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Feynman checkerboard
An intro to algorithmic quantum field theory
E. Akhmedova, M. Skopenkov, A. Ustinov, R. Valieva, A. Voropaev
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Basic model from ยง1
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Model with mass from ยง3
Figure 1: The probability tofind an electron at a given point(white depicts large oscillations)
Summary. It turns out that rainbow patterns on soapbubbles andunbelievable laws of motion of electrons can be explained by means ofone simple-minded model. It is a game, in which a checker moves on acheckerboard by certain simple rules, and we count the turnings. ThisโFeynman checkerboardโ can explain all phenomena in the world (witha serious proviso) except atomic nuclei and gravitation. We are goingto solve mathematical problems related to the game and discuss theirphysical meaning; no knowledge of physics is assumed.
Main results. The main results are explicit formulae for
โ the percentage of light of given color reflected by a glass plate ofgiven width (Problem 24);
โ the probability to find an electron at a given point, if it was emit-ted from the origin (Problem 15; see Figure 1).
Although the results are stated in physical terms, they are mathemati-cal theorems, because we provide mathematical models of the physicalphenomena in question, with all the objects defined rigorously. Moreprecisely, a sequence of models with increasing precision.
Plan. We start with a basic model and upgrade it step by stepin each subsequent section. Before each upgrade, we summarize whichphysical question does it address, which simplifying assumptions doesit resolve or impose additionally, and which experimental results doesit explain. Our aim is what is called 2-dimensional quantum electro-dynamics but the last steps on this way (sketched in Sections 8โ10)still have not been done. (A 4-dimensional one can already explain allphenomena โ with proviso and exceptions โ but we do not discuss it.)
The scheme of upgrades dependence might help to choose your way:
1. Basic model
))
6. External field // 8***. Interaction
**
4. Source
๏ฟฝ๏ฟฝ
2. Spin //oo
๏ฟฝ๏ฟฝ
OO
7. Identical particles & antiparticles
๏ฟฝ๏ฟฝ
OO
10***. QED
5. Medium 3. Mass //oo 9***. Creation & annihilation
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Conventions. If a problem is a statement of an assertion, then it is requested to prove the assertion.A puzzle is a problem, in which both a precise statement and a proof are requested. Hard problems aremarked with stars; you get first rank in Feynman checkerboard for solving one, and masters in Feynmancheckerboard for solving three from three different sections. Solutions are accepted in writing but you mayspend an earned star for 10 attempts to submit a solution in oral form (successful or not). If you cannotsolve a problem, proceed to the next ones: they may provide hints. Even if you do not reach the top(the proofs of main results), you learn much. You are encouraged to state and try to prove also your ownobservations and conjectures; you become a grand-master in Feynman checkerboard for discovering a newnontrivial one (and maybe even write your own scientific paper).
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1 Basic modelQuestion: what is the probability to find an electron at the point (๐ฅ, ๐ฆ), if it was emitted from the point (0, 0)?Assumptions: no self-interaction, no creation of electron-positron pairs, unit mass and lattice step, point source;no nuclear forces, no gravitation, electron moves uniformly along the ๐ฆ-axis and does not move along the ๐ง-axis.Results: double-slit experiment, charge conservation.
Figure 2: Checker paths
On an infinite checkerboard, a checker moves to the diagonal-neighboring squares, either upwards-right or upwards-left. To each path๐ of the checker, assign a vector ๏ฟฝ๏ฟฝ(๐ ) as follows. Start with a vector oflength 1 directed upwards. While the checker moves straightly, the vectoris not changed, but each time when the checker changes the direction, thevector is rotated through 90โ clockwise (independently of the direction thechecker turns). In addition, at the very end the vector is divided by 2(๐ฆโ1)/2,where ๐ฆ is the total number of moves. The final position of the vector iswhat we denote by ๏ฟฝ๏ฟฝ(๐ ). For instance, for the path in Figure 2 to the top,the vector ๏ฟฝ๏ฟฝ(๐ ) = (1/8, 0) is directed to the right and has length 1/8.
Denote ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) :=โ
๐ ๏ฟฝ๏ฟฝ(๐ ), where the sum is over all the paths of thechecker from the square (0, 0) to the square (๐ฅ, ๐ฆ), starting with the upwards-right move. Set ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) := 0, if there are no such paths. For instance,๏ฟฝ๏ฟฝ(1, 3) = (0,โ1/2)+(1/2, 0) = (1/2,โ1/2). The length square of the vector๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) is called the probability1 to find an electron in the square (๐ฅ, ๐ฆ), if itwas emitted from the square (0, 0). Notation: ๐ (๐ฅ, ๐ฆ) := |๐(๐ฅ, ๐ฆ)|2.
In Figure 1 to the top, the color of a point (๐ฅ, ๐ฆ) with even ๐ฅ+๐ฆ depictsthe value ๐ (๐ฅ, ๐ฆ). The sides of the apparent angle are not the lines ๐ฆ = ยฑ๐ฅ (and nobody knows why!).
In what follows squares (๐ฅ, ๐ฆ) with even and odd ๐ฅ + ๐ฆ are called black and white respectively.
1. Observations for small ๐ฆ. Answer the following questions for each ๐ฆ = 1, 2, 3, 4 (and state your ownquestions and conjectures for arbitrary ๐ฆ): Find the vector ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) and the probability ๐ (๐ฅ, ๐ฆ) for each ๐ฅ.When ๐ (๐ฅ, ๐ฆ) = 0? What is
โ๐ฅโZ ๐ (๐ฅ, ๐ฆ) for fixed ๐ฆ? What are the directions of ๏ฟฝ๏ฟฝ(1, ๐ฆ) and ๏ฟฝ๏ฟฝ(0, ๐ฆ)?
The probability2 to find an electron in the square (๐ฅ, ๐ฆ) subject to absorption in the square (๐ฅโฒ, ๐ฆโฒ) isdefined analogously to ๐ (๐ฅ, ๐ฆ), only the summation is over those paths ๐ that do not pass through (๐ฅโฒ, ๐ฆโฒ).The probability is denoted by ๐ (๐ฅ, ๐ฆ bypass ๐ฅโฒ, ๐ฆโฒ).
-1000 -500 500 1000
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Figure 3: ๐2(๐ฅ, 1000)2. Double-slit experiment. Is it true that ๐ (๐ฅ, ๐ฆ) = ๐ (๐ฅ, ๐ฆ bypass 0, 2)+๐ (๐ฅ, ๐ฆ bypass 2, 2)? Is it true that ๐ (๐ฅ, ๐ฆ) โฅ ๐ (๐ฅ, ๐ฆ bypass ๐ฅโฒ, ๐ฆโฒ)?
3. Find ๐ (0, 12). How to table the values ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) quickly without exhaustionof all paths? (The first solution grants first rank in Feynman checkers.)
Denote by ๐1(๐ฅ, ๐ฆ) and ๐2(๐ฅ, ๐ฆ) the coordinates of ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ); see Figure 3.
4. Diracโs equation. Express ๐1(๐ฅ, ๐ฆ) and ๐2(๐ฅ, ๐ฆ) through ๐1(๐ฅยฑ 1, ๐ฆ โ 1) and ๐2(๐ฅยฑ 1, ๐ฆ โ 1).
5. Probability/charge conservation. For each positive integer ๐ฆ we haveโ
๐ฅโZ ๐ (๐ฅ, ๐ฆ) = 1.
6. Symmetry. How the values ๐1(๐ฅ, 100) for ๐ฅ < 0 and for ๐ฅ โฅ 0 are related with each other? The samefor the values ๐1(๐ฅ, 100) + ๐2(๐ฅ, 100).
7. Huygensโ principle. What is a fast way to find ๏ฟฝ๏ฟฝ(๐ฅ, 199), if we know ๏ฟฝ๏ฟฝ(๐ฅ, 100) for all integers ๐ฅ?
8. Using a computer, plot the graphs of the functions ๐๐ฆ(๐ฅ) = ๐ (๐ฅ, ๐ฆ) for various ๐ฆ, joining each pair ofpoints (๐ฅ, ๐๐ฆ(๐ฅ)) and (๐ฅ + 2, ๐๐ฆ(๐ฅ + 2)) by a segment; cf. Figure 3. The same for the function ๐1(๐ฅ, ๐ฆ).
9.* Find an explicit formula for the vector ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) and the probability ๐ (๐ฅ, ๐ฆ) (it is allowed to use a sumwith at most ๐ฆ summands in the answer).
10.** (Skipable puzzle) Guess a simple โapproximate formulaโ for ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) and ๐ (๐ฅ, ๐ฆ), accurate for |๐ฅ| โช ๐ฆ.1One should think of the value ๐ฆ as fixed, and the squares (โ๐ฆ, ๐ฆ), (โ๐ฆ + 2, ๐ฆ), . . . , (๐ฆ, ๐ฆ) as all the possible outcomes of
an experiment. For instance, the ๐ฆ-th horizontal might be a photoplate detecting the electron.Familiarity with probability theory is not required for solving the presented problems.Beware that our rule for the probability computation is valid only for the basic model in question; we are going to changethe rule slightly in the upgrades. We make similar remarks each time we break some fundamental principles for simplicity.
2Thus an additional outcome of the experiment is that the electron has been absorbed and has not reached the photoplate.
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2 Spin
Question: what is the probability to find a right electron at (๐ฅ, ๐ฆ), if a right electron was emitted from (0, 0)?Assumptions: the same.Results: spin reversal.
The trick in the solution of the previous problems has a physical meaning: it is convenient to consideran electron as being in one of the two states: right-moving or left-moving. We write just โright โ or โleft โfor brevity3. This is not just a convenience but reflects an inalienable electronโs property called spin4.
Denote ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,+) :=โ
๐ ๏ฟฝ๏ฟฝ(๐ ), where the sum is over only those paths from (0, 0) to (๐ฅ, ๐ฆ), which bothstart and finish with an upwards-right move. Define ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,โ) to be an analogous sum over paths whichstart with an upwards-right move but finish with an upwards-left move.
The length square of the vector ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,+) (respectively, ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,โ)) is called the probability5 to find aright (respectively, left) electron in the square (๐ฅ, ๐ฆ), if a right electron was emitted from the square (0, 0).Denote by ๐ (๐ฅ, ๐ฆ,+) := |๐(๐ฅ, ๐ฆ,+)|2 and ๐ (๐ฅ, ๐ฆ,โ) := |๐(๐ฅ, ๐ฆ,โ)|2 these probabilities.
11. Express ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,+) and ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,โ) through ๐1(๐ฅ, ๐ฆ) and ๐2(๐ฅ, ๐ฆ); ๐ (๐ฅ, ๐ฆ) through ๐ (๐ฅ, ๐ฆ,+) and ๐ (๐ฅ, ๐ฆ,โ).
12.* Spin reversal. What is the probability ๐ (๐ฆ0,โ) :=โ
๐ฅโZ ๐ (๐ฅ, ๐ฆ0,โ) to find a left electron on theline ๐ฆ = ๐ฆ0 (it is allowed to use a sum with at most ๐ฆ0 summands in the answer)? Find the maximalelement and the limit of the sequence ๐ (1,โ), ๐ (2,โ), ๐ (3,โ), . . . .
A crash-course in calculus. For this and other problems with stars, the following formuli might be useful.Recall that
โ๐๐=0
12๐
= 2โ 12๐ . Clearly, as ๐ increases, the sum becomes closer and closer to 2. We would like
to writeโโ
๐=012๐
= 2; let us give a definition of such an infinite sum. A sequence ๐1, ๐2, ๐3, . . . has a limit ๐, iffor each real ๐ > 0 there is ๐ such that for each integer ๐ > ๐ we have |๐๐ โ ๐| < ๐. Notation: ๐ = lim๐โโ ๐๐.For instance, lim๐โโ
(2โ 1
2๐
)= 2. By definition, put
โโ๐=0 ๐๐ = lim๐โโ
โ๐๐=0 ๐๐. Then indeed
โโ๐=0
12๐
= 2.The following generalization is called Newtonโs binomial theorem (allowed to use without proof):
(1 + ๐ฅ)๐ =
โโ๐=0
๐(๐ โ 1) ยท ยท ยท (๐ โ ๐ + 1)
๐(๐ โ 1) ยท ยท ยท 1๐ฅ๐
for each complex ๐ฅ with |๐ฅ| < 1 and each real ๐, or for ๐ฅ = 1 and ๐ > โ1. In particular, for ๐ = โ1 and โ12 we get
1
1โ ๐ฅ=
โโ๐=0
๐ฅ๐ and1โ1โ ๐ฅ
=โโ๐=0
2๐(2๐ โ 1) ยท ยท ยท (๐ + 1)
๐(๐ โ 1) ยท ยท ยท 1๐ฅ๐
4๐.
The following Stirling formula allows to estimate the summands in Newtonโs binomial theorem:
โ2๐ ๐๐+1/2๐โ๐ โค ๐(๐ โ 1) ยท ยท ยท 1 โค ๐ ๐๐+1/2๐โ๐.
Here ๐ denotes lim๐โโ(1 + 1/๐)๐. It is an irrational number between 2.71 and 2.72.
3 MassQuestion: what is the probability to find a right electron of mass ๐ at (๐ฅ, ๐ฆ), if it was emitted from (0, 0)?Assumptions: the mass and the lattice step are now arbitrary.Results: a formula for the probability for small lattice step.
To check our model against experiment we need the following generalization.Fix ๐,๐ > 0 called lattice step and particle mass respectively. To each path ๐ of the checker, assign
a vector ๏ฟฝ๏ฟฝ(๐ ,๐๐) as follows. Start with the vector (0, 1). While the checker moves straightly, the vectoris not changed, but each time when the checker changes the direction, the vector is rotated through 90โ
3Beware that in 3 or more dimensions โright โ and โleft โ mean something very different from the movement direction.Although often visualized as the direction of the electron rotation, these states cannot be explained in nonquantum terms.
4And chirality ; beware that the term spin usually refers to a property, not related to the movement direction at all.5Thus an experiment outcome is a pair (final ๐ฅ-coordinate, last-move direction), whereas the final ๐ฆ-coordinate is fixed.
These are the fundamental probabilities, whereas ๐ (๐ฅ, ๐ฆ) should in general be defined by the formula from the solution ofProblem 11 rather than the above formula ๐ (๐ฅ, ๐ฆ) = |๐(๐ฅ, ๐ฆ)|2 (being a coincidence).
3
clockwise and multiplied by ๐๐. In addition, at the very end the vector is divided by (1 + ๐2๐2)(๐ฆโ1)/2,where ๐ฆ is the total number of moves. The final position of the vector is what we denote by ๏ฟฝ๏ฟฝ(๐ ,๐๐). Thevectors ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,๐๐,ยฑ) and the numbers ๐ (๐ฅ, ๐ฆ,๐๐,ยฑ) are defined analogously to ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,ยฑ) and ๐ (๐ฅ, ๐ฆ,ยฑ),only ๏ฟฝ๏ฟฝ(๐ ) is replaced by ๏ฟฝ๏ฟฝ(๐ ,๐๐). For instance, ๐ (๐ฅ, ๐ฆ, 1,+) = ๐ (๐ฅ, ๐ฆ,+). In Figure 1 to the bottom, thecolor of a point (๐ฅ, ๐ฆ) with even ๐ฅ + ๐ฆ depicts the value ๐ (๐ฅ, ๐ฆ, 0.02,+) + ๐ (๐ฅ, ๐ฆ, 0.02,โ).
13. (Puzzle) Massless and heavy particles. Find ๐ (๐ฅ, ๐ฆ, 0,+) and define ๐ (๐ฅ, ๐ฆ,โ,+) for each ๐ฅ, ๐ฆ.
14. Solve analogues of Problems 4, 5, and 9 for ๐๐ = 1.
In an experiment, we measure the probabilities to find the electron in intervals ๐ฅ0 โค ๐ฅ โค ๐ฅ0 + โ๐ฅ,๐ฆ = ๐ฆ0 rather than at particular points. Here ๐ฅ0, ๐ฆ0,โ๐ฅ are not integers but actual lengths measured inmeters. If all squares have small size 1
๐ร 1
๐, then the interval is approximated by the collection of black
squares(2โ๐๐ฅ2
โ, 2
โ๐๐ฆ2
โ)with ๐ฅ, ๐ฆ satisfying the above (in)equalities. This leads to the following problem.6
15.* (First main problem) Continuum limit. For each ๐ฅ, ๐ฆ,๐ find lim๐โโ ๐ ๏ฟฝ๏ฟฝ(2โ๐๐ฅ2
โ, 2โ๐๐ฆ2
โ, ๐๐,โ
)and
lim๐โโ ๐ ๏ฟฝ๏ฟฝ(2โ๐๐ฅ2
โ, 2โ๐๐ฆ2
โ, ๐๐,+
)In the answer, it is allowed to use the following expressions7
๐ฝ0(๐ง) :=โโ๐=0
(โ1)๐(๐ง/2)2๐
(๐!)2and ๐ฝ1(๐ง) :=
โโ๐=0
(โ1)๐(๐ง/2)2๐+1
๐!(๐ + 1)!.
4 SourceQuestion: what is the probability to find a right electron at (๐ฅ, ๐ฆ), if it was emitted by a source of wave length ๐?Assumptions: the source is now realistic.Results: wave propagation.
Figure 4: Checker paths maynow start at distinct squares
A realistic source does not produce electrons localized at ๐ฅ = 0 (asin our game) but a rather wide wave impulse instead. For our game,this means that the checker can start from an arbitrary black square onthe horizontal line ๐ฆ = 0 (not too far from the origin), but the initialdirection of the vector ๏ฟฝ๏ฟฝ(๐ ) is rotated through an angle proportional tothe distance from the starting square to the origin; see Figure 4.
Formally, fix real ๐, ๐ > 0 and odd โ called lattice step, wave length,and impulse width respectively. Denote by ๐ ๐ผ ๏ฟฝ๏ฟฝ the rotation of a vector ๏ฟฝ๏ฟฝthrough the angle |๐ผ|, which is counterclockwise for ๐ผ โฅ 0 and clockwisefor ๐ผ < 0. Define the vector
๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ,+) :=1โโ
ฮโ1โ๐ฅ0=1โฮ๐ฅ0 even
โ๐
๐ 2๐๐ฅ0๐/๐ ๏ฟฝ๏ฟฝ(๐ ),
where the second sum is over all checker paths ๐ from the square (๐ฅ0, 0) to the square (๐ฅ, ๐ฆ), starting andending with an upwards-right move. The length square of the vector is the probability to find a right electronat (๐ฅ, ๐ฆ), emitted by a source of wave length ๐ and impulse width โ. It is denoted by ๐ (๐ฅ, ๐ฆ, ๐/๐,โ,+).Define ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ,โ) and ๐ (๐ฅ, ๐ฆ, ๐/๐,โ,โ) analogously. For instance, ๐ (๐ฅ, ๐ฆ, ๐/๐, 1,+) = ๐ (๐ฅ, ๐ฆ,+)for each ๐, ๐, and ๏ฟฝ๏ฟฝ(๐ฅ + 1, 1, ๐/๐,โ,+) = 1โ
ฮ
(โ sin 2๐๐ฅ๐
๐, cos 2๐๐ฅ๐
๐
)for even |๐ฅ| < โ.
16. Let โ = 3, ๐/๐ = 4. Find the vector ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, 4, 3,+) and the probability ๐ (๐ฅ, ๐ฆ, 4, 3,+) for ๐ฆ = 1, 2, 3and each ๐ฅ. What is
โ๐ฅโZ(๐ (๐ฅ, ๐ฆ, 4, 3,+)+๐ (๐ฅ, ๐ฆ, 4, 3,โ)) for fixed ๐ฆ = 1, 2, 3? When ๐ (๐ฅ, 3, 4, 3,+) = 0?
17. Probability/charge conservation. Solve analogues of Problems 4 and 5 for ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ,โ),๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ,+), and ๐ (๐ฅ, ๐ฆ, ๐/๐,โ,+) + ๐ (๐ฅ, ๐ฆ, ๐/๐,โ,โ) instead of ๐1(๐ฅ, ๐ฆ), ๐2(๐ฅ, ๐ฆ), and ๐ (๐ฅ, ๐ฆ).
18. Causality. Bothโ
โ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ,+) and โ๐ (๐ฅ, ๐ฆ, ๐/๐,โ,+) do not depend on โ for โ > ๐ฆ + |๐ฅ|.Denote8 ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,ยฑ)=
โโ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ,ยฑ) and ๐ (๐ฅ, ๐ฆ, ๐/๐,ยฑ)=โ๐ (๐ฅ, ๐ฆ, ๐/๐,โ,ยฑ) for any โ>๐ฆ+|๐ฅ|.
19. Wave. How to find ๏ฟฝ๏ฟฝ(๐ฅ, 100, ๐/๐,+) for all even ๐ฅ, if we know it for just one even ๐ฅ?
20.* Wave propagation. For each ๐ฅ, ๐ฆ, ๐, ๐ find ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ), ๐ (๐ฅ, ๐ฆ, ๐/๐,โ), ๐ (๐ฅ, ๐ฆ, ๐/๐,โ)+๐ (๐ฅ, ๐ฆ, ๐/๐,+).6In the limits, the normalization factor of ๐ is a bit harder to explain; we not discuss it.7Called Bessel functions, which are almost as well-studied as sine and cosine; but familiarity with them is not required.8The notation ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,ยฑ) should not be confused with ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,๐๐,ยฑ).
4
5 MediumQuestion: which part of light of given color reflects from a glass plate of given width?Assumptions: right angle of incidence, no polarization of light; the mass now depends on ๐ฅ but not on the color.Results: thin-film reflection.
Our model can be applied also to describe propagation of light in transparent media such as glass9.Light propagates as if it had some nonzero mass inside the media, while the mass remains zero outside10.The wavelength determines the color of light.
21. (Puzzle) Define an analogue of ๏ฟฝ๏ฟฝ(๐ ,๐๐) in the case when the mass ๐ = ๐(๐ฅ) depends on ๐ฅ so thatanalogues of Problems 4 and 5 remain true.
Given a mass ๐ = ๐(๐ฅ), define ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,๐(๐ฅ)๐, ๐/๐,ยฑ) and ๐ (๐ฅ, ๐ฆ,๐(๐ฅ)๐, ๐/๐,ยฑ) analogously to๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,ยฑ) and ๐ (๐ฅ, ๐ฆ, ๐/๐,ยฑ), only replace ๏ฟฝ๏ฟฝ(๐ ) by ๏ฟฝ๏ฟฝ(๐ ,๐๐) in the definition. First we take ๐ = 1.
22. One-surface reflection. Find ๐ (๐ฅ, ๐ฆ,๐(๐ฅ), ๐,+) and ๐ (๐ฅ, ๐ฆ,๐(๐ฅ), ๐,โ) for
๐(๐ฅ) = ๐0(๐ฅ) โก 0 and ๐(๐ฅ) = ๐1(๐ฅ) =
{0.2, for ๐ฅ = 0;
0, for ๐ฅ = 0.
Now fix odd ๐ฟ > 1 called the width of a glass plate. First assume for simplicity that light is reflectedonly by the two surfaces of the plate and thus take11
๐2(๐ฅ)=
โงโชโจโชโฉโ0.2, for ๐ฅ = 1;
+0.2, for ๐ฅ = ๐ฟ;
0, otherwise.
The reflection/transmission probabilities12 of light of wavelength ๐ by glass plate of width ๐ฟ are respectively๐ (๐, ๐ฟ,โ) = lim
๐ฆโ+โ๐ฆ even
๐ (0, ๐ฆ,๐2(๐ฅ), ๐,โ);
๐ (๐, ๐ฟ,+) = lim๐ฆโ+โ๐ฆ even
๐ (๐ฟ + 1, ๐ฆ,๐2(๐ฅ), ๐,+).
23. Plot the graph of the function ๐(๐ฟ) = ๐ (0, 2๐ฟ,๐2(๐ฅ), 16,โ).
24.* (Second main problem)Two-surface reflection. Find ๐ (๐, ๐ฟ,โ), ๐ (๐, ๐ฟ,+), ๐ (๐, ๐ฟ,โ)+๐ (๐, ๐ฟ,+),and max๐ฟ ๐ (๐, ๐ฟ,โ).
In fact the light is reflected inside the plate; this should be taken into account for a more accuratecomputation of the reflection probability (there is no hope for an exact solution for complicated matter suchas glass). For this purpose we need to modify the model essentially. Take arbitrary ๐ > 0. Fix ๐ > 0 andlet ๐3(๐ฅ) = ๐ for 0 < ๐ฅ โค ๐ฟ/๐ and ๐3(๐ฅ) = 0 otherwise. Now for each move starting in a square insidethe glass, our vector is additionally rotated through the angle arctan๐๐ clockwise (independently on if thechecker does or does not turn in the square)13. In other words, set ๏ฟฝ๏ฟฝm(๐ ,๐(๐ฅ)๐) := ๐ โ๐ arctan๐๐๏ฟฝ๏ฟฝ(๐ ,๐(๐ฅ)๐),where ๐ is the number of moves starting in the strip 0 < ๐ฅ โค ๐ฟ/๐ in the path ๐ . Define ๐m(๐ฅ, ๐ฆ,๐(๐ฅ)๐, ๐
๐,โ)
analogously to ๐ (๐ฅ, ๐ฆ,๐(๐ฅ)๐, ๐๐,โ), only replace ๏ฟฝ๏ฟฝ(๐ ,๐(๐ฅ)๐) by ๏ฟฝ๏ฟฝm(๐ ,๐(๐ฅ)๐) in the definition.
25.** Thin-film reflection. Find lim๐โ0+
lim๐ฆโ+โ๐ฆ even
๐m(0,๐ฆ,๐3(๐ฅ)๐,๐๐,โ). For which ๐ maximum of the expres-
sion over ๐ฟ equals max๐ฟ ๐ (๐, ๐ฟ,โ)? (Use existence of lim๐ฆโ+โ๐ฆ+๐ฅ even
๐ 2๐๐ฆ๐/๐๏ฟฝ๏ฟฝ(๐ฅ,๐ฆ,๐3(๐ฅ)๐,๐๐,โ) without proof.)
9Beware: in general Feynman checkerboard is inappropriate to describe light; partial reflection is a remarkable exception.10The mass is proportional to (๐โ 1)/2
โ๐, where ๐ is the refractive index; for glass ๐ โ 1.5 and (๐โ 1)/2
โ๐ โ 0.2.
11This simplifying assumption requires negative mass for the left surface; the origin of that becomes clear after solving 25.12It is more conceptual to define ๐ (๐, ๐ฟ,โ) = lim๐โ0 limฮโ+โ lim๐ฆโ+โ
โ๐ฅโZ ๐ (๐ฅ, ๐ฆ,๐3(๐ฅ)๐,
๐๐ ,ฮ,โ) but we do not.
13This additional rotation is explained as follows. The light can be scattered in each square inside the glass several times.Each individual scattering gives a factor of โ๐๐๐ to our vector (viewed as a complex number) and may or may not changethe movement direction. Assume that ๐๐ < 1. Thus a move without changing the direction contributes a factor of
1 (no scattering)โ ๐๐๐ (1 scattering)+ (โ๐๐๐)2 (2 scatterings)+ ยท ยท ยท = 1
1 + ๐๐๐.
Without a scattering, the checker moves straightly. Thus a turn in a particular square inside the glass contributes a factor
โ๐๐๐ (1 scattering)+ (โ๐๐๐)2 (2 scatterings)+ (โ๐๐๐)3 (3 scatterings)+ ยท ยท ยท = โ๐๐๐
1 + ๐๐๐.
These are the same factors as in the model from ยง3 but additionally rotated through the angle arctan๐๐ clockwise.
5
6 External fieldQuestion: what is the probability to find a right electron at (๐ฅ, ๐ฆ), if it moves in a given magnetic field ๐ข?Assumptions: the magnetic field vanishes outside the ๐ฅ๐ฆ-plane, it is not affected by the electron.Results: deflection of electron and spin โprecessionโ in a magnetic field, charge conservation.
Figure 5: Paths in a field
An external magnetic field changes the motion as follows14.A common point of 4 squares of the checkerboard is called a ver-
tex. A magnetic field15 is a fixed assignment ๐ข of numbers +1 andโ1 to all the vertices. For instance, in Figure 5, the magnetic fieldis โ1 at the top-right vertex of each square (๐ฅ, ๐ฆ) with both ๐ฅ and๐ฆ even. Modify the definition of the vector ๏ฟฝ๏ฟฝ(๐ ) by reversing the di-rection each time when the checker passes through a vertex with themagnetic field โ1. Denote by ๏ฟฝ๏ฟฝ(๐ , ๐ข) the resulting vector. Formally,put ๏ฟฝ๏ฟฝ(๐ , ๐ข) = ๏ฟฝ๏ฟฝ(๐ )๐ข(๐ถ1)๐ข(๐ถ2) . . . ๐ข(๐ถ๐ฆ), where ๐ถ1, ๐ถ2, . . . , ๐ถ๐ฆ are allthe vertices passed by ๐ . Define ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐ข,ยฑ) and ๐ (๐ฅ, ๐ฆ, ๐ข,ยฑ) analogously to ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,ยฑ) and ๐ (๐ฅ, ๐ฆ,ยฑ) re-placing ๏ฟฝ๏ฟฝ(๐ ) by ๏ฟฝ๏ฟฝ(๐ , ๐ข) in the definition. For instance, if ๐ข(๐ถ) = +1 identically, then ๐ (๐ฅ, ๐ฆ, ๐ข) = ๐ (๐ฅ, ๐ฆ).
26. Homogeneous field. Let ๐ข(๐ถ) = โ1, if ๐ถ is the top-right vertex of a square (๐ฅ, ๐ฆ) with both ๐ฅand ๐ฆ even, and ๐ข(๐ถ) = +1 otherwise. Find the vector ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐ข,+) and the probability ๐ (๐ฅ, ๐ฆ, ๐ข,+) for๐ฆ = 1, 2, 3, 4 and each integer ๐ฅ. What is
โ๐ฅโZ(๐ (๐ฅ, ๐ฆ, ๐ข,+) + ๐ (๐ฅ, ๐ฆ, ๐ข,โ)) for fixed ๐ฆ = 1, 2, 3, or 4?
27. Spin โprecessionโ in a magnetic field. Plot the graph of the function ๐(๐ฆ) =โ
๐ฅโZ ๐ (๐ฅ, ๐ฆ, ๐ข,+)for the field ๐ข from the previous problem using a computer.
For a given magnetic field ๐ข, a white square is negative, if ๐ข equals โ1 at 1 or 3 vertices of the square.
28. Gauge transformations. Changing the signs of the values of ๐ข at the 4 vertices of one black squaresimultaneously does not change ๐ (๐ฅ, ๐ฆ, ๐ข,+).
29. Curvature. One can make ๐ข to be identically +1 in a rectangle formed by checkerboard squares usingthe transformations from Problem 28, if and only if there are no negative white squares in the rectangle.
30. Homology. The magnetic field ๐ข equals +1 on the boundary of a rectangle ๐ ร ๐ formed bycheckerboard squares. Which can be the number of negative white squares in the rectangle?
31. Probability/charge conservation. Solve analogues of Problems 4, 5 for ๐ข not being identically +1.
7 Identical particles and antiparticles
Question: what is the probability to find electrons (or electron+positron) at ๐น and ๐น โฒ, emitted from ๐ด and ๐ดโฒ?Assumptions: the same as in the basic model; ๐ฆ-coordinate is interpreted as time.Results: exclusion principle.
The motion of several electrons is described by a similar model as follows.To each pair of paths ๐ , ๐ โฒ of the checker, consisting of ๐ฆ moves each, assign a vector ๏ฟฝ๏ฟฝ(๐ , ๐ โฒ) as follows.
Start with the vector (0, 1). Move the checker consecutively along both paths, and rotate the vectoraccording to the same rule as in ยง1: each time when the checker changes the direction, the vector isrotated through 90โ clockwise. (Thus the vector is rotated totally ๐ก(๐ ) + ๐ก(๐ โฒ) times, where ๐ก(๐) is thenumber of turns in a path ๐.) In addition, at the very end the vector is divided by 2๐ฆโ1. The final positionof the vector is denoted by ๏ฟฝ๏ฟฝ(๐ , ๐ โฒ). For instance, in Figure 2 we have ๏ฟฝ๏ฟฝ(๐ , ๐ 0) = (โ1/4, 0).
Fix squares ๐ด = (0, 0), ๐ดโฒ = (๐ฅ0, 0), ๐น = (๐ฅ, ๐ฆ), ๐น โฒ = (๐ฅโฒ, ๐ฆ) and their diagonal neighbors ๐ต = (1, 1),๐ตโฒ = (๐ฅ0 + 1, 1), ๐ธ = (๐ฅโ 1, ๐ฆ โ 1), ๐ธ โฒ = (๐ฅโฒ โ 1, ๐ฆ โ 1), where ๐ฅ0 = 0. Denote16
๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธ โฒ๐น โฒ) :=โ
๐ :๐ด๐ตโ๐ธ๐น๐ โฒ:๐ดโฒ๐ตโฒโ๐ธโฒ๐น โฒ
๏ฟฝ๏ฟฝ(๐ , ๐ โฒ) โโ
๐ :๐ด๐ตโ๐ธโฒ๐น โฒ๐ โฒ:๐ดโฒ๐ตโฒโ๐ธ๐น
๏ฟฝ๏ฟฝ(๐ , ๐ โฒ),
14Beware that this method of adding the magnetic field, although well-known, is very different from the one from [Feynman].15Or electromagnetic vector-potential, to be precise. The field is interpreted as magnetic or electric depending on if the
๐ฆ-coordinate is interpreted as position or time.16Here it is essential that ๐ and ๐ โฒ are paths of particles of the same sort, e.g., two electrons. Otherwise the 2nd sum is
omitted. The sign before the 2nd sum is changed to plus for some other sorts of particles, e.g., photons (particles of light).
6
where the first sum is over all pairs consisting of a checker path ๐ starting with the move ๐ด๐ต and endingwith the move ๐ธ๐น , and a path ๐ โฒ starting with the move ๐ดโฒ๐ตโฒ and ending with the move ๐ธ โฒ๐น โฒ, whereasin the second sum the final moves are interchanged.
The length square ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธ โฒ๐น โฒ) = |๐(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธ โฒ๐น โฒ)|2 is called the probability17
to find right electrons at ๐น and ๐น โฒ, if they are emitted from ๐ด and ๐ดโฒ. In particular, ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ๐ธ๐น,๐ธ๐น ) = 0, i.e., two right electrons cannot be found at the same point; this is called exclusion principle.
32. Independence. For ๐ฅ0 โฅ 2๐ฆ and ๐ฅโฒ > ๐ฅ express ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธ โฒ๐น โฒ) through ๐ (๐ฅ, ๐ฆ,+) and๐ (๐ฅโฒ โ ๐ฅ0, ๐ฆ,+).
33. Exclusion principle (for intermediate states). Prove that ๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธ โฒ๐น โฒ) is notchanged, if the sums in the definition are over only those pairs of paths ๐ , ๐ โฒ which have no common moves.
Define ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธ โฒ๐น โฒ) analogously also for ๐ธ = (๐ฅยฑ 1, ๐ฆ โ 1), ๐ธ โฒ = (๐ฅโฒ ยฑ 1, ๐ฆ โ 1).
34. Probability/charge conservation. We haveโ
๐ธ,๐ธโฒ,๐น,๐น โฒ ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธ โฒ๐น โฒ) = 1, where thesum is over all quadruples ๐น = (๐ฅ, ๐ฆ), ๐น โฒ = (๐ฅโฒ, ๐ฆ), ๐ธ = (๐ฅยฑ1, ๐ฆโ1), ๐ธ โฒ = (๐ฅโฒยฑ1, ๐ฆโ1) with fixed ๐ฆ โฅ 1.
An electron has an antiparticle called positron. One can think of an antiparticle as a particle movingbackwards in time. The motion is described by a similar model as follows.
Add one more checker to the checkerboard, which now moves either downwards-right or downwards-leftto the diagonal neighbors. The added checker is called black, while the one studied before is called white.
For each pair of paths ๐ , ๐ โฒ of the white and black checker respectively, define a vector ๏ฟฝ๏ฟฝ(๐ , ๐ โฒ) analogouslyto the above, only for each turn of the black checker rotate the vector counterclockwise rather than clockwise(independently of the direction the checker turns). For instance, ๏ฟฝ๏ฟฝ(๐ , ๐ โฒ) = (0, 21โ๐ฆ), if ๐ and ๐ โฒ is the samepath with opposite directions. Denote18 by
๏ฟฝ๏ฟฝ(๐ด๐ต,๐ตโฒ๐ดโฒ โ ๐ธ๐น, ๐น โฒ๐ธ โฒ) :=โ๐ ,๐ โฒ
๏ฟฝ๏ฟฝ(๐ , ๐ โฒ)
the sum is over all the pairs of a white-checker path ๐ starting with the move ๐ด๐ต and ending with themove ๐ธ๐น , and a black-checker path ๐ โฒ starting with the move ๐น โฒ๐ธ โฒ and ending with the move ๐ตโฒ๐ดโฒ. Thelength square ๐ (๐ด๐ต,๐ตโฒ๐ดโฒ โ ๐ธ๐น, ๐น โฒ๐ธ โฒ) = |๐(๐ด๐ต,๐ตโฒ๐ดโฒ โ ๐ธ๐น, ๐น โฒ๐ธ โฒ)|2 is called the probability to find anelectron at ๐น and a positron at ๐น โฒ, if they are emitted from ๐ด and ๐ดโฒ.
35. Independence. Solve analogue of Problem 32 for ๐ (๐ด๐ต,๐ตโฒ๐ดโฒ โ ๐ธ๐น, ๐น โฒ๐ธ โฒ).
8 Interaction***Question: the same as in the basic model but the ๐ฆ-coordinate is now interpreted as time.Assumptions: the electron now generates a magnetic field affecting the motion; encircling by waveproof walls.Results: interaction with the walls changes the motion of an electron.
We have reached an unexplored area: a physically correct rigorous definition of the next upgrade isnot yet known. In this section we give a WRONG definition leading to paradoxical results:
โ the upgrades measures the interaction of the electron with the walls rather than self-interaction;
โ the interaction (with the walls) propagates at infinite speed rather than the speed of light.
But we hope that the introduced ideas still might be of interest.The moving electron itself generates an magnetic field, which in turn affects the motion. The generated
field is random, with the probability resulting from summation over all possible intermediate fields.Fix ๐ โฅ 0 called the interaction constant19 (it is related to the electron charge). Fix a black square
(๐ฅ, ๐ฆ) called the final position. Fix the rectangle ๐ formed by all the squares (๐ฅโฒ, ๐ฆโฒ) such that 1โ๐ฆ < ๐ฅโฒ < ๐ฆ
17One should think of the value ๐ฆ as fixed, and the quadruples (๐น, ๐น โฒ, ๐ธ,๐ธโฒ) as the possible outcomes of an experiment.18This definition makes sense only for ๐ฅ0 โฅ 2๐ฆ; otherwise annihilation of particles cannot be ignored, and the model
becomes inappropruate.19If the lattice step ๐ is not fixed, then ๐ might depend on it. This is called renormalization; we do not discuss it.
7
and 0 < ๐ฆโฒ < ๐ฆ. This waveproof box ๐ encircles any possible checker path ๐ from (0, 0) to (๐ฅ, ๐ฆ) excludingthe first and the last move; it is required to make the sum over all intermediate fields below finite.
Take any such path ๐ and any assignment ๐ข of the numbers ยฑ1 to all the vertices in the rectangle ๐ .Let ๐ be the number of negative white squares in ๐ for ๐ข (they play the role of turnings of the checker).Denote by
๏ฟฝ๏ฟฝ(๐ , ๐ข, ๐) =๐๐
2(๐ฆโ2)2(1 + ๐2)(๐ฆโ1)2/2๐ โ๐๐/2๏ฟฝ๏ฟฝ(๐ , ๐ข)
the vector20 ๏ฟฝ๏ฟฝ(๐ , ๐ข), rotated clockwise through ๐ยท90โ, multiplied by ๐๐ and divided by 2(๐ฆโ2)2(1+๐2)(๐ฆโ1)2/2.Let ๐ขfin be any assignment of the numbers ยฑ1 to all the vertices on the top side of ๐ such that ๐ขfin
equals +1 at the endpoints of the side. It is called the final magnetic field. Denote by
๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐ขfin, ๐,+) :=โ๐ ,๐ข
๏ฟฝ๏ฟฝ(๐ , ๐ข, ๐)
the sum over all the paths ๐ from (0, 0) to (๐ฅ, ๐ฆ) starting and ending with an upwards-right move and overall the assignments ๐ข of the numbers ยฑ1 to all the vertices in the rectangle ๐ such that
๐ข =
{๐ขfin on the top side of ๐ ;
+1 on all the other sides of ๐ .
The length square ๐ (๐ฅ, ๐ฆ, ๐ขfin, ๐,+) = |๐(๐ฅ, ๐ฆ, ๐ขfin, ๐,+)|2 is the probability that the final magnetic fieldequals ๐ขfin and a right electron is found at (๐ฅ, ๐ฆ). Define ๐ (๐ฅ, ๐ฆ, ๐ขfin, ๐,โ) analogously.
36. Find the probabilities ๐ (๐ฅ, ๐ฆ, ๐ขfin, 1,+) and ๐ (๐ฅ, ๐ฆ, ๐ขfin, 1,โ) for ๐=1, ๐ฆ=2, 3 and all possible ๐ฅ and ๐ขfin.
The sum21
๐ (๐ฅ, ๐ฆ, ๐,+) =โ๐ขfin
๐ (๐ฅ, ๐ฆ, ๐ขfin, ๐,+)
over all assignments ๐ขfin of the numbers ยฑ1 to the vertices on the top side of ๐ is the probability to find aright electron at (๐ฅ, ๐ฆ). Define ๐ (๐ฅ, ๐ฆ, ๐,โ) analogously. For instance, ๐ (๐ฅ, ๐ฆ, 0,+) = ๐ (๐ฅ, ๐ฆ,+) (why?).
37.* For a path ๐ from (0, 0) to (๐ฅ, ๐ฆ), consider the sumโ
๐ข ๏ฟฝ๏ฟฝ(๐ , ๐ข, ๐) over all assignments ๐ข of ยฑ1 to allthe vertices in the rectangle ๐ such that ๐ข = +1 on the boundary of ๐ . Express the sum through ๏ฟฝ๏ฟฝ(๐ )and the number of white squares in each of the two parts, into which ๐ is divided by the path ๐ .
38.* Let ๐ขยฑ be equal to ยฑ1 respectively at the top-right corner of the square (2 โ ๐ฆ, ๐ฆ โ 1), and equal to+1 at all the other vertices on the top side of ๐ . Solve analogues of Problems 4 and 5 for ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐ขยฑ, ๐,ยฑ)and ๐ (๐ฅ, ๐ฆ, ๐,+) + ๐ (๐ฅ, ๐ฆ, ๐,โ) instead of ๐1(๐ฅ, ๐ฆ), ๐2(๐ฅ, ๐ฆ), and ๐ (๐ฅ, ๐ฆ).
9 Creation and annihilation***Question: the same as in the model with identical particles and antiparticles.Assumptions: electron-positron pairs now created and annihilated, no interaction, encircling by reflecting walls.Results: no; this is only an ingredient for more realistic models with interaction.
Finally we have reached an unexplored area: we state an almost 40-years-old open problem.Start with mentioning what is not done in this section:
โ we do not give a definition of the new upgrade (it is unknown so far);
โ the upgrade (even if defined) would not explain any new experimental results.
But
โ we do give a precise statement of the problem: which exactly properties of the upgrade are requested;
20It is well-defined because the path ๐ is contained in ๐ except the first and the last move. We set ๐๐ = 1 for ๐ = ๐ = 0.21Here we sum probabilities rather than vectors. The notation ๐ (๐ฅ, ๐ฆ, ๐,+) should not be confused with ๐ (๐ฅ, ๐ฆ,๐๐,+).
8
โ the upgrade is an important ingredient of further ones fantastically agreeing with experiment.
Informally, our plan is as follows. Checker paths turning downwards or upwards or forming cycles meancreation and annihilation of electron-positron pairs. Even if we start with just one electron, we might endup with many electrons and positrons. To each possible configuration of the resulting particles, we wantto assign a complex number so that its length square is the probability of the configuration in a sense. Thenumber itself is the sum over all possible transitions from the initial configuration to the final one, that is,all possible paths configurations joining them. To make the sum finite, we put reflecting walls around.
Fix a rectangle ๐ formed by all the squares (๐ฅ, ๐ฆ) such that ๐ฅmin โค ๐ฅ โค ๐ฅmax and 0 โค ๐ฆ โค ๐ฆmax (thelines ๐ฅ = ๐ฅmin and ๐ฅ = ๐ฅmax are called reflecting walls). Fix ๐, ๐ > 0 called mass and lattice step.
An initial configuration is any assignment22 of one of the signs โ+โ or โโโ to some vertices inside ๐ lying between the lines ๐ฆ = 0 and ๐ฆ = 1. Physically the signs mean the initial positions of positrons andelectrons respectively (and points without a sign are vacant). For our game, this means that the checkerspass the vertices with the โโโ sign upwards-left or -right, and the vertices with the โ+โ sign โ downwards-left or -right (and do not pass through vertices without a sign). Analogously, a final configuration is anassignment to vertices between the lines ๐ฆ = ๐ฆmax and ๐ฆ = ๐ฆmax โ 1. An intermediate configuration is anyassignment of one of the signs โ+โ or โโโ to some vertices inside ๐ such that the difference between thenumber of โ+โ and โโโ signs on the 2 top vertices of each black square in the strip 1 โค ๐ฆ โค ๐ฆmaxโ1 equalsthe difference on the 2 bottom vertices. For our game, this means that the checkers start and finish motionin the lines ๐ฆ = 0 or ๐ฆ = ๐ฆmax only. In other words, the signs at the vertices of each black square are inone of the 19 positions23 shown in Figure 6 to the left. These 19 positions are called basic configurations.
Figure 6: Basic configurations
Suppose that one has assigned a complex number (depending on ๐๐) to each of the 19 basic configu-rations. A โrightโ choice of the numbers is unknown; think of them as fixed parameters of our model.
Then take any intermediate configuration. In each black square inside ๐ not having common pointswith the boundary, write the complex number assigned to the basic configuration in the black square.Assign the product of all the written numbers to the intermediate configuration.
Now to any pair of initial and final configurations, sum the complex numbers assigned to all possibleintermediate configurations between them. Assign the resulting complex number to the pair.
39. (Puzzle) Restrict to configurations without โ+โ signs. Let ๐ด,๐ดโฒ, ๐ต,๐ตโฒ, ๐ธ, ๐ธ โฒ, ๐น, ๐น โฒ be the black squaresdefined in ยง7. Take any ๐ฅmin < โ๐ฆ and ๐ฅmax > ๐ฅ0 + ๐ฆ. Fix the initial and final configurations with exactlytwo โโโ signs, located at the top-right vertices of the squares ๐ด, ๐ดโฒ, and ๐ธ, ๐ธ โฒ respectively. Assign complexnumbers to the 6 basic configurations without โ+โ signs so that the sum of the numbers assigned to allintermediate configurations without โ+โ signs equals to the vector ๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธ โฒ๐น โฒ) from ยง7.
A case of particular interest is when the initial configuration consists of just one โโโ sign at the top-right vertex of the square (0, 0) and the final configuration consists of just one โโโ sign at the bottom-right
22The assignment has nothing to do with the magnetic field from ยง6.23We consider positions of signs, but not possible checker paths in a particular black square like in Figure 6 to the right.
9
or bottom-left vertex of a black square (๐ฅ, ๐ฆmax). The complex numbers assigned to the pairs in questionare denoted by ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆmax,๐๐, ๐ฅmin, ๐ฅmax,โ) and ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆmax,๐๐, ๐ฅmin, ๐ฅmax,+) respectively.
The desired continuum limit of these complex numbers involves the following modified Bessel functionsand Hankel functions :
๐พ0(๐ฅ) =
โซ โ
1
๐โ๐ฅ๐ก
โ๐ก2 โ 1
๐๐ก ๐ป(1)0 (๐ฅ) =
2
๐๐
โซ โ
1
๐๐๐ฅ๐กโ๐ก2 โ 1
๐๐ก
๐พ1(๐ฅ) = ๐ฅ
โซ โ
1
๐โ๐ฅ๐กโ๐ก2 โ 1 ๐๐ก ๐ป
(1)1 (๐ฅ) = โ2๐ฅ
๐๐
โซ โ
1
๐๐๐ฅ๐กโ๐ก2 โ 1 ๐๐ก
40.*** Continuum limit. Assign complex numbers to the 19 basic configurations so that for each |๐ฆ| < |๐ฅ|
lim๐โโ
lim๐ฅmaxโโ
๐ ๏ฟฝ๏ฟฝ(
2โ๐๐ฅ
2
โ, 2โ๐๐ฆ
2
โ,๐
๐,โ๐ฅmax, ๐ฅmax,โ
)= ๐๐พ0(๐
โ๐ฅ2 โ ๐ฆ2);
lim๐โโ
lim๐ฅmaxโโ
๐ ๏ฟฝ๏ฟฝ(
2โ๐๐ฅ
2
โ, 2โ๐๐ฆ
2
โ,๐
๐,โ๐ฅmax, ๐ฅmax,+
)= โ๐๐
๐ฅ + ๐ฆโ๐ฅ2 โ ๐ฆ2
๐พ1(๐โ๐ฅ2 โ ๐ฆ2);
and for each |๐ฆ| > |๐ฅ| we have
lim๐โโ
lim๐ฅmaxโโ
๐ ๏ฟฝ๏ฟฝ(
2โ๐๐ฅ
2
โ, 2โ๐๐ฆ
2
โ,๐
๐,โ๐ฅmax, ๐ฅmax,โ
)= ๐๐
๐
2๐ป
(1)0 (๐
โ๐ฆ2 โ ๐ฅ2);
lim๐โโ
lim๐ฅmaxโโ
๐ ๏ฟฝ๏ฟฝ(
2โ๐๐ฅ
2
โ, 2โ๐๐ฆ
2
โ,๐
๐,โ๐ฅmax, ๐ฅmax,+
)= ๐
๐
2ยท ๐ฅ + ๐ฆโ
๐ฆ2 โ ๐ฅ2๐ป
(1)1 (๐
โ๐ฆ2 โ ๐ฅ2).
Let us discuss the physical meaning of the upgrade. For ๐ฆ โซ |๐ฅ| the right-hand sides of the lattertwo equations are very close to the right-hand sides of the answer to Problem 10. Thus one may wish tointerpret the upgrade as a more accurate approximation for the probability to find the electron in a square(๐ฅ, ๐ฆ). But this faces serious objections.
First, it is in principle impossible to measure the coordinates of an electron exactly 24. Such a prioriuncertainty has the same order of magnitude as the correction introduced by the upgrade. Thus theupgrade does not actually add anything to description of the electron motion.
Second, for fixed initial configuration, the squares of the absolute values of the numbers assigned toall the possible final configurations do not sum up to 1 (even in the continuum limit). The reason is thatdistinct configurations are not mutually exclusive: for a field in a given configuration, there is a positiveprobability to find it in a different configuration. We do not know any clear explanation of that.
To summarize, the upgrade lacks a direct physical interpretation, and should be considered as aningredient for further upgrades.
10 (1 + 1)-dimensional quantum electrodynamics***
Question: what is the probability to find electrons (or electron+positron) with momenta ๐ and ๐โฒ in the far future,if they were emitted with momenta ๐ and ๐โฒ in the far past?Assumptions: interaction now switched on; all simplifying assumptions removed except the default ones:no nuclear forces, no gravitation, electron moves only along the ๐ฅ-axis (and ๐ฆ-coordinate is interpreted as time).Results: quantum corrections.
Unifying the (so far unknown) upgrades discussed in the previous two sections would give an elementarydefinition of (1 + 1)-dimensional QED.
Future research
An algorithmic quantum field theory is a one which for each experimentally observable quantity and apositive number ๐ provides a precise statement of an algorithm giving the predicted value of the quantitywithin accuracy ๐. (Surely, the predicted value does not have to agree with the experiment for ๐ less thanaccuracy of theory itself.) This is an extension of constructive quantum field theory, the latter currentlybeing far away from algorithmic one.
24This should not be confused with uncertainty principle, which does not allow simultaneous measurement of the coordi-nates and momentum.
10
Epilogue (underwater rocks)
We hope that at least some of our readers have become interested in elementary particles and want tolearn more about them. As an epilogue, let us give a few warnings to such readers.
In popular science, theory of elementary particles is usually oversimplified. This sequence of problemsis not an exception. The toy models introduced here are very rough and should be considered with agrain of salt. Simplicity is their only advantage; if taken too seriously, the models could even give a wrongphysical intuition. Real understanding of particles theory requires excellent knowledge of both physics andmathematics.
We should also remark that nowadays there are almost no mathematical results in lattice quantumfield theory; what we have is usually just a numeric simulation. Finally, there are โtheories of NewPhysicsโ which are developed without any objective truth criterion: such theories are supported by neitherexperimental nor mathematical proofs (and some of them have experimental disproofs).
11
Hints, solutions, answers
For any vector ๏ฟฝ๏ฟฝ โ R2 denote by ๐1, ๐2 the coordinates of ๏ฟฝ๏ฟฝ. E.g., ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,๐๐,โ) = (๐1(๐ฅ, ๐ฆ,๐๐,โ), ๐2(๐ฅ, ๐ฆ,๐๐,โ)).Sometimes the vector ๏ฟฝ๏ฟฝ is considered as a complex number ๐1 + ๐๐2 (although complex numbers are not requiredfor the solution of most problems). In what follows assume that ๐ฅ and ๐ฆ have the same parity unless the oppositeis indicated.
1. Answer. In the table, the vector in the cell (๐ฅ, ๐ฆ) is ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ), and an empty cell (๐ฅ, ๐ฆ) means that ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) = (0, 0).
4(
12โ2, 0) (
0,โ 12โ2
) (1
2โ2,โ 1โ
2
) (0, 1
2โ2
)3
(12 , 0
) (12 ,โ
12
) (0, 12
)2
(1โ2, 0) (
0, 1โ2
)1 (0, 1)
y x โ2 โ1 0 1 2 3 4
In the following table, the number in the cell (๐ฅ, ๐ฆ) is ๐ (๐ฅ, ๐ฆ), and an empty cell (๐ฅ, ๐ฆ) means that ๐ (๐ฅ, ๐ฆ) = 0.
4 1/8 1/8 5/8 1/8
3 1/4 1/2 1/4
2 1/2 1/2
1 1
y x โ2 โ1 0 1 2 3 4
(The vectors ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) can be easily computed consecutively using Problem 4.) See also the following figure.
For any positive ๐ฆ we haveโ๐ฅโZ
๐ (๐ฅ, ๐ฆ) = 1. (This is Problem 5.)
For โ4 โค โ๐ฆ < ๐ฅ โค ๐ฆ โค 4 the probability ๐ (๐ฅ, ๐ฆ) equals 0 if and only if one of the numbers ๐ฅ, ๐ฆ is odd and theother one is even. For any ๐ฅ > ๐ฆ and ๐ฅ < โ๐ฆ + 1 we have ๐ (๐ฅ, ๐ฆ) = 0. (In general it is not known, if ๐ (๐ฅ, ๐ฆ) = 0can vanish for ๐ฅ and ๐ฆ of the same parity satisfying โ๐ฆ < ๐ฅ โค ๐ฆ.)
For each odd ๐ฆ we have ๏ฟฝ๏ฟฝ(0, ๐ฆ) = 0. For each nonnegative integer ๐ we have
the vector ๏ฟฝ๏ฟฝ(0, 8๐+ 2) is directed to the right,
the vector ๏ฟฝ๏ฟฝ(0, 8๐+ 4) is directed downwards,
the vector ๏ฟฝ๏ฟฝ(0, 8๐+ 6) is directed to the left,
the vector ๏ฟฝ๏ฟฝ(0, 8๐+ 8) is directed upwards.
Analogously, for each even ๐ฆ we have ๏ฟฝ๏ฟฝ(1, ๐ฆ) = 0. For each nonnegative integer ๐ we have
the vector ๏ฟฝ๏ฟฝ(1, 8๐+ 3) is directed downwards-right,
the vector ๏ฟฝ๏ฟฝ(1, 8๐+ 5) is directed downwards-left,
the vector ๏ฟฝ๏ฟฝ(1, 8๐+ 7) is directed upwards-left,
the vector ๏ฟฝ๏ฟฝ(1, 8๐+ 9) is directed upwards-right.
(See the remark after the answer to Problem 9.)
12
Remark. In the table, 2(๐+๐โ2)/2๏ฟฝ๏ฟฝ(๐โ๐+1, ๐+ ๐โ 1) stands at the intersection of ๐-th column and ๐-th row:
(0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (0, 1) (0, 1)(1, 0) (1,โ1) (1,โ2) (1,โ3) (1,โ4) (1,โ5) (1,โ6) (1,โ7)(1, 0) (0,โ1) (โ1,โ1) (โ2, 0) (โ3, 2) (โ4, 5) (โ5, 9) (โ6, 14)(1, 0) (โ1,โ1) (โ2, 0) (โ2, 2) (โ1, 4) (1, 5) (4, 4) (8, 0)(1, 0) (โ2,โ1) (โ2, 1) (0, 3) (3, 3) (6, 0) (8,โ6) (8,โ14)(1, 0) (โ3,โ1) (โ1, 2) (3, 3) (6, 0) (6,โ6) (2,โ12) (โ6,โ14)(1, 0) (โ4,โ1) (1, 3) (6, 2) (6,โ4) (0,โ10) (โ10,โ10) (โ20, 0)(1, 0) (โ5,โ1) (4, 4) (8, 0) (2,โ8) (โ10,โ10) (โ20, 0) (โ20, 20)
2. Answer: no for both questions. For example, ๐ (2, 4) = 5/8 = 1/8+1/4 = ๐ (2, 4 bypass 2, 2)+๐ (2, 4 bypass 0, 2)and ๐ (0, 4) = 1/8 < 1/4 = ๐ (0, 4 bypass 2, 2).
3. Answer: ๐ (0, 12) = 25/512. Hint: the answer is obtained immediately by means of Remark after the answerto Problem 9 or it can be quickly computed recursively by means of the answer to Problem 4.
4. Answer:
{๐1(๐ฅ, ๐ฆ) = 1โ
2๐2(๐ฅ+ 1, ๐ฆ โ 1) + 1โ
2๐1(๐ฅ+ 1, ๐ฆ โ 1);
๐2(๐ฅ, ๐ฆ) = 1โ2๐2(๐ฅโ 1, ๐ฆ โ 1)โ 1โ
2๐1(๐ฅโ 1, ๐ฆ โ 1).
Solution. Let us derive the formula for ๐2(๐ฅ, ๐ฆ); the derivation for ๐1(๐ฅ, ๐ฆ) is analogous. Consider anypath ๐ from (0, 0) to (๐ฅ, ๐ฆ). Denote by ๐ก(๐ ) the number of turnings in ๐ . Denote ๏ฟฝ๏ฟฝ(๐ ) = (๐1(๐ ), ๐2(๐ )). Noticethat the component ๐2(๐ ) = 0 if and only if ๐ก(๐ ) is even and ๐1(๐ ) = 0 if and only if ๐ก(๐ ) is odd. Therefore,๐2(๐ฅ, ๐ฆ) =
โ๐ :๐ก(๐ ) even
๐2(๐ ) and ๐1(๐ฅ, ๐ฆ) =โ
๐ :๐ก(๐ ) odd
๐1(๐ ).
The last move in the path ๐ is made either from (๐ฅโ 1, ๐ฆ) or from (๐ฅ+ 1, ๐ฆ). It is obvious that if ๐ก(๐ ) is even,then the last move is directed upwards-right, else it is directed upwards-left. Since we are interested in ๐2(๐ฅ, ๐ฆ),assume that last move is directed upwards-right.
Denote by ๐ โฒ the path ๐ without the last move. If the directions of the last moves in ๐ and ๐ โฒ coincide, then๏ฟฝ๏ฟฝ(๐ ) = 1โ
2๏ฟฝ๏ฟฝ(๐ โฒ), otherwise ๏ฟฝ๏ฟฝ(๐ ) = 1โ
2(๐2(๐
โฒ),โ๐1(๐ โฒ)). Therefore,
๐2(๐ฅ, ๐ฆ) =โ
๐ :๐ก(๐ ) even
๐2(๐ ) =1โ2
โโ โ๐ โฒ:๐ก(๐ โฒ) even
๐2(๐ โฒ)โ
โ๐ โฒ:๐ก(๐ โฒ) odd
๐1(๐ โฒ)
โโ =1โ2(๐2(๐ฅโ 1, ๐ฆ โ 1)โ ๐1(๐ฅโ 1, ๐ฆ โ 1)).
5. Let us prove by induction over ๐ฆ thatโ๐ฅโZ
๐ (๐ฅ, ๐ฆ) = 1 for all ๐ฆ โฅ 1: Obviously,โ๐ฅโZ
๐ (๐ฅ, 1) = 1. The step of
induction follows immediately from the following computation:โ๐ฅโZ
๐ (๐ฅ, ๐ฆ + 1) =โ๐ฅโZ
[๐1(๐ฅ, ๐ฆ + 1)2 + ๐2(๐ฅ, ๐ฆ + 1)2
]=
โ๐ฅโZ
๐1(๐ฅ, ๐ฆ + 1)2 +โ๐ฅโZ
๐2(๐ฅ, ๐ฆ + 1)2 =
=1
2
โ๐ฅโZ
(๐1(๐ฅ+ 1, ๐ฆ) + ๐2(๐ฅ+ 1, ๐ฆ))2 +1
2
โ๐ฅโZ
(๐2(๐ฅโ 1, ๐ฆ)โ ๐1(๐ฅโ 1, ๐ฆ))2 =
=1
2
โ๐ฅโZ
(๐1(๐ฅ, ๐ฆ) + ๐2(๐ฅ, ๐ฆ))2 +
1
2
โ๐ฅโZ
(๐2(๐ฅ, ๐ฆ)โ ๐1(๐ฅ, ๐ฆ))2 =
โ๐ฅโZ
[๐1(๐ฅ, ๐ฆ)
2 + ๐2(๐ฅ, ๐ฆ)2]=
โ๐ฅโZ
๐ (๐ฅ, ๐ฆ).
A generalization of conservation law by Gleb Minaev and Ivan Russkikh, participants of Summer conference of
Tournament of towns. Perform the change coordinates (๐ฅ, ๐ฆ) โฆโ (๐ก, ๐ข) = (๐ฅ+๐ฆ2 โ 1, ๐ฆโ๐ฅ
2 ), i.e., rotate the coordinatesystem through 45 degrees clockwise and shift it by the vector (โ1, 0).
Denote ๏ฟฝ๏ฟฝ(๐ก, ๐ข) := ๏ฟฝ๏ฟฝ(๐กโ ๐ข+ 1, ๐ก+ ๐ข+ 1), ๐(๐ก, ๐ข) := ๐ (๐กโ ๐ข+ 1, ๐ก+ ๐ข+ 1), ๏ฟฝ๏ฟฝ(๐ก, ๐ข) := (1 +๐2๐2)(๐ก+๐ข)/2 ๏ฟฝ๏ฟฝ(๐ก, ๐ข).The coordinates of the vectors ๏ฟฝ๏ฟฝ(๐ก, ๐ข) and ๏ฟฝ๏ฟฝ(๐ก, ๐ข) denote by ๐1(๐ก, ๐ข), ๐2(๐ก, ๐ข), ๐ต1(๐ก, ๐ข), ๐ต2(๐ก, ๐ข).
Remark. In the new coordinate system, a checker moves between neighbouring points of the integer latticerather than along the diagonals between black squares. Also we suppose that we start at (0, 0) and move to anypoint of (Nโช{0})2, and the additional โpre-moveโ from (โ1, 0) to (0, 0) is taken into account only to compute thenumber of turns.
For a subset ๐ โ (N โช {0})2 denote ๏ฟฝ๏ฟฝ(๐ก, ๐ข bypass ๐) :=โ
๐ ๏ฟฝ๏ฟฝ(๐ ), where we sum is over all paths ๐ from
(0, 0) to (๐ก, ๐ข), which bypass the points of the set ๐ . Analogously, define ๐(๐ก, ๐ข bypass ๐), ๏ฟฝ๏ฟฝ(๐ก, ๐ข bypass ๐,ยฑ),๐(๐ก, ๐ข bypass ๐,ยฑ). Also for a set ๐ denote
๐(๐) :=โ๐โ๐
๐(๐ bypass ๐ โ {๐}).
13
Remark. Note that for an infinite set ๐ the sum becomes infinite as well. The order of a summation isirrelevant because all the summands are positive.
Theorem 1. For each finite set ๐ โ (N โช {0})2 such that there are no infinite paths from (0; 0) bypassing the
points of ๐ we have ๐(๐) = 1.
Proof. Prove the theorem by induction over๐ := max(๐ก,๐ข)โ๐ (๐ก+๐ข), i.e., the maximal number of a downwards-rightdiagonal containing at least one point of the set ๐ . The diagonal is called maximal.
Base: ๐ = 0. In this case ๐ = (0, 0) and ๐(๐) = ๐(0, 0) = 1.Step. Let us prove a lemma.
Lemma 1. If a set ๐ด โ (Nโฉ{0})2, which does not contain the points (๐ก, ๐ข), (๐ก, ๐ข+1), (๐ก+1, ๐ข) โ (Nโฉ{0})2 , then
๐(๐ก, ๐ข bypass ๐ด) = ๐(๐ก, ๐ข+ 1 bypass ๐ด,โ) +๐(๐ก+ 1, ๐ข bypass ๐ด,+).
Proof. This is straightforward:
๐(๐ก, ๐ข+ 1 bypass ๐ด,โ) +๐(๐ก+ 1, ๐ข bypass ๐ด,+) = ๐1(๐ก, ๐ข+ 1 bypass ๐ด)2 + ๐2(๐ก+ 1, ๐ข bypass ๐ด)2 =
=(๐1(๐ก, ๐ข bypass ๐ด) + ๐2(๐ก, ๐ข bypass ๐ด))2 + (๐2(๐ก, ๐ข bypass ๐ด)โ ๐1(๐ก, ๐ข bypass ๐ด))2
2=
= ๐1(๐ก, ๐ข bypass ๐ด)2 + ๐2(๐ก, ๐ข bypass ๐ด)2 = ๐(๐ก, ๐ข bypass ๐ด).
Suppose that the point (๐ก, ๐ข) โ ๐ is such that ๐ก+ ๐ข is maximal. Suppose that there is a checker path startingat the point (0, 0) and ending at (๐ก, ๐ข) with the last move, say, in the upwards direction bypassing all the pointsof the set ๐ โ {(๐ก, ๐ข)}. Then there exists a path to the point (๐ก, ๐ข โ 1), bypassing all the points of the set๐ โ {(๐ก, ๐ข โ 1)}. Notice that in this case (๐ก, ๐ข โ 1) /โ ๐ , because the move to (๐ก;๐ข) is in the upwards direction.Hence there exists a path to the point (๐ก + 1, ๐ข โ 1), bypassing all the points of the set ๐ โ {(๐ก + 1, ๐ข โ 1)}. If(๐ก + 1, ๐ข โ 1) /โ ๐ then there exists a path going to infinity passing through (๐ก + 1, ๐ข โ 1) and bypassing all thepoints of the set ๐ . For example, the path turning right at the point (๐ก, ๐ข) and only going right from there passesthrough (๐ก, ๐ข) and bypasses all the points in ๐ because the diagonal is maximal. Therefore (๐ก + 1, ๐ข โ 1) โ ๐ .Notice that ๐(๐, ๐ bypass ๐ด) = ๐(๐, ๐ bypass ๐ด,+) + ๐(๐, ๐ bypass ๐ด,โ) for any set ๐ด โ (N โฉ {0})2. Denote๐พ := ๐ โ{(๐ก, ๐ข); (๐ก+1, ๐ขโ1)}. Then we have the following chain of equalities (where we use 1 for the set ๐ด = ๐พ):
๐(๐) = ๐(๐พ) +๐(๐ก, ๐ข bypass ๐ โ {(๐ก, ๐ข)}) +๐(๐ก+ 1, ๐ขโ 1 bypass ๐ โ {(๐ก+ 1, ๐ขโ 1)}) == ๐(๐พ) +๐(๐ก, ๐ข bypass ๐ โ {(๐ก, ๐ข)},+) +๐(๐ก, ๐ข bypass ๐ โ {(๐ก, ๐ข)},โ)+
+๐(๐ก+ 1, ๐ขโ 1 bypass ๐ โ {(๐ก+ 1, ๐ขโ 1)},+) +๐(๐ก+ 1, ๐ขโ 1 bypass ๐ โ {(๐ก+ 1, ๐ขโ 1)},โ) =
= ๐(๐พ) +๐(๐ก, ๐ข bypass ๐ โ {(๐ก, ๐ข)},+)+
+๐(๐ก, ๐ขโ 1 bypass ๐ โ {(๐ก, ๐ขโ 1)}) +๐(๐ก+ 1, ๐ขโ 1 bypass ๐ โ {(๐ก+ 1, ๐ขโ 1)},โ) = ๐(๐ โช {(๐ก, ๐ขโ 1)}).
Thus if the point (๐ก, ๐ขโ 1) is added to the set ๐ then the probability ๐(๐) is not changed.This way we put new points onto the diagonal ๐ก+ ๐ข = ๐โ 1, therefore, the maximal diagonal is not changed.
Thus if we add a few points to the set ๐ , then the paths bypassing other points of the set ๐ bypass the points of๐ in the maximal diagonal as well. Therefore, if we remove all the points from the maximal diagonal, then ๐(๐)is not changed and no infinite path bypassing the points of the set ๐ appears. This way we change ๐ keeping๐(๐) fixed but decreasing the number of a maximal diagonal. By the inductive hypothesis the new ๐(๐) equals1, hence the old one also equals 1.
6. Answer. For each ๐ฅ we have ๐1(๐ฅ, ๐ฆ) = ๐1(โ๐ฅ, ๐ฆ) and ๐2(๐ฅ, ๐ฆ) + ๐1(๐ฅ, ๐ฆ) = ๐2(2โ ๐ฅ, ๐ฆ) + ๐1(2โ ๐ฅ, ๐ฆ).Solution. For each path ๐ denote by ๐(๐ ) the reflection of ๐ with respect to the ๐ฆ axis. Then if ๐ is a path to
(๐ฅ, ๐ฆ), then ๐(๐ ) is a path to (โ๐ฅ, ๐ฆ).For each path ๐ denote by ๐(๐ ) the path consisting of the same moves as ๐ , but in the opposite order. Notice
that reordering of moves does not affect the endpoint.Now consider a path ๐ to (๐ฅ, ๐ฆ) such that ๐ก(๐ ) is odd. Then the last move in ๐ is upwards-left. Therefore,
the last move in ๐(๐ ) is upwards-right, hence the first move in ๐(๐(๐ )) is upwards-right. Thus ๐ โ ๐ is a bijectionbetween paths to (๐ฅ, ๐ฆ) with odd number of tunings beginning with an upwards-right move and paths to (โ๐ฅ, ๐ฆ)with odd number of tunings beginning with an upwards-right move.
To prove the second equation use the result of Problem 4:
๐1(๐ฅ, ๐ฆ) + ๐2(๐ฅ, ๐ฆ) =โ2 ยท ๐1(๐ฅโ 1, ๐ฆ + 1) =
โ2 ยท ๐1(1โ ๐ฅ, ๐ฆ + 1) = ๐1(2โ ๐ฅ, ๐ฆ) + ๐2(2โ ๐ฅ, ๐ฆ).
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Remark. These identities can also be proved simultaneously by induction over ๐ฆ using Problem 4. AlsoProblem 9 implies another identity (๐ฆ โ ๐ฅ) ๐2(๐ฅ, ๐ฆ) = (๐ฆ + ๐ฅโ 2) ๐2(2โ ๐ฅ, ๐ฆ).
7. Answer : for each 0 < ๐ฆโฒ < ๐ฆ we have
๐1(๐ฅ, ๐ฆ) =โ๐ฅโฒ
[๐2(๐ฅ
โฒ, ๐ฆโฒ)๐1(๐ฅโ ๐ฅโฒ + 1, ๐ฆ โ ๐ฆโฒ + 1) + ๐1(๐ฅโฒ, ๐ฆโฒ)๐2(๐ฅ
โฒ โ ๐ฅ+ 1, ๐ฆ โ ๐ฆโฒ + 1)],
๐2(๐ฅ, ๐ฆ) =โ๐ฅโฒ
[๐2(๐ฅ
โฒ, ๐ฆโฒ)๐2(๐ฅโ ๐ฅโฒ + 1, ๐ฆ โ ๐ฆโฒ + 1)โ ๐1(๐ฅโฒ, ๐ฆโฒ)๐1(๐ฅ
โฒ โ ๐ฅ+ 1, ๐ฆ โ ๐ฆโฒ + 1)].
The required expression for ๏ฟฝ๏ฟฝ(๐ฅ, 199) is obtained by taking ๐ฆ = 199 and ๐ฆโฒ = 100.Solution. Fix any positive integer ๐ฆโฒ < ๐ฆ. Consider a path ๐ from (0, 0) to (๐ฅ, ๐ฆ). Denote by (๐ฅโฒ, ๐ฆโฒ) the square
at which ๐ intersects the line ๐ฆ = ๐ฆโฒ. Denote by ๐ 1 the part of ๐ that joins (0, 0) with (๐ฅโฒ, ๐ฆโฒ) and by ๐ 2 the partstarting at the intersection square of ๐ with the line ๐ฆ = ๐ฆโฒ โ 1 and ending at (๐ฅ, ๐ฆ). Translate the path ๐ 2 so thatit starts at (0, 0).
Denote ๏ฟฝ๏ฟฝ(๐ ) = (๐1(๐ ), ๐2(๐ )). Then
๐1(๐ ) =
{๐1(๐ 1)๐2(๐ 2) if the move to (๐ฅโฒ, ๐ฆโฒ) is upwards-left,
๐2(๐ 1)๐1(๐ 2) if the move to (๐ฅโฒ, ๐ฆโฒ) is upwards-right,
๐2(๐ ) =
{โ๐1(๐ 1)๐1(๐ 2), if the move to (๐ฅโฒ, ๐ฆโฒ) is upwards-left,
๐2(๐ 1)๐2(๐ 2), if the move to (๐ฅโฒ, ๐ฆโฒ) is upwards-right.
Therefore,
๐1(๐ฅ, ๐ฆ) =โ๐
๐1(๐ ) =โ๐ฅโฒ
โ๐ through (๐ฅโฒ,๐ฆโฒ)
๐1(๐ ) =โ๐ฅโฒ
โกโขโขโขโขโขโฃโ
๐ through (๐ฅโฒ,๐ฆโฒ),move to (๐ฅโฒ,๐ฆโฒ)is upwards-right
๐2(๐ 1)๐1(๐ 2) +โ
๐ through(๐ฅโฒ,๐ฆโฒ),move to (๐ฅโฒ,๐ฆโฒ)is upwards-left
๐1(๐ 1)๐2(๐ 2)
โคโฅโฅโฅโฅโฅโฆ =
=โ๐ฅโฒ
[๐2(๐ฅ
โฒ, ๐ฆโฒ)๐1(๐ฅโ ๐ฅโฒ + 1, ๐ฆ โ ๐ฆโฒ + 1) + ๐1(๐ฅโฒ, ๐ฆโฒ)๐2(๐ฅ
โฒ โ ๐ฅ+ 1, ๐ฆ โ ๐ฆโฒ + 1)].
The formula for ๐2(๐ฅ, ๐ฆ) is proven analogously.Remark. Formulae illustrating the Huygens principle have the form of a convolution. In combinatorics the
convolution of two given sequences {๐๐} and {๐๐} (๐, ๐ โฅ 0) is the sequence {๐๐} (๐ โฅ 0), defined by the equality๐๐ =
โ๐+๐=๐ ๐๐๐๐. A convolution of two sequences corresponds to the product of their generating functions: if
๐ด(๐ก) =
โโ๐=0
๐๐๐ก๐, ๐ต(๐ก) =
โโ๐=0
๐๐๐ก๐, ๐ถ(๐ก) =
โโ๐=0
๐๐๐ก๐,
then ๐ถ(๐ก) = ๐ด(๐ก)๐ต(๐ก). Therefore, choosing the generating functions in a right way, we can write down the Huygensprinciple in a more compact form.
Define two series of Laurent polynomials (i.e., polynomials in variables ๐ก and ๐กโ1):
๐๐ = ๐๐(๐ก) =
๐โ๐=โ๐+2
๐1(๐,๐)๐ก๐, ๐๐ = ๐๐(๐ก) =
๐โ๐=โ๐+2
๐2(๐,๐)๐ก๐.
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These polynomials can be considered as generating fuctions of the sequences ๐1,2(๐,๐) with fixed ๐. From theanswer to the Problems 6 and 7 we have
๐2(๐ฅ, ๐ฆ) =โ๐ฅโฒ
[๐2(๐ฅ
โฒ, ๐ฆโฒ)๐2(๐ฅโ ๐ฅโฒ + 1, ๐ฆ โ ๐ฆโฒ + 1)โ ๐1(๐ฅโฒ, ๐ฆโฒ)๐1(๐ฅ
โฒ โ ๐ฅ+ 1, ๐ฆ โ ๐ฆโฒ + 1)]=
=โ๐ฅโฒ
[๐2(๐ฅ
โฒ, ๐ฆโฒ)๐2(๐ฅโ ๐ฅโฒ + 1, ๐ฆ โ ๐ฆโฒ + 1)โ ๐1(๐ฅโฒ, ๐ฆโฒ)๐1(๐ฅโ ๐ฅโฒ โ 1, ๐ฆ โ ๐ฆโฒ + 1)
].
This formula is equivalent to the following addition theorem for ๐โpolynomials:
๐๐+๐ =1
๐ก๐๐๐๐+1 โ ๐ก๐๐๐๐+1.
Also from the Problems 6 and 7 we have
๐1(๐ฅ, ๐ฆ) =โ๐ฅโฒ
[๐2(๐ฅ
โฒ, ๐ฆโฒ)๐1(๐ฅโ ๐ฅโฒ + 1, ๐ฆ โ ๐ฆโฒ + 1) + ๐1(๐ฅโฒ, ๐ฆโฒ)๐2(๐ฅ
โฒ โ ๐ฅ+ 1, ๐ฆ โ ๐ฆโฒ + 1)]=
=โ๐ฅโฒ
[๐2(๐ฅโฒ, ๐ฆโฒ)๐1(๐ฅโ ๐ฅโฒ + 1, ๐ฆ โ ๐ฆโฒ + 1)+
+ ๐1(๐ฅโฒ, ๐ฆโฒ)(๐2(๐ฅโ ๐ฅโฒ + 1, ๐ฆ โ ๐ฆโฒ + 1) + ๐1(๐ฅโ ๐ฅโฒ + 1, ๐ฆ โ ๐ฆโฒ + 1)โ ๐1(๐ฅโ ๐ฅโฒ โ 1, ๐ฆ โ ๐ฆโฒ + 1))].
This formula respectively is equivalent to the following addition theorem for ๐โpolynomials:
๐๐+๐ =1
๐ก(๐๐๐๐+1 + ๐๐๐๐+1) +
(1
๐กโ ๐ก
)๐๐๐๐+1.
In particular substitution of ๐ = 1 gives Diracโs equation written in terms of ๐โ and ๐โpolynomials:{๐๐+1 = 1
๐ก๐๐๐2 + ๐๐ ยท (1๐ก๐2 +1๐ก๐2 โ ๐ก๐2) =
1๐กโ2(๐๐ + ๐๐),
๐๐+1 = 1๐ก๐๐๐2 โ ๐ก๐๐๐2 =
๐กโ2(๐๐ โ ๐๐).
It follows immediately that{๐กโ2๐๐+1 โ๐๐ = ๐๐ = ๐๐ โ
โ2๐ก ๐๐+1
๐กโ2๐๐+1 โ ๐๐ = ๐๐ = ๐๐ +
โ2๐ก ๐๐+1
โ
{๐๐+1 =
โ2๐ก ๐๐ โ ๐๐+1
๐ก2
๐๐+1 = ๐ก2๐๐+1 โ ๐กโ2๐๐
โ
{๐๐+1 = 1โ
2(๐ก+ 1
๐ก )๐๐ โ ๐๐โ1
๐๐+1 = 1โ2(๐ก+ 1
๐ก )๐๐ โ๐๐โ1.
The latter formulae give us yet another recursive definitions for ๐๐ and ๐๐.
8. Answer : the graphs of ๐ (๐ฅ, 1000), ๐1(๐ฅ, 1000), and ๐2(๐ฅ, 1000) respectively as functions in ๐ฅ:
-1000 -500 500 1000
0.0005
0.0010
0.0015
0.0020
0.0025
0.0030
-1000 -500 500 1000
-0.06
-0.04
-0.02
0.02
0.04
0.06
-1000 -500 500 1000
-0.05
0.05
In the above graph of ๐ (๐ฅ, 1000), large values are cut out, to make the oscillations more visible. Compare withthe following graphs of ๐ (๐ฅ, 1000) and ๐ (๐ฅ, ๐ฆ) without cutting (kindly provided by A. Daniyarkhodzhaev andF. Kuyanov, participants of the Summer School in Contemporary Mathematics in Dubna, 2019):
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9. Answer 1: for each ๐ฆ > |๐ฅ| we have
๐1(๐ฅ, ๐ฆ) = 2(1โ๐ฆ)/2โ
๐คโก2๐ฆโ๐ฅโ2
2
(โ1)๐ฆโ๐ฅโ2+2๐คยทsign(๐ฅ)
4
(|๐ฅ|๐ค
)( ๐ฆโ|๐ฅ|โ22
๐ฆโ๐ฅโ2โ2๐ค4
),
๐2(๐ฅ, ๐ฆ) = 2(1โ๐ฆ)/2โ
๐คโก2๐ฆโ๐ฅ2
(โ1)๐ฆโ๐ฅ+2๐คยทsign(๐ฅ)
4
(|๐ฅ|๐ค
)( ๐ฆโ|๐ฅ|โ22
๐ฆโ๐ฅโ2๐ค4
);
for each ๐ฆ = ๐ฅ > 0 we have
๐1(๐ฆ, ๐ฆ) = 0,
๐2(๐ฆ, ๐ฆ) = 2(1โ๐ฆ)/2;
for each 0 < ๐ฆ < |๐ฅ| or ๐ฆ = โ๐ฅ > 0 we have
๐1(๐ฅ, ๐ฆ) = 0,
๐2(๐ฅ, ๐ฆ) = 0.
Answer 2: for each ๐ฆ > |๐ฅ| we have
๐1(๐ฅ, ๐ฆ) = 2(1โ๐ฆ)/2
โ๐ฆ/2โโ๐=0
(โ1)๐((๐ฅ+ ๐ฆ โ 2)/2
๐
)((๐ฆ โ ๐ฅโ 2)/2
๐
),
๐2(๐ฅ, ๐ฆ) = 2(1โ๐ฆ)/2
โ๐ฆ/2โโ๐=1
(โ1)๐((๐ฅ+ ๐ฆ โ 2)/2
๐
)((๐ฆ โ ๐ฅโ 2)/2
๐ โ 1
);
For other ๐ฆ the answer is the same as Answer 1.Answer 3. For even ๐+ ๐ and 2โ ๐ โค ๐ โค ๐ we have25
๐1(๐, ๐) = (โ1)๐2(๐โ3)/2 ยท๐โ2โ
๐=๐+|๐|โ2
2
(โ2)โ๐ ยท ๐!
(๐ โ 2โ ๐)!(๐โ๐2 + 1 + ๐)!(๐ โ ๐+๐
2 + 1)!
๐2(๐, ๐) = 2(1โ๐)/2 ยท(๐ โ 1๐โ๐2
)โ (โ1)๐2(๐โ3)/2
๐โ2โ๐=
๐+|๐|โ22
(โ2)โ๐ ยท ๐!(๐ + ๐+ 1)
(๐ โ 1โ ๐)!(๐โ๐2 + 1 + ๐)!(๐ โ ๐+๐
2 + 1)!
Remark. Answer 2 shows that ๐1(โ๐ฆ+2๐, ๐ฆ) and ๐2(โ๐ฆ+2๐, ๐ฆ) are the coefficients before ๐ก๐ฆโ๐โ1 and ๐ก๐ฆโ๐ inthe expansion of 2(1โ๐ฆ)/2(1 + ๐ก)๐ฆโ๐โ1(1โ ๐ก)๐โ1 for 1 โค ๐ โค ๐ฆ โ 1. In particular,
๐1(0, 4๐+ 2) =(โ1)๐
2(4๐+1)/2
(2๐
๐
),
๐2(0, 4๐+ 2) = 0,
and
๐1(0, 4๐) = 0,
๐2(0, 4๐) =(โ1)๐
2(4๐โ1)/2
(2๐โ 1
๐
).
The sums in the expressions for ๐1(๐ฅ, ๐ฆ) and ๐2(๐ฅ, ๐ฆ) for ๐ฆ > |๐ฅ| are particular cases of well-known Gaussianhypergeometric functions (but familiarity with them is not required for what follows):
๐1(๐ฅ, ๐ฆ) = 2(1โ๐ฆ)/22๐น1
(1โ ๐ฅ+ ๐ฆ
2, 1 +
๐ฅโ ๐ฆ
2, 1;โ1
),
๐2(๐ฅ, ๐ฆ) = 2(1โ๐ฆ)/2
(1โ ๐ฅ+ ๐ฆ
2
)2๐น1
(2โ ๐ฅ+ ๐ฆ
2, 1 +
๐ฅโ ๐ฆ
2, 2;โ1
).
Hint. Let us find ๐1(๐ฅ, ๐ฆ) for ๐ฆ > |๐ฅ|. Consider a path from the square (0, 0) to the square (๐ฅ, ๐ฆ) with anodd number of turns (the others do not contribute to ๐1(๐ฅ, ๐ฆ)). Denote by ๐ and ๐ฟ the number of upwards-right
25Fractions in those equalities are respectively(๐
๐ โ 2โ ๐, ๐โ๐2 + 1 + ๐, ๐ โ ๐+๐
2 + 1
)and
(๐
๐ โ 2โ ๐, ๐โ๐2 + 1 + ๐, ๐ โ ๐+๐
2 + 1
)ยท ๐ + ๐+ 1
๐ โ 1โ ๐.
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and upwards-left moves respectively. These numbers are uniquely determined by the conditions ๐ฟ + ๐ = ๐ฆ and๐ โ ๐ฟ = ๐ฅ. Denote by 2๐ + 1 the number of turns in the path. Let ๐ฅ1, ๐ฅ2, . . . , ๐ฅ๐+1 be the number of consecutiveupwards-right moves before the first, the third, . . . , the last turn respectively. Let ๐ฆ1, ๐ฆ2, . . . , ๐ฆ๐+1 be the numberof consecutive upwards-left moves after the first, the third, . . . , the last turn respectively. Then ๐ฅ๐, ๐ฆ๐ โฅ 1 and
๐ = ๐ฅ1 + ยท ยท ยท+ ๐ฅ๐+1;
๐ฟ = ๐ฆ1 + ยท ยท ยท+ ๐ฆ๐+1.
Conversely, each collection (๐ฅ1, ๐ฅ2, . . . , ๐ฅ๐+1, ๐ฆ1, ๐ฆ2, . . . , ๐ฆ๐+1) satisfying the resulting equations determines a pathfrom (0, 0) to (๐ฅ, ๐ฆ) with an odd number of turns.
The problem now reduces to a combinatorial one: find the number of positive integer solutions of the resultingequations. For the first equation, this number equals to the number of ways to put ๐ sticks between ๐ coins in arow, that is,
(๐ โ1๐
). Thus
๐1(๐ฅ, ๐ฆ) = 2(1โ๐ฆ)/2
โ๐ฆ/2โโ๐=0
(โ1)๐(๐ โ 1
๐
)(๐ฟโ 1
๐
).
Since ๐ฟ+๐ = ๐ฆ and ๐ โ ๐ฟ = ๐ฅ, the required formula from Answer 2 follows.The formula for ๐1(๐ฅ, ๐ฆ) from Answer 1 can be derived using the polynomial from the remark after the answer.
Depending on the sign of ๐ฅ we get
2(1โ๐ฆ)/2(1 + ๐ก)๐ฆโ๐โ1(1โ ๐ก)๐โ1 =
{2(1โ๐ฆ)/2(1โ ๐ก2)
๐ฆโ๐ฅโ22 (1โ ๐ก)๐ฅ, for ๐ฅ โฅ 0;
2(1โ๐ฆ)/2(1โ ๐ก2)๐ฆ+๐ฅโ2
2 (1 + ๐ก)โ๐ฅ, for ๐ฅ < 0.
For each |๐ฅ| < ๐ฆ the numbers ๐1(๐ฅ, ๐ฆ) and ๐2(๐ฅ, ๐ฆ) are equal to the coefficients before ๐ก๐ฅโ๐ฆโ2
2 and ๐ก๐ฅโ๐ฆ2 in the
expansion of the polynomials. Considering the cases ๐ฅ โฅ 0 and ๐ฅ < 0 separately we get the required formula.The proof of the formulae for ๐2(๐ฅ, ๐ฆ) for ๐ฆ > |๐ฅ| is analogous. The case ๐ฆ โค |๐ฅ| is obvious.
Solution by Alexander Kudryavtsev (participant of Summer conference of Tournament of Towns). Denote๐๐,๐ = ๐1(2๐ โ ๐ + 1, ๐ + 1) ยท 2๐/2, ๐๐,๐ = ๐2(2๐ โ ๐ + 1, ๐ + 1) ยท 2๐/2 (they are well-defined for 0 โค ๐ โค ๐ โ 1). Let๐ถ(๐ฅ, ๐ฆ), ๐ท(๐ฅ, ๐ฆ) be generating functions of ๐๐,๐ and ๐๐,๐ respectively. From the answer to Problem 4 we obtain thefollowing equations: โงโชโจโชโฉ
๐ถ(๐ฅ, ๐ฆ)โ ๐ถ(๐ฅ, 0)
๐ฆ= ๐ท(๐ฅ, ๐ฆ) + ๐ถ(๐ฅ, ๐ฆ)
๐ท(๐ฅ, ๐ฆ)โ๐ท(๐ฅ, 0)
๐ฆ= ๐ฅ ยท (๐ท(๐ฅ, ๐ฆ)โ ๐ถ(๐ฅ, ๐ฆ))
Since ๐ถ(๐ฅ, 0) = 0 and ๐ท(๐ฅ, 0) = 1, we get{๐ถ(๐ฅ, ๐ฆ) = ๐ฆ ยท (๐ท(๐ฅ, ๐ฆ) + ๐ถ(๐ฅ, ๐ฆ))
๐ท(๐ฅ, ๐ฆ)โ 1 = ๐ฅ๐ฆ ยท (๐ท(๐ฅ, ๐ฆ)โ ๐ถ(๐ฅ, ๐ฆ))
The solution of this system is ๐ถ(๐ฅ, ๐ฆ) =๐ฆ
1โ ๐ฆ โ ๐ฅ๐ฆ + 2๐ฅ๐ฆ2and ๐ท(๐ฅ, ๐ฆ) =
1โ ๐ฆ
1โ ๐ฆ โ ๐ฅ๐ฆ + 2๐ฅ๐ฆ2. Denote
๐ธ(๐ฅ, ๐ฆ) :=๐ถ(๐ฅ, ๐ฆ)
๐ฆ=
๐ท(๐ฅ, ๐ฆ)
1โ ๐ฆ= 1 + ๐ฆ(1 + ๐ฅโ 2๐ฅ๐ฆ) + ๐ฆ2(1 + ๐ฅโ 2๐ฅ๐ฆ)2 + . . .
The coefficient of ๐ฅ๐ ยท ๐ฆ๐ in the ๐ธ(๐ฅ, ๐ฆ) is equal to
๐โ๐=max(๐,๐โ๐)
(โ2)๐โ๐ ยท ๐!
(๐โ ๐)!(๐โ๐+ ๐)!(๐ โ ๐)!,
since from every summand of the form ๐ฆ๐(1 + ๐ฅโ 2๐ฅ๐ฆ)๐ we must take exactly one combination of factors:
โ for the power of ๐ฆ to be equal to ๐, the number of factors โ2๐ฅ๐ฆ must be exactly ๐โ ๐;
โ for the power of ๐ฅ to be equal to ๐, the number of factors ๐ฅ must be exactly ๐โ (๐โ ๐);
โ for the total number of factors to be ๐, the number of factors 1 must be ๐โ (๐โ (๐โ๐))โ (๐โ๐) = ๐โ๐.
18
And the number of ways to choose ๐ โ ๐, ๐ โ ๐ and ๐ โ (๐ โ ๐) factors (1 + ๐ฅ โ 2๐ฅ๐ฆ) equals to the fraction inthe formula for ๐ธ(๐ฅ, ๐ฆ).
Now Answer 3 can be easily obtained from the equations ๐ถ(๐ฅ, ๐ฆ) = ๐ฆ ยท๐ธ(๐ฅ, ๐ฆ) and ๐ท(๐ฅ, ๐ฆ) = ๐ธ(๐ฅ, ๐ฆ)โ๐ถ(๐ฅ, ๐ฆ).
10. Answer:
๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ) =
โ2(sin
(๐๐ฆ4 โ ๐ฅ2
2๐ฆ
), ๐๐ฅ/๐ฆ cos
(๐๐ฆ4 + ๐ฅ(2โ๐ฅ)
2๐ฆ
))โ๐๐ฆ
+O
(log2 ๐ฆ
๐ฆ
)for |๐ฅ| = O(
โ๐ฆ) .
Remark. Compare the graphs of ๐ (๐ฅ, 400), ๐1(๐ฅ, 400), and ๐2(๐ฅ, 400) (blue) with the graphs of their approxi-mations in the right-hand side for ๐ฆ = 400 (orange):
-400 -200 200 400
0.001
0.002
0.003
0.004
0.005
0.006
0.007
-400 -200 200 400
-0.05
0.05
0.10
-400 -200 200 400
-0.10
-0.05
0.05
0.10
Hint. Note that by Problem 9
๐2(โ๐+ 2๐ + 1, ๐+ 1) = 2โ๐/2[๐ฅ๐โ1](โ1)๐(1 + ๐ฅ)๐โ๐โ1(1โ ๐ฅ)๐;
๐1(โ๐+ 2๐ + 1, ๐+ 1) = 2โ๐/2[๐ฅ๐](โ1)๐(1 + ๐ฅ)๐โ๐โ1(1โ ๐ฅ)๐.
Thus
๐2(โ๐+ 2๐ + 1, ๐+ 1) = 2โ๐/2(โ1)๐โซ 1
0(1 + ๐2๐๐๐ก)๐โ๐โ1(1โ ๐2๐๐๐ก)๐๐โ2๐๐(๐โ1)๐ก๐๐ก.
For an odd ๐ denote ๐ = ๐ โ ๐โ12 . We have
๐2(โ๐+ 2๐ + 1, ๐+ 1) =๐๐โ2
โซ 1
0(sin(2๐๐ก))
๐โ12 (tan(๐๐ก))๐๐โ2๐๐(๐โ1)๐ก๐๐ก.
Since ๐ << ๐, we can calculate the integral in the areas (1/4โ ๐ฟ, 1/4 + ๐ฟ), (3/4โ ๐ฟ, 3/4 + ๐ฟ) with good accuracy.Denote ๐ โช
โ๐ log ๐, |๐| โค ๐.
In (1/4โ ๐ฟ, 1/4 + ๐ฟ) we have ๐ก = 1/4 + ๐, |๐ | โค ๐ฟ,
sin(2๐๐ก) = cos(2๐๐) = 1โ 2๐2๐2 +๐(๐4) = ๐โ2๐2๐2 +๐(๐4);
tan(๐๐ก) = 1 + 2๐๐ +๐(๐2) = ๐2๐๐ +๐(๐2);
๐โ2๐๐(๐โ1)๐ก = ๐1โ๐๐โ2๐๐(๐โ1)๐
and we get๐๐+1โ๐
โ2
โซ ๐ฟ
โ๐ฟ(๐โ2๐2๐2 +๐(๐4))
๐โ12 (๐2๐๐ +๐(๐2))๐๐โ2๐๐(๐โ1)๐๐๐. (*)
As ๐๐4 โช 1 and ๐๐2 โช 1, i.e. ๐๐ฟ4 โช 1 and ๐๐ฟ2 โช 1,
(*) = ๐๐+12
โ2
โซ ๐ฟ
โ๐ฟ(๐โ(๐โ1)๐2๐2 +๐(๐4))(๐2๐๐๐ +๐(๐2))๐โ2๐๐(๐โ1)๐๐๐.
If we open the brackets we get the summand ๐(๐๐4๐2๐๐๐ ) = ๐(๐๐ฟ4๐2๐๐๐ฟ) = ๐(๐โฮ), since we ask ๐๐ฟ โช log ๐.Then
(*) = ๐๐+12
โ2
โซ ๐ฟ
โ๐ฟ(๐โ(๐โ1)๐2๐2+2๐๐๐โ2๐๐(๐โ1)๐ +๐(๐โฮ) +๐(๐๐ฟ2))๐๐.
Now estimate the errorโซ โ
๐ฟ๐โ(๐โ1)๐2๐2+2๐๐๐๐๐ =
โซ โ
๐ฟ๐โ(๐โ1)๐2(๐โ ๐
๐(๐โ1))2+ ๐2
๐โ1๐๐ โชโซ โ
๐ฟ๐โ(๐โ1)๐2 ๐2
4+ ๐2
๐โ1๐๐ โค 1โ๐โ 1
๐โ(๐โ1)๐ฟ2 ,
since ๐2 โฅ ๐
๐(๐โ1) , i.e. ๐ฟ โฅ ๐/(๐โ 1).
19
So the necessary condition is ๐๐ฟ2 โซ log ๐. Then we can calculate
(*) = ๐๐+12
โ2
โซ โ
โโ๐โ(๐โ1)๐2๐2+2๐๐๐โ2๐๐(๐โ1)๐๐๐+๐(๐โฮ)+๐(๐๐ฟ3) =
๐๐+12
โ2
โซ โ
โโ๐โ(๐โ1)๐2๐2+2๐๐๐โ2๐๐(๐โ1)๐๐๐+๐(log2 ๐/๐),
since ๐๐ฟ3 โช log ๐ ยท ๐ฟ2 โช log2 ๐/๐.Finally, from the formula
โซโโโ ๐โ๐๐ก2+๐๐ก๐๐ก =
โ๐/๐๐๐
2/4๐ we get
(*) = ๐๐+12โ
2๐(๐โ 1)๐
(2๐โ1)โ2๐(๐โ1)๐โ1 +๐(log2 ๐/๐).
In the area of 3/4 the integral approximately equals ๐3๐+12โ2
๐(2๐โ1)+2๐(๐โ1)
๐โ1 and, since 3๐+12 โก โ๐+1
2 mod 4,
๐2(โ๐+ 2๐ + 1, ๐+ 1) = ๐2๐โ1
โ2
๐(๐โ 1)cos
(๐(๐+ 1)
4โ 2๐2
๐โ 1+โ 2๐
๐โ 1
)+๐(log2 ๐/๐) =
= โ๐2๐โ1
โ2
๐(๐โ 1)sin
(๐(๐โ 1)
4โ 2๐2
๐โ 1+
2๐
๐โ 1
)+๐(log2 ๐/๐).
Analogously,
๐1(โ๐+ 2๐ + 1, ๐+ 1) =
โ2
๐(๐โ 1)cos
(๐(๐โ 1)
4โ 2๐2
๐โ 1
)+๐(log2 ๐/๐).
For an even ๐ the answer is
๐2(โ๐+ 2๐ + 1, ๐+ 1) = ๐2๐
โ2
๐(๐โ 2)cos
(๐(๐+ 1)
4โ 2(๐2 โ 1/4)
๐โ 2
)+๐(log2 ๐/๐);
๐1(โ๐+ 2๐ + 1, ๐+ 1) =
โ2
๐(๐โ 2)sin
(๐(๐+ 1)
4โ 2(๐ + 1/2)2
๐โ 2
)+๐(log2 ๐/๐),
where ๐ = ๐ โ ๐/2.
11. Answer : ๐ (๐ฅ, ๐ฆ) = ๐ (๐ฅ, ๐ฆ,+) + ๐ (๐ฅ, ๐ฆ,โ); ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,+) = (0, ๐2(๐ฅ, ๐ฆ)); ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,โ) = (๐1(๐ฅ, ๐ฆ), 0).Solution. Note that if a path ๐ from (0, 0) to (๐ฅ, ๐ฆ) starts and finishes with an upward-left move, then the checker
changes the direction an odd number of times. Therefore, ๏ฟฝ๏ฟฝ(๐ ) = (ยฑ21โ๐ฆ2 , 0). Analogously, if a path ๐ finishes with
an upward-right move, then ๏ฟฝ๏ฟฝ(๐ ) = (0,ยฑ21โ๐ฆ2 ). Hence, ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,+) = (0, ๐2(๐ฅ, ๐ฆ)) and ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,โ) = (๐1(๐ฅ, ๐ฆ), 0).
For the probability, the following formula comprehensively explains the answer:
๐ (๐ฅ, ๐ฆ) = |๐(๐ฅ, ๐ฆ,+) + ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,โ)|2 = ๐21(๐ฅ, ๐ฆ) + ๐22(๐ฅ, ๐ฆ) = ๐ (๐ฅ, ๐ฆ,+) + ๐ (๐ฅ, ๐ฆ,โ).
12. Answer: for each integer ๐ฆ โฅ 2 we have ๐ (๐ฆ,โ) = 12
โโ๐ฆ/2โโ1๐=0
1(โ4)๐
(2๐๐
). We have also max๐ฆ ๐ (๐ฆ,โ) =
๐ (2,โ) = ๐ (3,โ) = 1/2 and lim๐ฆโโ ๐ (๐ฆ,โ) = 12โ2.
Remark. The graph of the sequence kindly provided by Gleb Minaev and Ivan Russkikh, participants ofSummer Conference of Tournament of Towns:
Solution. Denote
๐1(๐ฆ) =โ๐ฅ
๐21(๐ฅ, ๐ฆ); ๐2(๐ฆ) =โ๐ฅ
๐22(๐ฅ, ๐ฆ); ๐12(๐ฆ) =โ๐ฅ
๐1(๐ฅ, ๐ฆ)๐2(๐ฅ, ๐ฆ).
By Problem 5 for each ๐ฆ โฅ 1 we have๐1(๐ฆ) + ๐2(๐ฆ) = 1.
20
By the answers to Problem 4, Problem 6 and Problem 7 we have
๐1(0, 2๐ฆ) =1โ2
โ๐ฅ
๐1(๐ฅ, ๐ฆ)(๐2(๐ฅ, ๐ฆ)โ ๐1(๐ฅ, ๐ฆ)) + ๐2(๐ฅ, ๐ฆ)(๐2(๐ฅ, ๐ฆ) + ๐1(๐ฅ, ๐ฆ)) =1โ2(๐2(๐ฆ) + 2๐12(๐ฆ)โ ๐1(๐ฆ)).
By definition and the answer to Problem 4 we have
๐1(๐ฆ + 1)โ ๐2(๐ฆ + 1) = 2๐12(๐ฆ).
Hence,๐1(๐ฆ + 1)โ ๐2(๐ฆ + 1) = ๐1(๐ฆ)โ ๐2(๐ฆ) + ๐1(0, 2๐ฆ)
โ2.
Since ๐1(๐ฆ) + ๐2(๐ฆ) = 1, we have the recurrence relation ๐1(๐ฆ + 1) = ๐1(๐ฆ) +1โ2๐1(0, 2๐ฆ).
The remark from the solution of Problem 9 implies by induction that
๐1(๐ฆ) =1
2
โ๐ฆ/2โโ1โ๐=0
1
(โ4)๐
(2๐
๐
).
It only remains to recall that ๐ (๐ฆ,โ) = ๐1(๐ฆ) by Problem 11.Since the sequence
(2๐๐
)14๐
decreases, it follows that ๐ (๐ฆ,โ) < ๐ (3,โ) for each ๐ฆ > 3, thus max๐ฆ ๐ (๐ฆ,โ) =๐ (2,โ) = ๐ (3,โ) = 1/2.
By the Newton binomial theorem we getโโ
๐=0
(2๐๐
)๐ฅ๐ = 1โ
1โ4๐ฅfor each ๐ฅ โ
[โ1
4 ,14
). Setting ๐ฅ = โ1
4 we
obtain lim๐ฆโโ ๐ (๐ฆ,โ) = lim๐ฆโโ12
โโ๐=0
(2๐๐
) (โ1
4
)๐= 1
2โ2.
Remark. Using the Stirling formula (see ยง2) one can estimate the convergence rate:
|๐ (๐,โ)โ lim๐ฆโโ
๐ (๐ฆ,โ)| < 1
2 ยท 4โ๐/2โ
(2โ๐/2โโ๐/2โ
)<
๐
2๐โ
2โ๐/2โ<
1
2โ๐.
13. Answer :
๐ (๐ฅ, ๐ฆ, 0,+) =
{1, for ๐ฅ = ๐ฆ;
0, for ๐ฅ = ๐ฆ.
๐ (๐ฅ, ๐ฆ,โ,+) =
{1, for ๐ฅ = 1, ๐ฆ โก 1 mod 2;
0, otherwise.
Solution. By the definition, if the number of turns in a path ๐ is nonzero then ๏ฟฝ๏ฟฝ(๐ , 0) = 0. Therefore,๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, 0,+) = 0 if and only if ๐ฅ = ๐ฆ. Hence, we get the answer for ๐ (๐ฅ, ๐ฆ, 0,+). The case ๐ฅ = ๐ฆ is obvious.
One cannot prove the answer for ๐ (๐ฅ, ๐ฆ,โ,+) because it is a definition; but let us provide a motivation forthe definition. If ๐๐ is โvery largeโ, then ๏ฟฝ๏ฟฝ(๐ ,๐๐) is โsmallโ unless the checker turns at each move; this gives theformula for ๐ (๐ฅ, ๐ฆ,โ,+).
Remark. ๐ (๐ฅ, ๐ฆ,โ,+) = lim๐๐โโ ๐ (๐ฅ, ๐ฆ,๐๐,+). Indeed,
lim๐๐โโ
|๐(๐ ,๐๐)| = lim๐๐โโ
(๐๐)๐ก(๐ )
(โ1 + (๐๐)2)๐ฆโ1
= lim๐๐โโ
1
(โ1 + (๐๐)2)๐ฆโ1โ๐ก(๐ )
=
{1, for ๐ก(๐ ) = ๐ฆ โ 1;
0, otherwise;
where ๐ก(๐ ) is the number of turns in the path ๐ . To compute ๐ (๐ฅ, ๐ฆ,๐๐,+), we take only paths with ๐ก โก 0 mod 2,hence we get the answer.
14. Answer. Recall the notation introduced at the beginning of section โHints. . . โ. We have
๐1(๐ฅ, ๐ฆ,๐๐,โ) =1โ
1 +๐2๐2(๐1(๐ฅ+ 1, ๐ฆ โ 1,๐๐,โ) +๐๐๐2(๐ฅ+ 1, ๐ฆ โ 1,๐๐,+)),
๐2(๐ฅ, ๐ฆ,๐๐,+) =1โ
1 +๐2๐2(๐2(๐ฅโ 1, ๐ฆ โ 1,๐๐,+)โ๐๐๐1(๐ฅโ 1, ๐ฆ โ 1,๐๐,โ)),
๐2(๐ฅ, ๐ฆ,๐๐,โ) = ๐1(๐ฅ, ๐ฆ,๐๐,+) = 0.
For each ๐ฆ > |๐ฅ| we have
๐1(๐ฅ, ๐ฆ,๐๐) = (1 +๐2๐2)(1โ๐ฆ)/2
โ๐ฆ/2โโ๐=0
(โ1)๐((๐ฅ+ ๐ฆ โ 2)/2
๐
)((๐ฆ โ ๐ฅโ 2)/2
๐
)(๐๐)2๐+1,
๐2(๐ฅ, ๐ฆ,๐๐) = (1 +๐2๐2)(1โ๐ฆ)/2
โ๐ฆ/2โโ๐=1
(โ1)๐((๐ฅ+ ๐ฆ โ 2)/2
๐
)((๐ฆ โ ๐ฅโ 2)/2
๐ โ 1
)(๐๐)2๐;
21
For each ๐ฆ = ๐ฅ > 0 we have
๐1(๐ฆ, ๐ฆ,๐๐) = 0,
๐2(๐ฆ, ๐ฆ,๐๐) = (1 +๐2๐2)(1โ๐ฆ)/2;
For each 0 < ๐ฆ < |๐ฅ| or ๐ฆ = โ๐ฅ > 0 we have
๐1(๐ฅ, ๐ฆ,๐๐) = 0,
๐2(๐ฅ, ๐ฆ,๐๐) = 0.
Hint. First let us solve the analogue of Problem 4 for ๐๐ = 1. Let us derive the formula for ๐2(๐ฅ, ๐ฆ,๐๐,+).Consider any path ๐ from (0, 0) to (๐ฅ, ๐ฆ).
The last move in the path ๐ is made either from (๐ฅโ 1, ๐ฆ) or from (๐ฅ+ 1, ๐ฆ). Denote by ๐ โฒ the path ๐ withoutthe last move. If the directions of the last moves in ๐ and ๐ โฒ coincide, then ๏ฟฝ๏ฟฝ(๐ ,๐๐) = 1โ
1+๐2๐2๏ฟฝ๏ฟฝ(๐ โฒ,๐๐), otherwise
๏ฟฝ๏ฟฝ(๐ ,๐๐) = ๐๐โ1+๐2๐2
(๐2(๐ โฒ,๐๐),โ๐1(๐
โฒ,๐๐)).
Therefore, ๐2(๐ฅ, ๐ฆ,๐๐,+) = 1โ1+๐2๐2
(๐2(๐ฅโ 1, ๐ฆ โ 1,๐๐,+)โ๐๐๐1(๐ฅโ 1, ๐ฆ โ 1,๐๐,โ)). The formula for
๐1(๐ฅ, ๐ฆ,๐๐,โ) is proved analogously.Now let us solve the analogue of Problem 5: we prove that
โ๐ฅโZ
๐ (๐ฅ, ๐ฆ,๐๐) = 1 for all ๐ฆ โฅ 1 by induction over
๐ฆ using these results. Obviously,โ๐ฅโZ
๐ (๐ฅ, 1,๐๐) = 1. The step of induction follows immediately from the following
computation:โ๐ฅโZ
๐ (๐ฅ, ๐ฆ+1,๐๐) =โ๐ฅโZ
[๐1(๐ฅ, ๐ฆ + 1,๐๐,โ)2 + ๐2(๐ฅ, ๐ฆ + 1,๐๐,+)2
]=
โ๐ฅโZ
๐1(๐ฅ, ๐ฆ+1,๐๐,โ)2+โ๐ฅโZ
๐2(๐ฅ, ๐ฆ+1,๐๐,+)2 =
=1
1 +๐2๐2
โ๐ฅโZ
(๐1(๐ฅ+1, ๐ฆ,๐๐,โ)+๐๐๐2(๐ฅ+1, ๐ฆ,๐๐,+))2+1
1 +๐2๐2
โ๐ฅโZ
(๐2(๐ฅโ1, ๐ฆ,๐๐,+)โ๐๐๐1(๐ฅโ1, ๐ฆ,๐๐,โ))2 =
=1
1 +๐2๐2
โ๐ฅโZ
(๐1(๐ฅ, ๐ฆ,๐๐,โ) +๐๐๐2(๐ฅ, ๐ฆ,๐๐,+))2 +1
1 +๐2๐2
โ๐ฅโZ
(๐2(๐ฅ, ๐ฆ,๐๐,+)โ๐๐๐1(๐ฅ, ๐ฆ,๐๐,+))2 =
=โ๐ฅโZ
[๐1(๐ฅ, ๐ฆ,๐๐,โ)2 + ๐2(๐ฅ, ๐ฆ,๐๐,+)2
]=
โ๐ฅโZ
๐ (๐ฅ, ๐ฆ,๐๐).
Let us now solve the analogue of Problem 9 for ๐๐ = 1. Let us find ๐1(๐ฅ, ๐ฆ) for ๐ฆ > |๐ฅ|. We need to considerpaths with an odd number of turns only. Denote by 2๐+1 the number of turns in a path. Denote by ๐ and ๐ฟ thenumber of upwards-right and upwards-left moves respectively. Let ๐ฅ1, ๐ฅ2, . . . , ๐ฅ๐+1 be the number of upwards-rightmoves before the first, the third, . . . , the last turn respectively. Let ๐ฆ1, ๐ฆ2, . . . , ๐ฆ๐+1 be the number of upwards-leftmoves after the first, the third, . . . , the last turn respectively. Then ๐ฅ๐, ๐ฆ๐ โฅ 1 for 1 โค ๐ โค ๐ + 1 and
๐ = ๐ฅ1 + ยท ยท ยท+ ๐ฅ๐+1;
๐ฟ = ๐ฆ1 + ยท ยท ยท+ ๐ฆ๐+1.
The problem now reduces to a combinatorial one: find the number of positive integer solutions of the resultingequations. For the first equation, this number equals to the number of ways to put ๐ sticks between ๐ coins in arow, that is,
(๐ โ1๐
). Thus
๐1(๐ฅ, ๐ฆ,๐๐) = (1 +๐2๐2)(1โ๐ฆ)/2
โ๐ฆ/2โโ๐=0
(โ1)๐(๐ โ 1
๐
)(๐ฟโ 1
๐
)(๐๐)2๐+1.
Since ๐ฟ+๐ = ๐ฆ and ๐ โ ๐ฟ = ๐ฅ, the required formula follows. The proof of the formula for ๐2(๐ฅ, ๐ฆ) for ๐ฆ > |๐ฅ| isanalogous. The case ๐ฆ โค |๐ฅ| is obvious.15. Answer: for |๐ฅ| < ๐ฆ we have
lim๐โโ
๐ ๏ฟฝ๏ฟฝ(2โ๐๐ฅ2
โ, 2โ๐๐ฆ2
โ,๐
๐,โ
)= (๐๐ฝ0(๐
โ๐ฆ2 โ ๐ฅ2), 0);
lim๐โโ
๐ ๏ฟฝ๏ฟฝ(2โ๐๐ฅ2
โ, 2โ๐๐ฆ2
โ,๐
๐,+
)= (0,โ๐
๐ฅ+ ๐ฆโ๐ฆ2 โ ๐ฅ2
๐ฝ1(๐โ๐ฆ2 โ ๐ฅ2));
for |๐ฅ| โฅ ๐ฆ both limits vanish.
22
Remark. The limit coincides with the retarded fundamental solution of Diracโs equation, describing motion ofan electron in the plane.
Solution by Ivan Gaidai-Turlov, Timofey Kovalev and Alexey Lvov, participants of Summer conference of
Tournament of towns. Denote ๐ฅ๐ := 2โ๐๐ฅ2 โ, ๐ฆ๐ := 2โ๐๐ฆ2 โ, ๐ด := ๐ฅ๐+๐ฆ๐2 , ๐ต := ๐ฆ๐โ๐ฅ๐
2 .
Lemma 2. lim๐โโ
(1 + (๐๐ )2)
๐ฆ๐โ12 = 1.
Proof. We have 1 < (1 + (๐๐ )2)
๐ฆ๐โ12 < (1 + (๐๐ )
2)๐๐ฆ, therefore if ๐ โ โ then (1 + (๐๐ )2)๐๐ฆ = ๐
โ(1 + (๐๐ )
2)๐2๐ฆ โ๐โ๐๐2๐ฆ โ 1. Hence by the squeeze theorem we have lim
๐โโ(1 + (๐๐ )
2)๐ฆ๐โ1
2 = 1.
Lemma 3. lim๐โโ
๐(โ1)๐๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐๐ )
2๐+1 = (โ1)๐ ยท (๐ฅ+๐ฆ2
)๐( ๐ฆโ๐ฅ2
)๐ยท๐2๐+1
(๐!)2.
Proof. Clearly, ๐ ยท (โ1)๐๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐๐ )
2๐+1 = (โ1)๐ ยท (๐ดโ1)(๐ดโ2)...(๐ดโ๐)ยท(๐ตโ1)(๐ตโ2)...(๐ตโ๐)(๐!)2
ยท ๐2๐+1
๐2๐ .
Thus lim๐โโ
๐ดโ๐๐ = lim
๐โโ๐ด๐ = lim
๐โโโ๐๐ฅ
2โ+โ๐๐ฆ
2โ
๐ = lim๐โโ
๐๐ฅ2+๐๐ฆ
2+๐(๐)
๐ = ๐ฅ+๐ฆ2 . Analogously, lim
๐โโ๐ตโ๐๐ = ๐ฆโ๐ฅ
2 .
Therefore,
lim๐โโ
๐ ยท (โ1)๐๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐
๐)2๐+1 = lim
๐โโ(โ1)๐ ยท (๐ดโ1)...(๐ดโ๐)ยท(๐ตโ1)...(๐ตโ๐)
(๐!)2ยท ๐
2๐+1
๐2๐= (โ1)๐ ยท
(๐ฅ+๐ฆ2 )๐(๐ฆโ๐ฅ
2 )๐ ยท๐2๐+1
(๐!)2.
For each ๐ we have |(โ1)๐๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐๐ )
2๐+1| < |(โ1)๐ ยท (๐ฅ+๐ฆ2
)๐( ๐ฆโ๐ฅ2
)๐ยท๐2๐+1
(๐!)2| because ๐ดโ๐
๐ < ๐ด๐ < ๐ฆ+๐ฅ
2 and๐ตโ๐๐ < ๐ต
๐ < ๐ฆโ๐ฅ2 .
Lemma 4. The series ๐ ยทโโ๐=0
(โ1)๐๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐๐ )
2๐+1 converges absolutely for all ๐.
Proof. We have
๐ ยทโโ๐=0
|(โ1)๐๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐
๐)2๐+1| <
โโ๐=0
|(โ1)๐ ยท(๐ฅ+๐ฆ
2 )๐(๐ฆโ๐ฅ2 )๐ ยท๐2๐+1
(๐!)2| =
=
โโ๐=0
๐ ยท(๐ฅ+๐ฆ
2 )๐(๐ฆโ๐ฅ2 )๐ ยท๐2๐
(๐!)2<
โโ๐=0
๐ ยท(๐ฅ+๐ฆ
2 )๐(๐ฆโ๐ฅ2 )๐ ยท๐2๐
(๐!)= ๐ ยท ๐
๐ฆ2โ๐ฅ2
4๐2
.
Hence lim๐โโ
โโ๐=0
(โ1)๐๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐๐ )
2๐+1 = ( lim๐โโ
โโ๐=0;2|๐
๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐๐ )
2๐+1)โ ( lim๐โโ
โโ๐=0;2-๐
๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐๐ )
2๐+1).
Lemma 5. Suppose ({๐0(๐)}, {๐1(๐)} . . . ) is a sequence of nonnegative sequences such that lim๐โโ
๐๐(๐) = ๐๐ and
๐๐(๐) < ๐๐ for each ๐, ๐. Then lim๐โโ
โโ๐=0
๐๐(๐) =โโ๐=0
๐๐.
Proof. Denote ๐ :=โโ๐=0
๐๐. Then for each ๐ we haveโโ๐=0
๐๐(๐) < ๐. Take any ๐ > 0. Take such ๐ that๐โ๐=0
๐๐ > ๐โ ๐.
For each 0 โค ๐ โค ๐ take ๐๐ such that โ๐ก โฅ ๐๐ : ๐๐(๐ก) > ๐๐ โ ๐2๐+1 . Then for all ๐ > max(๐0,๐1, . . . ,๐๐ ) we
haveโโ๐=0
๐๐(๐) > ๐โ ๐. Therefore, lim๐โโ
โโ๐=0
๐๐(๐) = ๐.
From Lemmas 5 and 3 it follows that
lim๐โโ
๐ ยทโโ
๐=0;2|๐
๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐
๐)2๐+1 =
โโ๐=0;2|๐
(๐ฅ+๐ฆ2 )๐(๐ฆโ๐ฅ
2 )๐ ยท๐2๐+1
(๐!)2.
Therefore, by Problem 14 and Lemma 2 we have
lim๐โโ
๐๏ฟฝ๏ฟฝ(๐ฅ๐, ๐ฆ๐,๐
๐,โ) = lim
๐โโ๐ ยท
โโ๐=0
๐(โ1)๐๐ถ ๐๐ดโ1๐ถ
๐๐ตโ1(
๐
๐)2๐+1 =
= (
โโ๐=0;2|๐
ยท(๐ฅ+๐ฆ
2 )๐(๐ฆโ๐ฅ2 )๐ ยท๐2๐+1
(๐!)2)โ (
โโ๐=0;2-๐
ยท(๐ฅ+๐ฆ
2 )๐(๐ฆโ๐ฅ2 )๐ ยท๐2๐+1
(๐!)2) = ๐ ยท ๐ฝ0(๐
โ๐ฆ2 โ ๐ฅ2).
23
It can be proven analogously that lim๐โโ
๐๏ฟฝ๏ฟฝ(๐ฅ๐, ๐ฆ๐,๐๐ ,+) = โ๐๐ ๐ฅ+๐ฆโ
๐ฆ2โ๐ฅ2ยท ๐ฝ1(๐
โ๐ฆ2 โ ๐ฅ2).
Solution. Denote modified Bessel functions
๐ผ0(๐ง) :=โโ๐=0
(๐ง/2)2๐
(๐!)2and ๐ผ1(๐ง) :=
โโ๐=0
(๐ง/2)2๐+1
๐!(๐ + 1)!.
The answer follows from the following stronger theorem.
Theorem 2. Let 0 < ๐0 โค 1/2, min{๐ฆ + ๐ฅ, ๐ฆ โ ๐ฅ} โฅ ๐0, ๐ง := ๐โ๐ฆ2 โ ๐ฅ2, ๐ โฅ ๐0 = ๐0(๐ง, ๐0) = 4๐3๐ง/๐20. Then
๐1
(2โ๐๐ฅ2
โ, 2โ๐๐ฆ2
โ,๐
๐,โ
)=
๐
๐
(๐ฝ0(๐ง) +๐
(log2 ๐
๐0๐๐ผ0(๐ง)
)),
๐2
(2โ๐๐ฅ2
โ, 2โ๐๐ฆ2
โ,๐
๐,+
)= โ๐
๐
๐ฆ + ๐ฅโ๐ฆ2 โ ๐ฅ2
(๐ฝ1(๐ง) +๐
(log2 ๐
๐0๐๐ผ1(๐ง)
)).
Proof. Denote ๐ฅ๐ = 2โ๐๐ฅ2
โ, ๐ฆ๐ = 2
โ๐๐ฆ2
โ. By Problem 9 we have
๐1
(๐ฅ๐, ๐ฆ๐,
๐
๐,โ
)=
(1 +
๐2
๐2
) 1โ๐ฆ๐2 ๐
๐
[๐ฆ๐/2]โ๐=0
(โ1)๐((๐ฆ๐ + ๐ฅ๐ โ 2)/2
๐
)((๐ฆ๐ โ ๐ฅ๐ โ 2)/2
๐
)(๐๐
)2๐,
๐2
(๐ฅ๐, ๐ฆ๐,
๐
๐,+
)=
(1 +
๐2
๐2
) 1โ๐ฆ๐2 ๐
๐
[๐ฆ๐/2]โ๐=1
(โ1)๐((๐ฆ๐ + ๐ฅ๐ โ 2)/2
๐
)((๐ฆ๐ โ ๐ฅ๐ โ 2)/2
๐ โ 1
)(๐๐
)2๐โ1.
Since 1 + ๐2
๐2 โค ๐๐2/๐2
it follows that(1 + ๐2
๐2
) 1โ๐ฆ๐2 โฅ ๐
๐2
๐2 (1โ๐ฆ๐
2 ) because ๐ฆ๐ โฅ 1. Also 1โ๐ฆ๐2 โฅ โ๐ฆ๐
2 . Hence,
1 โฅ(1 +
๐2
๐2
) 1โ๐ฆ๐2
โฅ ๐๐2
๐2 (โ๐ฆ๐2 ) โฅ 1โ ๐2๐ฆ
2๐.
Thus(1 + ๐2
๐2
) 1โ๐ฆ๐2
= 1 +๐(๐2๐ฆ๐ ).
To find the asymptotic form for ๐1 and ๐2 consider the ๐ -th partial sums with ๐ = โlog ๐โ+1 summands. Theinequalities ๐ฆ โฅ ๐0 and ๐ โฅ 4
๐20guarantee that the total number of summands โ๐ฆ๐2 โ โฅ ๐ because ๐๐ฆ
2 โฅโ๐ โฅ log ๐+1
for ๐ > 0.For ๐ โฅ ๐ the ratios of consecutive summands in the expressions for ๐1 and ๐2 equal respectively(๐
๐
)2 ((๐ฆ๐ + ๐ฅ๐ โ 2)/2โ ๐)((๐ฆ๐ โ ๐ฅ๐ โ 2)/2โ ๐)
(๐ + 1)2<
(๐๐
)2ยท ๐(๐ฆ + ๐ฅ)
2๐ยท ๐(๐ฆ โ ๐ฅ)
2๐=
๐ง2
4๐ 2,(๐
๐
)2 ((๐ฆ๐ + ๐ฅ๐ โ 2)/2โ ๐)((๐ฆ๐ โ ๐ฅ๐ โ 2)/2โ ๐ + 1)
(๐ โ 1)๐<
(๐๐
)2ยท ๐(๐ฆ + ๐ฅ)
2๐ยท ๐(๐ฆ โ ๐ฅ)
2๐ โ 2=
๐ง2
4๐ (๐ โ 1).
From the condition ๐ > ๐3๐ง+1 it follows that ๐ = โlog ๐โ + 1 โฅ 3๐ง + 1 and ๐ง2
4๐ 2 < ๐ง2
4๐ (๐โ1) < 12 . Therefore, in
both cases the error term (i.e., the sum over ๐ โฅ ๐ ) is less then the sum of geometric series with ratio 12 . Hence,
๐1
(๐๐ฅ, ๐๐ฆ,
๐
๐,โ
)=
๐
๐
(1 +๐
(๐2๐ฆ
๐
))ยท
ยท
[๐โ1โ๐=0
(โ1)๐((๐ฆ๐ + ๐ฅ๐ โ 2)/2
๐
)((๐ฆ๐ โ ๐ฅ๐ โ 2)/2
๐
)(๐๐
)2๐+๐
((๐ฆ2 โ ๐ฅ2)๐
(๐ !)2ยท(๐2
)2๐)]
,
๐2
(๐๐ฅ, ๐๐ฆ,
๐
๐,+
)=
๐
๐
(1 +๐
(๐2๐ฆ
๐
))ยท
ยท
[๐โ1โ๐=1
(โ1)๐((๐ฆ๐ + ๐ฅ๐ โ 2)/2
๐
)((๐ฆ๐ โ ๐ฅ๐ โ 2)/2
๐ โ 1
)(๐๐
)2๐โ1+๐
((๐ฆ + ๐ฅ)๐ (๐ฆ โ ๐ฅ)๐โ1
(๐ โ 1)!๐ !ยท(๐2
)2๐โ1)]
.
The error term in the latter formulae are estimated as follows. Since ๐ ! โฅ (๐/3)๐ and ๐ โฅโ
๐ฆ2 โ ๐ฅ2 3๐โ๐
2 =3โ๐
2 ๐ง, it follows that
(๐ฆ2 โ ๐ฅ2)๐
(๐ !)2ยท(๐2
)2๐โค (๐ฆ2 โ ๐ฅ2)๐
(๐ )2๐ยท(3๐
2
)2๐
โค ๐โ๐ โค 1
๐.
24
Now transform binomial coefficients in the following way:(๐ผ๐โ 1
๐
)=
(๐ผ๐โ 1) ยท ยท ยท (๐ผ๐โ ๐)
๐!=
(๐ผ๐)๐
๐!
(1โ 1
๐ผ๐
)ยท ยท ยท
(1โ ๐
๐ผ๐
).
If ๐ผ โฅ ๐0 and ๐ < ๐ then ๐๐ผ๐ < ๐
๐ผ๐ <โ๐
๐๐0= 1โ
๐๐0โค ๐0
2 ยท 1๐0
= 12 . Hence from the inequalities ๐โ2๐ฅ โค 1 โ ๐ฅ < ๐โ๐ฅ
for 0 โค ๐ฅ โค 1/2 it follows that(1โ 1
๐ผ๐
)ยท ยท ยท
(1โ ๐
๐ผ๐
)โฅ ๐
โ2๐ผ๐ ๐
โ4๐ผ๐ ยท ยท ยท ๐
โ2๐๐ผ๐ = ๐
โ๐(๐+1)๐ผ๐ โฅ ๐
โ๐2
๐ผ๐ โฅ 1โ ๐ 2
๐ผ๐โฅ 1โ ๐ 2
๐0๐.
Therefore,(๐ผ๐โ1
๐
)= (๐ผ๐)๐
๐!
(1 +๐
(๐ 2
๐0๐
)).
Denote ๐ = ๐ 2
๐0๐๐ผ0(๐ง). Replacing
((๐ฆ๐+๐ฅ๐โ2)/2
๐
)and
((๐ฆ๐โ๐ฅ๐โ2)/2
๐
)in the previous expression for ๐1 we get
๐1
(๐ฅ๐, ๐ฆ๐,
๐
๐,โ
)=
๐
๐
(1 +๐
(๐2๐ฆ
๐
))ยท
[๐โ1โ๐=0
(โ1)๐(๐2
)2๐ (๐ฆ2 โ ๐ฅ2)๐
(๐!)2+๐(๐ )
]=
=๐
๐
(1 +๐
(๐2๐ฆ
๐
))ยท
[ โโ๐=0
(โ1)๐(๐2
)2๐ (๐ฆ2 โ ๐ฅ2)๐
(๐!)2+๐(๐ )
]=
๐
๐
(1 +๐
(๐2๐ฆ
๐
))ยท (๐ฝ0(๐ง) +๐(๐ )).
Here we can replace the sum with ๐ summands by an infinite sum because the โtailโ of alternating series withdecreasing absolute value of the summands can be estimated by the first summand:
โโ
๐=๐
(โ1)๐(๐2
)2๐ (๐ฆ2 โ ๐ฅ2)๐
(๐!)2
โค (๐ง/2)2๐
(๐ !)2โค 1
๐
(the latter inequality has been proved before).We have ๐2๐ฆ๐0 โค ๐2 ยท max{๐ฆ + ๐ฅ, ๐ฆ โ ๐ฅ} ยท min{๐ฆ + ๐ฅ, ๐ฆ โ ๐ฅ} = ๐2(๐ฆ2 โ ๐ฅ2) = ๐ง2 โค 9๐ง2 โค ๐ 2, thus
๐2๐ฆ๐ ๐ฝ0(๐ง) โค ๐ 2
๐0๐๐ผ0(๐ง), hence ๐1
(๐ฅ๐, ๐ฆ๐,
๐๐ ,โ
)= ๐
๐ (๐ฝ0(๐ง) +๐(๐ )) as required.Analogously,
๐2
(๐ฅ๐, ๐ฆ๐,
๐
๐,+
)=
๐
๐
(1 +๐
(๐2๐ฆ
๐
))ยท ๐ฆ + ๐ฅโ
๐ฆ2 โ ๐ฅ2ยท
[๐โ1โ๐=1
(โ1)๐(๐2
)2๐โ1 (๐ฆ2 โ ๐ฅ2)2๐โ1
2
(๐ โ 1)!๐!+๐(
๐ 2
๐0๐๐ผ1(๐ง))
]=
= โ๐
๐ยท ๐ฆ + ๐ฅโ
๐ฆ2 โ ๐ฅ2
(๐ฝ1(๐ง) +๐
(๐ 2
๐0๐๐ผ1(๐ง)
)).
16. Answer : see the following table, where the pair (๐, ๐)+ in the cell (๐ฅ, ๐ฆ) means that ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, 4, 3,+) = 1โ3(๐, ๐),
and (๐, ๐)โ means that ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, 4, 3,โ) = 1โ3(๐, ๐); an empty cell (๐ฅ, ๐ฆ) means that ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, 4, 3,+) = ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, 4, 3,โ) =
(0, 0).
3 (0, 0)+(โ1, 0)โ
(0, 1)+(0, 0)โ
(0,โ2)+(0, 0)โ
(0, 2)+(โ1, 0)โ
(0,โ1)+(0, 0)โ
2 (0, 0)+(โ1, 0)โ
(0,โ1)+(1, 0)โ
(0, 1)+(โ1, 0)โ
(0,โ1)+(0, 0)โ
1 (0,โ1)+(0, 0)โ
(0, 1)+(0, 0)โ
(0,โ1)+(0, 0)โ
y x โ3 โ2 โ1 0 1 2 3 4 5
Hint. Note that we are interested only in even ๐ฅ0, thus 2๐๐ฅ0๐/๐ = ๐๐ฅ02 is a multiple of ๐. Analogously to
Problem 1, that means that ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, 4, 3,+) is parallel to ๐๐ฆ and ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, 4, 3,โ) is parallel to ๐๐ฅ (be careful: this isnot true for an arbitrary ๐/๐).
17. Let us give a more general formula; the solution of the problem is the particular case ๐๐ = 1. Denote
๐โ(๐ฅ, ๐ฆ) := ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,๐๐, ๐/๐,ฮ,โ);
๐+(๐ฅ, ๐ฆ) := ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,๐๐, ๐/๐,ฮ,+),
25
where the vectors in the right-hand side are defined analogously to ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,ฮ,ยฑ) and ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ,๐๐,ยฑ). (Bewarethat ๐โ(๐ฅ, ๐ฆ) and ๐+(๐ฅ, ๐ฆ) are vector -valued functions; unlike ๐1(๐ฅ, ๐ฆ) and ๐2(๐ฅ, ๐ฆ) they are not coordinates of anyvector.) Then we have the Dirac equation (where ๐ := ๐ ๐/2 is the counterclockwise rotation through 90โ){
๐โ(๐ฅ, ๐ฆ) = 1โ1+๐2๐2
๐โ(๐ฅ+ 1, ๐ฆ โ 1) + โ๐๐๐โ1+๐2๐2
๐+(๐ฅ+ 1, ๐ฆ โ 1);
๐+(๐ฅ, ๐ฆ) = โ๐๐๐โ1+๐2๐2
๐โ(๐ฅโ 1, ๐ฆ โ 1) + 1โ1+๐2๐2
๐+(๐ฅโ 1, ๐ฆ โ 1).
From Problem 14 we know that this equation holds for ฮ = 1. It can be generalised for any greater ฮ becauseone can change the order of rotations ๐ 2๐๐ฅ0๐/๐ and ๐ = ๐ ๐/2.
Using this equation, analogously to Problem 14 we getโ๐ฅโZ
๐ (๐ฅ, ๐ฆ, ๐/๐,ฮ) = 1 for ๐ฆ โฅ 1 by induction over ๐ฆ.
18. Since ฮ๐ (๐ฅ, ๐ฆ, ๐/๐,ฮ,+) = |โฮ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,ฮ,+)|2, it suffices to prove that
โฮ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,ฮ,+) does not
depend on ฮ for ฮ > ๐ฆ+ |๐ฅ|. By definitionโฮ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,ฮ,+) =
๐ฅ0=ฮโ1โ๐ฅ0=1โฮ๐ฅ0 even
โ๐ ๐ 2๐๐ฅ0๐/๐๏ฟฝ๏ฟฝ(๐ ). But there are no paths
๐ starting at (๐ฅ0, 0) and ending at (๐ฅ, ๐ฆ) for |๐ฅโ ๐ฅ0| > ๐ฆ, thus increasing ฮ over ๐ฆ + |๐ฅ| does not change the sum.
19. Answer: if ๐ฅ1 and ๐ฅ2 are both even, then
๏ฟฝ๏ฟฝ(๐ฅ1, ๐ฆ, ๐/๐,+) = ๐ (๐ฅ1โ๐ฅ2)2๐๐/๐๏ฟฝ๏ฟฝ(๐ฅ2, ๐ฆ, ๐/๐,+),
๏ฟฝ๏ฟฝ(๐ฅ1, ๐ฆ, ๐/๐,โ) = ๐ (๐ฅ1โ๐ฅ2)2๐๐/๐๏ฟฝ๏ฟฝ(๐ฅ2, ๐ฆ, ๐/๐,โ).
Solution. From the solution of Problem 18 it follows that ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,+) =โ
๐ฅ0 even
โ๐ ๐ 2๐๐ฅ0๐/๐๏ฟฝ๏ฟฝ(๐ ), where the
second sum is over those paths from (๐ฅ0, 0) to (๐ฅ, ๐ฆ), which both start and finish with an upwards-right move. Toeach path ๐ 1 from (๐ฅ0, 0) to (๐ฅ1, ๐ฆ) assign the unique path ๐ 2 from (๐ฅ0 + (๐ฅ2 โ ๐ฅ1), 0) to (๐ฅ2, ๐ฆ), obtained from ๐ 1by translation by (๐ฅ1 โ ๐ฅ2, 0) and vice versa. Thus we sum over the same vectors, but for ๐ฅ = ๐ฅ1 they are rotatedthrough 2๐๐ฅ0๐/๐ while for ๐ฅ = ๐ฅ2 they are rotated through 2๐(๐ฅ0 + (๐ฅ2 โ ๐ฅ1))๐/๐, and we are done.
20. Let us give a more general formula; the solution of the problem is the particular case ๐๐ = 1.
Answer. Denote ๐ธ = arccos(cos(2๐๐/๐)โ
1+๐2๐2
). Then for even ๐ฅ+ ๐ฆ we have:
๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ) =(๐๐ cos(2๐๐๐ฅ/๐) sin((๐ฆ โ 1)๐ธ)โ
1 +๐2๐2 sin๐ธ,๐๐ sin(2๐๐๐ฅ/๐) sin((๐ฆ โ 1)๐ธ)โ
1 +๐2๐2 sin๐ธ
);
๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,+) =(cos(2๐๐(๐ฅโ 2)/๐) sin(๐ฆ๐ธ)โ cos(2๐๐๐ฅ/๐) sin((๐ฆ โ 2)๐ธ)
2 cos(2๐๐/๐) sin๐ธ,
โ sin(2๐๐(๐ฅโ 2)/๐) sin(๐ฆ๐ธ) + sin(2๐๐๐ฅ/๐) sin((๐ฆ โ 2)๐ธ)
2 cos(2๐๐/๐) sin๐ธ
);
๐ (๐ฅ, ๐ฆ, ๐/๐,โ) =๐2๐2 sin2((๐ฆ โ 1)๐ธ)
๐2๐2 + sin2(2๐๐/๐);
๐ (๐ฅ, ๐ฆ, ๐/๐,+) =๐2๐2 cos2((๐ฆ โ 1)๐ธ) + sin2(2๐๐/๐)
๐2๐2 + sin2(2๐๐/๐);
๐ (๐ฅ, ๐ฆ, ๐/๐,โ) + ๐ (๐ฅ, ๐ฆ, ๐/๐,+) =1.
Remark. In complex form,
๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ) = ๐๐ฅ๐๐ sin((๐ฆ โ 1)๐ธ)โ1 +๐2๐2 sin๐ธ
;
๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,+) = ๐๐ฅโ1(sin((๐ฆ โ 1)๐ธ) sin(2๐๐/๐)โ
1 +๐2๐2 sin๐ธ+ ๐ cos((๐ฆ โ 1)๐ธ)
),
where ๐ = cos(2๐๐/๐) + ๐ sin(2๐๐/๐). This is the solution of the Dirac equation (see the solution of Problem 17)with the (quasi)periodic initial condition ๐โ(๐ฅ, 0) = 0 and ๐+(๐ฅ, 0) = (โ sin 2๐๐ฅ๐
๐ , cos 2๐๐ฅ๐๐ ). The number ๐ธ has
physical meaning of energy.Hint. Formulas for ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,+) and ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ) can be proven by induction over ๐ฆ. The base is obvious;
the inductive step follows from Problem 17. Then ๐ (๐ฅ, ๐ฆ, ๐/๐,+) and ๐ (๐ฅ, ๐ฆ, ๐/๐,โ) are computed directly.Path to solution. Let us view vectors as complex numbers. Then ๐ ๐ผ๏ฟฝ๏ฟฝ = (cos๐ผ + ๐ sin๐ผ)๐. Denote ๐ =
cos(2๐๐/๐) + ๐ sin(2๐๐/๐); then ๐โ1 = cos(2๐๐/๐) โ ๐ sin(2๐๐/๐). Denote ๐(๐ฆ) = ๐โ๐ฆ๏ฟฝ๏ฟฝ(๐ฆ, ๐ฆ, ๐/๐,+) and ๐(๐ฆ) =๐โ๐ฆ๏ฟฝ๏ฟฝ(๐ฆ, ๐ฆ, ๐/๐,โ). Denote๐ = ๐๐โ
1+๐2๐2and ๐ = 1โ
1+๐2๐2. By Problem 19 for even ๐ฅ+๐ฆ we have ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,+) =
๐๐ฅ๐(๐ฆ) and ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ) = ๐๐ฅ๐(๐ฆ). By Problem 17 we have ๐๐ฅ๐(๐ฆ) = ๐๐๐ฅโ1๐(๐ฆ โ 1) โ ๐๐๐๐ฅโ1๐(๐ฆ โ 1) and
26
๐๐ฅ๐(๐ฆ) = ๐๐๐ฅ+1๐(๐ฆ โ 1) โ ๐๐๐๐ฅ+1๐(๐ฆ โ 1). Since ๐ = 0, we can divide each side by ๐๐ฅ and obtain the followingsystem: โงโชโชโชโจโชโชโชโฉ
๐(๐ฆ) = ๐๐โ1๐(๐ฆ โ 1)โ ๐๐๐โ1๐(๐ฆ โ 1) (1)
๐(๐ฆ) = ๐๐๐(๐ฆ โ 1)โ ๐๐๐๐(๐ฆ โ 1) (2)
๐(1) = ๐๐โ1 (3)
๐(1) = 0 (4)
There are multiple ways of solving this system (we have thought about generating functions and about diagonalisingthe โtransfer matrixโ) but the one which is probably most accessible (to math high-school students in Russia) istransforming this system into a linear recurrence relation of degree 2. From (1) we have ๐(๐ฆ โ 1) = 1
๐ ๐๐๐(๐ฆ) โ๐๐ ๐๐(๐ฆ โ 1). From this and (2) we have ๐(๐ฆ) = ๐
๐ ๐๐2๐(๐ฆ) โ 1๐ ๐๐๐(๐ฆ โ 1). From this and (1) we have ๐(๐ฆ) =
๐(๐ + ๐โ1)๐(๐ฆ โ 1)โ ๐(๐ฆ โ 2).The solution of this recurrence equation is ๐(๐) = ๐1๐
๐+ + ๐2๐
๐โ, where
๐+ =๐(๐ + ๐โ1) +
โ๐2(๐ + ๐โ1)2 โ 4
2= ๐ cos(2๐๐/๐) + ๐
โ๐2 sin2(2๐๐/๐) +๐2
๐โ =๐(๐ + ๐โ1)โ
โ๐2(๐ + ๐โ1)2 โ 4
2= ๐ cos(2๐๐/๐)โ ๐
โ๐2 sin2(2๐๐/๐) +๐2
are the roots of the equation ๐ก2โ๐(๐+๐โ1)๐ก+1 = 0, and ๐1 and ๐2 are constants that are found using (3) and (4).Since we are dealing with complex numbers, the notation
โ๐2(๐ + ๐โ1)2 โ 4 is ambiguous. But we can choose
any of the two roots;โ
๐2(๐ + ๐โ1)2 โ 4 denotes 2๐โ
๐2 sin2(2๐๐/๐) +๐2. It follows that
๐1 = ๐๐โ1 โ ๐๐โ๐+ โ ๐โ
and ๐2 = ๐๐โ1 โ ๐๐+๐โ โ ๐+
.
Note that ๐ธ = arccos(๐ cos(2๐๐/๐)). We have ๐+ = cos๐ธ + ๐ sin๐ธ and ๐โ = cos๐ธ โ ๐ sin๐ธ. Thus we have
๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,+) = ๐๐ฅ๐(๐ฆ) = ๐๐๐ฅ(๐โ1โ๐๐โ)๐๐ฆ+โ(๐โ1โ๐๐+)๐๐ฆโ
๐+โ๐โ= ๐๐๐ฅโ1 sin(๐ฆ๐ธ)
sin๐ธ โ ๐๐๐๐ฅ sin((๐ฆโ1)๐ธ)sin๐ธ .
Now we compute ๐ (๐ฅ, ๐ฆ, ๐/๐,+) = ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,+) ยท ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,+). It follows from formulas above that ๐ = ๐โ1,๐+ = ๐โ, ๐โ = ๐+, ๐1 = ๐ ๐โ๐๐+
๐+โ๐โand ๐2 = ๐ ๐โ๐๐โ
๐โโ๐+. From this (with some simplification) we finally have
๐ (๐ฅ, ๐ฆ, ๐/๐,+) =(๐+โ๐โ)2โ๐2(๐๐ฆโ1
+ โ๐๐ฆโ1โ )2
(๐+โ๐โ)2= 1โ ๐2 sin2((๐ฆโ1)๐ธ)
sin2 ๐ธ.
Analogously we get ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐/๐,โ) = ๐๐ฅ๐ sin((๐ฆโ1)๐ธ)sin๐ธ and ๐ (๐ฅ, ๐ฆ, ๐/๐,โ) = ๐2 sin2((๐ฆโ1)๐ธ)
sin2 ๐ธ.
21. Solution. Let (0, 0), (1, 1), (๐ฅ2, 2), ยท ยท ยท , (๐ฅ๐ฆ, ๐ฆ) be the squares passed by the path ๐ . To define the vector๏ฟฝ๏ฟฝ(๐ ,๐(๐ฅ)๐), start with the vector (0, 1). While the checker moves straightly, the vector is not changed, but if thechecker changes the direction after the ๐-th move, the vector is rotated through 90โ clockwise and multiplied by๐(๐ฅ๐)๐. In addition, at the very end the vector is divided by
โ๐ฆโ1๐=1
โ1 +๐2(๐ฅ๐)๐2, where ๐ฆ is the total number
of moves. The final position of the vector is what we denote by ๏ฟฝ๏ฟฝ(๐ ,๐(๐ฅ)๐). The numbers ๐1(๐ฅ, ๐ฆ,๐(๐ฅ)๐) and๐2(๐ฅ, ๐ฆ,๐(๐ฅ)๐) are defined analogously to ๐1(๐ฅ, ๐ฆ) and ๐2(๐ฅ, ๐ฆ), only ๏ฟฝ๏ฟฝ(๐ ) is replaced by ๏ฟฝ๏ฟฝ(๐ ,๐(๐ฅ)๐).
The formulas analogous to the ones from Problem 4 are the following:
๐1(๐ฅ0, ๐ฆ,๐(๐ฅ)๐) =1โ
1 +๐2(๐ฅ0 + 1)๐2(๐1(๐ฅ0 + 1, ๐ฆ,๐(๐ฅ)๐) +๐(๐ฅ0 + 1)๐ ยท ๐2(๐ฅ0 + 1, ๐ฆ,๐(๐ฅ)๐)),
๐2(๐ฅ0, ๐ฆ,๐(๐ฅ)๐) =1โ
1 +๐2(๐ฅ0 โ 1)๐2(๐2(๐ฅ0 โ 1, ๐ฆ,๐(๐ฅ)๐)โ๐(๐ฅ0 โ 1)๐ ยท ๐1(๐ฅ0 โ 1, ๐ฆ,๐(๐ฅ)๐)).
22. Answer : For each ๐ฅ, ๐ฆ we have๐ (๐ฅ, ๐ฆ,๐0(๐ฅ), ๐/๐,+) = 1;๐ (๐ฅ, ๐ฆ,๐0(๐ฅ), ๐/๐,โ) = 0.
๐ (๐ฅ, ๐ฆ,๐1(๐ฅ), ๐/๐,+) =
{1
1.04 = 2526 , for 0 < ๐ฅ < ๐ฆ;
1, otherwise;
๐ (๐ฅ, ๐ฆ,๐1(๐ฅ), ๐/๐,โ) =
{0.041.04 = 1
26 , for โ ๐ฆ < ๐ฅ < 0;
0, otherwise;
Solution. Analogously to Problem 13 we have ๐ (๐ฅ, ๐ฆ,๐0(๐ฅ), ๐/๐,+) = 1, ๐ (๐ฅ, ๐ฆ,๐0(๐ฅ), ๐/๐,โ) = 0 for each๐ฅ, ๐ฆ.
27
The condition ๏ฟฝ๏ฟฝ(๐ ,๐1(๐ฅ)) = 0 means that the checker, if at all changed the direction, turned in one ofthe squares with the coordinates (0, 2), (0, 4), ยท ยท ยท , (0, 2 ยท
[๐ฆ2
]). Thus the number of turns in any path ๐ with
๏ฟฝ๏ฟฝ(๐ ,๐1(๐ฅ)) = 0 is not greater than 1.To compute ๐ (๐ฅ, ๐ฆ,๐1(๐ฅ), ๐/๐,+) we need to consider only the paths ending with an upwards-right move โ
i.e. without turns. If the path ๐ does not contain any of the squares with coordinates (0, 1), (0, 2), . . . , (0, ๐ฆ โ 1)(amounting to ๐ฅ โค 0 or ๐ฅ โฅ ๐ฆ) then ๐ (๐ฅ, ๐ฆ,๐1(๐ฅ), ๐/๐,+) = 1. Otherwise, the length of the vector ๏ฟฝ๏ฟฝ(๐ ,๐(๐ฅ))equals 1โ
1+0.22.
To compute ๐ (๐ฅ, ๐ฆ,๐1(๐ฅ), ๐/๐,โ), consider paths having exactly one turn in the square with zero ๐ฅ-coordinate.If there is such a path ๐ (i.e. โ๐ฆ < ๐ฅ < 0), then the length of the vector ๏ฟฝ๏ฟฝ(๐ ,๐(๐ฅ)) equals 0.2โ
1+0.22. Otherwise, it
equals zero.
23. Solution. If ๏ฟฝ๏ฟฝ(๐ ,๐2(๐ฅ)) = 0 then the checker changed the direction only in the squares with the ๐ฅ-coordinateequal to 1 or ๐ฟ. To compute ๐ (0, 2๐ฟ,๐2(๐ฅ), 16,โ) we need to consider only the paths with an odd number ofturns. Since for ๐ฟ > 1 we have ๐ฆ = 2๐ฟ < 3(๐ฟโ 1) (for ๐ฟ = 1 the function ๐(๐ฟ) is undefined), it follows that thenumber of turns must be exactly 1. Therefore, either the path starts in the square (0, 0) and changes the directionin the square (๐ฟ,๐ฟ), or the path starts in the square (2 โ 2๐ฟ, 0) and changes the direction in (1, 2๐ฟ โ 1). The
vectors for these two paths are related by the rotation through the angle ๐ โ ๐(๐ฟโ1)4 and division by 1.04. Since
the vector for the latter path has square 126 and the one for the former one has square 1
26ยท1.042 , by the law of cosineswe get
๐ (0, 2๐ฟ,๐2(๐ฅ), 16,โ) =1
26
(1 +
1
1.042โ 2
1.04cos
๐(๐ฟโ 1)
4
)โ 0.074
(1โ cos
๐(๐ฟโ 1)
4
).
The graph of this function is shown in the following figure (kindly provided by Punnawith Thuwajit, a participantof Summer Conference of Tournament of Towns).
24. Answer :
๐ (๐, ๐ฟ,โ) =0.1664
1.1664 + cot2 2๐(๐ฟโ1)๐
; max๐ฟ
๐ (๐, ๐ฟ,โ) =0.1664
1.1664โ 0.14;
๐ (๐, ๐ฟ,+) =1
1 + 0.1664 sin2 2๐(๐ฟโ1)๐
; ๐ (๐, ๐ฟ,โ) + ๐ (๐, ๐ฟ,+) = 1.
Hint. Denote ๐ = 0.2 and ๐ = 2๐๐ (๐ฟโ 1). For ๐ฆ > 2๐ฟโ 2 we have
๐ (0, ๐ฆ,๐2(๐ฅ), ๐,โ) =
=
๐ 2๐๐
๐(2โ๐ฆ) ๐โ
1 +๐2
โโโ1 +1
1 +๐2๐2๐๐ +
๐2
(1 +๐2)2(๐2๐๐)2 + ยท ยท ยท+ 1
๐2
(๐2
1 +๐2๐2๐๐
)[๐ฆโ2
2(๐ฟโ1)
]2(๐ฟโ1)
โโ 2
,
Then, since cos 2๐ = 1โ 2 sin2 ๐, we have
๐ (๐, ๐ฟ,โ) =๐2
1 +๐2ยท
1โ 1
1 +๐2๐2๐๐ ยท
โโ๐=0
(๐2
1 +๐2๐2๐๐
)๐2
=
=๐2
1 +๐2ยท
1โ 1
1 +๐2๐2๐๐
1
1โ ๐2
1+๐2 ๐2๐๐
2
=๐2
1 +๐2ยท |1โ ๐2๐๐|2
|1โ ๐2
1+๐2 ๐2๐๐|2=
=๐2(1 +๐2)(2โ 2 cos(2๐))
(1 +๐2)2 +๐4 โ 2๐2(1 +๐2) cos(2๐)=
4๐2(1 +๐2)
(1 + 2๐2)2 + cot2 ๐=
0.1664
1.1664 + cot2 2๐๐ (๐ฟโ 1)
.
28
Therefore, max๐ฟ ๐ (๐, ๐ฟ,โ) = 0.16641.1664 โ 0.14. Analogously,
๐ (๐, ๐ฟ,+) =
1
1 +๐2ยท
โโ๐=0
(๐2
1 +๐2๐2๐๐
)๐2
=
1
1 +๐2ยท 1
1โ ๐2
1+๐2 ๐2๐๐
2
=
=1
(1 +๐2)2 +๐4 โ 2(1 +๐2)๐2 cos(2๐)=
1
1 + 4๐2(1 +๐2) sin2 ๐=
1
1 + 0.1664 sin2 2๐๐ (๐ฟโ 1)
.
Note that
1โ 1
1 + 0.1664 sin2 2๐๐ (๐ฟโ 1)
=0.1664
1sin2 2๐
๐(๐ฟโ1)
+ 0.1664=
0.1664
1.1664 + cot2 2๐๐ (๐ฟโ 1)
.
Hence, ๐ (๐, ๐ฟ,โ) + ๐ (๐, ๐ฟ,+) = 1.
25. Answer :(๐2 โ 1)2
(๐2 + 1)2 + 4๐2 cot2 2๐๐ฟ๐๐
and ๐ =
(0.3328 + 0.864
โ1.04
)ยท ๐
๐,
where ๐ =โ1 +๐๐/๐ (the refractive index ; notice that the relation between ๐ and ๐ has changed when we
passed to reflection inside the glass). For the latter ๐ the limit equals
0.1664
1.1664 + cot22๐๐ฟ(1.08+0.08
โ1.04)
๐
.
Remark. The first formula in the answer is well-known and confirmed by experiment. A remarkable detailedpopular science discussion of Problems 23โ25 can be found in [Feynman], and a derivation of the equations on aphysical level of rigor โ in [Landafshitz, Chapter X, ยง86, Problem 4].
Path to solution. Consider Diracโs equation from the solution of Problem 17. Search for a solution of the form
(๐+(๐ฅ, ๐ฆ), ๐โ(๐ฅ, ๐ฆ)) = (๐+(๐ฅ), ๐โ(๐ฅ))๐โ2๐๐๐๐ฆ/๐.
Use Problem 20 to guess ๐ยฑ(๐ฅ) separately for ๐ฅ โค 0, 0 โค ๐ฅ โค ๐ฟ/๐, and ๐ฅ โฅ ๐ฟ/๐. Assume for simplicity that 1/๐is an odd integer. Abbreviate ๐3(๐ฅ) =: ๐(๐ฅ).
First consider the half-plane ๐ฅ โค 0 (to the left from glass). We need to specify what is meant by a solutionof Diracโs equation in the half-plane. Analogously to Problem 17, by Diracโs equation in a black square (๐ฅ, ๐ฆ) wemean the system โงโจโฉ๐โ(๐ฅโ 1, ๐ฆ + 1) = 1โ
1+๐(๐ฅ)2๐2๐โ(๐ฅ, ๐ฆ) +
โ๐๐(๐ฅ)๐โ1+๐(๐ฅ)2๐2
๐+(๐ฅ, ๐ฆ);
๐+(๐ฅ+ 1, ๐ฆ + 1) = โ๐๐(๐ฅ)๐โ1+๐(๐ฅ)2๐2
๐โ(๐ฅ, ๐ฆ) +1โ
1+๐(๐ฅ)2๐2๐+(๐ฅ, ๐ฆ).
A solution of Diracโs equation in the half-plane ๐ฅ โค 0 is a pair (๐+(๐ฅ, ๐ฆ), ๐โ(๐ฅ, ๐ฆ)) satisfying the equation in eachblack square in the half-plane. In particular, ๐โ(๐ฅ, ๐ฆ) is defined for ๐ฅ โค 0, whereas ๐+(๐ฅ, ๐ฆ) is defined for ๐ฅ โค 1.Search for a solution of the form (๐+(๐ฅ), ๐โ(๐ฅ)) = (๐ ๐2๐๐๐ฅ๐/๐, ๐(๐) ๐โ2๐๐๐ฅ๐/๐) for some ๐(๐) โ C to be determinedlater, where the formulae for ๐+(๐ฅ) and ๐โ(๐ฅ) hold for ๐ฅ โค 1 and ๐ฅ โค 0 respectively. (Physically, ๐+(๐ฅ, ๐ฆ) is theincident wave, ๐โ(๐ฅ, ๐ฆ) is the reflected wave, and | lim๐โ0 ๐(๐)|2 is the reflection probability.)
For the half-plane ๐ฅ โฅ ๐ฟ/๐ take (๐+(๐ฅ), ๐โ(๐ฅ)) = (๐ก(๐) ๐2๐๐๐ฅ๐/๐, 0) for some ๐ก(๐) โ C to be determined later,where the formulae for ๐+(๐ฅ) and ๐โ(๐ฅ) hold for ๐ฅ โฅ ๐ฟ/๐+ 1 and ๐ฅ โฅ ๐ฟ/๐ respectively. (Physically, ๐+(๐ฅ) is thetransmitted wave, and | lim๐โ0 ๐ก(๐)|2 is the transmission probability.)
For 0 โค ๐ฅ โค ๐ฟ/๐ search for a solution of the form
๐+(๐ฅ) = ๐(๐) ๐2๐๐๐ฅ๐/๐โฒ(๐) + ๐(๐) ๐โ2๐๐๐ฅ๐/๐โฒ(๐),
๐โ(๐ฅ) = ๐(๐) ๐2๐๐๐ฅ๐/๐โฒ(๐) + ๐(๐) ๐โ2๐๐๐ฅ๐/๐โฒ(๐).
for some coefficients ๐, ๐, ๐, ๐ โ C and ๐โฒ > 0 depending on ๐ (we have a combination of refracted and reflectedwaves). In what follows ๐, ๐, ๐, ๐, ๐โฒ, ๐, ๐ก denote the limits of the above numbers as ๐ โ 0.
The limit ๐โฒ (wavelength of refracted wave) is determined by Diracโs equation as follows. Substituting theexpressions for ๐ยฑ(๐ฅ) into Diracโs equation, canceling common factors, and taking the limit ๐ โ 0 we get
๐๐ =
(๐โ 2๐
๐โฒ +2๐
๐
)๐,
๐๐ =
(๐+
2๐
๐โฒ +2๐
๐
)๐.
29
We also get an analogous equation with ๐ and ๐ instead of ๐ and ๐. This implies that ๐โฒ = ๐/๐, where ๐ =โ1 +๐๐/๐ (refractive index ).Sewing together the solutions (namely, ๐+(๐ฅ) at ๐ฅ = 1 and ๐ฅ = ๐ฟ/๐+1, whereas ๐โ(๐ฅ) at ๐ฅ = 0 and ๐ฅ = ๐ฟ/๐)
gives a system of linear equations in ๐, ๐, ๐, ๐, ๐, ๐ก:
๐โ(0) = ๐(๐) = ๐(๐) + ๐(๐), ๐โ(๐ฟ/๐) = 0 = ๐(๐) ๐2๐๐๐ฟ/๐โฒ+ ๐(๐) ๐โ2๐๐๐ฟ/๐โฒ
,
๐+(1) = ๐๐2๐๐๐ฟ/๐ = ๐(๐) ๐2๐๐๐ฟ/๐โฒ+ ๐(๐) ๐โ2๐๐๐ฟ/๐โฒ
, ๐+(๐ฟ/๐+ 1) = ๐ก(๐) ๐2๐๐(๐ฟ+๐)/๐ = ๐(๐) ๐2๐๐(๐ฟ+๐)/๐โฒ+ ๐(๐) ๐โ2๐๐(๐ฟ+๐)/๐โฒ
.
Passing to the limit ๐ โ 0 and solving the resulting system of linear equations we find ๐, ๐, ๐, ๐, ๐, ๐ก Surely, one stillneeds to check that even before passing to the limit, the expressions for ๐ยฑ(๐ฅ) determined by our equations indeedsatisfy Diracโs equation in each black square. We omit that.
It remains to prove that (๐+(๐ฅ), ๐โ(๐ฅ)) = lim๐ฆโ+โ๐ฆ+๐ฅ even
๐ 2๐๐ฆ๐/๐๏ฟฝ๏ฟฝ(๐ฅ,๐ฆ,๐3(๐ฅ)๐,๐๐ ,โ); then the number |๐|2 is the required
limit. We do not know any elementary proof of that assertion and thus omit the details.Informally, the idea is to show that the limit does not depend on the initial values ๐+(๐ฅ, 0) for ๐ฅ > 0 and
๐โ(๐ฅ, 0) for ๐ฅ < ๐ฟ/๐; then one can replace these initial values by the solution we have found above and get theresult. Independence on ๐+(๐ฅ, 0) for ๐ฅ > ๐ฟ/๐ and ๐โ(๐ฅ, 0) for ๐ฅ < 0 is obvious. To prove independence on๐ยฑ(๐ฅ, 0) for 0 โค ๐ฅ โค ๐ฟ/๐, assume that ๐ยฑ(๐ฅ, 0) vanishes outside the segment 0 โค ๐ฅ โค ๐ฟ/๐ and consider the lineartransformation
๐ : (๐โ(0, 0), ๐โ(2, 0), . . . , ๐โ(๐ฟ/๐โ 1, 0), ๐+(0, 0), . . . , ๐+(๐ฟ/๐โ 1, 0))
โฆโ (๐โ(0, 2), ๐โ(2, 2), . . . , ๐โ(๐ฟ/๐โ 1, 2), ๐+(0, 2), . . . , ๐+(๐ฟ/๐โ 1, 2)).
Let us show that the absolute values of all the eigenvalues of ๐ are less than 1. Consider an eigenvector ๐ฃ withthe coordinates ๐ฃ1 = ๐โ(0, 0), . . . , ๐ฃ๐ฟ/๐+1 = ๐+(๐ฟ/๐ โ 1, 0). Let ๐ฃ๐ be the leftmost nonzero coordinate. If ๐ > 1then ๐๐ฃ has nonzero (๐ โ 1)-th coordinate; but ๐๐ฃ is proportional to ๐ฃ, a contradiction. Thus ๐ = 1. Hence๐โ(โ2, 2) = 0 by Diracโs equation. Then by the probability conservation
โ๐ฃโ = |๐โ(0, 0)|2 + |๐โ(2, 0)|2 + ยท ยท ยท+ |๐+(๐ฟ/๐โ 1, 0)|2
= |๐โ(โ2, 2)|2 + |๐โ(0, 2)|2 + ยท ยท ยท+ |๐+(๐ฟ/๐โ 1, 2)|2
> |๐โ(0, 2)|2 + ยท ยท ยท+ |๐+(๐ฟ/๐โ 1, 2)|2
= โ๐๐ฃโ.
Thus the absolute value of the eigenvalue ๐ฃ is strictly less than 1. Thus โ๐๐ฆโ โ 0 as ๐ฆ โ โ, which proves thatthe values ๐ยฑ(๐ฅ, 0) for 0 โค ๐ฅ โค ๐ฟ do not affect the limit.
26. Answer : In the table, the vector in the cell (๐ฅ, ๐ฆ) is ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐ข,+), and an empty cell (๐ฅ, ๐ฆ) means that๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐ข,+) = (0, 0).
4(0, 1
2โ2
)(0, 0)
(0, 1
2โ2
)3
(0,โ1
2
) (0, 12
)2
(0,โ 1โ
2
)1 (0,โ1)
y x โ2 โ1 0 1 2 3 4
In the following table the number in the cell (๐ฅ, ๐ฆ) is ๐ (๐ฅ, ๐ฆ, ๐ข,+), and an empty cell (๐ฅ, ๐ฆ) means that๐ (๐ฅ, ๐ฆ, ๐ข,+) = 0.
4 1/8 0 1/8
3 1/4 1/4
2 1/2
1 1
y x โ2 โ1 0 1 2 3 4
(The vectors ๏ฟฝ๏ฟฝ(๐ฅ, ๐ฆ, ๐ข,+) can be easily computed using Problem 31.)For any positive ๐ฆ we have
โ๐ฅโZ
(๐ (๐ฅ, ๐ฆ, ๐ข,+)+๐ (๐ฅ, ๐ฆ, ๐ข,โ)) = 1. (This is one of the statements of Problem 31.)
27. Answer (kindly provided by Gleb Minaev and Ivan Russkikh, participants of Summer Conference of Tourna-ment of Towns):
30
28. Solution. Suppose that ๐ขโฒ is the result of changing the values of ๐ข in all the vertices of a black square (๐, ๐).Notice that if a path ๐ neither starts nor ends in the square (๐, ๐), then it either passes through two vertices witha changed value of ๐ข or does not pass through such vertices at all. Therefore, ๏ฟฝ๏ฟฝ(๐ , ๐ข) = ๏ฟฝ๏ฟฝ(๐ , ๐ขโฒ). The equation๐ (๐ฅ, ๐ฆ, ๐ขโฒ,+) = ๐ (๐ฅ, ๐ฆ, ๐ข,+) for any (๐ฅ, ๐ฆ) = (๐, ๐) follows immediately.
The only case left to prove is (๐, ๐) = (๐ฅ, ๐ฆ) or (๐, ๐) = (0, 0). For each path ๐ starting or ending in (๐, ๐)we have ๏ฟฝ๏ฟฝ(๐ , ๐ขโฒ) = โ๏ฟฝ๏ฟฝ(๐ , ๐ข) because ๐ passes through exactly one vertex with a changed value of ๐ข. Therefore,๐ (๐, ๐, ๐ขโฒ,+) = ๐ (๐, ๐, ๐ข,+).29. Solution. The top-right vertex of the square (๐ฅ, ๐ฆ) is called vertex (๐ฅ, ๐ฆ). Denote by ๐ข(๐ฅ, ๐ฆ) the value of ๐ข atthe vertex (๐ฅ, ๐ฆ). Reversing a black square (๐ฅ, ๐ฆ) means changing the values of ๐ข at all the four vertices of thesquare.
Notice that reversing a black square does not change the sign of any white squares, because the number ofchanged vertices in each white square is even. If all values of a field in a rectangle are positive, then all the whitesquares inside the rectangle are positive. Therefore, it is impossible to make ๐ข identically +1 by reversing blacksquares, if initially there is a negative white square.
Suppose that there are no negative white squares in a rectangle ๐ ร๐ , where ๐ is the height of the rectangle.Let the vertex (0, 0) be at the bottom-left of the rectangle. Let us prove that ๐ข can be made identically +1 throughreversal of black squares by induction over ๐ .
Base: ๐ = 0. Let us describe the algorithm for making ๐ข identically +1 in the rectangle ๐ ร 0 (i.e in vertices(0, 0), (1, 0), . . . , (๐, 0)).
Note that for all ๐ we can reverse either square (๐+ 1, 0) or square (๐+ 1, 1), thereby we can change the signsof the values of ๐ข in vertices (๐, 0) and (๐ + 1, 0). Denote this operation reversing ๐.
The algorithm is as follows: Find the minimum 0 โค ๐ โค ๐ such that ๐ข(๐, 0) = โ1, and reverse ๐. If after thatthere still exists 0 โค ๐ โค ๐ such that ๐ข(๐, 0) = โ1, than the minimum such ๐ is at least 1 more than before usingthe operation. Therefore, after at most ๐ + 1 operations we make ๐ข to be identically +1 in the rectangle ๐ ร 0.
Step: Suppose that the statement is true for ๐ and prove it for ๐ + 1 by giving an explicit algorithm.First make ๐ข identically +1 in the lower rectangle ๐ ร ๐ by applying the algorithm for ๐ . Then for each
white square (๐,๐ + 1) in the rectangle reverse the square (๐,๐ + 2) if ๐ข(๐ โ 1, ๐ + 1) = ๐ข(๐,๐ + 1) = โ1,otherwise do nothing. Finally, if the square (0, ๐ + 1) is black and ๐ข(โ1, ๐ + 1) = โ1 then reverse the square(โ1, ๐ +2), and if the square (๐,๐ +1) is black and ๐ข(๐,๐ +1) = โ1 then reverse the square (๐ +1, ๐ +2).The result is a field, positive in the whole rectangle (๐ + 1)ร๐ .30. Answer: Zero, if one of the sides of the rectangle is 1, and any even number no greater than the number ofwhite squares in the rectangle, if both sides are greater than 1.
Hint. Denote by ๐(๐ฅ, ๐ฆ, ๐ข) the number of vertices of the square (๐ฅ, ๐ฆ) where the value of ๐ข is negative. Noticethat any vertex that is strictly inside the rectangle (not on the boundary) is a vertex of exactly four squares inthat rectangle, two of which are white. Therefore, the sum of ๐(๐ฅ, ๐ฆ, ๐ข) over all white squares in the rectangle iseven. Hence the number of negative white squares is even.
Let us show that any even number no greater than the number of white squares in the rectangle is indeedpossible.
If both the length and the width of the rectangle are even, then for any field that is positive everywhere in therectangle except for ๐ vertices with both coordinates even there are exactly 2๐ negative white squares.
For odd length or width (greater than 1) see the figure below. The identically +1 field has no negative whitesquares. The field ๐ข2๐+2 with 2๐ + 2 negative white squares is obtained from the field ๐ข2๐ by changing all thevalues of ๐ข2๐ to the opposite in any single encircled area where they have not been changed before.
31
31. Answer. Recall that ๐ข(๐ฅ, ๐ฆ) is the value of ๐ข in the top-right vertex of the square (๐ฅ, ๐ฆ). Then
๐1(๐ฅ, ๐ฆ, ๐ข) =1โ2๐ข(๐ฅ, ๐ฆ โ 1)(๐1(๐ฅ+ 1, ๐ฆ โ 1, ๐ข) + ๐2(๐ฅ+ 1, ๐ฆ โ 1, ๐ข)),
๐2(๐ฅ, ๐ฆ, ๐ข) =1โ2๐ข(๐ฅโ 1, ๐ฆ โ 1)(โ๐1(๐ฅโ 1, ๐ฆ โ 1, ๐ข) + ๐2(๐ฅโ 1, ๐ฆ โ 1, ๐ข)).
Hint. The solution of the analogue of Problem 4 is very similar to the original one, only the factors ๐ข(๐ฅ, ๐ฆโ 1)and ๐ข(๐ฅ โ 1, ๐ฆ โ 1) are added due to the last step of the path ๐ passing through the top-right vertex of either(๐ฅ, ๐ฆ โ 1) or (๐ฅโ 1, ๐ฆ โ 1).
Let us prove the analogue of Problem 5 by induction over ๐ฆ. The step of induction follows immediately fromthe following computation:โ
๐ฅโZ๐ (๐ฅ, ๐ฆ + 1, ๐ข) =
โ๐ฅโZ
[๐1(๐ฅ, ๐ฆ + 1, ๐ข)2 + ๐2(๐ฅ, ๐ฆ + 1, ๐ข)2
]=
โ๐ฅโZ
๐1(๐ฅ, ๐ฆ + 1, ๐ข)2 +โ๐ฅโZ
๐2(๐ฅ, ๐ฆ + 1, ๐ข)2 =
=1
2
โ๐ฅโZ
๐ข(๐ฅ, ๐ฆ)2(๐1(๐ฅ+ 1, ๐ฆ, ๐ข) + ๐2(๐ฅ+ 1, ๐ฆ, ๐ข))2 +1
2
โ๐ฅโZ
๐ข(๐ฅโ 1, ๐ฆ)2(๐2(๐ฅโ 1, ๐ฆ, ๐ข)โ ๐1(๐ฅโ 1, ๐ฆ, ๐ข))2 =
=โ๐ฅโZ
(๐1(๐ฅ, ๐ฆ, ๐ข) + ๐2(๐ฅ, ๐ฆ, ๐ข))2
2+โ๐ฅโZ
(๐2(๐ฅ, ๐ฆ, ๐ข)โ ๐1(๐ฅ, ๐ฆ, ๐ข))2
2=
โ๐ฅโZ
[๐1(๐ฅ, ๐ฆ, ๐ข)
2 + ๐2(๐ฅ, ๐ฆ, ๐ข)2]=
โ๐ฅโZ
๐ (๐ฅ, ๐ฆ, ๐ข).
32. Answer: ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) =
โงโชโชโชโชโจโชโชโชโชโฉ๐ (๐ฅ, ๐ฆ,โ)๐ (๐ฅโฒ โ ๐ฅ0, ๐ฆ,โ), if ๐ธ = (๐ฅ+ 1, ๐ฆ), ๐ธโฒ = (๐ฅโฒ + 1, ๐ฆ),
๐ (๐ฅ, ๐ฆ,โ)๐ (๐ฅโฒ โ ๐ฅ0, ๐ฆ,+), if ๐ธ = (๐ฅ+ 1, ๐ฆ), ๐ธโฒ = (๐ฅโฒ โ 1, ๐ฆ),
๐ (๐ฅ, ๐ฆ,+)๐ (๐ฅโฒ โ ๐ฅ0, ๐ฆ,โ), if ๐ธ = (๐ฅโ 1, ๐ฆ), ๐ธโฒ = (๐ฅโฒ + 1, ๐ฆ),
๐ (๐ฅ, ๐ฆ,+)๐ (๐ฅโฒ โ ๐ฅ0, ๐ฆ,+), if ๐ธ = (๐ฅโ 1, ๐ฆ), ๐ธโฒ = (๐ฅโฒ โ 1, ๐ฆ).
Solution. Notice that ๏ฟฝ๏ฟฝ(๐ , ๐ โฒ) = โ๐๐(๐ )๐(๐ โฒ), where ๏ฟฝ๏ฟฝ๐ denotes the product of complex numbers ๏ฟฝ๏ฟฝ and ๏ฟฝ๏ฟฝ.Without loss of generality suppose that ๐ฅโฒ > ๐ฅ. Due to the condition ๐ฅ0 โฅ 2๐ฆ there are no paths starting at ๐ด andending in ๐น โฒ and no paths starting at ๐ดโฒ and ending in ๐น .
Thereforeโ
๐ :๐ด๐ตโ๐ธโฒ๐น โฒ๐ โฒ:๐ดโฒ๐ตโฒโ๐ธ๐น
๏ฟฝ๏ฟฝ(๐ , ๐ โฒ) = 0 and ๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) =โ
๐ :๐ด๐ตโ๐ธ๐น๐ โฒ:๐ดโฒ๐ตโฒโ๐ธโฒ๐น โฒ
๏ฟฝ๏ฟฝ(๐ , ๐ โฒ).
If ๐ธ = (๐ฅโ 1, ๐ฆ) and ๐ธโฒ = (๐ฅโฒ โ 1, ๐ฆ) then ๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) = โ๐โ๐ ,๐ โฒ
๏ฟฝ๏ฟฝ(๐ )๐(๐ โฒ), where the sum is over
๐ and ๐ โฒ ending with an upwards-right move. Therefore, ๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) = โ๐๐(๐ฅ, ๐ฆ,+)๐(๐ฅโฒ โ ๐ฅ0, ๐ฆ,+).Hence ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) = ๐ (๐ฅ, ๐ฆ,+)๐ (๐ฅโฒ โ ๐ฅ0, ๐ฆ,+).
If ๐ธ = (๐ฅ+1, ๐ฆ) and ๐ธโฒ = (๐ฅโฒโ 1, ๐ฆ) then ๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) = โ๐โ๐ ,๐ โฒ
๏ฟฝ๏ฟฝ(๐ )๐(๐ โฒ), where the sum is over ๐
ending with an upwards-left move and ๐ โฒ ending with an upwards right move. Hence ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) =๐ (๐ฅ, ๐ฆ,โ)๐ (๐ฅโฒ โ ๐ฅ0, ๐ฆ,+).
If ๐ธ = (๐ฅโ 1, ๐ฆ) and ๐ธโฒ = (๐ฅโฒ+1, ๐ฆ) then ๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) = โ๐โ๐ ,๐ โฒ
๏ฟฝ๏ฟฝ(๐ )๐(๐ โฒ), where the sum is over ๐
ending with an upwards-right move and ๐ โฒ ending with an upwards left move. Hence ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) =๐ (๐ฅ, ๐ฆ,+)๐ (๐ฅโฒ โ ๐ฅ0, ๐ฆ,โ).
32
If ๐ธ = (๐ฅ+ 1, ๐ฆ) and ๐ธโฒ = (๐ฅโฒ + 1, ๐ฆ) then ๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) = โ๐โ๐ ,๐ โฒ
๏ฟฝ๏ฟฝ(๐ )๐(๐ โฒ), where the sum is over
๐ and ๐ โฒ ending with an upwards-left move. Hence ๐ (๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) = ๐ (๐ฅ, ๐ฆ,โ)๐ (๐ฅโฒ โ ๐ฅ0, ๐ฆ,โ).
33. Hint. Suppose that paths ๐ : ๐ด๐ต โ ๐ธ๐น and ๐ โฒ : ๐ดโฒ๐ตโฒ โ ๐ธโฒ๐น โฒ have a common move. Suppose that ๐ถ๐ท is thefirst common move of ๐ and ๐ โฒ. Denote by ๐ the path that coincides with ๐ up to the square ๐ถ and with ๐ โฒ after ๐ถ.and ๐ โฒ the path that coincides with ๐ โฒ before ๐ถ and with ๐ after. The paths ๐ , ๐ โฒ start with the moves ๐ด๐ต,๐ดโฒ๐ตโฒ
and end with moves ๐ธโฒ๐น โฒ and ๐ธ๐น respectively.The total number of turns in the paths ๐ and ๐ โฒ equals the total number of turns in ๐ and ๐ โฒ , because ๐ถ๐ท is a
common move for all of them. Therefore, ๏ฟฝ๏ฟฝ(๐ , ๐ โฒ) = ๏ฟฝ๏ฟฝ(๐ , ๐ โฒ). Notice that ๐ : (๐ , ๐ โฒ) โฆโ (๐ , ๐ โฒ) is a bijection, because๐ถ๐ท is exactly the first common move. One of (๐ , ๐ โฒ) and (๐ , ๐ โฒ) contributes to ๏ฟฝ๏ฟฝ(๐ธ๐น,๐ธโฒ๐น โฒ) with the coefficient(+1) and the other one with the the coefficient (โ1). Therefore, these pairs cancel each other.
34. Hint. Let us prove this equality by induction over ๐ฆ. The base is obvious.Note that by analogy with Problem 11, ๏ฟฝ๏ฟฝ(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) is parallel to either the line ๐ฅ = 0 or the line
๐ฆ = 0. Denote by ๐1(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) and ๐2(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) its coordinates.Fix the starting moves ๐ด๐ต and ๐ดโฒ๐ตโฒ of the paths. Denote by ๐(๐ธ,๐ธโฒ, โ,+,โ) = ๐1(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ),
where ๐ธ๐น is an upwards-right move, ๐ธโฒ๐น โฒ is an upwards-left move, and ๐1(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ธ๐น,๐ธโฒ๐น โฒ) is the (possibly)non-zero coordinate of the corresponding vector. ๐(๐ธ,๐ธโฒ, โ,+,+), ๐(๐ธ,๐ธโฒ, โ,โ,+) and ๐(๐ธ,๐ธโฒ, โ,โ,โ) are definedanalogously. Denote ๐(๐ธ,๐ธโฒ, โ,+,โ) = ๐1(๐ด๐ต,๐ดโฒ๐ตโฒ โ ๐ท๐ธ,๐ทโฒ๐ธโฒ), where ๐ท๐ธ is an upwards-right move and ๐ทโฒ๐ธโฒ
is an upwards-left move. ๐(๐ธ,๐ธโฒ, โ,+,+), ๐(๐ธ,๐ธโฒ, โ,โ,+) and ๐(๐ธ,๐ธโฒ, โ,โ,โ) are defined analogously.Note that the induction step is a proof of the equalityโ
๐ธ,๐ธโฒ
((๐(๐ธ,๐ธโฒ, โ,+,+))2 + (๐(๐ธ,๐ธโฒ, โ,+,โ))2 + (๐(๐ธ,๐ธโฒ, โ,โ,+))2 + (๐(๐ธ,๐ธโฒ, โ,โ,โ))2) =
=โ๐ธ,๐ธโฒ
((๐(๐ธ,๐ธโฒ, โ,+,+))2 + (๐(๐ธ,๐ธโฒ, โ,+,โ))2 + (๐(๐ธ,๐ธโฒ, โ,โ,+))2 + (๐(๐ธ,๐ธโฒ, โ,โ,โ))2).
We will prove this equation by proving that for every pair (๐ธ,๐ธโฒ), where ๐ธ and ๐ธโฒ have the same ๐ฆ-coordinate,
(๐(๐ธ,๐ธโฒ, โ,+,+))2 + (๐(๐ธ,๐ธโฒ, โ,+,โ))2 + (๐(๐ธ,๐ธโฒ, โ,โ,+))2 + (๐(๐ธ,๐ธโฒ, โ,โ,โ))2 =
=(๐(๐ธ,๐ธโฒ, โ,+,+))2 + (๐(๐ธ,๐ธโฒ, โ,+,โ))2 + (๐(๐ธ,๐ธโฒ, โ,โ,+))2 + (๐(๐ธ,๐ธโฒ, โ,โ,โ))2.
Analogously to Problem 4 we obtain
๐(๐ธ,๐ธโฒ, โ,+,+) =1
2(๐(๐ธ,๐ธโฒ, โ,+,+)โ ๐(๐ธ,๐ธโฒ, โ,+,โ)โ ๐(๐ธ,๐ธโฒ, โ,โ,+)โ ๐(๐ธ,๐ธโฒ, โ,โ,โ)),
๐(๐ธ,๐ธโฒ, โ,+,โ) =1
2(๐(๐ธ,๐ธโฒ, โ,+,+)โ ๐(๐ธ,๐ธโฒ, โ,+,โ) + ๐(๐ธ,๐ธโฒ, โ,โ,+) + ๐(๐ธ,๐ธโฒ, โ,โ,โ)),
๐(๐ธ,๐ธโฒ, โ,โ,+) =1
2(๐(๐ธ,๐ธโฒ, โ,+,+) + ๐(๐ธ,๐ธโฒ, โ,+,โ)โ ๐(๐ธ,๐ธโฒ, โ,โ,+) + ๐(๐ธ,๐ธโฒ, โ,โ,โ)),
๐(๐ธ,๐ธโฒ, โ,โ,โ) =1
2(โ๐(๐ธ,๐ธโฒ, โ,+,+)โ ๐(๐ธ,๐ธโฒ, โ,+,โ)โ ๐(๐ธ,๐ธโฒ, โ,โ,+) + ๐(๐ธ,๐ธโฒ, โ,โ,โ)).
With this, we can prove our equation by substitution and expansion.
35. Answer: ๐ (๐ด๐ต,๐ตโฒ๐ดโฒ โ ๐ธ๐น,๐น โฒ๐ธโฒ) =
โงโชโชโชโชโจโชโชโชโชโฉ๐ (๐ฅ, ๐ฆ,โ)๐ (๐ฅ0 โ ๐ฅโฒ, ๐ฆ,โ), if ๐ธ = (๐ฅ+ 1, ๐ฆ), ๐ธโฒ = (๐ฅโฒ + 1, ๐ฆ),
๐ (๐ฅ, ๐ฆ,โ)๐ (๐ฅ0 โ ๐ฅโฒ, ๐ฆ,+), if ๐ธ = (๐ฅ+ 1, ๐ฆ), ๐ธโฒ = (๐ฅโฒ โ 1, ๐ฆ),
๐ (๐ฅ, ๐ฆ,+)๐ (๐ฅ0 โ ๐ฅโฒ, ๐ฆ,โ), if ๐ธ = (๐ฅโ 1, ๐ฆ), ๐ธโฒ = (๐ฅโฒ + 1, ๐ฆ),
๐ (๐ฅ, ๐ฆ,+)๐ (๐ฅ0 โ ๐ฅโฒ, ๐ฆ,+), if ๐ธ = (๐ฅโ 1, ๐ฆ), ๐ธโฒ = (๐ฅโฒ โ 1, ๐ฆ).
Hint. Notice that ๏ฟฝ๏ฟฝ(๐ , ๐ โฒ) = ๐๐(๐ )๐*(๐ โฒ), where ๏ฟฝ๏ฟฝ๐ denotes the product of complex numbers ๏ฟฝ๏ฟฝ and ๏ฟฝ๏ฟฝ, and ๏ฟฝ๏ฟฝ*
denotes conjugation.For the remaining problems, we only give preliminary draft solutions.
36. Answer: for each ๐ขfin we have
๐ (2, 2, ๐ขfin, 1,+) = 1/4, ๐ (0, 2, ๐ขfin, 1,โ) = 1/4,
๐ (3, 3, ๐ขfin, 1,+) = 1/32, ๐ (1, 3, ๐ขfin, 1,โ) = 1/32,
๐ (1, 3, ๐ขfin, 1,+) = 1/32, ๐ (โ1, 3, ๐ขfin, 1,โ) = 1/32;
all the other probabilities vanish for ๐ฆ = 2, 3.
37. Answer :โ๐ข๏ฟฝ๏ฟฝ(๐, ๐ , ๐ข) = 23๐ฆโ4(1 + ๐2)
๐ฆโ12 cos((๐ โ ๐) arctan(๐))๐(๐ ).
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Hint. For brevity, denote ๐ = 1
2(๐ฆโ2)2 (1+๐2)(๐ฆโ1)2
2
.
Suppose that ๐ and ๐ are the numbers of white squares in ๐ to the right and to the left from ๐ respectively.Notice that ๐ and ๐ are odd if and only if the number of โโโ signs on the path ๐ is odd (that is, ๏ฟฝ๏ฟฝ(๐ , ๐ข) = โ๏ฟฝ๏ฟฝ(๐ )).
Easy-to-prove lemma: if ๐ is the number of black squares in ๐ not touching the boundary, then the numberof fields ๐ข with exactly 2๐ negative white squares equals 2๐ times the number of ways to assign โโโ to 2๐ whitesquares in ๐ . The factor 2๐ is the number of gauge transformations from the Problem 28.
By the lemma we have
โ๐ข
๏ฟฝ๏ฟฝ(๐, ๐ , ๐ข) = ๐โ๐ข
(โ๐๐)๐๏ฟฝ๏ฟฝ(๐ , ๐ข) = 2๐๐
โกโฃ โ๐,๐ ๐๐ฃ๐๐
(โ๐๐)๐+๐
(๐
๐
)(๐
๐
)โ
โ๐,๐ ๐๐๐
(โ๐๐)๐+๐
(๐
๐
)(๐
๐
)โคโฆ ๏ฟฝ๏ฟฝ(๐ ) =
= 2๐โ2๐[((1 + ๐๐)๐ + (1โ ๐๐)๐)
((1 + ๐๐)๐ + (1โ ๐๐)๐
)โ ((1 + ๐๐)๐ โ (1โ ๐๐)๐)
((1 + ๐๐)๐ โ (1โ ๐๐)๐
)]๏ฟฝ๏ฟฝ(๐ ) =
= 2๐๐๐ ๐((1 + ๐๐)๐(1โ ๐๐)๐
)๏ฟฝ๏ฟฝ(๐ ) = 2๐(1 + ๐2)
๐+๐2 ๐ cos((๐ โ ๐) arctan(๐))๐(๐ ).
Since ๐ = ๐ + ๐ = ๐ฆ2 โ ๐ฆ, it follows thatโ๐ข๏ฟฝ๏ฟฝ(๐, ๐ , ๐ข) = 23๐ฆโ4(1 + ๐2)
๐ฆโ12 cos((๐ โ ๐) arctan(๐))๐(๐ ).
38. Answer. Denote
๐โโ(๐ฅ, ๐ฆ) = ๐2(๐ฅ, ๐ฆ, ๐ขโ, ๐,โ), ๐โ+(๐ฅ, ๐ฆ) = ๐1(๐ฅ, ๐ฆ, ๐ขโ, ๐,+),
๐+โ(๐ฅ, ๐ฆ) = ๐1(๐ฅ, ๐ฆ, ๐ข+, ๐,โ), ๐++(๐ฅ, ๐ฆ) = ๐2(๐ฅ, ๐ฆ, ๐ข+, ๐,+).
Denote by ๐ (๐ฆ) = ๐ the rectangle formed by all the squares (๐ฅโฒ, ๐ฆโฒ) such that 1 โ ๐ฆ < ๐ฅโฒ < ๐ฆ and 0 < ๐ฆโฒ < ๐ฆ.Denote ๐ผ = ๐ผ(๐ฅ, ๐ฆ) = (ฮ๐ โ ฮ๐) arctan ๐, where ฮ๐ and ฮ๐ are the numbers of white squares in ๐ (๐ฆ) outside๐ (๐ฆ โ 1) lying to the left and to the right from the square (๐ฅ+ 1, ๐ฆ โ 1). Denote ๐ฝ = ๐ผ(๐ฅโ 2, ๐ฆ). Thenโงโชโชโชโชโชโจโชโชโชโชโชโฉ
๐++(๐ฅ, ๐ฆ) = cos๐ฝโ2(๐++(๐ฅโ 1, ๐ฆ โ 1)โ ๐+โ(๐ฅโ 1, ๐ฆ โ 1))โ sin๐ฝโ
2(๐โ+(๐ฅโ 1, ๐ฆ โ 1)โ ๐โโ(๐ฅโ 1, ๐ฆ โ 1)) ;
๐+โ(๐ฅ, ๐ฆ) = cos๐ผโ2(๐++(๐ฅ+ 1, ๐ฆ โ 1) + ๐+โ(๐ฅ+ 1, ๐ฆ โ 1))โ sin๐ผโ
2(๐โ+(๐ฅ+ 1, ๐ฆ โ 1) + ๐โโ(๐ฅ+ 1, ๐ฆ โ 1)) ;
๐โ+(๐ฅ, ๐ฆ) = sin๐ฝโ2(๐++(๐ฅโ 1, ๐ฆ โ 1)โ ๐+โ(๐ฅโ 1, ๐ฆ โ 1)) + cos๐ฝโ
2(๐โ+(๐ฅโ 1, ๐ฆ โ 1)โ ๐โโ(๐ฅโ 1, ๐ฆ โ 1)) ;
๐โโ(๐ฅ, ๐ฆ) = sin๐ผโ2(๐++(๐ฅ+ 1, ๐ฆ โ 1) + ๐+โ(๐ฅ+ 1, ๐ฆ โ 1)) + cos๐ผโ
2(๐โ+(๐ฅ+ 1, ๐ฆ โ 1) + ๐โโ(๐ฅ+ 1, ๐ฆ โ 1)) .
Hint. The latter formula defines an orthogonal transformation, which implies thatโ
๐ฅโZ(๐ (๐ฅ, ๐ฆ, ๐,+) +๐ (๐ฅ, ๐ฆ, ๐,โ)) = 1 by induction over ๐ฆ.
39. Answer. The number assigned to a basis configuration is shown below it:
1 โ๐๐๐โ1+๐2๐2
1โ1+๐2๐2
1โ1+๐2๐2
โ๐๐๐โ1+๐2๐2
โ 1.
Hint. Let us prove that the sum over all pairs of paths passing through vertices with โโโ signs only equals theproduct of all the numbers assigned to the black squares in the rectangle.
Notice that the squares of type (1) do not contribute to the result at all. Squares of types (2) and (5) canonly be passed through by a path with a turn in that square, which means contributing a factor of โ๐๐๐โ
1+๐2๐2to
๏ฟฝ๏ฟฝ(๐ , ๐ โฒ). Squares of types (3) and (4) can only be passed through by a path with a turn in that square which meanscontributing a factor of 1โ
1+๐2๐2to ๏ฟฝ๏ฟฝ(๐ , ๐ โฒ). Squares of type (6) can be passed through in two different ways: by
two turning or two not turning paths. When the paths do turn, the square contributes(
โ๐๐๐โ1+๐2๐2
)2and when they
do not, it contributes โ11+๐2๐2
, which sums up to โ1.
Acknowledgements
The authors are grateful to V. Skopenkova for some of the figures, G. Chelnokov, I. Ivanov for useful discus-sions, and all participants of Summer Conference of Tournament of Towns in Arandelovac and Summer School inContemporary Mathematics in Dubna for their contribution.
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