analysis iii integration

8/9/2019 Analysis III Integration

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Analysis III: Integration

Easter 2007

Professor Terry Lyons

F19, Mathematical Institute

[email protected]

March 27, 2008

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1 Integrating continuous functions

The space of functions mapping an interval into the reals

f : J Ris huge! We are obviously familiar with those such as x

x2 which are given

by a formula but in some sense these are a minority. Even simple functions suchas

(x) = [x] = max{n | n x}(which is always defined because of the standard axioms for the real numbers)generated considerable controversy when they were first introduced.

Many functions cannot be graphed very easily if at all. For example

f

p

q

=

1

q, (p,q) = 1, p, q Z, q > 0

f(x) = 0, x R\Qf(0) = 0.

Our goal in this course is to identify a sensible class of functions that can be

integrated. Having done this, we must make it happen, make sense ofba

f(x) dx

and prove that it has all the obvious properties, such as linearity. I hope thatyou will be surprised by the way that we are able to move relatively smoothlyfrom rather primitive notions to quite sophisticated information about the basicfunctions, such as the exponential, logarithm, and the trigonometric functionssin, cos etc..

Integration will provide one of the first serious contexts you will meet wherewe examine, in a rigorous way, functions on functions!

In order to defined the integral we are going to consider two basic classes of

functions1. Step Functions: The functions f defined on an interval I, for which there

is a finite partition Pf I so that f is constant on each remaining (open)interval.

2. The continuous functions on a closed interval J.

If we fix J then both of these spaces of functions are real vector spaces.

1.1 Step functions and intervals

Definition 1.1 A non-empty subset I ofR is an interval if, whenever x I,y I, and

x < z < y, z

R

then z I.

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Definition 1.2 The endpoints of interval I are defined to be b = sup I anda = infI. It is allowed and can happen that b = + or a = .

Because an interval I is always nonempty we may choose x I and, recallingthe definition of infI as the greatest lower bound, concluded that a x andsimilarly that x b. It follows that a b. With similar effort (exercise) andusing the definition of infimum and the basic property of an interval, one mayprove that (a, b) I.Lemma 1.3 Any interval is of the form

[a, b) , (a, b] , [a, b] , (a, b)

[a, ) , (a, ) , (, b] , (, b) ,Rfor some a b R. In any case where one of these sets is non-empty, and inparticular when a < b, any such set is an interval.

Exercise 1.4 If I and J are intervals, and J contains no endpoint of I theneither J I or J and I are disjoint.

Exercise 1.5 The intersection of two intervals is an interval or it is empty.

Definition 1.6 A partitionP of an intervalI with endpoints a and b (possibly

) is a finite sequence (ai)ki=0 so and thata0 = a < a1 < a2 < < ak = b.

The compliment of a partition P in the interval I , and written I\P, is theset

I\P = x | x kj=1 (aj1, aj) ,and is a finite union of disjoint open intervals.

Definition 1.7 A function defined on an interval I is a step function if there

is a partition P = (ai)ki=0 of the interval so that is constant on each interval(aj1, aj), j k. Any particular choice P = (ai)ki=0 of partition for the intervalI with this property is called a witness.

If is a step function then there are, in general, infinitely many partitionswitnessing this fact. The class of Roe step functions and the class we havedefined here correspond.

Remark 1.8 By considering the intervals [aj , aj ] as well as the (aj1, aj) onesees that a function defined on an interval I is a step function if and only ifone can decompose the interval into finitely many disjoint subintervals so that is constant on each of them. The decomposition is not unique.

Lemma 1.9 Any step function defined on an interval I can be extended to astep function defined on the intervalR by making it zero off I.

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Proof. We treat the case of a bounded interval; the other cases are similar. Itis enough to add the points to a partition

P = (a0 = a < a1 < a2 < < ak = b)

witnessing the fact that is a step function on I. By definition is zero on

(b, +) and (, a) and so is constant there.P = ( < a0 = a < a1 < a2 < < ak = b < )

is a partition demonstrating that is a step function.

Definition 1.10 If P is a partition of an interval I and P is a finite set ofpoints satisfying P P I then we call P refinement of P.

Proposition 1.11 If is a step function on I and P is a partition that giveswitness to this, and if P is a refinement of P, then P is also a witness to the

fact that is a step function.

Suppose is a step function on I witnessed by the partition P, and that

a second partition P is a refinement of P so that P P. Then J\P J\P.Consider one of the disjoint open intervals N which together comprise J\P.It must intersect one of the disjoint open intervals N that make up J\P. Byconstruction N does not contain any end point of any such an interval. So usingexercise 1.4 it must be properly contained in N since it is not disjoint. But isconstant on N and so constant on N .

Definition 1.12 If A R is a set, then the indicator function A of A is thefunction

A (x) = 1, x AA (x) = 0, x = A.

If I is an interval then it is clear that its indicator function I (x) is a stepfunction.

If f is a step function on J and R then it is obvious that f is also astep function on J.

Proposition 1.13 The sum of two step functions f and g on an interval I isa step function on I.

Proof. Let Pf and Pg be finite partitions witnessing that f and g are stepfunctions. Now let P = Pf Pg . Then P is finite and a refinement of Pf andPg. In view of Proposition 1.11 P is witness to the fact that f and g are bothstep functions. We can order the points of P: (a0 < a1 < a2 < < ak) . Sincef and g are both constant on (aj1, aj) their sum is as well. As this holds for

all j k the function f + g is constant on every interval in I\P and so a stepfunction with witness P.

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Exercise 1.14 Prove that if I J are intervals then the restriction of a stepfunction on J to I is also a step function.

Exercise 1.15 Any step function can be written in the form of a finite sum

=

k

j=1

Ij

where Ir are intervals. Conversely any such sum is a step function onR.

A Priestley step function is any function that can be expressed in the formkj=1 Ij (x) where Ij are bounded intervals.

Exercise 1.16 The Priestley step functions are the step functions on theintervalR that are zero on the initial and final intervals

(, a1) , (ak1, )

of any partition that is witness to being a step function.

Remark 1.17 In this way, we have reconciled the different notations used indifferent publications from the Institute. We will stick to our version, which isequivalent to the definition used in Roe. However, the differences between theclasses of functions described by these definitions are minor, and the key issueslay elsewhere!

1.2 The Integral of a Step Function

Suppose that is a step function on a fixed closed interval J = [a, b] R, andthat P = (a0 = a < a1 < ak = b) is a partition ofJ that witnesses this; then,for i k, there are constants ci R so that

(x) = ci, x

(a

i1, a

i) .

We might define I(, P) =k

i=1 ci (ai ai1), and perhaps this makes a goodstarting point, as the definition for the integral of . However, it might bethe case that a second partition P witnessing as a step function could give adifferent integral.

Lemma 1.18 Suppose that is a step function on J witnessed by the partitionP, and that a second partition P is a refinement of P, then P also witnessesthat is a step function, and

I(, P) = I

, P

.

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Proof. Let |P| = n andP = n. We proceed by induction. Suppose the

proposition is false, then among all counterexamples, we can choose a one: ,J, P, P with

I(, P) = I

, P

.

which minimises n and among these choose an example that maximises n. If

n < n 1, then, by dropping one point from P\P we can find a partition Pwith

P = n 1 that is a refinement of P and for which P is a refinement. Byinduction I(, P) = I

, P

and so I

, P

= I

, P

, but

P = n 1 >|P| which contradicts the assumption that our (counter) example was maximalamong all those with

P = n.Hence, we may assume that |P| =

P 1. LetP = (a0 = a < a1 < an = b)

and let a be the unique element of P\P. Then a J\P so there is a j n sothat

a (aj1, aj) .Since is a step function witnessed by P there are ci so that (x) = cj forx (aj1, aj), j n and

I(, P) =ni=1

ci (ai ai1)

while, using the definition again we see that

I

, P

=

j1i=1

ci (ai ai1)

+cj (a aj1)+cj (aj a)

+ni=j

ci (ai ai1)

and becausecj (aj aj1) = cj (a aj1) + cj (aj a)

We have I

, P

= I

, P

. This contradicts the existence of a counterexam-

ple as required.

Corollary 1.19 The integral I(, P) of a step function does not depend on the

choice of partition.

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Proof. Let P and P be two partitions witnessing that is a step function.Let P = P P then P is a refinement of both of these partitions and

I(, P) = I

, P

= I(, P) .

Definition 1.20 We define the integralba of a step function to be I(, P)

for any partition witnessing that is a step function.

Theorem 1.21 (Linearity and Positivity on step functions) Let andbe two step functions on the intervalJ = [a, b] and let and be real numbers;then the integrals of the step functions + , and satisfyb

a

+ =

ba

+

ba

.

If on the interval J = [a, b] thenba

ba

.

If, for some non-empty open interval (c, d) [a, b] one has (a) > (a), a (c, d), then b

a

>

ba

.

Proof. Fix partitions of the interval J = [a, b] witnessing that and areboth step functions on J and, if neccessary refine the partition to include {c, d},by taking the union of these two partitions, choose a single partition P =(a0 = a < a1 < an1 = b) that is a refinement of both. Then, by assumption,there are constants ci and di so that

(x) = ci, x (ai1, ai) (x) = di, x

(ai1, ai)

and for each i

( + ) (x) = ici + di, x (ai1, ai) .So, from the definition of I and the usual distributive laws of arithmetic

I(( + ), P) =ni=1

(ici + di) (ai ai1)

=

ni=1

ici (ai ai1)

+n

i=1

di (ai ai1)

= I(, P) + I(, P) .

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The case of inequality can be proved by following a virtually identical argu-ment. The inequality between the functions becomes an inequality between theterms in a sum and noting that if there is a strict inequality on one of the termsin the sum there is a strict inequality on the sums.

Alternatively, one can use the linearity established in the first part of thisproof to reduce the problem to a special case: if

on the interval J = [a, b],

then ( ) 0 on the interval J = [a, b] .and is also a step function. Theintegral of a positive step function is clearly (from the definition) positive. Sob

a

( ) 0

and so ba

ba

0ba

ba

For strict inequality between the integrals of and it is enough that the stepfunction ( ) > 0 on a non-empty open interval.

1.3 The Integral of a Continuous Function - details

We have defined step functions, shown that they are a vector space, definedan integral on this space, and observed that the integral is linear and positive.Although these results simple and extremely intuitive they are not quite obvious,one must validate our model for the real numbers by developing these conceptswithin this context without fuss or complexity. In this section our goal is tointegrate continuous functions on bounded intervals that contain their endpoints.These are exactly the intervals on which continuous functions are bounded anduniformly continuous. Both of these properties will play an important role in

our discussion. First was set up some notation. The space of step functions on[a, b] will be denoted by Lstep [a, b] and the space of continuous functions will bedenoted by C[a, b].

Definition 1.22 If f is a real valued function on [a, b] then we may define

ba

f = sup

ba

Lstep [a, b] , f

ba

f = inf

| Lstep [a, b] , fRecall the following a standard result from last term:

Lemma 1.23 If f C[a, b] then there exists m, M (, ) so that for allx [a, b] m f(x) M.

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From this and the positivity of the integral on Lstep [a, b] we see that

Lemma 1.24 If f C[a, b] then

< m (b a) 0 so that if|x y| and x, y [a, b] then |f(x) f(y)| < .

Proof. The result is obvious if b = a as in this case both sides are zero.Fixf and > 0. Choose > 0 so that if |x y| then |f(x) f(y)| < ba .Choose a partition

P = (a0 = a < a1 < an = b)so that

|ar

ar1

| for every r

n.

Define

cr = inf {f(x) | x [ar1, ar]}dr = sup {f(x) | x [ar1, ar]}

and define , by:

= cr on (ar1, ar)

= dr on (ar1, ar)

and (ar) = (ar) = f(ar) .

Then and are in Lstep

[a, b] .Clearly

f

and our choice of ensures that

+ b a .

Thus, and in view of Lemma 1.24,it follows that

ba

ba

f ba

f ba

.

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2 The Fundamental Theorem of Calculus and

the link with differentiation

We now wish to develop some basic consequences that follow from our ability tointegrate continuous functions. We start by proving that the integral is positiveand linear.

Theorem 2.1 (Strict positivity of the integral) Suppose that a < b, f [a, b], f(x) 0 for allx [a, b], and that there is an x [a, b] so thatf(x) > 0.Then

ba

f > 0.

Proof. Suppose that x [a, b] and f(x) > 0. Choose (0, f(x) /2) ;theinterval is nonempty by hypothesis. Since f is continuous there is a > 0 sothat if y (x , x + ) [a, b] then |f(y) f(x)| < . In particular f(y) >. Suppose that = on (x , x + ) [a, b] and zero off it. Then Lstep [a, b] .Since a < b and x [a, b] it follows that there is a > 0 so that oneof the following holds:

x , x (x , x + ) [a, b]x, x + (x , x + ) [a, b]

Then, one of

(x,x) (x,x+)

and in either case, 0 < ba . By definitionba

ba

f ba

f.

Theorem 2.2 (Linearity of the integral) Suppose that f, g C[a, b] andthat , R then b

a

f + g =

ba

f +

ba

g.

Proof. It is enough to treat the cases

ba

f =

ba

f

and

ba (f + g) =

ba f +

ba g.

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separately. If 0 then is a step function less than f if is a step functionless than f.From the linearity of the integral on step functions one can conclude

ba

ba

af

and taking the supremum over the step functions less than f one has

ba

f ba

af.

and similarly ba

af aba

f

and recalling that, because f is continuous, f is integrable and

a

ba

f ba

af. ba

af aba

f

and the result follows. The case where 0 is equally simple.If b = a we are finished as all the integrals will be zero. Fix > 0, applying

corollary 1.27 identify and in Lstep [a, b] so that

f + 2 (b a)

g + 2 (b a)

+ .f + g + + (b a)

Then

ba

(f + g) ba

+ +

(b a)

=

ba

+

ba

+

ba

f +

ba

g + .

Because was arbitrary, we haveba (f + g)

ba

f +ba

g. Applying this tof and g using the relation in the first part of the proof we conclude thatba (f + g) =

ba

f +ba

g.

Theorem 2.3 (additivity on disjoint intervals) Suppose thatb

[a, c] and

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Proof. The result is obvious for step functions;

= [a,b) + [b,c]

so that

c

a

= c

a

[a,b) + c

a

[b,c]

but, adding the point b to any partition P witnessing that [a,b) is a stepfunction on [a, c] one quickly sees that

ba

=

ca

[a,b)cb

=

ca

[b,c]

and yields the required additivity.The result for continuous functions then follows from corollary 1.27. Choose

so that

f

+

(c a)then c

a

f ca

+

(c a)

=

ca

+

=

ba

+

cb

+

ba

f +

cb

f +

and was arbitrary soca f

ba f+

cb f ; using this inequality with f as well

we deduce there is equality.

Corollary 2.4 If f and g are in C[a, b] and f g thenba

f ba

g.

The proof is an excercise! using the previous results.

2.1 The main theorems

Theorem 2.5 (The first mean value theorem (for integrals) FMVT) Supposethat f C[a, b] then for some c [a, b]

ba f = f(c) (b a) .

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Proof. Let

M = sup {f(x) | x [a, b]}m = inf {f(x) | x [a, b]} .

Then f attains the values m and M. Fix e, E

[a, b] so that

f(e) = m

f(E) = M.

Of course m[a,b] f M[a,b] so that using the definition of integral,

m 1b a

ba

f M

(assuming b > a) and by the intermediate value theorem there exists c [e, E] [a, b] so that

1

b

a

b

a

f = f(c) .

If b = a the result is immediate with c = b.

Exercise 2.6 Suppose f, andg are continuous functions on the closed bounded

interval [a, b] and thatba

f =ba

g then there exists a point c [a, b] such thatf(c) = g (c).

Note that the first mean and value theorem (FMVT) follows from this remark

by taking g (x) 1baba

f .

Definition 2.7 If b < a then that we defineba

f to be -ab

f.

We can immediately,and easily, extend theorem 2.3

Theorem 2.8 (additivity) suppose f is a continuous function on some inter-val J and that a, b, c J then

ca

f =

ba

f +

cb

f

Proof. We only treat the new case a < c < b. Using theorem2.3 we see that

ba

f =

ca

f +

bc

f

and so

ca

f = ba

f

b

c

f

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and by definition this

=

ba

f +

cb

f.

Suppose again that f is a continuous function on some interval J and thata is in J.

Definition 2.9 We say F(x) is the (an) indefinite integral of f on J if thereis a constant C R so that for all x J

F(x) =

xa

f + C.

Theorem 2.10 (the fundamental theorem of calculus) Iff C[a, b] wherea < b, and for some fixed C

F(x) =

xa

f + C

thenF is differentiable on(a, b) with the left derivative atb and a right derivative

at a and F (x) = f(x) , x [a, b] .Proof. If we prove that

limh0h>0

F(x + h) F(x)h

= f(x) .

for all x [a, b). Then a similar argument (with the additivity of the integralon disjoint intervals) will prove the existence of a right derivative for x (a, b] .Taken together these give the theorem.

Fix > 0, because f is continuous, we may choose > 0 so that

|f(x)

f(x + h)

|<

whenever h (0, ) and by reducing we can always ensure that in additionx + [a, b]. Now there is a C so that F(x + h) is defined by

F(x + h) =

x+ha

f + C

=

x+hx

f +

xa

f + C

=

x+hx

f + F(x) .

and applying the first mean value theorem there exists [x, x + h] so that

F(x + h) F(x)h

= f() .

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If h (0, ) then |x | < and soF(x + h) F(x)h f(x) < .

which completes the proof.

This allows us to construct an enormous number of differentiable functions!There are many versions of the mean value theorem - and we set the followingas an easty excercise:

Theorem 2.11 (the second mean value theorem for integrals) Supposef, g C[a, b], and that g 0 then there exists c [a, b] so that

ba

fg = f(c)

ba

g

2.1.1 Applications

Example 2.12 (Logarithm) For y R>0 define

log y =

y

1

1x

then log y this differentiable and its derivative is 1y .

The following corrolary is also frequently referred to as the FundamentalTheorem of Calculus

Corollary 2.13 If F is differentiable on the interval J with derivative f andif f is continuous then there exists a constant C R so that for all x J

F(x) =

xa

f + C.

Proof. Consider H(x) defined by

H(x) = F(x) xa

f.

then we have just proved that H is differentiable and H 0 So applying themean value theorem for functions, H is a constant and the result follows.

Example 2.14 We will prove that if f is continuous then

In :=

10

nf(x)

1 + n2x2dx 1

2f (0)

Putg (x) = n

1 + n2x2

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then g is the derivative of tan1 nx. Hence10

g (x) dx =

tan1 nx10

12

The graph of 31+32x2

We know there is a choice ofcn so that10

nf(x)1+n2x2 dx = f(cn)

10

n1+n2x2 dx.If

we can show that cn 0 we will have finished. For this we need to be to becleverer.1

0

nf(x)

1 + n2x2dx =

1/n0

nf(x)

1 + n2x2dx +

11/n

nf(x)

1 + n2x2dx

= f(cn)

tan1 nx1/n0

+

11/n

nf(x)

1 + n2x2dx

where cn [0, 1/

n]. Now

1

1/n

nf(x)

1 + n2x2dx

M

1

1/n

n

1 + n2x2dx

= M

tan1 n tan1 n M

2

2

= 0

as n .Meanwhile, tan1 nx1/n0

= tan1 (

n) 2 , and f(cn) f(0)as |cn|

n.

3 Integration by parts, substitution and meth-

ods of integration

The methods we have introduced are practically useful for explicitly calculating

integrals and derivatives. Unfortunately, any subject that has existed for 400years in some form develops multiple notations and slight variants of definitions.

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Definition 3.1 A function F is the primitive (or anti-derivative) off on [a, b]if F = f on [a, b].

In view of Cor. 2.13 F is the indefininte integral of f if the latter is contin-uous.

Remark 3.2 There are good and bad notations; one often adds a dx to indicatethe argument in the function:

xdx

and sometimes wrongly writes:xdx =

1

2x2 + C

which is not really correct. The following is much better:xtdt =

1

2x2 + C

or even more informatively x

a

tdt =1

2

x2 a2

We can (easily) use our methods to do integration of some complicatedalgebraic functions:

Example 3.3 the function

x2 + 1

(x + 2)2=

(x + 2)2 4 (x + 2) + 5(x + 2)2

= 1 4x + 2

+5

(x + 2)2

so has primitive x 4log(x + 2) 5

x+2 and sox0

t2 + 1

(t + 2)2dt = x 4log(x + 2) 5

x + 2+ 4log2 +

5

2

As you probably learnt at school - anti-derivatives/primitives allow us totranslate known properties of differentiation to properties of integration. Thefollowing is one of the most important:

Proposition 3.4 (Integration by parts) If u, v are differentiable on [a, b]and u, v C[a, b] then

b

a

(uv) + b

a

(uv) = u (b) v (b) u (a) v (a)

= [uv]ba

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Proof. The function uv is differentable, its derivative is (uv) + (uv)Substitution

Proposition 3.5 (Substitution) Suppose that g C[c, d] is monotone in-creasing, and continously differentiable so g C[c, d] , g > 0 and suppose thatg (c) = a, g (d) = b.Suppose that f C[a, b] . Then

d

c

f gg

=

ba

f

Proof. We can apply the fundamental theorem of calculus. Let F(x) :=xa

fand define G (u) = F(g (u)) then

G (c) = F(g (c))

= F(a)

= 0

b

a

f = G (d)

= G (d) G (c)and providing G is continuously differentiable one has

G (d) G (c) =dc

G

so recalling that (FTC again) that F = f and applying the chain rule

G (u) = f g (u) g (u)and hence

ba f(x) dx =

dc f g (u) g (u) du

Example 3.6 The integral of log :x1

log t =

x1

(t)log t

= x1

(t)1

t+ [t log t]

x1

= x log x x + 1Example 3.7 The integral

I = 2

0

1

1 + sin xdx

and use substitutiont = tan (x/2)

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http://math.berkeley.edu/sassaf/math1b/fa03/tech.pdf

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4 Power series and limits - The trancendental

mathematical functions exp, sin etc.

4.1 The Interchange of Limits and Integration

In general, Real numbers (even interesting ones such as ) lack descriptions

in terms of simple formulae, and instead are defined in terms of limits. In ananalogous way, most of the interesting mathematical functions are constructedas limits; for example, as an infinite series. Operations such as differentiationand integration are often well understood when applied to the the intermediatefunctions used in the construction and we need understand how to inherit someof these properties for the limit.

For example, it is a simple calculation to see that

u (x, t) = a0 +N1

en2t (an cos nx + bn sin nx)

is a solution to the heat equation

t

u = 2

2xu

and most of you will know that it is possible to write a reasonably a generalfunction F as the sum of an infinite Fourier series

F(x) = a0 +

1

(an cos nx + bn sin nx) .

With this in mind it would certainly make sense to ask about the convergenceof

a0 +

1en

2t (an cos nx + bn sin nx)

and particularly to ask whether the limit is a solution to the same differentialequation. If it were, then one would have constructed a solution to the heatequation on the unit interval having periodic boundary conditions at the end ofthe intervals, and having as its initial condition a temperature F.

Problem 4.1 It is a simple calculation and an application of the FTC to seethat

1 +

x0

N0

tn

n!dt =

N+10

xn

n!

Can we deduce from this that:

1 +x0 e

t

dt = ex

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and hence conclude from a second application of the FTC that

d

dxex = ex

Remark 4.2 In the previous two examples, we hoped (and we shall see ourhope fulfilled in this context) that the derivative of the limit (or the integral of

the limit) the limit of the derivatives (integrals) of the approximating sequences -which can be computed computed term by term. Our wish is not always fulfilled!There are even really important examples where it works in our favour that thederivative of the limit is not the limit of the derivatives. So it is importantto understand the sort of hypotheses involved in proving the conditions we givebelow.

Example 4.3 Consider the function defined for x > 0 by

f(x) = n2x, x [0, 1/n]f(x) = 2n n2x, x [1/n, 2/n]f(x) = 0, [2/n, ]

then20 fn = 1 for all n . But for each fixed x, fn (x) 0.

Recall the following definition:

Definition 4.4 Let fn, f be functions on the set [a, b]. we say thatfn funiformly on [a, b] if

limn

supx[a,b]

|fn (x) f(x)| = 0.

and one of the basic theorems about uniform convergence:

Theorem 4.5 Suppose that fn f uniformly on [a, b], where fn C[a, b].Then f C[a, b].Remark 4.6 If you have too show that a sequence of functions converges uni-

formly then one should always be careful. It often happens that the locationwhere

|fn (x) f(x)|takes its maximum will frequently depend on n and so moves. It pays to usecalculus to determine the critical points offn (x) f(x).Exercise 4.7 The functions

fn (x) = nxe

n2n2

2 , x > 0

fn (x) = 0, x 0

converge to 0 uniformly iff < 1. (Hint - compute f(x) = lim fn (x). Thentake logs of |fn (x) f(x)| and differentiate to identify the maximum.)

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4.2 Power series

Now recall form last term that a power series

n=0 anxn always has a radius

of convergence R, and that if |x| < R then the series is absolutely convergentwith

n=0 |an| |x|n < and moreover, for |z| |x| one has

|anz

n

| |an

| |x|n

and applying the M-test, one sees that

n=0

anzn

converges uniformly for |z| |x|.

Lemma 4.11 For anyan R, the power series

n=0 anxn,

n=0 nanxn1and

n=0 anxn+1

n+1 have the same radius of convergence.

Proof. Let R be the radius of convergence of

n=1 nant

n1. Then if |x| < Rone has for n

1

|anxn| Rnanxn1

and so

n=0 anxn converge absolutely by comparison with

|a0x| + Rn=0

nanxn1 .

Hence the radius of convergence R of

n=0 antn satisfies R R. On the other

hand if |x| < R then there is a > 1 so that |x| < R. Then for x = 0

nanxn1 n|x| n |anxnn|

Now, since > 1 the quantity

C(x, ) = supn

n

|x| n

is strictly finite for each and x. So, by comparison with C(x, ) |anxnn| onededuces that

n=0 nanx

n1 is also absolutely convergent. So R R.The other comparison can be achieved by simply relabelling the co-efficients

- put

b0 = 0

bn+1 =an

n + 1

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then

n=0 bnxn,

n=0 nbnxn1 have the same radius of convergence so that

n=1 bnxn and

n=1 nbnx

n1 have the same radius of convergence. Substi-tuting in the definitions of the bn we get

n=1bnx

n =

n=1an1

nxn

=

n=1

an1n

xn

=

n=0

ann + 1

xn+1.

n=1

nbnxn1 =

n=1

an1xn1

=

n=0

anxn

Theorem 4.12 (term by term integration of power series) Suppose thatf(t) =

n=0 ant

n has a radius of convergence R. Then f is continuous on(R, R) and x

0

f(t) dt =

n=0

anxn+1

n + 1

for anyx with |x| < R.Proof. If |x| < R then we know the series n=0 antn converges uniformly on[ |x| , |x|]. By the Theorem 4.8 the integral

x0

f(t) dt = limN

x0

Nn=0

antndt

and by linearity this

= limN

Nn=0

an

x0

tndt

= limN

Nn=0

anxn+1

n + 1

=

n=0anx

n+1

n + 1.

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Example 4.13 The series

1

1 x =0

xn, |x| < 1

leads one to

log (1 x) =n=0

xn+1

n + 1

=

n=1

xn

n

for x (1, 1) .Note: although it is true that log2 = n=1 (1)nn it does not follow from the above arguments and is much more delicate.

Theorem 4.14 Suppose thatf(t) =

n=0 antn has a radius of convergence R

thenf can be differentiated term by term on(R, R) andf (t) = n=0 an1tn1for t (R, R).

Proof. Put g (t) =n=0 an1tn1 for t (R, R) and

G (x) =

x0

g (t) dt, x (R, R) .

Applying term by term integration one has that G (x) + a0 = f(x). Since g iscontinuous we may apply the fundamental theorem of calculus to deduce thatf is differentiable and its derivative is g.

4.3 Those famous functions

Rudin starts his (famous) text-book on Real and Complex Analysis with thestatement that exp is the most important function in mathematics!

We define it by

exp(z) =n=0

xn

n!

and this series converges absolutely for every (complex) number z and, amongother things, is a homomorphism from the abelian group (R, +) to the abeliengroup (R>0, ) . There are a number of important functions that are very closelyrelated to it.

Definition 4.15 We define sin and cos to be the series

sin z =

n=0

(1)n z2n+1(2n + 1)!

cos z =

n=0

(

1)n z2n

(2n)!

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and so z Cexp iz = cos z + i sin z

and if z R then

cos z = R (exp iz)

sin z =I

(exp iz)

Definition 4.16 Similarly we define

sinh z =1

2(exp z exp(z))

cosh z =1

2(exp z + exp (z))

tanh z =sinh z

cosh z.

It is not (at all) obvious that these functions satisfy our intuitive axioms.The logical sequence of events is that we started with axioms for real numbersand slowly but effectively we have created a rigorous machine in which these

functions can be defined and studied. To justify this approach we need to provethat these objects do correpond to our intuitive picture.

One basic property of sin and cos is that

Proposition 4.17 If t R then

(sin t)2 + (cos t)2 = 1

Proof. Using term by term differentiation one observes that ddt sin t = cos t andddt cos t = sin t so applying the product rule the derivative of

I(t) := (sin t)2 + (cos t)2

is identically 0. By the mean value theorem the function I is constant and wehave I(0) = 1 so I 1.It is not quite obvious that sin and cos are periodic! What is

Definition 4.18 If x is the smallest positive solution of cos x = 0 then wedefine = 2x.

Proposition 4.19 exists and is strictly between 2

2 and 4.

Proof. Since cos 0 = 1 and cos is continous, there exists a > 0 so that cos>12on [0, ] so if x exists then x > 0. But we can be more precise! Recall thebasic estimate of alternating series: if an are positive numbers decreasing tozero and if

SN =

Nn=0

(1)n

an

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then the series converges to some limit l and in fact one can be much moreprecise

S0 S2 S2n l S2n+1 S3 S1|S2n S2n+1| a2n+1

and we can apply this to the series for cos .

Lemma 4.20 If z > 0 then the terms

an :=z2n

(2n)!

are monotone decreasing for all n such that 0 z a2 > a3 etc.

Proof. Note thatan+1

an=

z2

(2n + 2) (2n + 1).

Corollary 4.21 For 0 x 2 the function cos x satisfies

1 x2

2+

x4

4! cos x

and if x [0, 3] then

cos x > 1 x2

2+

x4

4! x

6

6!

and in particular cos2 38 while for x

0,

2

the functioncos x > 0.

Corollary 4.22 The number exists and satisfies 22 < 4.Proof. Now cos0 = 1 and cos2 < 0 so that the continuity of cos,togetherwith the intermediate value theorem ensure that the set of x (0, 2) for whichcos x = 0 is closed and nonempty. It is therefore also compact and has a smallestelement which we can (and do) call . By using higher approximations we canof course get arbirarily good numerical approximations - but at the moment weare happy that is exists.

Corollary 4.23 One also has sin2

= 1

Proof. The derivative of sin is cos and cos 0 on [0, /2] so that sin isincreasing on [0, /2]. Since (sin t)2+(cos t)2 = 1 and cos

2 = 0 one concludes

that sin2

= 1 Putting the two remarks together gives the result.

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Suppose that f(t) =N

0 an cos2nt +N

1 bn sin2nt then the following inte-grals make sense and

a0 =

10

f(t) dt, and for n 1

an = 210 cos(2nt) f(t) dt

bn = 2

10

sin(2nt) f(t) dt

Definition 4.26 If f C[0, 1], with f(0) = f(1) then we define the fourierseries of f to be

a0 =

10

f(t) dt, and for n 1

an = 2

10

cos(2nt) f(t) dt

bn = 210

sin(2nt) f(t) dt

Which leads to a basic question - can two different continuous functionshave the same fourier co-efficients. The answer is no - the proof is only a littlemore difficult than one of the earlier excercises. Indeed we could prove a muchstronger theorem and explain how to recover f:

Theorem 4.27 Suppose that f C[a, b] with f(a) = f(b) and with fouriercoefficients an, bn. Define

SN =N0

an cos2nt +N1

bn sin2nt,

then the average CN =1

N+1

Nn=0 Sn of the Sn converges uniformly to f on

[0, 1]!

Unfortunately time runs out. However the situation is delicate. In general,the SN themselves do not converge uniformly or even pointwise.

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analysis iii integration

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