88886220 street fighting mathematics
TRANSCRIPT
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SANJOY MAHAJAN
FOREWORD BY CARVER A. MEAD
THE ART OF EDUCATED GUESSING AND
OPPORTUNISTIC PROBLEM SOLVING
STREET-FIGHTING
MATHEMATICS
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Street-Fighting Mathematics
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Street-Fighting Mathematics
The Art of Educated Guessing andOpportunistic Problem Solving
Sanjoy Mahajan
Fwd Cv A. Md
Th MIT P
Cmd, Mh
Ld, Ed
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C 2010 Sj MhjFwd C 2010 Cv A. Md
Street-Fighting Mathematics: The Art of Educated Guessing and Opportunistic
Problem Solving Sj Mhj (h), Cv A. Md (wd),
d MIT P (ph) d d h Cv Cmm
A–Nmm–Sh Ak 3.0 Ud S L.
A p h v
hp://vmm.//--/3.0//
F m p q d, p m
Tp P d E h h CTEX d PDFTEX
Library of Congress Cataloging-in-Publication DataMhj, Sj, 1969–
S-fih mhm : h dd dpp pm v / Sj Mhj ; wd
Cv A. Md.
p. m.
Id ph d dx.
ISBN 978-0-262-51429-3 (pk. : k. pp) 1. Pm v.
2. Hph. 3. Em h. I. T.
QA63.M34 2010510—d22
2009028867
Pd d d h Ud S Am
10 9 8 7 6 5 4 3 2 1
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For Juliet
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Brief contents
Fwd x
P x
1 Dm 1
2 E 13
3 Lmp 31
4 P p 57
5 Tk h p 77
6 A 99
Bph 123
Idx 127
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Contents
Foreword xi
Preface xiii
1 Dimensions 1
1.1 Em: Th pw m p 1
1.2 Nw mh: F 3
1.3 G 7
1.4 Summary and further problems 11
2 Easy cases 13
2.1 G vd 132.2 P m: Th p 16
2.3 Sd m: Th vm d pmd 17
2.4 Fd mh: D 21
2.5 Summary and further problems 29
3 Lumping 31
3.1 Em pp: Hw m ? 32
3.2 Em 33
3.3 Em dvv 37
3.4 Az dff q: Th p–m m 423.5 Pd h pd pdm 46
3.6 Summary and further problems 54
4 Pictorial proofs 574.1 Add dd m 58
4.2 Ahm d m m 60
4.3 Appxm h hm 66
4.4 B 70
4.5 Smm 734.6 Summary and further problems 75
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x
5 Taking out the big part 77
5.1 Mp d w 77
5.2 F h d w-p xp 79
5.3 F h wh xp 845.4 Sv ppxm: Hw dp h w? 91
5.5 D m 94
5.6 Summary and further problems 97
6 Analogy 99
6.1 Sp m: Th d mh 99
6.2 Tp: Hw m ? 103
6.3 Op: E–ML mm 107
6.4 T : A d d m 113
6.5 Bon voyage 121
Bibliography 123
Index 127
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Foreword
M k mhm m mhm—Bd Id!
Mhm mhm d w h.Th mhm p xp -
hp m q h wd, d dv
qv m h hp. F h pp, mh-m , h h d, dm hp d
dwh dv.
A d, I pmd m h I v m h, I wdv p d hh h kd h. I hv p m
fid d d p w d
xp h h qv, d I hv v kw k
m pm.
Wh xp, h mhm h I hv d m w
d d , m w, m h k.Street-Fighting Mathematics h h . Sj Mhj h
, h m d w, h wk h wd. J wh
w hk h p v, h p h v. M
p v h pph h Nv–Sk q:
h I wd v v mp . B h d hh ,
m h h w.
I h k h v . I hv p
dpd v h hq h w fid h. I mmd
hh v .
—Carver Mead
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Preface
T mh mhm h rigor mortis: h mk
jfid p v wh d . Id
p, hv —h fi d k q . Ahh
w p p, v pm-v phph, d
h hm h k: hw w wh p
x .Edd d pp pm v q x.
A , pph G P, k I w. Th k
d, hp, d dm dv fid
hm kwd. Th dv xmp hp p h —h
pp—m h p pp h p
d h pm p .
Th xmp d h h d wh-
, mm m h md, x
ph pp m dff q, m d wh v h Nv–Sk q, fid h h
ph h , d , d mm fi
wh v m kw d d.
Th k mpm wk h How to Solve It [37], Mathematics
and Plausible Reasoning [35, 36], d The Art and Craft of Problem Solving[49]. Th h hw v x d pm x, wh
hd p dfid pm d md
. A 2 m hw h
ppd d wd v d v wk.Th ff vd d h p p v
pm w d.
Th k w h h m m h I h
v MIT. Th d vd wd xp: m
fi- dd d d d -
h d h. Th d vd wd pz:
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xv Preface
m ph, mhm, d mm ,
mp , d . Dp h dv, h
d md fi m h d j h dv
d pp. I wh h m .
How to use this book
A w h Axd Md (, Axd
h G). A kw, kd d kwd
h m ffv h [8]. A kd mk w m dk m q, h kw h q, wd, d
d pm - . Th, q w
p pd hh h k.
Questions marked with a in the margin: Th q wh mh k d , d k wk h x p
. Th wd h q x, wh
hk d m .
Numbered problems: Th pm, mkd wh hdd kd,
wh mh v k hm . Th k
p h , xd xmp, v h,
d v v (pp) pdx.
T m q h p!
Copyright license
Th k d d h m MIT’ OpCW:
Cv Cmm A-Nmm-Sh Ak . Th
ph d I , mpv, d h h wk -
mm, d w w d v d .
Acknowledgments
I hk h w dvd d z.
For the title: C M.
For editorial guidance: Kh Amd d R P.
For sweeping, thorough reviews of the manuscript: Mh G, DvdH, Dvd MK, d Cv Md.
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Preface xv
For being inspiring teachers: Jh Am, Ah Ek, P Gd-
h, Jh Hpfid, J K, Gff Ld, Dd Kh, Cv
Md, Dvd Mddk, S Ph, d Edw T.
For many valuable suggestions and discussions: Shh Adm, DC, D Fm, Mh Gd, H H, Jz H, T
Hkw, Sph H, K J, Ad Mhj, H M,
Eh M, H Phm, Bjm Rpp, Rh Sphk,
Md Shd-D, Edw T, Tdh Tkd, Mk W,
d Jh Zk.
For advice on the process of writing: Cv Md d H R.
For advice on the book design: Y Ih.
For advice on free licensing: D Rvh d Rhd Sm.
For the free software used for calculations: Fdk Jh (mpmh), h
Mxm pj, d h Ph mm.
For the free software used for typesetting: H H d T Hkw
(CTEX); H Th Thh (PDFTEX); Dd Kh (TEX); Jh H
(MP); Jh Bwm, Ad Hmmd, d Tm P (Amp-); M Mk (M); Rhd Sm (Em); d h D
GNU/Lx pj.
For supporting my work in science and mathematics teaching: Th Whk
Fd Bmd E; h Hz Fd; h Md Fw Cp Ch C, Cmd; h MIT Thd L L d h Offi h D Udd
Ed; d p R Bk, Jh Wm, d h T
h G Ch Fd.
Bon voyage
A fi , ’ wm v m ph d :
h mhd dm .
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1
Dimensions
1.1 Em: Th pw m p 11.2 Nw mh: F 31.3 G 7
1.4 Summary and further problems 11
O fi -fih dm , wh vd,
dm. T hw dv pp, h ddwh m xmp d hpd xmp m Nw
mh d .
1.1 Economics: The power of multinational corporations
C z mk h w mp [25] pv
h xv pw m p:
I N, v m , h GDP [ dmpd] $99 . Th wh Exx $119 . “Wh m- hv wh hh h h GDP h whh hp, wh kd pw hp w k ?” k LM.
B , xp h w q:
What is the most egregious fault in the comparison between Exxon and Nigeria?
Th fid mpv, d . I m vd
pk h m GDP. A GDP $99 hhd
m flw $99 p . A , whh h m h h v d h , m phm h
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2 1 Dimensions
h h m phm—m,
m flw.
Spp d h m hd h h dd h
m m GDP. Th N’ GDP (m h flw md m ) wd h $1 p dd d
pd $1 . Nw N w v Exx, wh p
m -h N’ GDP. T dd h pp
, pp h wk w h m m GDP.
N’ GDP m $2 p wk, pd $2 . Nwp N d hp h mh Exx, 50-d h
N.
A vd m m h h dpd
h m phm h m m. Th mk mp mp q. N wh m: I hdm m d p md d. GDP,
hwv, flw : I h dm m p m d
p d p . (A dm d dpd
h m mm, wh h hw h dm
md p m.) Cmp wh GDP mp
m m m flw. B h dm dff,
h mp mk [39] d h d
.
Problem 1.1 Units or dimensions?
A m, km, d d dm? Wh ,h, pw, d ?
A m flwd mp h p m (pd) v h:
“I wk 1.5 m −1—mh m h h Emp S d Nw
Yk, whh 300 m hh.” I . T pd h pp
, m m h: “I wk 5400 m/h—
mh h h Emp S d, whh 300 m hh.”
I mp p d pw m
N–Exx xmp. I w h xp h I
mphzd wh h h h m d
dm mk. H pd h I hd md p
h h m mp hw h ’ wk w
h hd w , h w v hd!
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1.2 Newtonian mechanics: Free fall 3
A dm vd mp wd mp k wh k: h
N’ GDP wh Exx’ v, Exx’ wh wh N-
’ wh. B wh d,
wh p v wd v, mp Exx’
v wh N’ GDP. B 2006, Exx hd m ExxM wh v h $350 —m w N-
’ 2006 GDP $200 . Th vd mp h h
flwd , h flwd mp w v xpd!
Th mpd q m hv d dm d mk vd mp, ffi. A
h 1999 M Cm O (MCO), whh hd
h M h h pp d . Th ,
d h Mhp Iv Bd (MIB), w mmh -
w Eh d m [26, p. 6]:
Th MCO MIB h dmd h h h h MCOp w h m h d dw fi, Sm F, d j md. Spfi, hpm d Eh d m w d hw pp d d SM_FORCES (m ). A fi d A- Mmm D (AMD) d h p d m hSM_FORCES w. Th d h AMD fi w qd m p x w dm, d h j md- md h d w pvdd m p h qm.
Mk md dm d .
Problem 1.2 Finding bad comparisons
Lk vd mp— xmp, h w, h wpp, h I—h dm .
1.2 Newtonian mechanics: Free fall
Dm j dk m
. T d , h q pm d hv
dm. A xmp hw wh d, h hw
m xk d pm m:
A m hh h feet d h h d pd v feet per second. Fd v m v g
feet per second squared d .
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4 1 Dimensions
Th h p d hhhd d
h q hw p , d
h fi pm. B h hh h
, h v h d h hh: h h
dm. (F h hv dm, h pm wd d mp h h m hh h ; h h dm
h wd h .) A m xp pfi m
h h v g d v dm. B g, h , d v
dm, mp v wh q dvd m g
d h mp w dm q. I h
w dm vd, dm hp
h mp pd.
Gv p h v dm k fih wh hd
d hd k. Th d, w m d v hw dff q wh d:
d2 y
dt2= −g, wh y(0) = h d dy/dt = 0 t = 0, (1.1)
wh y(t) h ’ hh, dy/dt h ’ v, d g h
v .
Problem 1.3 Calculus solution
U hw h h - dff qd2 y/dt2 = −g
wh d y(0) = h d dy/dt = 0 t = 0 h h w :
dy
dt= −gt d y = −
1
2gt2 + h. (1.2)
Using the solutions for the ball’s position and velocity in Problem 1.3, what isthe impact speed?
Wh y(t) = 0, h m h d. Th h mp m t0
2h/g. Th mp v −gt0 −
√ 2gh . Th h mp
pd (h d v) √
2gh .
Th v v mk: k q
wh v t0, dvd h h mp g wh
fid h mp v. P— h wd, mk d -
m mk—d h pv mp pm,
mpx pm wh m p m mfid. W wd k -p mhd.
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1.2 Newtonian mechanics: Free fall 5
O v h mhd dm . B h
q h q m v, g, d h hv dm.
Ohw, v dd mp pd, m hw d, q
dm q d h h vd dm.
Th, ’ h - pm h h q
h dm:
A m hh h d h h d pd v.Fd v m v g d .
Th m , fi, h d p h h ph:
A m hh h d h h d pd v p d. Fd v m v g
p d qd d .
Sd, h m m . I mk mp h m , v m, , h h. M mp, h m v dm
h , g, d v. Th dm w m q dm h mp
pd—wh d v dff q.
Th dm hh h mp h , h, L. Th dm-
v g h p m qd LT−2,
wh T p h dm m. A pd h dm
LT−1, v g d h wh dm LT−1.
Problem 1.4 Dimensions of familiar quantities
I m h dm h L, m M, d m T, wh hdm , pw, d q?
What combination of g and h has dimensions of speed?
Th m√
gh h dm pd.
LT−2
g
× L
h 1/2
=√
L2T−2 = LT−1
pd
. (1.3)
Is√
gh the only combination of g and h with dimensions of speed?
I d dd whh√
gh h p,
pp [43]. Th h h m g d
h , pd, hd hv dm v m (T−1). B
h dm m, hp T −1. B
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6 1 Dimensions
g T −2, h T −1 m m m√
g. Th d
h h m L1. Th√
g d L1/2, h
m L1/2 m m m√
h . Th w h dm
q hw g d h pp h mp pd v.
Th x xp v , hwv, q. I d √
gh ,√
2gh ,
, ,√
gh ×dm . Th dm mp
dm q d dv mp
k h q :
v ∼
gh. (1.4)
Id h ∼ , w hv v p q:
∝q xp php wh dm,
∼ q xp php wh dm,
≈ q xp php 1.
(1.5)
Th x mp pd √
2gh , h dm √
gh h dpd! I k h dm √
2, d h mp. I h xmp, h hh
mh v m w m ( fl hpp) w m (
jmp m d). Th --100 v hh
--10 v mp pd. Sm, h v -
mh v m 0.27 m −2 ( h d C) 25 m −2 ( Jp). Th --100 v g h --10
v mp pd. Mh v h mp pd, h,m m h dm
√ 2 h m h m
—whh mpd x dm .
Fhm, h x w dv. Ex
w hv d m, pm mp m,
h h dm √
2, mp m
h √
gh . A Wm Jm dvd, “Th w h
kw wh vk” [19, Chp 22].
Problem 1.5 Vertical throw
Y hw d pwd wh pd v0. U dm m hw h k hd ( ).Th fid h x m v h - dff q. Whdm w m m h dm- ?
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1.3 Guessing integrals 7
1.3 Guessing integrals
Th (S 1.2) hw h v p
dmd q m h . Hwv, wh h q
dm, h h 5 d x h w G : ∞−∞
e−5x2
dx ? (1.6)
Av, h dm mh pfid— mm
mhm v . F xmp, p
h h G x2
x1
e−x2/2σ2
dx, (1.7)
wh x d hh, d , mh . Thm ph
h m e− 1
2 mv2/kT dv, (1.8)
wh v m pd. Mhm, h mm ,
d h mm m
e−αx2wh p h dm
α d x. Th k pfi v mhm pw ,
mk dm dffi.
How can dimensional analysis be applied without losing the benefits of mathe-matical abstraction?
Th w fid h q wh pfid dm d h
hm dm. T h pph,
’ pp h dfi G ∞−∞
e−αx2
dx. (1.9)
Uk pfi wh α = 5, whh h ∞−∞
e−5x2
dx,
h m d p h dm x α —d h
p pvd h dm dd h mhd dm
.
Th mhd q h q dm vd. Th, h w q, h d h d m hv d
dm:
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8 1 Dimensions
∞−∞
e−αx2
dx = mh. (1.10)
Is the right side a function of x? Is it a function of α ? Does it contain a constant
of integration?
Th d m q h h x d α . B
x h v d h v dfi ,
x dpp p (d pp).
Th, h h d—h “mh”— α . I
m, ∞−∞
e−αx2
dx = f(α ). (1.11)
Th f mh d dm m h 2/3 √ π ,
α p wh dm.
F h q dm vd, h m hv h
m dm f(α ), d h dm f(α ) dpd hdm α . Ad, h dm- pd h
h w h p:
Sp 1. A dm α (S 1.3.1).
Sp 2. Fd h dm h (S 1.3.2).
Sp 3. Mk f(α ) wh h dm (S 1.3.3).
1.3.1 Assigning dimensions to α
Th pm α pp xp. A xp pfi hw
m m mp q . F xmp, h 2n:
2n = 2 × 2 × · · · × 2
n m
. (1.12)
Th “hw m m” p m, xp
dm.
H h xp −αx2 h G dm. F
v, d h dm α [α ] d x [x ]. Th
[α ] [x ]2
= 1, (1.13)
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1.3 Guessing integrals 9
[α ] = [x ]−2 . (1.14)
Th , pfid dm q , d h k h
.
Th mp v mk x dm. Th h mk α
d f(α ) dm, dd f(α ) wd dm
vd, mk dm . Th mp ffvv v x mp dm— xmp, h. (Th
h m h x x h fl.) Th
[α ] = L−2.
1.3.2 Dimensions of the integral
Th m [x ] = L d [α ] = L−2 dm h dm h
G . H h : ∞−∞
e−αx2
dx. (1.15)
Th dm dpd h dm h
p: h , h d e−αx2, d h dff dx.
Th d d S Summe, h Gmwd m. I vd m, m hv d dm: Th
dm pp dm q h pp ddd
pp. F h m , h m h h m dm
m. Th, h mm —d h h
—d ff dm: Th dm.
Problem 1.6 Integrating velocity
P h v. Hwv, p d v hv dff- dm. Hw h dff wh h h h
dm?
B h dm, h dm h -
h dm h xp e−αx2mpd h
dm dx. Th xp, dp fi xp −αx2,
m v p e mpd h. B e dm, e−αx2
.
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10 1 Dimensions
What are the dimensions of dx?
T fid h dm dx, w h dv Sv Thmp
[45, p. 1]: Rd d “ .” Th dx “ x.” A
h h, dx h. I , dx h h mdm x. Eqv, d—h v
— dm.
Am h p, h wh h dm h: e−αx2
dx
=
e−αx2
1
× [dx ] L
= L. (1.16)
Problem 1.7 Don’t integrals compute areas?
A mm h mp . A hv dm
L2. Hw h h G hv dm L?
1.3.3 Making an f(α ) with correct dimensions
Th hd d fi p h dm f(α )
wh h m dm h . B h dm α
L−2, h w α h m α −1/2. Th,
f(α ) ∼ α −1/2. (1.17)
Th , whh k dm , w dwh .
T dm h dm , α = 1 d v
f(1) =
∞−∞
e−x2
dx. (1.18)
Th w ppxmd S 2.1 d d √ π . Th w f(1) =
√ π d f(α ) ∼ α −1/2 q h f(α ) =
π/α ,
whh d ∞−∞
e−αx2
dx =
π
α . (1.19)
W mmz h dm h pw α .
D d h. Th α mh m mp h h
dm . Cv, h α wh dm mp.
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1.4 Summary and further problems 11
Problem 1.8 Change of variable
Rwd k p 8 d pd h d kw f(α ). Wh ddm , hw h f(α ) ∼ α −1/2.
Problem 1.9 Easy case α = 1
S α = 1, whh xmp - (Chp 2), vh mp h x h d α h dm L−2. Wh k α = 1?
Problem 1.10 Integrating a difficult exponential
U dm v
∞0
e−αt3 dt.
1.4 Summary and further problems
D dd pp : Ev m q m m
hv d dm! Th pw . I hp v wh d pd h
dff q. H h pm p h .
Problem 1.11 Integrals using dimensions
U dm fid
∞0
e−ax dx d
dx
x2 + a2. A
dxx2 + 1
= arctan x + C. (1.20)
Problem 1.12 Stefan–Boltzmann law
Bkd d m phm, h d - dpd h pd h c. I hm phm, dpd h hm k BT , wh T h j’ mp d k B Bzm’ . Ad qm phm, dpd Pk’ h . Th h kd-d I dpd c, k BT ,d h . U dm hw h I ∝ T 4 d fid h pp σ. Th k p h m dm . (Th
d S 5.3.3.)
Problem 1.13 Arcsine integral
U dm fid
1 − 3x2 dx. A
1 − x2 dx =
arcsin x
2+
x
1 − x2
2+ C, (1.21)
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12 1 Dimensions
Problem 1.14 Related rates
h
W pd vd (wh 90◦ p- ) dV/dt = 10 m3 −1. Wh h wdph h = 5 m, m h whh h dph
. Th fid h x .
Problem 1.15 Kepler’s third law
Nw’ w v v—h m v-q w— hh v w w m
F = −Gm 1m 2
r2, (1.22)
wh G Nw’ , m 1 d m 2 h w m, d r hp. F p h , v v h whNw’ d w v
m d2r
dt2= −
GMm
r2r̂, (1.23)
wh M h m h , m h m h p, r h v mh h p, d r̂ h v h r d.
Hw d h pd τ dpd d r? Lk p Kp’hd w d mp .
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2
Easy cases
2.1 G vd 132.2 P m: Th p 162.3 Sd m: Th vm d pmd 17
2.4 Fd mh: D 212.5 Summary and further problems 29
A wk , d h . Th mxm
d h d —h mhd . I w hp
, dd vm, d v x dff q.
2.1 Gaussian integral revisitedA h fi pp, ’ v h G m S 1.3, ∞
−∞e−αx2
dx. (2.1)
Is the integral√
πα or
π/α ?
Th h m wk α 0. A h ’ dp
(α = ∞ d α = 0), h v.
What is the integral when α = ∞?
e−10x2
0 1
A h fi , α ∞. Th −αx2 -
m v v, v wh x . Th xp-
v m , h v
w v, d hk z. Th-, α → ∞ h hk z. Th h p
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14 2 Easy cases
√ πα , whh fi wh α = ∞; d pp h p
π/α ,
whh z wh α = ∞.
What is the integral when α = 0?
e−x2/10
0 1
I h α = 0 xm, h v fl
hz wh hh. I , d
v h fi , fi. Th
h√
πα p, whh z wh α = 0; d
pp h
π/α p, whh fi wh α =
0. Th h√
πα p h - , d h
π/α p
p h - .
I h w p w h p, w wd h π/α . Hw-
v, hd p w
2/α , hw d dd w d π/α ? Bh p p h - ; h hv d
dm. Th h k dffi.
T h, hd : α = 1. Th h mpfi ∞−∞
e−x2
dx. (2.2)
Th vd d m p -
d, h mhd q k wh w h pp
(xk mv v h d). A m pph v h m d
h ppxm v h d m.
Th, p h mh v e−x2wh v
hv n m. Th pw- ppx-
m h m n pzd. A
n pph fi, h h pzd m d m
pph h d h mh v.
n A
10 2.07326300569564
20 1.77263720482665
30 1.77245385170978
40 1.77245385090552
50 1.77245385090552
Th v h d h v h
x = −1 0 . . . 1 0, dvd h v n m. Th
v, d k m. I
wh 1.7, whh mh m√
3. Hwv,
1.77, whh √
3.
F, π h h 3, h mh v
√ π .
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2.1 Gaussian integral revisited 15
L’ hk mp h qd π :
1.772453850905522 ≈ 3.14159265358980,
π
≈3.14159265358979.
(2.3)
Th mh h h α = 1 G dd√
π : ∞−∞
e−x2
dx =√
π. (2.4)
Th h G ∞−∞
e−αx2
dx (2.5)
m d √
π wh α = 1. I m hv h h
w α = 0 d α = ∞.
Am h h h
2/α ,
π/α , d√
πα ,
π/α p
h α = 0, 1, d ∞. Th, ∞−∞
e−αx2
dx =
π
α . (2.6)
E h w jd h h. Dm -, xmp, h p (S 1.3). I v
m h k√
π/α h p h - . Hwv,
, d, mp. Th d q v
dd dm x, α , d dx (h xv S 1.3).
E , k dm , m h k 2/α wh dm. Eh h h.
Problem 2.1 Testing several alternatives
F h G ∞−∞
e−αx2 dx, (2.7)
h h - v h w dd v.
()√
π/α () 1 + (√
π − 1)/α () 1/α 2 + (√
π − 1)/α .
Problem 2.2 Plausible, incorrect alternative
I h v
π/α h h vd dm d p h h- ?
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16 2 Easy cases
Problem 2.3 Guessing a closed form
U h v hw h
∞
0
dx
1 + x2
= 2 1
0
dx
1 + x2
. (2.8)
Th d h fi , h h fi v m. Em h d h pzdppxm d mp pmm . Th dm h fi .
2.2 Plane geometry: The area of an ellipse
b
a
Th d pp m p
m: h p. Th p hmmj x a d mm x b. F A
d h w dd:
() ab2 () a2 + b2 () a3/b (d) 2ab () πab.
What are the merits or drawbacks of each candidate?
Th dd A = ab2 h dm L3, wh m hv
dm L2. Th ab2 m w.
Th dd A = a
2
+ b
2
h dm ( d h mdd), h x h h d a d b. F a,
h w xm a = 0 pd fim h p wh z. Hwv, wh a = 0 h dd A = a2 + b2 d A = b2
h h 0; a2 + b2 h a = 0 .
Th dd A = a3/b pd z wh a = 0. B
a = 0 w , d h x a d b m
h, mm p b = 0 hd
. I pd fim h p wh z ;
, h dd a3/b pd fi , h b = 0 .
Tw dd m.
Th dd A = 2ab hw pm. Wh a = 0 b = 0, h
d pdd z, A = 2ab p h -
. Fh q h hd : a = b. Th h p
m wh d a d πa2. Th dd 2ab, hwv,d A = 2a2, h a = b .
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2.3 Solid geometry: The volume of a truncated pyramid 17
Th dd A = πab p h : a = 0, b = 0, d a = b.
Wh h p , fid h dd ; d
πab dd h (Pm 2.4).
Problem 2.4 Area by calculus
U hw h A = πab.
Problem 2.5 Inventing a passing candidate
C v d dd h h h dm dp h a = 0, b = 0, d a = b ?
Problem 2.6 Generalization
G h vm pd wh pp d a, b, d c.
2.3 Solid geometry: The volume of a truncated pyramid
Th G- xmp (S 2.1) d h p- xmp
(S 2.2) hwd mhd : hk
whh m . Th x v ph
mhd h: m.
h
b
aA xmp, k pmd wh q d
p m p k p h
. Th d pmd (d h m) h q d q p p h . L h
v hh, b h d h , d a
h d h p.
What is the volume of the truncated pyramid?
L’ hz h m h vm. I h h
h h , a, d b. Th h p w kd: hh d
h. F xmp, flpp h d hd h
h m a d b pv h ; d mp p -
h hh wh a b. Th h vm p h w ,
h h h kd:
V (h,a,b) = f(h ) × g(a, b). (2.9)
Pp w dm f; dm d - w dm g.
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18 2 Easy cases
What is f : How should the volume depend on the height?
T fid f, pp- hh xp-
m. Chp h d v v, h k
-d ; h m d h . Th hd h vm h v d h d
h wh vm V . Th f ∼ h d V ∝ h :
V = h × g(a, b). (2.10)
What is g : How should the volume depend on a and b?
B V h dm L3, h g(a, b) h dm
L2. Th h dm . Fh
dd hz g, d h pvdd h mhd .
2.3.1 Easy cases
What are the easy cases of a and b?
Th h xm a = 0 ( d pmd). Th
mm w a d b w h , m a = b
d h xm b = 0. Th h hd:
h
b
h
a
h
a
a = 0 b = 0 a = b
Wh a = 0, h d d pmd, d g
h d h b. B g h dm L2, h p g g ∼ b2; dd, V ∝ h ; , V ∼ hb2. Wh b = 0,
h d pd-dw v h b = 0 pmd d h
h vm V ∼ ha2. Wh a = b, h d pm hv
vm V = ha2 ( hb2).
Is there a volume formula that satisfies the three easy-cases constraints?
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2.3 Solid geometry: The volume of a truncated pyramid 19
Th a = 0 d b = 0 fid h mm m
V ∼ h (a2 + b2). I h m dm 1/2, mk
V = h (a2 + b2)/2, h h vm fi h a = b , d
h vm d pmd (a = 0) wd hb2/2.
When a = 0, is the prediction V = hb2/2 correct?
T h pd q fid h x dm
V ∼ hb2. Th k k k pm: S pmd
h hz d dd () h vm. Hwv,
mp v pp .
b
h = b
Th w v
m mp pm: fid h
wh b d hh h . Th fiA ∼ hb, wh h dm ? T
fid , h b d h mk :
h wh h = b. Tw h mk hp: q wh b2. Th h h h
A = b2/2; h dm 1/2. Nw xd h
h dm—fid d pmd (wh q ) h
m wh mk d.
What is the easy solid?
A v d d h pmd’ q
: Php h . Th h
q x pmd wh p m h h; h h pmd hv h p h = b/2. F
m mp, ’ m h d wh b = 2
d h = 1.
Sx h pmd m wh vm b3 = 8, h vm
pmd 4/3. B h pmd h vm V ∼ hb2, d hb2 = 4
h pmd, h dm V ∼ hb2 m 1/3.
Th vm d pmd ( pmd wh a = 0) h
V = hb2/3.
Problem 2.7 Triangular base
G h vm pmd wh hh h d A.Am h h p vx d v h d h . Th Pm 2.8.
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20 2 Easy cases
Problem 2.8 Vertex location
Th x pmd d mk h pmd’ p vx d v h h . Th h V = hb2/3 mh pp wh h . I d h p vx v h
v, wh h vm?
Th pd m h fi h - w V = hb2/2 (wh
a = 0), wh h h h = b/2 d a = 0 j hwd
h V = hb2/3. Th w mhd mk d pd.
How can this contradiction be resolved?
Th d m hv k d h p.
T fid h p, v h p . Th m V ∝ h k
. Th h - qm—h V ∼ hb
2
wh a = 0, hV ∼ ha2 wh b = 0, d h V = h (a2 + b2)/2 wh a = b— k
. Th mk w p m h h pd
V ∼ h (a2 + b2) a b.
Id ’ h w m h d ab m:
V = h (αa2 + βab + γb2). (2.11)
Th v h ffi α , β, d γ pp h -
qm.
Th b = 0 wh h h = b/2 , whh hwd hV = hb2/3 d pmd, q h α = 1/3. Th a = 0
m q h γ = 1/3. Ad h a = b q h
α + β + γ = 1. Th β = 1/3 d và,
V =1
3h (a2 + ab + b2). (2.12)
Th m, h pp , dm ,
d h mhd , x (Pm 2.9)!
Problem 2.9 Integration
U hw h V = h (a2 + ab + b2)/3.
Problem 2.10 Truncated triangular pyramid
Id pmd wh q , wh pmd wh q d h b . Th mk h d d p m h p k p h . I m h hh h
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2.4 Fluid mechanics: Drag 21
d h p d m d h a d b, wh h vm h d?(S Pm 2.7.)
Problem 2.11 Truncated cone
Wh h vm d wh d r1 d p d r2 (wh h p p h )? Gz -m h vm d pmd wh hh h , hp d A , d pd p Ap.
2.4 Fluid mechanics: Drag
Th pd xmp hwd h hk d
m, h xmp d wh ( xmp,
wh ). F h x q, m fld mh, x
kw , d h -fih
m h w mk p.
H h h Nv–Sk q fld mh:
∂v
∂t+ (v·∇)v = −
1
ρ∇ p + ν∇2v, (2.13)
wh v h v h fld ( p d m),
ρ d, p h p, d ν h km v. Th
q d mz v phm d flh,
d, d v pd.
O xmp h w hm xpm d. Php h
p wh m 2; h h w w
mp:
1
2
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22 2 Easy cases
Wh h mp, p h h hdd
mk . Th w hv h m
hp, h h w h hh d wdh
h m .
When the cones are dropped point downward, what is theapproximate ratio of their terminal speeds (the speeds at which drag balancesweight)?
Th Nv–Sk q h w h q. Fd
h m pd vv p.
Sp 1. Imp d d. Th d d h m
h d h qm h fld h pp.
Sp 2. Sv h q, h wh h q ∇·v =
0, d fid h p d v h h
.
Sp 3. U h p d v fid h p d v
d h h ; h h
fid h d q h .
Sp 4. U h d q fid h m h . Thp dffi h m m
wh h m md p 1. I , k p 1, m dff m, d hp k
p h h p.
U, h Nv–Sk q pd d
p-dff q. Th kw v
mp : xmp, ph mv v w v fld,
ph mv pd z-v fld. Th
hp v h mpd flw d , qv
hp h flx pp .
Problem 2.12 Checking dimensions in the Navier–Stokes equations
Chk h h fi h m h Nv–Sk q hv ddm.
Problem 2.13 Dimensions of kinematic viscosity
Fm h Nv–Sk q, fid h dm km v ν.
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2.4 Fluid mechanics: Drag 23
2.4.1 Using dimensions
B d h Nv–Sk q h
q, ’ h mhd dm d . A
d pph hm dd h m v . Ad pph dd h d pd
d h fid h pd whh h d h wh h . Th w-p pph mpfi h pm. I d
w q (h d ) m w q: h
v d h m h .
Problem 2.14 Explaining the simplification
Wh h d dpd h v g d h’ m m ( h dpd h ’ hp d z)?
Th pp dm h m vd q hv
d dm. Appd h d F, m h h
q F = f(q h ff F) h d hv dm
. Th, h fid h q h ff F, fidh dm, d h m h q q wh
dm .
On what quantities does the drag depend, and what are their dimensions?
v pd h LT−1
r z h L
ρ d ML−3
ν v L2T−1
Th d dpd q-: w pm h d
w pm h fld (). (Fh dm ν, Pm 2.13.)
Do any combinations of the four parameters v, r, ρ, and ν have dimensions of force?
Th x p m v, r, ρ, d ν q wh dm
. U, h p m— xmp,
F1 = ρv2r2,
F2 = ρνvr,(2.14)
h pd m√
F1F2 d F21/F2. A m h
pd , h d F d √
F1F2 + F21/F2,
3√
F1F2 − 2F21/F2, mh w.
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24 2 Easy cases
Nw h p q mhd m phd h
mp m wh dm. T dvp h
phd pph, h fi pp dm: A
m q hv d dm. Th pp pp m d h
A + B = C (2.15)
wh h A, B, d C F, v, r, ρ, d ν.
Ahh h d mpx , h hv ddm. Th, dvd h m A, whh pd h
q
A
A +
B
A =
C
A , (2.16)
mk h m dm. Th m mhd vd q-
dm q. Th, () q d
h wd w dm m.
A dm m m dm p: m
dm pd h v. B q d
h wd w dm m, d dmm w dm p, q d
h wd w dm p.
Is the free-fall example (Section 1.2) consistent with this principle?
B pp h pp h mpd pm d,
h mp xmp (S 1.2). Th x mp pd
j dppd m hh h v =√
2gh , wh g h v
. Th dd w h dm m
v/√
gh =√
2, whh h dm p v/√
gh . Th
w pp p fi .
Th dm-p m, wh vd, m mhd h. L’ wm p hz h mp pd v.
F, h q h pm; h, h v, g, d h . Sd,
m h q dm p. H, dm-
p d j m p. F h p, ’
h v2/gh (h p h d ff h ). Thh p dm m
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2.4 Fluid mechanics: Drag 25
v2
gh = dm . (2.17)
(Th h d dm d p
v h.) I h wd, v2/gh ∼ 1 v ∼ √ gh .
Th pd h h phd dm
S 1.2. Idd, wh dm p, h
d h m . Hwv, hd pm—
xmp, fid h d —h phd mhd d
pvd m; h h mhd dm
p .
Problem 2.15 Fall time
Shz ppxm m h - m t m g d h .
Problem 2.16 Kepler’s third law
Shz Kp’ hd w h pd p d. (S Pm 1.15.)
What dimensionless groups can be constructed for the drag problem?
O dm p d F/ρv2r2; d p d rv/ν.A h p d m h p (Pm 2.17),
h pm dd w independent dm p. Thm dm m h
p = f(d p), (2.18)
wh f -kw ( dm) .
Which dimensionless group belongs on the left side?
Th hz m F, d F pp h fi
p F/ρv2r2. Wh h md, p h fi p h
d h h wpp h -m f. Whh h, h m m d
F
ρv2r2= f
rv
ν
. (2.19)
Th ph h (d-) d h d h dm f.
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26 2 Easy cases
Problem 2.17 Only two groups
Shw h F, v, r, ρ, d ν pd w dpd dm p.
Problem 2.18 How many groups in general?
I h mhd pd h m dpd dmp? (Th w w v 1914 Bkhm [9].)
Th pd mh m p, hv pdd d h
dpd h kw f. B h mpvd
h fid f. Th pm m qd - h -v h F = h ( v, r, ρ, ν), wh dm
dd h pm v
(h vr/ν). Th v h mpfi w q dd
h d ph Hd Jff (qd [34, p. 82]):
A d v m q p; h w v vm; h h v k;d h v .
Problem 2.19 Dimensionless groups for the truncated pyramid
Th d pmd S 2.3 h vm
V =1
3h (a2 + ab + b2). (2.20)
Mk dm p m V , h , a, d b, d w h vm
h p. (Th m w d .)
2.4.2 Using easy cases
Ahh mpvd, h d k hh: Ev h -v
d pm h x . B mh hv x
. B h xm , k fi
h xm .
Extreme cases of what?
Th kw f dpd rv/ν,
F
ρv2r2= f
rv
ν
, (2.21)
xm rv/ν. Hwv, vd p md m- ph, fi dm h m rv/ν. Th m rv/ν,
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2.4 Fluid mechanics: Drag 27
dd R, h m Rd m. (I ph p-
q h hq mp d xpd S 3.4.3.)
Th Rd m ff h d v h kw f:
Fρv2r2
= f (R) . (2.22)
Wh k, f ddd xm h Rd m; wh
h k, h xmp xm.
Are the falling cones an extreme of the Reynolds number?
Th Rd m dpd r, v, d ν. F h pd v, vd
xp h h h 1 m −1 (wh, ,
2). Th z r h 0.1 m ( wh 2). Adh km v ν ∼ 10−5 m2 −1. Th Rd m
r 0.1 m ×
v 1 m −1
10−5 m2 −1 ν
∼ 10 4. (2.23)
I fi h 1, h xm
hh Rd m. (F w Rd m, Pm 2.27d [38].)
Problem 2.20 Reynolds numbers in everyday flows
Em R m dw, p , dp, d 747 h A.
Th hh-Rd-m m hd m w. O w
hk h v ν 0, ν v h dm
h Rd m. Th, h m hh Rd m,v dpp m h pm d h d hd d-
pd v. Th v h,
m . (C h qd w - p mhm, m p
d h h d [12, 46].)
V ff h d hh h Rd m:
F
ρv2r2= f
rv
ν
. (2.24)
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28 2 Easy cases
T mk F dpd v, F m dpd Rd
m! Th pm h dpd dm
p, F/ρv2r2, h m m d
Fρv2r2
= dm . (2.25)
Th d h F ∼ ρv2r2. B r2 pp h
’ - A, h d mm w
F ∼ ρv2A. (2.26)
Ahh h dv w , h pp
j h Rd m hh. Th hp ff
h m dm . F ph, h 1/4;
d mv ppd x, h 1/2; d fl p mv ppd , h 1.
2.4.3 Terminal velocities
Fd
W = mg
Th F ∼ ρv2A h pd h m v-
h . Tm v m z ,
h d m h wh. Th wh
W = σppAppg, wh σpp h d pp
(m p ) d App h h mp h q . B App mp
h - A, h wh h v
W ∼ σppAg. (2.27)
Th,
ρv2A
d
∼ σppAg
wh
. (2.28)
Th dvd d h m v m
v ∼
gσpp
ρ. (2.29)
A d m h m pp d hv h m hp,
whatever their size, h m pd!
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2.5 Summary and further problems 29
T h pd, I d h m d dd
p 21, hd h hd v m hd, d hm . Th
2 m d h 2 , d h dd wh 0.1 h.
Chp xpm d hp h !
Problem 2.21 Home experiment of a small versus a large cone
T h hm xpm (p 21).
Problem 2.22 Home experiment of four stacked cones versus one cone
Pd h
m v m kd d h h
m v m . (2.30)
T pd. C fid mhd q m qpm?
Problem 2.23 Estimating the terminal speed
Em k p h d pp; pd h ’ m pd;d h mp h pd h h hm xpm.
2.5 Summary and further problems
A wk , d h . Th-
, hk ppd m h , d m
xp h p - . T pp d
xd h d, h w pm d h d
v k Cp [10].
Problem 2.24 Fencepost errors
A d h 10 m hz h wd k dvd 1 mm v p. D d 10 11 v p (dh p dd h d)?
Problem 2.25 Odd sum
H h m h fi n dd :
Sn = 1 + 3 + 5 + · · · + ln n m
(2.31)
. D h m ln q 2n + 1 2n − 1?
. U Sn ( n).
A v dd S 4.1.
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30 2 Easy cases
Problem 2.26 Free fall with initial velocity
Th S 1.2 w d m . Nw m h v v v0 (wh pv v0 m pwd hw). G h mpv v.
Th v h - dff q fid h x v, d mph x .
Problem 2.27 Low Reynolds number
I h m R 1, h m f
F
ρv2r2= f
rv
ν
. (2.32)
Th , wh md wh h dm , kw Sk d [12].
Problem 2.28 Range formula
v
R
θ
Hw d k v hz ( )?U dm d m h R h h v v, h h θ, d h v g.
Problem 2.29 Spring equation
Th q d m–p m (S 3.4.2)
k/m ,wh k h p d m h m. Th xp h h p k h m. U xm k m dd whh h
pm .
Problem 2.30 Taping the cone templates
Th p mk h mp (p 21) w wd h pmk h m mp. I h wd, h p h , , 6 mm wd, h p h m hd 3 mm wd. Wh?
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3
Lumping
3.1 Em pp: Hw m ? 323.2 Em 333.3 Em dvv 37
3.4 Az dff q: Th p–m m 423.5 Pd h pd pdm 463.6 Summary and further problems 54
Wh w p 6 mh m w? T pd w
, w mp mp h 6 mh h p’
v, v v. Sh h
h w vd. I dm d dvd h m
v v whh h v , mp h m h pd v mp d, d h
dd h d.
Amz, h mp d x, v wh h
v hv fim wdh d h fi m.
Hwv, h m mp h d, w,
dd mp v m h h pm. U
mhd, xmp, w x h dh G e−x2
w x = 0 d ∞; h m v
xp z fi, x m mp.I , ppxm mhd : Th m w pvd
w. Ad h m mhd
mp. Id dvd h p m p,
p mp w p. Th mp ppxm d
dv d xmp m dmph(S 3.1) dff q (S 3.5).
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32 3 Lumping
3.1 Estimating populations: How many babies?
Th fi xmp m h m h Ud S.
F dfi, hd h h 2 d. A
x q h h d v p h UdS. Th, m, m d v dd
h US C B.
()
106
0 50
0
4
N(t)
A ppxm h vmd, h C B [47] ph
h m pp h . Th
d 1991 p
w N(t), wh t . Th
N = 2
0N(t) dt. (3.1)
Problem 3.1 Dimensions of the vertical axis
Wh h v x d pp p h h pp? Eqv, wh d h x hv dm T−1?
Th mhd h v pm. F, dpd h h
h US C B, d d k-
-h-vp . Sd, q v wh
m, h m d m. Thd, h d pfi h pm, wh mhm hd
. A x , h, pvd h
d h mm v. Id h pp
v x, ppxm —mp h v .
What are the height and width of this rectangle?
Th ’ wdh m, d p m d pp
h xp. I h 80 , mk 80 h wdh
pd h v d p h h 80h hd.Th ’ hh mpd m h ’ , whh h US pp—v 300 m 2008. Th,
hh =
wdh∼
3 × 108
75 . (3.2)
Why did the life expectancy drop from 80 to 75 years?
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3.2 Estimating integrals 33
mpd
()
106
0 750
4
d
Fd h xp mpfi h
m dv: 75 dvd 3 d
300. Th h h
md mp, d mh v h mp . U 75
h wdh mk h hh ppxm
4 × 106 −1.
I h pp v v h t = 0 . . . 2 m j
mp:
N ∼ 4 × 106 −1
hh
× 2
= 8 × 106. (3.3)
Th C B’ fi v : 7.980 × 106. Th mmp d h m d h xp 75 !
Problem 3.2 Landfill volume
Em h US dfi vm d dp dp.
Problem 3.3 Industry revenues
Em h v h US dp d.
3.2 Estimating integrals
Th US pp v (S 3.1) w dffi p -
w kw. B v w-kw dffi
. I h , w mp mhd p : h
1/e h (S 3.2.1) d h wdh h mxmm (FWHM)
h (S 3.2.2).
3.2.1 1/e heuristic
0
1
0 1t
. . .
e−t
E , mph p, d d-
v d h q xp d
(v h dm m) ∞0
e−t dt. (3.4)
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34 3 Lumping
T ppxm v, ’ mp h e−t v .
What values should be chosen for the width and height of the rectangle?
mpd
0
1
0 1t
e−t
A hh h h mxmm e−t, m 1. T h wdh, fi
h h ( mhd d S-
3.3.3): Ch fi h e−t; h
fid h wdh Δt h pd h h. I xp d, mp d fi
h wh e−t m e
fi v (whh 0 h t ∞). Wh h ,
Δt = 1. Th mp h h —whh h x
v h !
e−x2
0 1−1
Ed h , ’ h h
h dffi ∞−∞
e−x2
dx. (3.5)
0 1−1
A mp h . I hh
h mxmm e−x2, whh 1. I wdh
h h e−x2 e. Th dp hp-
p x =
±1, h wdh Δx = 2 d
1 × 2. Th x √ π ≈ 1.77 (S 2.1), mp mk 13%: F h h dv, h
xm hh.
Problem 3.4 General exponential decay
U mp m h ∞0
e−at dt. (3.6)
U dm d hk h w mk .
Problem 3.5 Atmospheric pressure
Amph d ρ d h xp wh hh z :
ρ ∼ ρ0e−z/H, (3.7)
wh ρ0 h d v, d H h -d hh (hhh whh h d e). U vd xp m H.
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3.2 Estimating integrals 35
Th m h mph p v m h wh fi hh d .
Problem 3.6 Cone free-fall distance
Rh hw d S 2.4 h fi m v? Hw h d mpd hdp hh 2 m? Hint: Skh (v h) h ’ vm d mk mp ppxm.
3.2.2 Full width at half maximum
Ah mp h h d p-
p. A pp wp hh wvh,
h d wd p hw m d d
h wvh. Th v m pk wh d v h h m (d w dvp
qm h [14]). B dd d h d xd,
hw d h h pk mpd?
Th w mpd mp h pk wh hh
h hh h pk d wh wdh h wdh h mxmm
(FWHM). Wh h 1/e h e h fi
h, h FWHM h 2.
T h p h G ∞−∞ e−x2
dx.
√ ln 2−
√ ln 2
FWHM
Th mxmm hh e−x2 1, h h mxm
x = ±√
ln 2 d h wdh 2√
ln 2. Th
mpd h h 2√
ln 2 ≈ 1.665.
Th x √
π ≈ 1.77 (S 2.1): Th FWHM
h mk 6%, whh h
-h h h 1/e h.
Problem 3.7 Trying the FWHM heuristic
Mk - mp m h w . Ch hhh d wdh h h FWHM h. Hw h m?
.
∞−∞
1
1 + x2dx [x v: π ].
.
∞−∞
e−x4 dx [x v: Γ (1/4)/2 ≈ 1.813].
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36 3 Lumping
3.2.3 Stirling’s approximation
Th 1/e d FWHM mp h x hp ppxm h
q n!; h ’ m p-
h mh d h hm. Fpv , n! dfid n × (n − 1) × (n − 2) × · · · × 2 × 1. I
h d m, dffi ppxm. Hwv, h
p n!,
n! ≡ ∞
0
tne−t dt, (3.8)
pvd dfi v wh n pv —d h
ppxmd mp.
Th mp w m S’ m ppx-m m
n! ≈ nne−n√
2πn. (3.9)
Lumping requires a peak, but does the integrand tne−t have a peak?
T dd h d tne−t tn/et, xm h xm
t. Wh t = 0, h d 0. I h pp xm, t → ∞,
h pm tn mk h pd fi wh h xp
e−t mk z. Wh w h ? Th T
et
v pw t (d wh pv ffi), , fi-d pm. Th, t fi, et
pm tn d mk h d tn/et q 0 h
t → ∞ xm. B z h xm, h d m hv
pk w. I , h x pk. (C hw h?)
1
te−t
2
t2e−t
3
t3e−tI n h h pm
tn, tn vv hh t et
. Th, h pk tn/et h h
n . Th ph fim h pd
d h h pk t = n
. L’hk mxmz tne−t ,
m mp, mxmz hm f(t) =
n l n t − t. A pk, h z p.
B df/dt = n/t−1, h pk tpk = n, wh h d
tne−t nne−n—h pd h d m mp
S’ m.
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3.3 Estimating derivatives 37
tne−
t
2Δt
nn/en
What is a reasonable lumping rectangle?
Th ’ hh h pk hh nne−n.
F h ’ wdh, h h 1/e
h FWHM h. B h h -q ppxm tne−t, xpd hm
f(t) T d pk t = n:
f(n + Δt) = f(n) + Δtdf
dt
t=n
+(Δt)2
2
d2f
dt2
t=n
+ · · · . (3.10)
Th d m h T xp vh f(t) h z
p h pk. I h hd m, h d dvv d2f/dt2
t = n −n/t2 −1/n. Th,
f(n + Δt) ≈ f(n) − (Δt)2
2n. (3.11)
T d tne−t F q d f(t) ln F. Th
h m Δt =√
2nlnF. B h ’ wdh 2Δt, h
mpd- m n!
n! ∼ nne−n√
n ×√
8 (1/e : F = e)√ 8 l n 2 (FWHM : F = 2).
(3.12)
F mp, S’ m n!
≈nne−n
√ 2πn. Lmp h
xpd m v . Th nne−n h hh h -, d h
√ n m h wdh h . Ahh
h x√
2π m m (Pm 3.9), ppxmd
wh 13% (h 1/e h) 6% (h FWHM h).
Problem 3.8 Coincidence?
Th FWHM ppxm h d G (S 3.2.2) w 6%. Cd?
Problem 3.9 Exact constant in Stirling’s formula
Wh d h m √ 2π m m?
3.3 Estimating derivatives
I h pd xmp, mp hpd m . B d dff d, mp pvd
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38 3 Lumping
mhd m dvv. Th mhd wh dm-
v dvv. A dvv dff;
xmp, df/dx h df dx. B d dm
(S 1.3.2), h dm df/dx h dm f/x. Th
, p wh wh m xmp:Dff hh y wh p m t pd v dy/dt,
wh dm LT−1 dd h dm y/t.
Problem 3.10 Dimensions of a second derivative
Wh h dm d2f/dx2?
3.3.1 Secant approximation
x
x2
A df/dx d f/x hv d dm,
php h md m:
df
dx∼
f
x. (3.13)
Gm, h dvv df/dx h p h , wh h ppxm
f/x h p h . B p-
h v wh h , w mk
mp ppxm.L’ h ppxm h f(x) = x2. Gd
w—h d p dff 2:
df
dx= 2x d
f(x)
x= x. (3.14)
Problem 3.11 Higher powers
Iv h ppxm f(x) = xn.
Problem 3.12 Second derivatives
U h ppxm m d2f/dx2 wh f(x) = x2. Hw dh ppxm mp h x d dvv?
How accurate is the secant approximation for f(x) = x2 + 100?
Th ppxm qk d mk .Wh f(x) = x2 + 100, xmp, h d x = 1
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3.3 Estimating derivatives 39
hv dm dff p. Th p df/dx 2, wh
h p f(1)/1 101. Th h w p, hh
dm, d .
Problem 3.13 Investigating the discrepancy
Wh f(x) = x2 + 100, kh h
p
p(3.15)
x. Th ! Wh h dm ? (Th q k.)
Th dp p h dvv df/dx, whh
lim Δx→0
f(x) − f(x − Δx)
Δx , (3.16)
wh h p f(x)/x d w ppxm. Th fi
ppxm k Δx = x h h Δx = 0. Th df/dx ≈(f(x) − f(0))/x. Th fi ppxm pd h p h
m (0, f(0)) (x, f(x)). Th d ppxm p f(0) wh
0, whh pd df/dx ≈ f/x; h h p h m
(0, 0) (x, f(x)).
3.3.2 Improved secant approximation
x
x2 +C
x = 0
Th d ppxm fixd -
h (0, f(0)) d (0, 0).
With that change, what are the secant and tan-
gent slopes when f(x) = x2 + C?
C h (0, 0) h
; h w h x = 0 .
Th h x = 0 w h -h
h p h , m h C. Th x = 0
ppxm — ffd —v .
How robust is the x = 0 secant approximation against horizontal translation?
T v hw h x = 0 hd hz , - f(x) = x2 hwd 100 mk f(x) = (x−100)2. A h p’
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40 3 Lumping
vx x = 100, h x = 0 , m (0,10 4) (100,0), h p −100;
hwv, h h z p. Th h x = 0 , hh
mpvm h , ffd hz .
3.3.3 Significant-change approximation
Th dvv ffd hz d v ,
dvv ppxmd mh v. A
ppxm dvv
df
dx≈ f(x + Δx) − f(x)
Δx, (3.17)
wh Δx z m.
How small should Δx be? Is Δx = 0.01 small enough?
Th h Δx = 0.01 h w d. F, wk wh x h
dm. I x h, wh h m h? Ch Δx =
1 mm p m h mp dvv d h
m, p mp dvv d
dp. Sd, fixd h v.
Ahh Δx = 0.01 pd dvv wh f(x) = sinx, wh f(x) = sin 1000x, h mp x 1000x.
Th pm h w fi-h ppx-m:
df
dx∼
fi Δf (h f) x
Δx h pd fi Δf. (3.18)
B h Δx h dfid h pp h v h p
, wh v p d v v Δx,
h ppxm d v.
cosx
(0, 1)(0, 1)
(2π,1)(2π,1)
x = 0
T h ppxm, ’
f(x) = cosx d m df/dx x =3π/2 wh h h ppxm: h
, h x = 0 , d h
fi-h ppxm. Th
m (0, 0) (3π/2, 0),
h z p. I p ppx-m h x p 1. Th x = 0
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3.3 Estimating derivatives 41
m (0, 1) (3π/2, 0), h p −2/3π , whh
w h pd z p v h w!
cosx
(2π,1)(2π,1)
(3π2 , 0)(3π2 , 0)
(5π3 , 1
2)(5π3 , 1
2)
Th fi-h ppxm mh p-
vd m . Wh fi h f(x) = cosx? B h h 2
(m −1 1), 1/2 fi h f(x).
Th h hpp wh x h m 3π/2,
wh f(x) = 0, 3π/2 + π/6, wh f(x) = 1/2.
I h wd, Δx π/6. Th ppxm d-vv h
df
dx∼
fi Δf x
Δx∼
1/2
π/6=
3
π . (3.19)
Th m ppxm 0.955—mz h dv-v 1.
Problem 3.14 Derivative of a quadratic
Wh f(x) = x2, m df/dx x = 5 h ppxm: h , h x = 0 , d h fi-h ppxm. Cmph m h p.
Problem 3.15 Derivative of the logarithm
U h fi-h ppxm m h dvv ln x
x = 10. Cmp h m h p.
Problem 3.16 Lennard–Jones potential
Th Ld–J p md h w wp m h N2 CH4. I h h m
V (r) = 4
σ
r
12−σ
r
6
, (3.20)
wh r h d w h m, d d σ hdpd h m. U h m r0, h p r
whh V (r) mmm. Cmp h m h r0 d
.
Problem 3.17 Approximate maxima and minima
L f(x) d g(x) d . U h hw, ppxm, h h (x) = f(x) + g(x) h mmm whf(x) = g(x). Th hm, whh z Pm 3.16, d h h.
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42 3 Lumping
3.4 Analyzing differential equations: The spring–mass system
Em dvv d dff dv (S 3.3);
h d dff q q.
k
m
x0
T pd xmp q z, -
k m m d p wh
p (ff) k , p h k d-
x0 h h v h qm
p x = 0, d m t = 0. Th k k d
h, p x dd h d-p dff q
m d2x
dt2+ kx = 0. (3.21)
L’ ppxm h q d h m h -q.
3.4.1 Checking dimensions
Up q, fi hk dm (Chp 1). I
m d hv d dm, h q wh
v— v ff. I h dm mh, h hk h
pmpd fl h m h m; h fl hp
pp v h q d dd .
What are the dimensions of the two terms in the spring equation?
Lk fi h mp d m kx. I m Hk’ w, whh
h d p x kx wh x h x h
p v qm h. Th h d m kx
. I h fi m ?
Th fi m m (d2x/dt2) h d dvv d2x/dt2, whh
m . M dff q, hwv,
m dvv. Th Nv–Sk q fld mh(S 2.4),
∂v
∂t+ (v·∇)v = −
1
ρ∇ p + ν∇2v, (3.22)
w dvv: (v·∇)v d ∇2v. Wh h dm- h m?
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3.4 Analyzing differential equations: The spring–mass system 43
T p hd h mpd m, ’ w fid h
dm d2x/dt2 hd. B d2x/dt2 w xp
2, d x h d t m, d2x/dt2 mh p hv dm-
L2T−2.
Are L2T−2 the correct dimensions?
T dd, h d m S 1.3.2 h h dff m d
m “ .” Th m d2x, m d dx, “
x.” Th, h. Th dm dt2 dp m (dt)2 d(t2). [I m (dt)2.] I h ,
dm T2. Th, h dm h d dvv
LT−2:
d
2
xdt2
= LT−2. (3.23)
Th m , h p q’ fi m
m (d2x/dt2) m m —v h m dm
h kx m.
Problem 3.18 Dimensions of spring constant
Wh h dm h p k ?
3.4.2 Estimating the magnitudes of the terms
Th p q p h dm , wh z fid h q. Th mhd p h m wh
ppxm md. Th pm w mpd
dff q mp q h q.
T ppxm h fi m m (d2x/dt2), h fi-h p-
pxm (S 3.3.3) m h md h
d2x/dt2.
d2x
dt2∼
fi Δx
(Δt h pd fi Δx)2. (3.24)
Problem 3.19 Explaining the exponents
Th m h fi pw Δx, wh h dm h d pw Δt. Hw h dp ?
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44 3 Lumping
T v h ppxm , fi dd fi
Δx— wh fi h h m’ p. Th
m mv w h p x = −x0 d x = +x0, fi
h p hd fi h pk--pkmpd 2x0. Th mp h Δx = x0.
Nw m Δt: h m h k mv d mp
Δx. Th m—d h h m h m— d
h pd T . D pd, h m mv k
d h d v d 4x0—mh h h x0. I Δt w,, T/4 T/2π , h h m Δt h m wd v d
mp x0. Th h Δt hv p
ppxm 1/ω, wh h q ω d
h pd h dfi ω≡
2π/T . Wh h pd h
Δx d Δt, h m (d2x/dt2) m h mx0ω2.
What does “is roughly” mean?
Th ph m h mx0ω2 d m (d2x/dt2) wh, , 2, m (d2x/dt2) v d mx0/τ 2 . Rh, “
h” m h p h md m (d2x/dt2)—
xmp, -m-q v— mp mx0ω2. L’
d h m wh h wdd ∼. Th h p-
md m w
m d2x
dt2∼ mx0ω2. (3.25)
Wh h m m “ h”, m h h p m-
d mp, h p q’ d m kx h kx0.
Th w m m dd z— q h p q
m d2x
dt2+ kx = 0. (3.26)
Th, h md h w m mp:
mx0ω2 ∼ kx0. (3.27)
Th mpd x0 dvd ! Wh x0 , h q ω d -
pd T = 2π/ω dpd mpd. [Th
v ppxm, h x (Pm 3.20).]Th ppxmd q ω h
k/m .
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3.4 Analyzing differential equations: The spring–mass system 45
F mp, h x h p dff q ,
m Pm 3.22,
x = x0 cos ωt, (3.28)
wh ω
k/m . Th ppxmd q x!
Problem 3.20 Amplitude independence
U dm hw h h q ω dpd h mpd x0.
Problem 3.21 Checking dimensions in the alleged solution
Wh h dm ωt? Wh h dm cos ωt? Chk hdm h ppd x = x0 cos ωt, d h dm hppd pd 2π m/k .
Problem 3.22 Verification
Shw h x = x0 cos ωt wh ω =
k/m v h p dff q
m d2x
dt2+ kx = 0. (3.29)
3.4.3 Meaning of the Reynolds number
A h xmp mp— p, h fi-h
ppxm—’ z h Nv–Sk q dd
S 2.4,
∂v
∂t+ (v·∇)v = −
1
ρ∇ p + ν∇2v, (3.30)
d x m hm ph m h Rd m rv/ν.
T d , w m h p md h m (v·∇)v
d h v m ν∇2v.
What is the typical magnitude of the inertial term?
Th m (v·∇)v h p dvv ∇v. Ad
h fi-h ppxm (S 3.3.3), h dvv ∇v
h h
fi h flw v
d v whh flw v h fi. (3.31)
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46 3 Lumping
Th flw v (h v h ) z m h
d mp v h (whh mv pd v).
Th, v, v, fi h
flw v. Th pd h hpp v d mp h z h : Sv h w, h hd kw
h . Th ∇v ∼ v/r. Th m (v·∇)v
d v, (v·∇)v h v2/r.
What is the typical magnitude of the viscous term?
Th v m ν∇2v w p dvv v. B
h p dvv 1/r h p md,
ν∇2v h νv/r2. Th h m h v m h h ( v2/r)/(νv/r2). Th mpfi rv/ν—h m,
dm, Rd m.
Th, h Rd m m h mp v. Wh
R 1, h v m m, d v h ff. I
pv p fld m q fi dffv, d h flw m . Wh R 1, h v
m , d v h dm ph ff. Th flw
z, wh p d h.
3.5 Predicting the period of a pendulum
Lmp mp, - dff q. O xmp h h
pd pdm, h W mkp.
How does the period of a pendulum depend on its amplitude?
m
l
θ
Th mpd θ0 h mxmm h w; -
pdm d m , h .
Th ff mpd d h h pd-m dff q ( [24] h q’ dv):
d2θ
dt2+
g
lsinθ = 0. (3.32)
Th w : dm (S 3.5.2),
(S 3.5.1 d S 3.5.3), d mp (S 3.5.4).
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3.5 Predicting the period of a pendulum 47
Problem 3.23 Angles
Exp wh dm.
Problem 3.24 Checking and using dimensions
D h pdm q hv dm? U dm - hw h h q h m h (xp mm h dvd ).
3.5.1 Small amplitudes: Applying extreme cases
θ1 sinθ
θ
Th pdm q dffi
sinθ. F, h
h m-mpd xm θ
→0. I h
m, h hh h , whh sinθ, m x h h θ. Th, m
, sinθ ≈ θ.
Problem 3.25 Chord approximation
Th sinθ ≈ θ ppxm p h wh h, v . Tmk m ppxm, p h wh h hd ( h v ). Wh h ppxm sinθ?
I h m-mpd xm, h pdm q m :
d2θ
dt2+
g
lθ = 0. (3.33)
Cmp h q h p–m q (S 3.4)
d2x
dt2+
k
m x = 0. (3.34)
Th q pd wh x θ d k/m
g/l. Th q h p–m m ω =
k/m , d
pd T = 2π/ω = 2π m/k . F h pdm q, h
pd pd
T = 2π
l
g( m mpd). (3.35)
(Th pvw h mhd , whh h j
Chp 6.)
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48 3 Lumping
Problem 3.26 Checking dimensions
D h pd 2π
l/g hv dm?
Problem 3.27 Checking extreme cases
D h pd T = 2π
l/g mk h xm g → ∞ dg → 0?
Problem 3.28 Possible coincidence
I d h g ≈ π 2 m −2? (F xv h dh vv h pdm, [1] d m d [4, 27, 42].)
Problem 3.29 Conical pendulum for the constant
m
lθ
Th dm 2π dvd -h m H [15, p. 79]: z h m
pdm mv hz ( pd-m). Pj w-dm m v- pd -dm pdm m, h pd h w-dm m h m h pd -dm pdm m! Uh d wh Nw’ w m xph 2π .
3.5.2 Arbitrary amplitudes: Applying dimensional analysis
Th pd mh h h mpd θ0 m.
As θ0 increases, does the period increase, remain constant, or decrease?
A m xpd dm p
(S 2.4.1). Th pm vv h pd T , h l, v
h g, d mpd θ0. Th, T h dm-
p T
l/g. B dm, θ0
dm p. Th w p T
l/g d θ0 dpd
d d h pm (Pm 3.30).
k
m
x0
A v h d p–m
m. Th pd T , p k , d m
m m h dm p T
m/k ;
h mpd x0, h q
h, p dm p (Pm 3.20) d
h ff h pd h p–m m. I ,
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3.5 Predicting the period of a pendulum 49
h pdm’ mpd θ0 d dm p,
ff h pd h m.
Problem 3.30 Choosing dimensionless groups
Chk h pd T , h l, v h g, d mpd θ0 p-d w dpd dm p. I p z h pd, wh hd T pp p? Ad wh hdθ0 pp h m p T ?
Tw dm p pd h dm m
p = h h p, (3.36)
T l/g = θ0. (3.37)
B T
l/g = 2π wh θ0 = 0 (h m-mpd m),
h 2π mp h q q, d dfi dm
pd h w:
T l/g
= 2π h (θ0). (3.38)
Th h m hw mpd ff h
pd pdm. U h , h q h pd -m h w: I h , , d
mpd? Th q wd h w .
3.5.3 Large amplitudes: Extreme cases again
F h hv h mpd, m m v h w mpd. O mpd
h xm z mpd, wh h (0) = 1. A d mpd
h pp xm mpd. How does the period behave at large amplitudes? As part of that question, whatis a large amplitude?
A mpd π/2, whh m h pd-
m m hz. Hwv, π/2 h x h h w wxp (Pm 3.31):
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50 3 Lumping
h (π/2) =
√ 2
π
π/2
0
dθ√ cosθ
. (3.39)
I h h, q , m h 1? Wh kw? Th -
k hv d m d q m v(Pm 3.32).
Problem 3.31 General expression for h
U v hw h h pd
T (θ0) = 2√
2
l
g
θ00
dθ√ cosθ − cosθ0
. (3.40)
Cfim h h qv dm m
h (θ0) = √ 2π θ0
0dθ√
cos θ − cosθ0. (3.41)
F hz , θ0 = π/2, d
h (π/2) =
√ 2
π
π/2
0
dθ√ cosθ
. (3.42)
Problem 3.32 Numerical evaluation for horizontal release
Wh d h mp p (S 3.2) h Pm 3.31?Cmp h (π/2) m .
B θ0 = π/2 hp xm, v m xm. Tθ0 = π , whh m h pdm m v. I h
d h pv p , hwv, v
wd m h h h dw d . Th
v hv h dd dd h pdm
dff q.
θ0
h(θ0)
π
11
F, hh xpm hp m-pv: Rp h wh m
d. Bd p θ0 = π , h pdm
h pd dw v, T (π ) = ∞ dh (π ) = ∞. Th, h (π ) > 1 d h (0) = 1. Fm
h d, h m k j h h -
m wh mpd. Ahh
h d fi d d h , h w d wd
p hv m h dff q. (F h hv h θ0 = π , Pm 3.34).
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3.5 Predicting the period of a pendulum 51
Problem 3.33 Small but nonzero amplitude
θ0
h
1 AB
A h mpd pph π , h dm pd h
dv fi; z mpd, h = 1. B wh h dvv h ? A z mpd (θ0 = 0), d h (θ0)
hv z p (v A) pv p (v B)?
Problem 3.34 Nearly vertical release
β h (π − β)
10−1 2.791297
10−2 4.255581
10−3 5.721428
10−4 7.187298
Im h pdm m m v: π − β wh β . A β,h hw d h pdm k fi —, 1 d? U h m pd hw h (θ0) hv wh θ0 ≈ π . Chk dfi j h d v. Thpd h (π − 10−5).
3.5.4 Moderate amplitudes: Applying lumping
Th j h h m w dvd h x-
m z d v mpd, hd pp md
mpd. B k h m h, pv mm- : “T, v.”
At moderate (small but nonzero) amplitudes, does the period, or its dimensionlesscousin h , increase with amplitude?
I h z-mpd xm, sinθ θ. Th ppxm
d h pdm q
d2θ
dt2+
g
lsinθ = 0 (3.43)
h , d-p q— whh h pd dpd
mpd.
A z mpd, hwv, θ d sinθ dff d h dff
ff h pd. T h dff d pd h pd,p sinθ h θ d djm f(θ). Th
q
d2θ
dt2+
g
lθ
sinθ
θ f(θ)
= 0. (3.44)
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52 3 Lumping
0
1
0 θ0
f(θ)Th f(θ) p h
h pdm q. Wh θ , f(θ) ≈ 1: Th
pdm hv k , d-p m.
B wh θ , f(θ) fi w 1,
mk h d-p ppxm fi. A h , h p
dffi z— xmp, h w Pm 3.31.
A m, mk mp ppxm p h
h f(θ) wh .
0
1
0 θ0
f(0)Th mp f(0). Th h pd-
m dff q m
d2θ
dt2
+g
l
θ = 0. (3.45)
Th q , , h d-p q.
I h ppxm, pd d dpd mpd, h = 1
mpd. F dm hw h pd ppxmd
pdm dpd mpd, h f(θ) → f(0) mp ppxm-
dd mh m.
0
1
0 θ0
f(θ0)
Th, p f(θ) wh h h xm
f(θ0). Th h pdm q m
d2θ
dt2 +
g
l θf(θ0) = 0. (3.46)
Is this equation linear? What physical system doesit describe?
B f(θ0) , h q ! I d z-mpd pdm p wh v gff h h wk
h h v— hw h w h p:
d2θ
dt2 +
gff
gf(θ0)
l θ = 0. (3.47)
B h z-mpd pdm h pd T = 2π
l/g, h z-
mpd, w-v pdm h pd
T (θ0) ≈ 2π
l
gff = 2π
l
gf(θ0). (3.48)
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3.5 Predicting the period of a pendulum 53
θ0π
1
h
f−1/2U h dm pd h vd w
h 2π , l, d g, d d h mp
pd
h (θ0) ≈ f(θ0)−1/2 =
sinθ0
θ0
−1/2
. (3.49)
A md mpd h ppxm
w h x dm pd (dk v). A ,
pd h (π ) = ∞, wh h hh xpm h pdm m ph (S 3.5.3).
How much larger than the period at zero amplitude is the period at 10◦ amplitude?
A 10◦ mpd h 0.17 d, md , h ppxmpd h ppxmd T .
Th T sinθ θ − θ3/6,
f(θ0) =sinθ0
θ0≈ 1 −
θ20
6. (3.50)
Th h (θ0), whh h f(θ0)−1/2, m
h (θ0) ≈
1 −θ2
0
6 −1/2
. (3.51)
Ah T d (1 + x)−1/2 ≈ 1 − x/2 ( m x). Th,
h (θ0) ≈ 1 +θ2
0
12. (3.52)
R h dmd q v h pd .
T ≈ 2π
l
g
1 +
θ20
12
. (3.53)
Cmpd h pd z mpd, 10◦ mpd pd h θ2
0/12 ≈ 0.0025 0.25%. Ev md
mpd, h pd dpd mpd!
Problem 3.35 Slope revisited
U h pd h (θ0) hk Pm 3.33 h p h (θ0) θ0 = 0.
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54 3 Lumping
Does our lumping approximation underestimate or overestimate the period?
Th mp ppxm mpfid h pdm dff q-
p f(θ) wh f(θ0). Eqv, md h h m
w md h dp h m wh |θ| = θ0. Id,h pdm pd mh m md p wh
|θ| < θ0 d f(θ) > f(θ0). Th, h v f h f(θ0).
B h v d f (h = f−1/2), h f(θ) → f(θ0) mpppxm vm h d h pd.
Th f(θ) → f(0) mp ppxm, whh pd T = 2π
l/g,
dm h pd. Th, h ffi h θ20 m
h pd ppxm
T ≈ 2π
lg
1 + θ
2
012
(3.54)
w 0 d 1/12. A h h ffi hw
w h xm—m, 1/24. Hwv, h pdm pd
m m wd h xm (wh f(θ) = f(θ0)) h pd h qm p (wh f(θ) = f(0)). Th, h -
fi p 1/12—h pd h f(θ) → f(θ0)
ppxm—h 0. A mpvd mh w-hd
h w m 0 1/12, m 1/18.
I mp, v-ppxm h pdm
dff q v h w pd [13, 33]:
T = 2π
l
g
1 +
1
16θ2
0 +11
3072θ 4
0 + · · ·
. (3.55)
O dd 1/18 v h ffi 1/16!
3.6 Summary and further problems
Lmp hd. Wh z h-
p dvd v fi v, mp mpfi
h p m h p. I
v h , dffi mp, d md
dff q dff q.
. . . the crooked shall be made straight, and the rough places plain. (Ih 40:4)
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3.6 Summary and further problems 55
Problem 3.36 FWHM for another decaying function
U h FWHM h m
∞
−∞
dx
1 + x4. (3.56)
Th mp h m wh h x v π/√
2. F jdd pm, dv h x v.
Problem 3.37 Hypothetical pendulum equation
Spp h pdm q hd
d2θ
dθ2+
g
ltanθ = 0. (3.57)
Hw wd h pd T dpd mpd θ0? I p, θ0 ,wd T d, m , ? Wh h p dT/dθ0
z mpd? Cmp wh h Pm 3.33.F m z θ0, fid ppxm xp h dmpd h (θ0) d hk pv .
Problem 3.38 Gaussian 1-sigma tail
Th G p d wh z m d v
p(x) =e−x2/2
√ 2π
. (3.58)
Th mp q , h d m.I h pm m h h 1-m ∞
1
e−x2/2√
2π dx. (3.59)
. Skh h v G d hd h 1-m .
. U h 1/e mp h (S 3.2.1) m h .
. U h FWHM h m h .
d. Cmp h w mp m wh h m : ∞1
e−x2/2
√ 2π
dx =1 − (1/
√ 2)
2≈ 0.159, (3.60)
wh (z ) h .
Problem 3.39 Distant Gaussian tails
F h p G, m h n-m ( n). I h wd, m ∞
n
e−x2/2
√ 2π
dx. (3.61)
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4
Pictorial proofs
4.1 Add dd m 584.2 Ahm d m m 604.3 Appxm h hm 664.4 B 704.5 Smm 734.6 Summary and further problems 75
Hv v wkd hh p, dd d fimd h
p, vd h hm? Y z that h hm
, why .
T h m m xmp, m h
hd h v d h h mp Fhh Cd, whhv m. I m vd xp, mp-
m Fhh. Wh I h mp 40◦C,
I h w :
1. I v 40◦ C Fhh: 40 × 1.8 + 32 = 104.
2. I : “Ww, 104◦ F. Th’ d! G h d!”
Th C mp, hh m qv h Fh-h mp, . M d v
h mp v h mp m xp.
A m dp, whh p m mp,
v mpd m h pk pp -
m. Th hw qd h p m
. (S Evolving Brains [2] d, h h h .) Sm, q q , whh h
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58 4 Pictorial proofs
vvd 105 . Ahh 105 p m hm m,
v k. I p, h mpd h m
p v whh pp hdw h vvd: F v h-
dd m , m hv fid h p h,m, , h, d .
Ev h wkd 1000 m pp h
m- . Cmpd pp hd-
w, m, q hdw -dvpd m.
N p, pp p m -. Ev pp hh-v m v h p
dm h m pp hdw [16]. Seeing d
v dph dd h m dp
mh.
Problem 4.1 Computers versus people
A k k xpd (x + 2y)50, mp mh h pp. Ak k z m, v hd mh h mp. Hw d xp h ?
Problem 4.2 Linguistic evidence for the importance of perception
I v (), hk h m m d-d ( xmp, p).
4.1 Adding odd numbers
T h v p, ’ fid h m h fi n dd
m ( h j Pm 2.25):
Sn = 1 + 3 + 5 + · · · + (2n − 1) n m
. (4.1)
E h n = 1, 2, 3 d h j h Sn = n2.B hw h j pvd? Th dd m mhd
p d:
1. V h Sn = n2 h base case n = 1. I h , S1 1,
n2, h vfid.
2. Mk h induction hypothesis: Am h Sm = m 2 m h
q mxmm v n. F h p, h w, wkd hph ffi:
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4.1 Adding odd numbers 59
n1
(2k − 1) = n2. (4.2)
I h wd, w m h hm h h m = n.3. Pm h induction step: U h d hph hw h
Sn+1 = (n + 1)2. Th m Sn+1 p w p:
Sn+1 =
n+11
(2k − 1) = (2n + 1) +
n1
(2k − 1). (4.3)
Thk h d hph, h m h h n2. Th
Sn+1 = (2n + 1) + n2, (4.4)
whh (n + 1)2; d h hm pvd.
Ahh h p pv h hm, why h m Sn d p n2
v.
Th m dd—h kd h dd
Whm [48]—q p p. S dw h dd
m L-hpd pzz p:
1
3
5
(4.5)
How do these pieces fit together?
Th mp Sn fi h h pzz p w:
S2 = 1 +3
= 1
3
S3 = 1 +
3
+
5
= 1
3
5
(4.6)
Eh v dd m—h p—xd h q 1
hh d wdh, h n m d n × n q. [O
(n − 1) × (n − 1) q?] Th, h m n2. A p h
p p, wh dd p h fi n dd mpd n2.
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60 4 Pictorial proofs
Problem 4.3 Triangular numbers
Dw p p hw h
1 + 2 + 3 + · · · + n + · · · + 3 + 2 + 1 = n2. (4.7)
Th hw h
1 + 2 + 3 + · · · + n =n(n + 1)
2. (4.8)
Problem 4.4 Three dimensions
Dw p hw h
n0
(3k 2 + 3k + 1) = (n + 1)3. (4.9)
Gv p xp h 1 h mmd 3k 2
+ 3k + 1; h 3 dh k 2 3k 2; d h 3 d h k 3k .
4.2 Arithmetic and geometric means
Th x p p wh w v m— xm-
p, 3 d 4—d mp h w w v:
hm m ≡ 3 + 4
2= 3.5; (4.10)
m m ≡ √ 3 × 4 ≈ 3.464. (4.11)
T h p m— xmp, 1 d 2. Th hm m
1.5; h m m √
2 ≈ 1.414. F h p, h mm m h h hm m. Th p ;
h m hm-m–m-m (AM–GM) q [18]:
a + b
2
AM
√
ab
GM
. (4.12)
(Th q q h a, b 0.)
Problem 4.5 More numerical examples
T h AM–GM q vd m xmp. Wh d wh a d b h h? C mz h p?(S Pm 4.16.)
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4.2 Arithmetic and geometric means 61
4.2.1 Symbolic proof
Th AM–GM q h p d m p. Th m- p wh (a − b)2— p h h q-
a + b h h a − b. Th d dd h m(a − b)2. I v, a2 − 2ab + b2 0. Nw m dd
dd 4ab h d. Th
a2 + 2ab + b2 (a+b)2
4ab. (4.13)
Th d (a + b)2, a + b 2√
ab d
a + b
2
√ ab. (4.14)
Ahh h p mp, h wh h m k m d v
h why m. I h hd dd wh (a + b)/4 √
ab,
wd k v w. I , v p wd
v h h q hp .
4.2.2 Pictorial proof
Th pvdd p p.
What is pictorial, or geometric, about the geometric mean?
x
a b
A m p h m m
wh h . L wh hp
hz; h wh h d x
h h d dk . Th hp
p w h a d b, d h d
x h m m√
ab.
Why is the altitude x equal to√
ab?
b
x
T hw h x =√
ab, mp h m, dk
h , h h m
d h . Th w
m! Th, h p (h h
h h d) d. I m, x/a =
b/x: Th d x h h m m√
ab.
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62 4 Pictorial proofs
Th h p h m-m p h
AM–GM q. Th hm m (a + b)/2 h p,
-h h hp. Th, h q m h
hp2
d. (4.15)
A, h m p v.
Can you find an alternative geometric interpretation of the arithmetic mean thatmakes the AM–GM inequality pictorially obvious?
√ aba+ b
2
a b
Th hm m h d
wh dm a + b. Th-
, m m d
h , mh h ’ dm- wh h hp a + b (P-
m 4.7). Th d xd h
d; h,
a + b
2
√ ab. (4.16)
Fhm, h w d q wh h d h
d h m—m wh a = b. Th ph h q d q d -
-p j. (A v p p h AM–GM q
dvpd Pm 4.33.)
Problem 4.6 Circumscribing a circle around a triangle
H w xmp hw md d .
Dw p hw h h q dmd h .
Problem 4.7 Finding the right semicircle
A q dm m (Pm 4.6). Hwv,h ’ dm mh wh d h . C m- w md d h wh h ’dm h hp?
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4.2 Arithmetic and geometric means 63
Problem 4.8 Geometric mean of three numbers
F h v m, h AM–GM q
a + b + c
3 (abc)1/3. (4.17)
Wh h q, w-m , k hv m p? (I fid p, m kw.)
4.2.3 Applications
Ahm d m m hv wd mhm pp.
Th fi pp pm m vd wh dvv:Fd fixd h h d.
What shape of rectangle maximizes the area?
a
b
d
Th pm vv w q: pm h
fixd d mxmz. I h pm -
d h hm m d h h m
m, h h AM–GM q mh hp mxmz
h . Th pm P = 2(a + b) m hhm m, d h A = ab h q h
m m. Th, m h AM–GM q,
P 4
AM
√ A GM
(4.18)
wh q wh a = b. Th d fixd h m .
Th h h d, whh v dpd a d b, h mxmm
P/4 wh a = b. Th mxm- q.
Problem 4.9 Direct pictorial proof
Th AM–GM h mxm d d p
. I m p h p p h AM–GM q. C dw p hw d h h q hpm hp?
Problem 4.10 Three-part product
Fd h mxmm v f(x) = x2(1 − 2x) x 0, wh .Skh f(x) fim w.
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64 4 Pictorial proofs
Problem 4.11 Unrestricted maximal area
I h d d , wh h mxm- hp?
Problem 4.12 Volume maximization
flp x
x
Bd p-ppd x w: S wh q, d , d d h flp. Th xh vm V = x(1 − 2x)2, wh x h d h . Wh h x mxmz h vm h x?
H p mdd h h d. S a = x, b = 1 − 2x, d c = 1 − 2x. Th abc hvm V , d V 1/3 =
3√
abc h m m (Pm 4.8). B hm m v xd h hm m d h w m q wh a = b = c, h mxmm vm d wh x = 1 − 2x.Th, h x = 1/3 hd mxmz h vm h x.
Nw hw h h h w ph V (x) dV/dx = 0;xp wh w wh h pd ; d mk v.
Problem 4.13 Trigonometric minimum
Fd h mmm v
9x2 sin2 x + 4
xsinx(4.19)
h x ∈ (0, π ).
Problem 4.14 Trigonometric maximum
I h t ∈ [0, π/2 ], mxmz sin 2t , qv, 2sintcost.
Th d pp hm d m m md,
mz pd mhd mp π [5, 6]. A mhd
mp π dd h pm m-dd
p d pvdd w dm p .
R mp hv d Lz’
arctan x = x −x3
3+
x5
5−
x7
7+ · · · . (4.20)
Im h w mp π 109 d, php h
hdw w pmp d whh h d π
dm ( hm C S’ v Contact [40]). S x = 1 h
Lz pd π/4, h v xm w.
O 109 d q h 10109m— m m h
m h v.
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4.2 Arithmetic and geometric means 65
F, p m d d Jh Mh (1686–
1751)
arctan 1 = 4 arctan1
5
− arctan1
239
(4.21)
h v d x:
π
4= 4 ×
1 −
1
3 × 53+ · · ·
arctan (1/5)
−
1 −
1
3 × 2393+ · · ·
arctan (1/239)
. (4.22)
Ev wh h pdp, 109-d q h
109 m.
I , h md B–Sm hm [3, 41], whh
hm d m m, v π xm pd. Th
hm d mz mhd h pm p (Pm 4.15) d m
d [23]. Th hm v q
wh a0 = 1 d g0 = 1/√
2; h mp v hm m
an, m m gn, d h qd dff dn.
an+1 =an + gn
2, gn+1 =
√ angn, dn = a2
n − g2n. (4.23)
Th a d g q pd v m M(a0, g0) dh hm–m m a0 d g0. Th M(a0, g0) d h
dff q d dm π .
π =4M(a0, g0)2
1 −∞
j=1 2 j+1d j. (4.24)
Th d q pph z qd; h wd, dn+1 ∼ d2n
(Pm 4.16). Th, h h mp π d
h d . A -d π q
30 — w h h 10109m h
wh x = 1 v h h 109 m Mh’ pdp.
Problem 4.15 Perimeter of an ellipse
T mp h pm p wh mmj x a0 d mmx g0, mp h a, g, d d q d h mm m M(a0, g0) h a d g q, h mp π . Th h pm P mpd wh h w m:
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66 4 Pictorial proofs
P =A
M(a0, g0)
⎛⎝a2
0 − B
∞j=0
2jdj
⎞⎠ , (4.25)
wh A d B dm. U h mhd (Chp 2) dm h v. (S [3] hk v d p h mpd m.)
Problem 4.16 Quadratic convergence
S wh a0 = 1 d g0 = 1/√
2 ( h pv p) d w v h AM–GM q
an+1 =an + gn
2d gn+1 =
√ angn. (4.26)
Th dn = a2n − g2
n d log10 dn hk h dn+1 ∼ d2n (qd
v).
Problem 4.17 Rapidity of convergence
Pk pv x0; h q h
xn+1 =1
2
xn +
2
xn
(n 0). (4.27)
T wh d hw pd d h q v? Wh x0 < 0?
4.3 Approximating the logarithm
θ1sinθ
θ
A ppxmd T
f(x) = f(0) + xdf
dx
x=0
+x2
2
d2f
dx2
x=0
+ · · · , (4.28)
whh k k v q m.
F, p xp h fi d m
mp m ppxm. F xmp, h -m
ppxm sinθ ≈ θ, whh p h d h
h h , h pdm dff q , q (S 3.5).
Ah T- h v p m m h
h hm :
ln(1 + x) = x −x2
2+
x3
3− · · · . (4.29)
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4.3 Approximating the logarithm 67
I fi m, x, w d h wd ppxm (1 + x)n ≈ enx
m x d n (S 5.3.4). I d m, −x2/2, hp
v h h ppxm. Th fi w m
h m m—d h hv p xp.
11+t
ln(1 + x)
0 x
1
t
Th p h p
ln(1 + x) =
x0
dt
1 + t. (4.30)
What is the simplest approximation for the shaded area?
11+t
x
0 x
1
t
A fi ppxm, h hdd hh md — xmp mp-
. Th h x:
= hh 1
× wdh x
= x. (4.31)
Th pd h fi m h T . B
md , h vm ln(1 + x).
11+t
0 x
1
t
Th ppxmd dw -d . I wdh x, hh
1 h 1/(1 + x), whh ppxm
1 − x (Pm 4.18). Th h d
h h ppxm x(1 − x) = x − x2. Th h dm ln(1 + x).
Problem 4.18 Picture for approximating the reciprocal function
Cfim h ppxm
1
1 + x≈ 1 − x ( m x) (4.32)
x = 0.1 x = 0.2. Th dw p h qvppxm (1 − x)(1 + x) ≈ 1.
W w hv w ppxm ln(1 + x). Th fi d hmp ppxm m m dw h md .
Th d ppxm m m dw h d .
Bh d d h x v.
How can the inscribed- and circumscribed-rectangle approximations be combinedto make an improved approximation?
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68 4 Pictorial proofs
11+t
0 x
1
t
O ppxm vm h , d h
h dm h ; h v h
mpv h ppxm. Th v
pzd wh
x + (x − x2)
2= x −
x2
2. (4.33)
Th pd h fi w m h T
ln(1 + x) = x −x2
2+
x3
3− · · · . (4.34)
Problem 4.19 Cubic term
Em h m h T m h dff w
h pzd d h .
F h hm ppxm, h hd pm ln 2.
ln(1 + 1) ≈
1 ( m)
1 −1
2(w m).
(4.35)
Bh ppxm dff fi m h v (h
0.693). Ev md ln 2 q m m h T
, d wh p xp (Pm 4.20). Th pm
h x ln(1 + x) 1, h xn
h m h T d hk h hh-n m.
Th m pm hpp wh mp π Lz’
(S 4.2.3)
arctan x = x −x3
3+
x5
5−
x7
7+ · · · . (4.36)
B x = 1, h d ppxm π/4 q m m
v md . F, h m d
arctan 1 = 4 arctan 1/5 − arctan 1/239 w h x 1/5 d
h pd h v.
Is there an analogous that helps estimate ln 2?
B 2 ( 4/3)/(2/3), w ln 2
ln 2 = ln4
3− ln
2
3. (4.37)
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4.3 Approximating the logarithm 69
Eh h h m 1 + x wh x = ±1/3. B x m,
m h hm mh pvd . L’
h ln(1 + x) ≈ x ppxm h w hm:
ln 2 ≈ 13
−
− 13
= 2
3. (4.38)
Th m wh 5%!
Th w k h hpd mp π ( w h arctan x
) d m ln(1 + x) ( w x ). Th d h-
m mhd— k h I w (h dfi
d P).
Problem 4.20 How many terms?
Th T h hm
ln(1 + x) =
∞1
(−1)n+1 xn
n. (4.39)
I x = 1 h , hw m m qd m ln 2 wh 5%?
Problem 4.21 Second rewriting
Rp h w mhd w 4/3 d 2/3; h m ln 2 m h hm . Hw h vd m?
Problem 4.22 Two terms of the Taylor series
A w ln 2 ln( 4/3) − ln(2/3), h w-m ppxm hln(1+x) ≈ x−x2/2 m ln 2. Cmp h ppxm h -mm, m 2/3. (Pm 4.24 v p xp.)
Problem 4.23 Rational-function approximation for the logarithm
Th pm ln 2 = ln( 4/3) − ln(2/3) h h m
ln(1 + x) = ln1 + y
1 − y, (4.40)
wh y = x/(2 + x).
U h xp y d h -m ln(1+x) ≈ x xp ln(1+x)
x ( pm x). Wh h fi wm T ?
Cmp h m h fi w m h ln(1 + x) T , dh xp wh h - ppxm m hv h w-m ln(1 + x) ≈ x − x2/2.
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70 4 Pictorial proofs
Problem 4.24 Pictorial interpretation of the rewriting
11+t
ln 2
−1/3 1/3
1
t
. U h p ln(1 + x) xpwh h hdd ln 2.
. O h h p
ln4
3− ln
2
3(4.41)
wh h md- ppxm h hm.
. O h m wh h pzd p-
pxm ln(1+x) = x−x2/2. Shw p hh , hh dff hp, h h m h h dw m .
4.4 Bisecting a triangle
P p k m pm:
What is the shortest path that bisects an equilateral triangle into two regions of equal area?
Th p ph m fi . T mh mpx, (Chp 2)—dw w q -
d hm wh ph. P, d, v mh m.
What are a few easy paths?
l =√
3/2
1
l
Th mp ph v m h p
h w h h wh 1/2. Th
ph h ’ d, d h h
l =
12 − (1/2)2 =
√ 3
2≈ 0.866. (4.42)
l = 1/√
2
A v h ph p h pzd
d m .
What is the shape of the smaller triangle, and how long is the path?
Th m h , q.Fhm, h -h h h , h
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4.4 Bisecting a triangle 71
d, whh h ph, √
2 m h h
d h . Th h ph h h 1/√
2 ≈ 0.707—
mpvm h v ph wh h√
3/2.
Problem 4.25 All one-segment paths
A q h fi m -m ph.A w hm hw h fi. Whh -m ph h h?
l = 1
Nw ’ v w-m ph. O p
ph dmd d xd w m .
Th w m p -h h .
Eh m h p -h h
d h d h 1/2. B h ph -
w h d, h h 1. Th ph , , h w -m dd, wh h 1/
√ 2 d
√ 3/2.
Th, j h h h ph h h w
m. Th j dv d (Pm 4.26).
Problem 4.26 All two-segment paths
Dw fi hw h v w-m ph. Fd h h ph,hw h h h
l = 2
×31/4
×sin15◦
≈0.681. (4.43)
Problem 4.27 Bisecting with closed paths
Th ph d d d h . Tw xmp d h:
D xp d ph h h h h-m ph? Gv m j, d hk h
j fid h h h w v d ph.
Does using fewer segments produce shorter paths?
Th h -m ph h ppxm h 0.707; h
h w-m ph h ppxm h 0.681. Th h
d xm ph: ph wh fi m
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72 4 Pictorial proofs
m. I h wd, vd ph. Th vd ph
p p .
What is a likely candidate for the shortest circle or piece of a circle that bisects
the triangle?
Whh h ph p , d .
Hwv, p h d h d
pd ph (Pm 4.27). Th
h p vx h , m
d vx.
How long is this arc?
Th d -xh (60
◦) h , h l = πr/3
,wh r d h . T fid h d, h qm
h h m h . Th, h -h
h ’ . Th d r h πr2 = 3√
3/4:
1
6× h
πr2
=1
2× h √
3/4
. (4.44)
Th d h (3√
3/4π )1/2; h h h πr/3, whh
ppxm 0.673. Th vd ph h h h h
w-m ph. I mh h h p ph.T h j, w mm. B q -xh hx, d hx p h d
q . H h hx m h d
hz :
Th x ph m hx wh -h
h h hx.
What happens when replicating the triangle bisected by the circular arc?
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4.5 Summing series 73
Wh h pd, x p mk
wh q -h h h hx.
F fixd , h h h pm (h
pm hm [30] d Pm 4.11); h,-xh h h h ph.
Problem 4.28 Replicating the vertical bisection
Th d v , pd d d, pd md d h h vx p. Hw h pd h h x ph m p?
Problem 4.29 Bisecting the cube
O h w q vm, whh h hm ?
4.5 Summing series
F h fi xmp wh p xp, h
. O fi ppxm n! wh p-
d h d mp (S 3.2.3).
ln 2
ln 3
ln 4ln 5
ln k
1 2 3 4 5k
Lmp, p v wh
wh mpd,
d p . A dp n! wh h mm- p
ln n! =
n1
lnk. (4.45)
Th m q h md h m .
Problem 4.30 Drawing the smooth curve
S h hh h q dw h ln k v—whhd h p d h wh h d. I hpd fi d h h , h v h hdp h d. A d h , d h w h:
. Th v h dp h d.
. Th v h mdp h d.
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74 4 Pictorial proofs
n1
l nkdk
ln k
1 · · · nk
Th md ppxm
h d h ln k v,
ln n!
≈ n
1
lnkdk = n l n n − n + 1.
(4.46)
Eh m h ln n! ppxm
n!:
n! ≈ nn × e−n × e. (4.47)
Eh h p m S’ ppxm
(S 3.2.3). I dd d mp, h S’
ppxm
n! ≈ nn × e−n × √ n × √ 2π. (4.48)
Th ppxm pd h w m mp
d m pd h h : e d√
2π dff 8%.
Th xpd √
n.
ln k
1 · · · nk
From where does the√
n factor come?
Th√
n m m m h m
v h ln k v. Th m
d wd dd h w .Th, dw h ln k v h-
m (h mp).
ln k
1 · · · nk
ln k
1 · · · nk
Th wd dd
h w . Th, ’ d h
mk .
What is the sum of these rectangular pieces?
T m h p, h hd h
k = n v . Wh hd, hv hp h h h h h hd.
Th p h k m h ln n .
B h p d h pd
p, h p
m (ln n)/2. Th mpv h ppx-m. Th ppxm ln n! w h m m:
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4.6 Summary and further problems 75
ln n! ≈ n l n n − n + 1
+ln n
2
. (4.49)
Up xp n!, h √ n.
n! ≈ nn × e−n × e × √ n. (4.50)
Cmpd S’ ppxm, h m dff
h e h hd √
2π , 8%— m d
d dw w p.
Problem 4.31 Underestimate or overestimate?
D h ppxm wh h dm
vm n!? U p ; h hk h m.
Problem 4.32 Next correction
Th h fi fi . Th - d m pp n−2, n−3, . . ., d h dffi dv p. B h n−1 dvd wh p.
. Dw h hw h md p h mh ln k vwh pw- v ( v md h m).
. Eh dd v v h m p, wh v Ahmd’ m (Pm 4.34)
=2
3× h m . (4.51)
U h pp ppxm h h .
. Shw h wh v ln n! =n
1 ln k , h m ppx-
m (1 − n−1)/12.
d. Wh h , mpvd m (m e) h ppxm-
n! d hw √
2π ? Wh d h n−1 m hln n! ppxm h n! ppxm?
Th d q dvd S 6.3.2 h hq
.
4.6 Summary and further problems
F m , v h fid pp .A m hd z p m d qk h d
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76 4 Pictorial proofs
h pmp. P , h, p h md’
v mp pw. I mk m hp
dd d d .
F xv d j p p, h wk N [31, 32]. H h pm dvp p .
Problem 4.33 Another picture for the AM–GM inequality
Skh y = ln x hw h h hm m a d b w h q h m m, wh q wh a = b.
Problem 4.34 Archimedes’ formula for the area of a parabola
Ahmd hwd ( !) h h d p w-hd m . Pv h .
Shw h h d p w-hd h m- pm wh v d. Th p p wh ppxm ( xmp, Pm 4.32).
Problem 4.35 Ancient picture for the area of a circle
Th Gk kw h h m wh d r w2πr. Th h d h w p hw h πr2. C h m?
=
Problem 4.36 Volume of a sphere
Exd h m Pm 4.35 fid h vm ph d r,v h 4πr2. I h m wh kh.
Problem 4.37 A famous sum
U p ppxm h m B m
∞1
n−2.
Problem 4.38 Newton–Raphson method
I , v f(t) = 0 q ppxm. O mhd wh t0 d mpv v h Nw–Rph mhd
tn+1 = tn −f(tn)
f(tn), (4.52)
wh f(tn) h dvv df/dt vd t = tn. Dw p j h p; h h p m
√ 2. (Th Pm 4.17.)
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5
Taking out the big part
5.1 Mp d w 775.2 F h d w-p xp 795.3 F h wh xp 845.4 Sv ppxm: Hw dp h w? 915.5 D m 945.6 Summary and further problems 97
I m v qv pm, h mpfi wh
w h pv dv d fi h fi. F ppxmd dd h m mp ff—h p—h fi
d dd. Th pd v ppxm
“k h p” m, mm, d xp. Th w xmp d h d d w-
p xp (S 5.2) d z m mp (S-
5.1), xp (S 5.3), qd q (S 5.4),
d dffi m (S 5.5).
5.1 Multiplication using one and few
Th fi mhd m mp d h,
k--h-vp m. Th p h
p d CD-ROM. A d CD-ROM h h m m d p m CD, wh p md h
pd h :
1 h × 3600
1 h p m
× 4.4 × 10 4 mp
1 mp
× 2 h × 16
1 mp mp z
. (5.1)
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78 5 Taking out the big part
(I h mp-z , h w h ph d.)
Problem 5.1 Sample rate
Lk p h Sh–Nq mp hm [22], d xp wh h
mp (h whh h d p md) h 40 kHz.
Problem 5.2 Bits per sample
B 216 ∼ 105, 16- mp— h h CD m—q h 0.001%. Wh dd’ h d h CDm h mh mp z, 32 (p h)?
Problem 5.3 Checking units
Chk h h h m dvd —xp h dd .
Bk--h-vp h m h h p
m d mp h h dvd d- d . I h d m h m, mp wh
3 dm p wd vk. A ppxm
d ppxm mhd .
What is the data capacity to within a factor of 2?
Th (h p!) (Pm 5.3), d h h m-
3600×
4.4×
10 4
×32. T m h pd, p
p d .
The big part: Th m mp k--h-vp pd-
m m h pw 10, v h p fi:
3600 h pw 10, 4.4 × 10 4 , d 32
. Th h pw 10 pd 108.
The correction: A k h p, h m p -
3.6 × 4.4 × 3.2. Th pd mpfid k
p. Rd h h m m h h:
1, w, 10. Th vd m w mdw w 1 d 10:I h m m 1 d 10, (w)2 = 10 d w ≈ 3. I hpd 3.6× 4.4×3.2, h d w, 3.6× 4.4×3.2 ≈ (w)3
h 30.
Th , h pw 10, d h m v
p ∼ 108 × 30 = 3 × 109 . (5.2)
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5.2 Fractional changes and low-entropy expressions 79
Th m wh 2 h x pd (Pm 5.4),
whh h p 5.6 × 109 .
Problem 5.4 Underestimate or overestimate?
D 3 × 109 vm dm 3600 × 4.4 × 104 × 32? Chk mp h x pd.
Problem 5.5 More practice
U h --w mhd mp pm h w - m; h mp h ppxm d pd.
. 161 × 294 × 280 × 438. Th pd h 5.8 × 109.
. Eh’ A = 4πR2, wh h d R ∼ 6 × 106 m. Th
h 5.1 × 1014 m2.
5.2 Fractional changes and low-entropy expressions
U h --w mhd m mp . F xm-
p, 3.15 × 7.21 qk m w × 101 ∼ 30, whh wh 50% h x pd 22.7115. T m m, d 3.15
3 d 7.21 7. Th pd 21 8%. T d h
h, d p 3.15 × 7.21 p d ddv
. Th dmp pd
(3 + 0.15)(7 + 0.21) = 3 × 7 p
+ 0.15 × 7 + 3 × 0.21 + 0.15 × 0.21 ddv
. (5.3)
Th pph d, h pp k h
p pd m h hd mm d d-d. Sh mdfid, hwv, k h p pvd
d v . A v, dvp h mpvd -
d w mp -fih d: h
(S 5.2.1) d w-p xp (S 5.2.2). Th mpvd
w h, fi m , hp m h vd hhw pd m (S 5.2.3).
5.2.1 Fractional changes
Th h v ddv p h pd p d multiplicative :
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80 5 Taking out the big part
3.15 × 7.21 = 3 × 7 p
× (1 + 0.05) × (1 + 0.03)
. (5.4)
Can you find a picture for the correction factor?
1
1
0.05
0.03
1 0.05
0.03 ≈ 0Th h wh
wdh 1 + 0.05 d hh 1 + 0.03. Th
h m h x-
p (1 + 0.05) × (1 + 0.03). Th md
h 1 + 0.05 + 0.03 p 8%
v h p. Th p
21, d 8% 1.68, 3.15 × 7.21 = 22.68,whh wh 0.14% h x pd.
Problem 5.6 Picture for the fractional error
Wh h p xp h h 0.15%?
Problem 5.7 Try it yourself
Em 245× 42 d h mp 10, d mph p wh h x pd. Th dw h , m , d h p.
5.2.2 Low-entropy expressions
Th 3.15 × 7.21 w mpd ddvh mp h. Th . U
h ddv , w- pd m
(x + Δx)( y + Δy) = xy + xΔy + yΔx + ΔxΔy ddv
. (5.5)
Problem 5.8 Rectangle picture
Dw p h xp
(x + Δx)( y + Δy) = xy + xΔy + yΔx + ΔxΔy. (5.6)
Wh h h Δx d Δy m (x Δx d y Δy),
h mpfi xΔy + yΔx, v hd mm
h m p v. F xmp, d p m h ΔxΔy, xΔx, yΔy. Th x
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5.2 Fractional changes and low-entropy expressions 81
h p v m h p w d
; h h p, h hd h m wk fi
, d h hd w m wk mm h .
Sh p h j mh d m h[20, 21], whh dfi h p h hm h m p
v d h hm q h p. Th hm
d h p h xp dff h m
p v d h hh-p xp [28]— whm p v— hd mm d dd.
I , w-p xp w w p v,
d , “Y! Hw d hw?!” Mh mhm dfi p fid w hk h hh-
p xp --dd, w-p xp.
What is a low-entropy expression for the correction to the product xy?
A mpv , dm, m h w
p h h ddv : Th p dm
xp mh m h h p xp.
Th mpv (x + Δx)( y + Δy)/xy. A w, h
p. I w dmd m
x + Δx d y + Δy, mp hm, d fi dvd h pd xy.
Ahh h dm, m h p.A mhd p d mk dm
q h w:
(x + Δx)( y + Δy)
xy=
x + Δx
x
y + Δy
y=
1 +
Δx
x
1 +
Δy
y
. (5.7)
Th h d m h dm dm q 1
d m m dm : (Δx)/x h h
x, d (Δy)/y h h y.
Th p m m mx x + Δx, y + Δy, x, d y w, d w mvd p mx. Umx d-
fi wh ph m. T, xmp, mv dp d
mxd w. Th pm h
w h 1025 m. F, m mhm
xp hv w . W p d mxh md p d h d h p h xp.
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82 5 Taking out the big part
Problem 5.9 Rectangle for the correction factor
Dw p h w-p
1 +Δx
x1 +
Δy
y . (5.8)
A w-p pd w-p h:
Δ (xy)
xy=
1 +
Δx
x
1 +
Δy
y
− 1 =
Δx
x+
Δy
y+
Δx
x
Δy
y, (5.9)
wh Δ(xy)/xy h h m xy (x + Δx)( y + Δy).
Th hm m h pd w m , m
mpd h pd w m. Wh h m, qd m,
Δ (xy)
xy ≈Δx
x +
Δy
y . (5.10)
Sm h mp dd!
Th -h mp h h pd ppx-
m h h h xΔy + yΔx. Smp d
w p; dd, h p v h ppd
h p h h mp. Ad h j
k: Wh Δy = 0, pd h Δ(xy) = 0 m h v Δx (h pd xpd Pm 5.12).
Problem 5.10 Thermal expansion
I, d hm xp, m h xpd h dm 4%,wh hpp ?
Problem 5.11 Price rise with a discount
Im h fl, ph w, h p k 10%mpd . F, q k , d 15%. Wh h p h h ?
5.2.3 Squaring
I z h d d wd, mm p q— p mp. Sqd h , d
qd pd pp h d m j (S 2.4):
Fd ∼ ρv2A, (5.11)
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5.2 Fractional changes and low-entropy expressions 83
wh v h pd h j, A - , d ρ
h d h fld. A q, dv hhw pd
d d m E = Fdd ∼ ρAv2d. E mp
h dd dv m w. Th p mmp W h 1970 wh p pd
( [7] ). A , h Ud S d hhw
pd m 55 mph (90 kph).
By what fraction does gasoline consumption fall due to driving 55 mph insteadof 65 mph?
A w pd m d mp d h d
ρAv2 d d h dv d d: Pp m d
h mm m m h d. B fid
w hm j w p. Th, z fi h fi—m h h h dv d d fixd (h
Pm 5.14).
Wh h mp, E pp v2, d
ΔE
E= 2 × Δv
v. (5.12)
G m 65 mph 55 mph h 15% dp v, h
mp dp h 30%. Hhw dv fi
h md m vh, whh h Ud Sm fi md. Th h 30% dp
dd US mp.
Problem 5.12 A tempting error
I A d x d A = x2, mp j h
ΔA
A≈
Δx
x
2
. (5.13)
Dpv h j (Chp 2).
Problem 5.13 Numerical estimates
U h m 6.33. Hw h m?
Problem 5.14 Time limit on commuting
Am h dv m, h h d, fixd hhw dvpd 15%. Wh h h h -md hhw dv?
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84 5 Taking out the big part
Problem 5.15 Wind power
Th pw d d wd pp v3 (wh?). I wd pd m 10%, wh h ff h d pw?Th q wd h wd pd ff
hp .
5.3 Fractional changes with general exponents
Th -h ppxm h x2 (S 5.2.3) d
x3 (Pm 5.13) p h ppxm xn
Δ (xn)
xn≈ n × Δx
x. (5.14)
Th ff mhd m dv (S 5.3.1), m
q (S 5.3.2), d jd mm xp h
(S 5.3.3). Th q h h h
m d h h xp n (S 5.3.4).
5.3.1 Rapid mental division
Th p n = −1 pvd h mhd pd m dv.
A xmp, ’ m 1/13. Rw (x + Δx)−1 wh x = 10
d Δx = 3. Th p x−1 = 0.1. B (Δx)/x = 30%, h x−1 h −30%. Th 0.07.
1
13≈ 1
10− 30% = 0.07, (5.15)
wh h “−30%” , m “d h pv j
30%,” hhd 1 − 0.3.
How accurate is the estimate, and what is the source of the error?
Th m 9%. Th h ppxm
Δ
x−1
x−1≈ −1 × Δx
x(5.16)
d d h q ( hh pw) h h
(Δx)/x (Pm 5.17 k fid h qd m).
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5.3 Fractional changes with general exponents 85
How can the error in the linear approximation be reduced?
T d h , d h h. B h h dmd h p, ’ h h
p. Ad, mp 1/13 8/8, v m 1, 8/104. I p 0.08 ppxm 1/13 d wh 4%.
T mpv , w 1/104 (x + Δx)−1 wh x = 100 d Δx = 4. Th
h (Δx)/x w 0.04 (h h 0.3); d h
1/x d 8/x m − 4%. Th d m 0.0768:
1
13≈ 0.08 − 4% = 0.08 − 0.0032 = 0.0768. (5.17)
Th m d m d d 0.13%!
Problem 5.16 Next approximation
Mp 1/13 v m 1 mk dm 1000; hm 1/13. Hw h ppxm?
Problem 5.17 Quadratic approximation
Fd A, h ffi h qd m h mpvd -hppxm
Δ
x−1
x−1≈ −1 × Δx
x+ A ×
Δx
x
2
. (5.18)
U h ppxm mpv h m 1/13.
Problem 5.18 Fuel efficiency
F ffi v pp mp. I 55 mphpd m d mp 30%, wh h w ffi h m 30 m p US (12.8 km p )?
5.3.2 Square roots
Th xp n = 1/2 pvd h mhd m
q . A xmp, ’ m √ 10. Rw (x + Δx)1/2
wh x = 9 d Δx = 1. Th p x1/2 3. B (Δx)/x = 1/9 d
n = 1/2, h 1/18. Th d m
√ 10 ≈ 3 ×
1 +
1
18
≈ 3.1667. (5.19)
Th x v 3.1622..., h m 0.14%.
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86 5 Taking out the big part
Problem 5.19 Overestimate or underestimate?
D h -h ppxm vm q ( vmd
√ 10)? I , xp wh; , v xmp.
Problem 5.20 Cosine approximation
U h m- ppxm sinθ ≈ θ hw h cosθ ≈ 1 − θ2/2.
Problem 5.21 Reducing the fractional change
T d h h wh m√
10, w √
360/6 dh m
√ 360. Hw h m
√ 10?
Problem 5.22 Another method to reduce the fractional change
B√
2 d m h q √
1 d√ 4, h d v d d m
√ 2. A
m pm d m ln 2 (S 4.3); h, w 2 ( 4/3)/(2/3) mpvd h . D h w hp m
√ 2?
Problem 5.23 Cube root
Em 21/3 wh 10%.
5.3.3 A reason for the seasons?
Smm wm h w, d, h h
h h mm h h w. Th mm xp- w . F, mm h h hmph
hpp d w h h hmph, dp m
dff h pv d h . Sd, w w
w m, h v h– d pd m m-p dff. Th h—h h d dm h
d d h h dm h
mp— m zd h.
Intensity of solar radiation: Th h pw dvdd h
v whh pd. Th pw hd h v (h h xd v ); hwv, d r
m h , h h pd v ph wh
∼ r2. Th I h v d I ∝ r−2. Th
h d d d
ΔI
I≈ −2 × Δr
r. (5.20)
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5.3 Fractional changes with general exponents 87
Surface temperature: Th m m d
p kd d. I dpd
h h’ mp T d h S–Bzm
w I = σT 4 (Pm 1.12), wh σ h S–Bzm .Th T ∝ I1/4. U h,
ΔT
T ≈ 1
4× ΔI
I. (5.21)
Th d mp. Th mp d
d d (ΔI)/I = −2 × (Δr)/r. Wh jd, h w
d d mp w:
− 21
4
ΔT
T ≈ −
1
2× Δr
r
Δr
r
ΔII ≈ −2× Δr
r
I ∝ r−2 T ∝ I1/4
l
rmx rm0◦
θr
Th x p h mp m
h p (Δr)/r—m, h h
h h– d. Th h h
p; d
r =l
1 + cosθ, (5.22)
wh h h , θ h
p , d l h m m. Th r v m rm =l/(1 + ) (wh θ = 0◦) rmx = l/(1 − ) (wh θ = 180◦). Th
m rm l h h . Th
m l rmx h h h . Th,
r v h 2. F h h’ , = 0.016, h h–
d v 0.032 3.2% (mk h v 6.4%).
Problem 5.24 Where is the sun?
rmx
rmTh pd dm h h’ pd h wm h h p. Th dm h h hwh v d php m : h h p. Wh ph w, , pvh m h h p?
Problem 5.25 Check the fractional change
Lk p h mmm d mxmm h– d d hk h hd d v 3.2% m mmm mxmm.
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88 5 Taking out the big part
A 3.2% d h dp mp:
ΔT
T ≈ −
1
2× Δr
r= −1.6%. (5.23)
Hwv, m d v h d xph mp h ΔT .
ΔT = −1.6% × T. (5.24)
In winter T ≈ 0◦ C, so is ΔT ≈ 0◦ C?
I pd h ΔT ≈ 0◦ C, m w. A v p m m T Fhh d,
whh mk T v p h h hmph. Y
ΔT flp j T md Fhh d!F, h mp d h S–Bzm
w. F kd flx pp T 4, mp m md v wh z hm : z.
Nh h C h Fhh fi h qm.
I , h Kv d m mp v
z. O h Kv , h v mp T ≈ 300 K;
h, 1.6% h T mk ΔT ≈ 5 K. A 5 K h 5◦ C
h—Kv d C d h m z, hh h
hv dff z p. (S Pm 5.26.) A p mp- h w mm d w mp d 20◦ C—
mh h h pdd 5◦ C h, v w
h m. A v h– d d xp
h h .
Problem 5.26 Converting to Fahrenheit
Th v w Fhh d C mp
F = 1.8C + 32, (5.25)
h 5◦ C hd h 41◦ F—ffi xp h! Wh w wh h ?
Problem 5.27 Alternative explanation
I v d h xp h , wh ? Ypp hd, p, xp wh h h d h hmphhv mm 6 mh p.
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5.3 Fractional changes with general exponents 89
5.3.4 Limits of validity
Th -h ppxm
Δ (xn)
xn ≈ n ×Δx
x (5.26)
h . B wh vd? T v wh dw
, w z Δx; h h x = 1 mk z h
d h h. Th h d m nz , d h -h ppxm qv
(1 + z )n ≈ 1 + nz. (5.27)
Th ppxm m wh z : xmp,
wh v √ 1 + z wh z = 1 (Pm 5.22). I h xp n d? Th pd xmp d md-zd
xp: n = 2 mp (S 5.2.3), −2
ffi (Pm 5.18), −1 p (S 5.3.1), 1/2 q
(S 5.3.2), d −2 d 1/4 h (S 5.3.3). W
d h d.
What happens in the extreme case of large exponents?
Wh xp h n = 100 d, , z = 0.001, h ppx-
m pd h 1.001100
≈ 1.1— h v 1.105...Hwv, h h m n d z = 0.1 ( h 0.001
m) pd h pd
1.1100 (1+z )n
= 1 + 100 × 0.1 nz
= 11; (5.28)
1.1100 h 14,000, m h 1000 m h h pd.
Bh pd d n d m z , pd w
; h, h pm n z . Php h p
h dm pd nz . T h d, hd nz wh v n. F nz , 1—h mp
dm m. H v xmp.
1.110 ≈ 2.59374,
1.01100 ≈ 2.70481,
1.0011000 ≈ 2.71692.
(5.29)
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90 5 Taking out the big part
I h xmp, h ppxm pd h (1 + z )n = 2.
What is the cause of the error?
k
1 + 10−k
10k
1 2.5937425
2 2.7048138
3 2.7169239
4 2.7181459
5 2.7182682
6 2.7182805
7 2.7182817
T fid h , h q d1.0011000 d hp h p w m: Th
v m pph e = 2.718281828 . . ., h h hm. Th, k h
hm h wh ppxm.
ln(1 + z )n = n ln(1 + z ). (5.30)
P hwd h ln(1 + z ) ≈ z wh
z 1 (S 4.3). Th, n ln(1 + z ) ≈ nz , mk-
(1 + z )
n
≈ e
nz
. Th mpvd ppxmxp wh h ppxm (1 + z )n ≈ 1 + nz d wh nz :
O wh nz 1 enz ppxm 1 + nz . Th, wh z 1
h w mp ppxm
(1 + z )n ≈
1 + nz (z 1 d nz 1),
enz (z 1 d nz d).(5.31)
n
z
n z =
1 n / z
= 1
z
=
1
n = 11+nz
enz znen/z
zn
zn
1+n l n z
Th dm hw, h wh
n–z p, h mp ppxm
h . Th x h-
m d n d z md pv:Th h h p hw z 1, dh pp h p hw n 1. O
h w h, h d v
n l n z = 1. Exp h d
d xd h ppxm
v x (Pm 5.28).
Problem 5.28 Explaining the approximation plane
I h h h p, xp h n/z = 1 d n l n z = 1 d. F hwh p, x h mp pv n d z p.
Problem 5.29 Binomial-theorem derivation
T h w v dv (1+z )n ≈ enz (wh n 1). Expd(1 + z )n h m hm, mp h pd h mffi ppxm n − k n, d mp h xp h T enz.
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5.4 Successive approximation: How deep is the well? 91
5.4 Successive approximation: How deep is the well?
Th x k h p mphz v
ppxm d dd ph pm.
Y dp dw w kw dph h d h h ph 4 . N , fid h wh 5%. U c = 340 m −1 h pd d d g = 10 m −2 h h v.
Appxm d x v m h m w dph,
ff fi dff dd.
5.4.1 Exact depth
Th dph dmd h h h 4 w p w
m: h k dw h w d h d v ph w. Th - m
2h/g (Pm 1.3), h m
T =
2h
g k
+h
c d
. (5.32)
T v h x, h h q d d q
h d qd q h (Pm 5.30); ,
-p mhd, w h qd q
w v z = √ h .
Problem 5.30 Other quadratic
Sv h h q d d q h d.Wh h dv d ddv h mhd mp whh mhd w h qd z =
√ h ?
A qd q z =√
h , h
1
c
z 2 + 2g
z − T = 0. (5.33)
U h qd m d h h pv d
z =−
2/g +
2/g + 4T/c
2/c. (5.34)
B z 2 = h ,
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92 5 Taking out the big part
h =
−
2/g +
2/g + 4T/c
2/c
2
. (5.35)
S g = 10 m −2
d c = 340 m −1
v h ≈ 71.56 m.Ev h dph , h x m m. Sh hh-p h q m h qd m;
h mph m mp v hh. Ex w,
w w fid, m h ppxm w.
5.4.2 Approximate depth
T fid w-p, ppxm dph, d h p—h
m mp ff. H, m h m h k’ : Th k’ mxmm pd, v h 4 ,
gT = 40 m −1, whh w c. Th, h m mp ff
hd h xm fi d pd.
If c = ∞, how deep is the well?
I h zh ppxm, h - m t0 h m T = 4 , h w dph h 0 m
h 0 =
1
2 gt
2
0 = 80 m. (5.36)
Is this approximate depth an overestimate or underestimate? How accurate is it?
Th ppxm h d-v m, vm
h - m d h h dph. Cmpd h dph
h 71.56 m, vm h dph 11%—
qk mhd ff ph h. Fhm, h
ppxm w fim.
How can this approximation be improved?
T t h
12 gt2
T − hc
T mpv , h ppxm dph h 0 ppx-m h d-v m.
td ≈ h 0
c≈ 0.24 . (5.37)
Th m m h x ppxm h - m.
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5.4 Successive approximation: How deep is the well? 93
t1 = T −h 0c
≈ 3.76 . (5.38)
I h m, h k d gt21/2, h x ppxm
h dph
h 1 =1
2gt2
1 ≈ 70.87 m. (5.39)
Is this approximate depth an overestimate or underestimate? How accurate is it?
Th h 1 d h 0 m h d-v m. B
h 0 vm h dph, h pd vm h d-v
m d, h m m, dm h - m. Th
h 1 dm h dph. Idd, h 1 h m h h
dph h 71.56 m— 1.3%.
Th mhd v ppxm h v dv v v- h qd m x. F, hp dvp ph
dd h m; w z, xmp, h m h
T = 4 p , h dph h gT 2/2. Sd,
h p xp (Pm 5.34). Thd, v ffi
w qk. I w kw whh jmp
h w, wh h dph h dm p?
F, h mhd hd m h h md. M h
pd d v wh dph, m mp
(Pm 5.32). Th h -, qd-m mhd . Thqd m d h v m d h q m
d-m mpd q. M q
hv d-m . Th, m h v
md pd md— w dmd x
w. Th mhd v ppxm v
h pd w-p, mph .
Problem 5.31 Parameter-value inaccuraciesWh h 2, h d ppxm h dph? Cmp h h 1d h 2 wh h md g = 10 m −2.
Problem 5.32 Effect of air resistance
Rh wh h dph pdd - (S 2.4.2)? Cmp h h h fi ppxmh 1 d h d ppxm h 2 (Pm 5.31).
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94 5 Taking out the big part
Problem 5.33 Dimensionless form of the well-depth analysis
Ev h m d hv w p dmm. Th q h , g, T , d c pd w dpd dmp (S 2.4.1). A v p
h ≡ h
gT 2d T ≡ gT
c. (5.40)
. Wh ph p T ?
. Wh w p, h dm m h = f(T ). Wh h
h T → 0?
. Rw h qd-m
h =
−
2/g +
2/g + 4T/c
2/c
2
(5.41)
h = f(T ). Th hk h f(T ) hv h T → 0.
Problem 5.34 Spacetime diagram of the well depth
dph
t
4 k
dwv
Hw d h pm dm [44] h v ppxm h w dph?O h dm, mk h 0 (h zh ppx-m h dph), h 1, d h x dphh . Mk t0, h zh ppxm h- m. Wh p h k dd-wv v dd? Hw wd
dw h dm h pd ddd? I g dd?
5.5 Daunting trigonometric integral
Th fi xmp k h p m d
m h I d dd. M m
d I p m h h ph v hmwk
pm; h d d, d h m h , wd
wh h v mhm d ph pm.Th ppd h mhm-pm xm h Ld I Th Ph h m USSR. Th
pm v π/2
−π/2
(cost)100 dt (5.42)
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5.5 Daunting trigonometric integral 95
wh 5% h 5 m wh mp!
Th (cost)100 k h. M m d d
hp. Th hp d (cost)2 = (cos 2t − 1)/2 pd
(cost)100 =
cos 2t − 1
2
50
, (5.43)
whh m m m p xpd h 50h pw.
A p mp mhd h 5% ffi—,
fid h p! Th d wh t z. Th,
cost ≈ 1 − t2/2 (Pm 5.20), h d h
(cost)100
≈ 1 −t2
2
100
. (5.44)
I h h m m (1 + z )n, wh h z = −t2/2 d
xp n = 100. Wh t m, z = −t2/2 , (1 + z )n m
ppxmd h S 5.3.4:
(1 + z )n ≈
1 + nz (z 1 d nz 1)
enz (z 1 d nz d).(5.45)
B h xp n , nz v wh t d z
m. Th, h ppxm (1 + z )n ≈ enz ; h
(cost)100 ≈
1 −t2
2
100
≈ e−50t2
. (5.46)
costA d hh pw m G!
A hk h p , mp-
d p (cost)n n = 1 . . . 5 hw
G hp k m n .
Ev wh h ph vd, p (cost)100 G
p. I h , t m −π/2 π/2, d
h dp d h wh cost ≈ 1 − t2/2 ppxm. F, h
(Pm 5.35). I h h G wh fi m: π/2
−π/2
(cost)100 dt ≈ π/2
−π/2
e−50t2
dt. (5.47)
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96 5 Taking out the big part
U, wh fi m h h d m. B
xd h m fi pd d m wh
m (Pm 5.36). Th ppxm h w
π/2
−π/2
(cost)100 dt ≈ π/2
−π/2
e−50t2dt ≈ ∞
−∞e−50t2
dt. (5.48)
Problem 5.35 Using the original limits
Th ppxm cost ≈ 1 − t2/2 q h t m. Wh d’ h ppxm d h m-t fi ?
Problem 5.36 Extending the limits
Wh d’ xd h m m ±π/2 ±∞ fi ?
Th d d (S 2.1):∞
−∞ e−αt2
dt =
π/α . Wh
α = 50, h m
π/50. Cv, 50 h 16π ,
h q —d 5% m— h 0.25.
F mp, h x (Pm 5.41) π/2
−π/2
(cost)n dt = 2−n
n
n/2
π. (5.49)
Wh n = 100, h m ffi d pw w pd
12611418068195524166851562157158456325028528675187087900672
π ≈ 0.25003696348037. (5.50)
O 5-m, wh-5% m 0.25 m 0.01%!
Problem 5.37 Sketching the approximations
P (cost)100 d w ppxm e−50t2 d 1 − 50t2.
Problem 5.38 Simplest approximation
U h -h ppxm (1 − t2/2)100 ≈ 1 − 50t2 ppxm h d; h v h wh 1 − 50t2 pv. Hw h h 1-m mhd h x v0.2500...?
Problem 5.39 Huge exponent
Em π/2
−π/2(cost)10000 dt. (5.51)
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5.6 Summary and further problems 97
Problem 5.40 How low can you go?
Iv h h ppxm
π/2
−π/2
(cost)n dt
≈ π
n
, (5.52)
m n, d n = 1.
Problem 5.41 Closed form
T v h π/2
−π/2(cost)100 dt (5.53)
d m, h w p:
. Rp cost wh (eit + e−it)2.
. U h m hm xpd h 100h pw.
. P h m k eikt wh p e−ikt; h h mm −π/2 π/2. Wh v v k pd m wh z?
5.6 Summary and further problems
Up m mpd pm, dvd p—h m
mp ff—d . Az h p fi, d w h wd. Th v-ppxm pph,
p dvd-d-q , v m
w-p m. Lw-p xp dm w p
v; h h mm d mph. I h,
ppxm m h x .
Problem 5.42 Large logarithm
Wh h p ln(1+ e2)? Gv h m ln(1 +e2)
wh 2%.
Problem 5.43 Bacterial mutations
I xpm dd Ch m h 1990, hpd dd pp d m. Ih d d, 5% h md. A 140 d,h wh w md? (Th m pkv h d 3 mk , hd h m v fid .)
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98 5 Taking out the big part
Problem 5.44 Quadratic equations revisited
Th w qd q, pd [29], d v dmpd m.
s2
+ 109
s + 1 = 0. (5.54)
. U h qd m d dd fid h hqd. Wh w d wh?
. Em h k h p. ( Hint: Appxm d vh q ppp xm .) Th mpv h m v ppxm.
. Wh h dv d ddv h qd-m v v ppxm?
Problem 5.45 Normal approximation to the binomial distribution
Th m xp1
2+
1
2
2n
(5.55)
m h m
f(k ) ≡
2n
n − k
2−2n, (5.56)
wh k = −n . . . n. Eh m f(k ) h p n − k hd(d n + k ) 2n flp; f(k ) h -d m dwh pm p = q = 1/2. Appxm h d w h
w q:
. I f(k ) v dd k ? F wh k d f(k ) hv mxmm?
. Appxm f(k ) wh k n d kh f(k ). Th, dv d xph m ppxm h m d.
. U h m ppxm hw h h v h md n/2.
Problem 5.46 Beta function
Th w pp B :
f(a, b) =
10
xa(1 − x)b dx, (5.57)
wh f(a − 1, b − 1) h E . U -fih mhd j m f(a, 0), f(a, a), d, fi, f(a, b). Chk j wh hh-q mp- mh Mxm.
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6
Analogy
6.1 Sp m: Th d mh 996.2 Tp: Hw m ? 1036.3 Op: E–ML mm 1076.4 T : A d d m 1136.5 Bon voyage 121
Wh h h, h h w h dd. Th d,
h hm h wh k, d h fi -fih
. I dv mp: Fd wh dffi pm,
d v m mp pm— pm.
P dvp fl. Th dd p m
(S 6.1); hpd d m d p (S 6.2);h ppd d mhm (S 6.3) d, h w
xmp, fi d m (S 6.4).
6.1 Spatial trigonometry: The bond angle in methane
θ
Th fi m m p m. Imh (hm m CH 4), m
h hd, d hd
m h vx. Wh h θ
ww –hd d?
A h dm hd vz. T,
xmp, m d h w w
hd. B w-dm vz, ’
d z p m. Kw d mh hp mh’ d .
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100 6 Analogy
Should the analogous planar molecule have four or three hydrogens?
F hd pd d whh, wh pd
p, pd w dff d . I
, mh d . Th, hd h
pm. Th k h p m h hd.
θTh hd d p
d : θ = 120◦. Php h h d
mh! O d p, hwv, h d
whh h pd hh dm. Th
d p w dm (d = 2) wh m
j— xmp, h d dm h d 120◦ (60d)◦ mh .
θS j q h h
d. E v d m m v mp pm: h -dm, m
CH2. I w hd pp h, h
w C–H d m θ = 180◦.
Based on the accumulated data, what are reasonable conjectures for the three-dimensional angle θ3?
d θd
1 180◦
2 120
3 ?
Th -dm m m h j h
θd = (60d)◦. I w j— xmp,
h θd = (240 − 60d)◦ θd = 360◦/(d + 1). T h
j d k h mhd .
Th - hh dm (hh d) hj h θd = (240 − 60d)◦. F hh d, pd
mp d —m, θ = 0 d = 4 d θ < 0 d > 4.
F, h d , θd = 360◦/(d + 1), p h m
- . L’ v pd mh—m, θ3 = 90◦. Im h h mh:
CH6 m wh h d x hd h
. I m d 90◦. (Th h d 180◦.)
Nw mv w hd CH 6 CH 4, v pd
h m hd. Rd h wd h m d v 90◦—d h pd h θ3 = 90◦.
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6.1 Spatial trigonometry: The bond angle in methane 101
Problem 6.1 How many hydrogens?
Hw m hd dd h - d fiv-dm d- pm? U h m hw h θ4 > 90◦. I θd > 90◦
d?
Th d hv d h mp - j
(240−60d)◦ d 360◦/(d+1). Ahh h - j
mh vv, wh w d p h p v.W, θd mh v d.
P q w d: Th d mh h mp
v d. A dffi wh j h
x m h 3, 5, 11, 29, . . .
What is the next term in the series?A fi , h m m m dm. Y 2 m
h m pd 1, 3, 9, 27, . . . Th, h h x
m k 83. Sm, mp m h θd dmh hp j p θd.
What transformation of the θd data produces simple patterns?
Th dd m hd pd mp p d hv -
h jfi. O jfi h h
h d , whh mpd d pd w C–H v (Pm 6.3). B d pd vv ,
whwh m θd cosθd.
d θd cosθd
1 180◦ −1
2 120 −1/2
3 ? ?
Th m mpfi h d: Th cosθd
mp −1, −1/2, . . . Tw p −1/4 −1/3; h pd,
pv, h m −1/2d−1 −1/d.
Which continuation and conjecture is the more plausible?
Bh j pd cosθ < 0 d h θd > 90◦ ( d). Th
hd pd (Pm 6.1); hwv, hd
m h d dh w h j.
HHCC
HH1 1D h j mh h m m?
A mp m , p m h d, h p h . I dm, hw
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102 6 Analogy
w h w hd, p h H–H m w
p hv 1 : 1 h .
HH HH
HH
CC1
2
I w dm, h h d h
hd h mdp h hw hd. Th p h d w
p hv 1 : 2 h .
How does the carbon split the analogous altitude of methane?
CC
I mh, h d m h p
vx h h . Th h
m p d h h m hh h
hd. B h h hd hv
z hh, h m hh h hd h/4, wh h h hh h p hd. Th,
h dm, h p h d
w p hv h h/4 : 3h/4 1 : 3. I d dm,h, h p p h d w p hv
h 1 : d (Pm 6.2).
109.47◦
B 1 : d h m, cosθd
m k 1/d h h 1/2d−1. Th, h
m k h w cosθd j h
cosθd = − 1d
. (6.1)
F mh, wh d = 3, h pdd d
arccos(−1/3) ppxm 109.47◦. Th pd
wh xpm d wh h
m (Pm 6.3).
Problem 6.2 Carbon’s position in higher dimensions
J j h h p h d w p hv
h 1 : d.
Problem 6.3 Analytic-geometry solution
I d hk h , m w fid h d . F, d (xn, yn, z n) h n hd,wh n = 1 . . . 4, d v h d. (U mm mk hd mp .) Th h w C–H v d mph h h d.
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6.2 Topology: How many regions? 103
Problem 6.4 Extreme case of high dimensionality
Dw p xp h m- ppxm arccos x ≈ π/2 − x.Wh h ppxm d hh dm ( d)? C fid v xp h ppxm d ?
6.2 Topology: How many regions?
Th d mh (S 6.1) d d wh
m (Pm 6.3), d hw
pw. Th, h w pm.
Into how many regions do five planes divide space?
Th m pm d m h fiv pp, p m p, h p m . T
m h d h d, ’ p d h p
dm, h mxmz h m . Th pm h fid h mxmm m pdd fiv p.
Fv p hd m, h mhd
— w p—mh pd ph z fiv p. Th z
p: Sp m wh R(0) = 1 (wh R(n)
d h m pdd n p).
Th fi p dvd p w hv, v
R(1) = 2. T dd h d p, m
w pd wd: R(2) = 4.
What pattern(s) appear in the data?
A j h R(n) = 2n. T , h n = 3 h hd m d
h h p w m p;
h, R(3) dd 8. Php h p
wh R( 4) = 16 d R(5) = 32. I h w
R(n), h w xp mkd
dh hm m h vfid .
n 0 1 2 3 4 5
R 1 2 4 8 16 32
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104 6 Analogy
How can the R(n) = 2n conjecture be tested further?
A d dffi h hd
vz h dm. A w-dm p-
m wd v, d m hp h h-dm j. A w-dm p pd ,
h q h w:
What is the maximum number of regions into which n lines divide the plane?
Th mhd mh p. I h p 2n,
h h R(n) = 2n j k pp h dm.
What happens in a few easy cases?
Z v h p wh, v R(0) = 1. Th x h
w (hh Pm 6.5):
R(1)=2 R(2)= 4 R(3)=7
Problem 6.5 Three lines again
Th R(3) = 7 hwd h pd v .H h xmp wh h , dm -m, m pd x . Wh, wh, h vh ? O R(3) = 6?
Problem 6.6 Convexity
M h d h vx? (A vx d m w p d h d h .) Wh h h-dm d pp p?
U R(3) d 7, h j R(n) = 2n kdd. Hwv, dd h mp j,
dw h d h . F
mk 11 h h h pdd 16, h 2n
j dd.
A w j mh m h w-dm d R2(n)
d h h-dm d R3(n).
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6.2 Topology: How many regions? 105
n 0 1 2 3 4
R2 1 2 4 7 11
R3 1 2 4 8
I h , v m mk . F xmp,
R2(1) d R3(1)—h w h n = 1 m—m R2(2)
R3(2). Th w m h R3(3) . B h
h m m m wh m w m hm; dd
h d q h h d—d h mp d
h -dm pm.
What is the maximum number of segments into which n points divide a line?
A mp w h n p mk n m. Hwv,
—h p pd w m—d h mp.
Rh, n p mk n + 1 m. Th h R1 w
h w .
n 0 1 2 3 4 5 n
R1 1 2 3 4 5 6 n + 1
R2 1 2 4 7 11
R3 1 2 4 8
What patterns are in these data?
Th 2n j vv p. I h R1 w,
n = 2. I h R2 w, n = 3. Th h R3 w,
p n = 4, mk h j R3( 4) = 16 d
R3(5) = 32 mp. M p m h, h
, h p h R3( 4) = 16 j w 0.5; w
m 0.01. (F m m d pd h p-
j, h mp wk p
Cfid [11], J [21], d P [36].)
I w, h pp d p:
n 0 1 2 3 4 5 n
R1 1 2 3 4 5 6 n + 1
R2 1 2 4 7 11
R3 1 2 4 8
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106 6 Analogy
If the pattern continues, into how many regions can five planes divide space?
Ad h p,
R3( 4) = R2(3) 7
+ R3(3) 8
= 15 (6.2)
d h
R3(5) = R2( 4) 11
+ R3( 4) 15
= 26. (6.3)
Th, fiv p dvd p mxmm 26 .
Th m hd dd dw fiv p d h
. Fhm, h - pph wd v h v R3(5), wh d v mhd mp h . Th h pvd h d j
xp R2(n) (Pm 6.9), R3(n) (Pm 6.10), d h
Rd(n) (Pm 6.12).
Problem 6.7 Checking the pattern in two dimensions
Th jd p pd R2(5) = 16: h fiv dvd h p 16 . Chk h j dw fiv d h.
Problem 6.8 Free data from zero dimensions
B h -dm pm v d, h z-dmpm. Exd h p h R3, R2, d R1 w pwd R0 w. I v h m z-dm (p) pdd p p wh n j ( dm −1). Wh R0 h w w h vd p? I h wh h mm dvd p?
Problem 6.9 General result in two dimensions
Th R0 d fi R0(n) = 1 (Pm 6.8), whh zh-d pm.
Th R1 d fi R1(n) = n + 1, whh fi-d pm. Th,h R2 d p fi qd.
T h j fi h d n = 0 . . . 2 h qdAn2 + Bn + C, pd k h p (Chp 5) w.
. G v h qd ffi A. Th k (-
) h p An2 d h v, R2(n) − An2, n = 0 . . . 2.
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6.3 Operators: Euler–MacLaurin summation 107
I h v n, h qd m m mhw mvd. I h , dj A.
. O h qd ffi A , pd fid h ffi B.
. Sm v h ffi C.
d. Chk qd fi w d (R2(n) n 3).
Problem 6.10 General result in three dimensions
A j h h R3 w mh (Pm 6.9). Uk h p fi h n = 0 . . . 3 d. D pd hjd v R3( 4) = 15 d R3(5) = 26?
Problem 6.11 Geometric explanation
Fd m xp h vd p. Hint: Exp fi wh
h p h R2 w m h R1 w; h z h xp h R3 w.
Problem 6.12 General solution in arbitrary dimension
Th p h h Rd(n) h ph P’ [17]. B P’ pd mffi, h xp Rd(n) hd m ffi.
Th, m ffi xp R0(n) (Pm 6.8), R1(n), dR2(n) (Pm 6.9). Th j m-ffi m R3(n) dRd(n), hk h Pm 6.10.
Problem 6.13 Power-of-2 conjecture
O fi j h m w Rd(n) = 2n. I h dm-, wkd n = 4. I d dm, hw h Rd(n) = 2n n d
(php h Pm 6.12).
6.3 Operators: Euler–MacLaurin summation
Th x d . M m
h m, p kd —p— -
h . A m xmp h dvv pD. I h h , h hp
h hp . I p ,
D(sin) = cos d D(sinh ) = cosh ; m h ph v h
d xp Dsin = cos d Dsinh = cosh . T dd
d hw p, :Op hv mh k d v k m.
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108 6 Analogy
6.3.1 Left shift
Lk m, h dvv p D qd mk D2 (h
d-dvv p) mk pw D. Sm,
h dvv p d pm. I h , d pm h P(x) = x2 + x/10 + 1 pd h p
pm P(D) = D2 + D/10 + 1 (h dff p h
dmpd p–m m).
Hw d h m xd? F xmp, d cosh D
sinD hv m? B h w
h xp , ’ v h p xp eD.
What does eD mean?
Th d p eD h f eDf.
D expf eDfDf
Hwv, h p d . I 2f e2Df,
whh h q eDf, wh p h pd eDf
m f wd pd 2eDf m 2f. T p,
T — D w m— d eD p.
eD = 1 + D +1
2D2 +
1
6D3 +
· · ·. (6.4)
What does this eD do to simple functions?
Th mp z h f = 1. H h
d eD:
(1 + D + · · ·) eD
1 f
= 1. (6.5)
Th x mp x x + 1.1 + D +
D2
2+ · · ·
x = x + 1. (6.6)
M , x2 (x + 1)2.1 + D +
D2
2+
D3
6· · ·
x2 = x2 + 2x + 1 = (x + 1)2. (6.7)
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6.3 Operators: Euler–MacLaurin summation 109
Problem 6.14 Continue the pattern
Wh eDx3 d, , eDxn?
What does eD
do in general?
Th pd xmp w h p eDxn = (x+1)n. B m
x xpdd pw x, d eD h xn m
(x + 1)n, h h eD f(x) f(x + 1). Amz,
eD mp L, h -h p.
Problem 6.15 Right or left shift
Dw ph hw h f(x) → f(x + 1) h h h h.App e−D w mp hz hv.
Problem 6.16 Operating on a harder function
App h T xp eD sinx hw h eD sinx = sin(x + 1).
Problem 6.17 General shift operator
I x h dm, h h dvv p D = d/dx dm,d eD xp. T mk h xp eaD , whm h dm a ? Wh d eaD d?
6.3.2 Summation
J h dvv p p h -h p ( L =
eD), h -h p p h p mm. Th
p p w d pw mhd ppxm
m wh d m.
Smm h m m p .
I dfi d dfi flv: Dfi
qv dfi wd v h m
. A xmp, h h dfi f(x) = 2x.
ba
b2 −a22x
x2 +C
m
I , h w p g d h
dfi DG = g, wh D h dvv p d
G =
g h dfi . Th D d
v
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110 6 Analogy
h—D
= 1 D = 1/
— pd h
p h dm. (
D = 1 p .)
ba
D
G(b)−G(a)gG
What is the analogous picture for summation?
f(k)
k
f(2)
2
f(3)
3
f(4)
4 5
A , dfi dfimm dfi mm d
h v h m. B pp h
wh vd ff--
p (Pm 2.24). Th m 42 f(k ) d h —f(2), f(3), d f( 4)—wh h dfi-
4
2f(k ) dk d d h f( 4) . Rh
h h dp dfi h m p
, p dfi mm xd h .
Th dfi mm wd v h m a d b
pd m wh dx m a b − 1.
A xmp, k f(k ) = k . Th h dfi m f h
F dfid F(k ) = k (k −1)/2+C (wh C h mm).
Ev F w 0 d n v n(n − 1)/2, whh n−1
0 k . I h
w dm, h p h wd ph.
ba
Δ
F(b) − F(a) =
b−1k=a
f(k)fF
Δ
I h v ph, h w Δ p v Σ j dff
v . Th, p p Δ pvd
Σ. B Δ d h dvv p D , h
p p . A dvv h m
df
dx= lim
h →0
f(x + h ) − f(x)
h . (6.8)
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6.3 Operators: Euler–MacLaurin summation 111
Th dvv p D h h p m
D = lim h →0
Lh − 1
h , (6.9)
wh h Lh p f(x) f(x + h )—h , Lh h h .
Problem 6.18 Operator limit
Exp wh Lh ≈ 1 + hD m h . Shw h h L = eD.
What is an analogous representation of Δ?
Th p m D fim h; pd,
h v p m fim wdh.
B mm Σ m wdh, v Δ hd h—m, Lh wh h = 1. A j,
Δ = lim h →1
Lh − 1
h = L − 1. (6.10)
Th Δ—d h fi-dff p— d 1/Σ. I
h , h (L − 1)Σ h d p 1. I
h wd, (L − 1)Σ hd hmv.
How well does this conjecture work in various easy cases?
T h j, pp h p (L−1)Σ fi h
g = 1. Th Σg w d m, d (Σg)(k )
h d k . Wh h , (Σg)(k ) = k + C. Fd
h h L − 1 p pd g.(L − 1)Σg
(k ) = (k + 1 + C)
(LΣg)(k )
− (k + C) (1Σg)(k )
= 1 g(k )
. (6.11)
Wh h x- —dfid g(k ) = k —h dfi m
(Σg)(k ) k (k − 1)/2 + C. P Σg hh L − 1 pd g.
(L − 1)Σg
(k ) =
(k + 1)k
2+ C
(LΣg)(k )
−
k (k − 1)
2+ C
(1Σg)(k )
= k g(k )
. (6.12)
I mm, h g(k ) = 1 d g(k ) = k , h ppd (L − 1)Σ k g k , k h d p.
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112 6 Analogy
Th hv —(L−1)Σ1 dd 1, d Σ = 1/(L−1). B
L = eD, w hv Σ = 1/(eD − 1). Expd h h d T
v mz p h mm p.
= 1
eD − 1= 1
D− 1
2+ D
12− D
3
720+ D
5
30240− · · · . (6.13)
B D
= 1, h d m 1/D . Th, mm
ppxm — p d h h
p p .
App h p f d h v h
m a d b pd h E–ML mm m
b−1
a
f(k ) = b
a
f(k ) dk −f(b) − f(a)
2+
f(1)(b) − f(1)(a)
12
−f(3)(b) − f(3)(a)
720+
f(5)(b) − f(5)(a)
30240− · · · ,
(6.14)
wh f(n) d h nh dvv f.
Th m k h fi m f(b). Id h m v h
vb
a
f(k ) = b
a
f(k ) dk +f(b) + f(a)
2+
f(1)(b) − f(1)(a)
12
−f(3)(b) − f(3)(a)
720+
f(5)(b) − f(5)(a)
30240− · · · .
(6.15)
A hk, :n
0 k . U E–ML mm,
f(k ) = k , a = 0, d b = n. Th m h n2/2;
h m
f(b) + f(a)
2 n/2; d m vh.
Th m d :n
0
k =n2
2+
n
2+ 0 =
n(n + 1)
2. (6.16)
A m E–ML mm ppxm
ln n!, whh h mn
1 ln k (S 4.5). Th, m f(k ) = ln k
w h (v) m a = 1 d b = n. Th
n1
ln k =
n1
lnkdk +ln n
2+ · · · . (6.17)
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6.4 Tangent roots: A daunting transcendental sum 113
ln k
1 · · · n k
Th , m h 1/D p,
h d h ln k v. Th ,
m h 1/2 p, p h
p (Pm 6.20). Th p dh hh-d (Pm 6.21)—hd
v p (Pm 4.32) m-
p E–ML mm (Pm 6.21).
Problem 6.19 Integer sums
U E–ML mm fid d m h w m:
()
n0
k 2 ()
n0
(2k + 1) ()
n0
k 3.
Problem 6.20 Boundary cases
I E–ML mm, h m
f(b) + f(a)
2—-h h fi m p -h h m. Th p mm ln k
(S 4.5) hwd h h p ppxm -h h m, m ln n. Wh, p, hppd -h h fi m?
Problem 6.21 Higher-order terms
Appxm ln 5! E–ML mm.
Problem 6.22 Basel sum
Th B m ∞1
n−2 m ppxmd wh p (Pm 4.37).
Hwv, h ppxm d hp h d m. AE dd, E–ML mm mpv h fid h d m. Hint: Sm h fi w m xp.
6.4 Tangent roots: A daunting transcendental sum
O w xmp, h m dv -
fih , dffi fi m.
Fd S ≡
x−2n wh h xn h pv tanx = x.
Th tanx = x , qv, h tanx − x,
d d hv d m, d m qd
m v mm mhd. S-fih mhd w m .
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114 6 Analogy
6.4.1 Pictures and easy cases
B h wh hp .
What is the first root x1?
y = x
π2
3π2
11
5π2
22
7π2
33
x
Th tanx − x v h
y = x d y = tanx.Sp,
h h tanx wh 0 < x < π/2
(Pm 6.23); h fi
j h mp x = 3π/2.
Th, x1 ≈ 3π/2.
Problem 6.23 No intersection with the main branchShw m h tanx = x h 0 < x < π/2. (Th k p p wh hk d dw h p.)
Where, approximately, are the subsequent intersections?
A x w, h y = x h y = tanx ph v hh
d h v h v mp. Th, mk h
w mp ppxm h p xn:
xn ≈ n + 1
2
π. (6.18)
6.4.2 Taking out the big part
Th ppxm, w-p xp xn v h p S
(h zh ppxm).
S ≈ n +1
2π ≈xn
−2
=4
π 2
∞
1
1
(2n + 1)2. (6.19)
Th m∞
1 (2n + 1)−2 , m p (S 4.5) m E–
ML mm (S 6.3.2), h h w .
∞1
(2n + 1)−2 ≈ ∞
1
(2n + 1)−2 dn = −1
2× 1
2n + 1
∞
1
=1
6. (6.20)
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6.4 Tangent roots: A daunting transcendental sum 115
Th,
S ≈ 4
π 2× 1
6= 0.067547... (6.21)
(2k+ 1)−2
1 2 3 4k
Th hdd p h ,d h m -h h fi .
Th h 1/9,
∞1
(2n + 1)−2 ≈ 1
6+
1
2× 1
9=
2
9. (6.22)
Th, m m S
S ≈ 4
π 2× 2
9= 0.090063..., (6.23)
whh h hh h h fi m.
Is the new approximation an overestimate or an underestimate?
Th w ppxm d w dm. F, h mp-
ppxm xn ≈ (n + 0.5)π vm h xn d h
dm h qd p h m
x−2n . Sd,
mk h mp ppxm, h p ppxm h
m
∞1 (2n + 1)−2 p h p wh d
d h dm h p (Pm 6.24).
Problem 6.24 Picture for the second underestimate
Dw p h dm h p ppxm
∞1
1
(2n + 1)2≈ 1
6+
1
2× 1
9. (6.24)
How can these two underestimates be remedied?
Th d dm (h p) md mm∞1 (2n + 1)−2 x. Th m m p fi m h 1/9—wh h p h m.
Id h n = 0 m, whh 1, d h v qd p
1/(2n)2 pd mp d m w-p m.
∞1
1
(2n + 1)2+ 1 +
∞1
1
(2n)2=
∞1
1
n2. (6.25)
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116 6 Analogy
Th fi, w-p m h m B m (hh-p
m). I v B = π 2/6 (Pm 6.22).
How does knowing B = π 2/6 help evaluate the original sum ∞1 (2n + 1)−2?
Th mj mdfi m h m w d h v
qd p. Th m B/4.
∞1
1
(2n)2=
1
4
∞1
1
n2. (6.26)
Th d mdfi w d h n = 0 m. Th, ∞1 (2n + 1)−2, dj h B v B B/4 d h h
n = 0 m. Th , B = π 2/6,
∞1
1
(2n + 1)2= B −
1
4B − 1 =
π 2
8− 1. (6.27)
Th x m, d h mp ppxm xn, pd
h w m S.
S ≈ 4
π 2
∞1
1
(2n + 1)2=
4
π 2
π 2
8− 1
. (6.28)
Smp xpd h pd v
S ≈ 1
2−
4
π 2= 0.094715... (6.29)
Problem 6.25 Check the earlier reasoning
Chk h p (Pm 6.24) h 1/6 + 1/18 = 2/9
dm∞
1 (2n + 1)−2. Hw w h m?
Th m S h hd h h mp ppxm
xn ≈ (n + 0.5)π . Amd h, h m
S ≈⎧⎨⎩ 0.067547 ( ppxm ∞1 (2n + 1)−2),
0.090063 ( ppxm d vh),
0.094715 (x m ∞
1 (2n + 1)−2).
B h hd m pd h x v ∞
1 (2n + 1)−2,
m h m S m h mp
ppxm .
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6.4 Tangent roots: A daunting transcendental sum 117
For which term of
x−2n is the asymptote approximation most inaccurate?
A x w, h ph x d tanx v h vmp. Th, h mp ppxm mk
wh n = 1. B x1 h m , h xn , v h xn, v m d
n = 1. Th x−2n , −2 m h
xn (S 5.3), q d n = 1. B x−2n h
n = 1, h x−2n (h m x−2
n
) , , h n = 1.
Problem 6.26 Absolute error in the early terms
Em, n, h x−2n h pdd h
mp ppxm.
Wh h d n = 1, h mpvm h
m S m m p h ppxm x1 = (n + 0.5)π
wh m v. A mp m pph vppxm h Nw–Rph mhd (Pm 4.38). T
fid wh h mhd, mk x d pd
mpv h pm
x −→ x −tanx − x
sec2 x − 1. (6.30)
Wh h x h w h fi mp 1.5π ,
h pd pd v x1 = 4.4934...
Th, mpv h m S ≈ 0.094715, whh w d h
mp ppxm, ppxm fi m ( p)
d dd h d fi m.
S ≈ Sd −1
(1.5π )2+
1
4.49342≈ 0.09921. (6.31)
U h Nw–Rph mhd fi, dd, h 1/x22 m
v S ≈ 0.09978 (Pm 6.27). Th, hh dd
S =1
10. (6.32)
Th fi m kw d m m h
d ! Th mp d p mdv mp xp.
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118 6 Analogy
Problem 6.27 Continuing the corrections
Ch m N, 4. Th h Nw–Rph mhd mp v xn n = 1 . . . N; d h v fi h m S. A xd h mp v N, d h fid
m S pph dd 1/10?
6.4.3 Analogy with polynomials
I h q tanx − x = 0 hd j w d-m !
Th h m S wd mp. Th wh fid
p tanx − x wh pm q wh mp . Thmp pm h qd, xpm wh
mp qd— xmp, x2 − 3x + 2.
Th pm h w , x1 = 1 d x2 = 2; h
x−2n , h
pm- m h - m, h w m.x−2
n =1
12+
1
22=
5
4. (6.33)
Th - mhd mp h m q
h qd q. Hwv, mhd h hq tanx − x = 0, whh h d-m ,
h hmv. I m h qd—
m, w ffi 2 d −3. U, p mhd m 2 d −3 pd h
x−2
n = 5/4.
Where did the polynomial analogy go wrong?
Th pm h h qd x2 − 3x + 2 ffi m
tanx − x. Th qd h pv ; hwv, tanx − x,
dd , h mm pv d v d h
x = 0. Idd, h T tanx x + x3/3 + 2x5/15 + · · ·(Pm 6.28); h,
tanx − x =x3
3+
2x5
15+ · · · . (6.34)
Th mm x3 m h tanx − x h p x = 0.
A pm—h, wh p x = 0, pv
, d mm v — (x+2)x3(x−2) , xp,
x5 − 4x3. Th m
x−2n ( h pv ) m
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6.4 Tangent roots: A daunting transcendental sum 119
d mp 1/4. Th v d p h (v)
h w ffi h pm.
T dd whh h p d, h pm:
wh −2, −1, 0 (hd), 1, d 2. O h pm
(x + 2)(x + 1)x3(x − 1)(x − 2) = x7 − 5x5 + 4x3. (6.35)
Th pm- m h w pv 1 d 2 d
1/12 + 1/22, whh 5/4—h (v) h w ffi.
A fi h p, d −3 d 3 m h . Th
pm
(x7 − 5x5 + 4x3)(x + 3)(x − 3) = x9 − 14x7 + 49x5 − 36x3. (6.36)
Th pm- m h h pv 1, 2, d 3 d 1/12 + 1/22 + 1/32, whh 49/36— h (v) h
w ffi h xpdd pm.
What is the origin of the pattern, and how can it be extended to tanx − x?
T xp h p, d h pm w:
x9 − 14x7 + 49x5 − 36x3 = −36x3
1 −
49
36x2 +
14
36x 4 −
1
36x6
. (6.37)
I h m, h m 49/36 pp h v h fi ffi. L’ z. P k x = 0 d
±x1, ±x2, . . ., ±xn v h pm
Axk
1 −
x2
x21
1 −
x2
x22
1 −
x2
x23
· · ·
1 −x2
x2n
, (6.38)
wh A . Wh xpd h pd h
ph, h ffi h x2 m h xp v
m h x2/x2k m . Th, h xp
Axk
1 −
1x2
1
+ 1x2
2
+ 1x2
3
+ · · · + 1x2
n
x2 + · · · . (6.39)
Th ffi x2 ph
x−2n , whh h pm
h - m.
L’ pp h mhd tanx − x. Ahh pm, T k fi-d pm. Th T
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120 6 Analogy
x3
3+
2x5
15+
17x7
315+ · · · =
x3
3
1 +
2
5x2 +
17
105x 4 + · · ·
. (6.40)
Th v h x2 ffi hd − x−2n . F h -
m pm, x−2n hd h −2/5. U, h m
pv q v!
What went wrong with the analogy?
O pm h tanx − x mh hv m mpx
wh q v m S. F,
(Pm 6.29). A hd--v pm h tanx − x
fi fi v x, d d fi , wh
pm d v .
sinx − xc osx
0x1
x2
x3
Th hv
fi hv h m tanx−x. Th
fi tanx − x wh tanx w p,
whh wh cosx = 0. T mv h fi
wh d , mp
tanx − x cosx. Th pm-k xpd h sinx − xcosx.
I T xp x − x3
6+ x5
120− · · ·
sinx
−
x − x3
2+ x5
24− · · ·
xcosx
. (6.41)
Th dff h w
sinx − xcosx =x3
3
1 −
1
10x2 + · · ·
. (6.42)
Th x3/3 d h p x = 0. Ad h , hv h x2 ffi, - m S = 1/10.
Problem 6.28 Taylor series for the tangent
U h T sinx d cosx hw h
tanx = x +x3
3+
2x5
15+ · · · . (6.43)
Hint: U k h p.
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6.5 Bon voyage 121
Problem 6.29 Only real roots
Shw h tanx − x .
Problem 6.30 Exact Basel sum
U h pm v h B m
∞1
1
n2. (6.44)
Cmp wh Pm 6.22.
Problem 6.31 Misleading alternative expansions
Sq d k h p tanx = x v cot2 x = x−2; qv,cot2 x − x−2 = 0. Th, x tanx − x, cot2 x − x−2.Th T xp cot2 x − x−2
−2
3
1 −
1
10x2 −
1
63x4 − · · ·
. (6.45)
B h ffi x2 −1/10, h - m S— cotx = x−2
d h tanx = x—hd 1/10. A w d xpm d tanx = x, h . Hwv, wh wwh h ?
Problem 6.32 Fourth powers of the reciprocals
Th T sinx − xcosx
x3
3
1 − x2
10+ x4
280− · · ·
. (6.46)
Th fid
x−4n h pv tanx = x. Chk m
h p.
Problem 6.33 Other source equations for the roots
Fd
x−2n , wh h xn h pv cosx.
6.5 Bon voyageI hp h hv jd p -fih mhd
pm-v x. M fid dv pp
dm , , mp, p , k
h p, d . A pp h , w hphm—d v d w .
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Index
An italic page number refers to a problem on that page.
ν
km v
1 w w
≈ (ppxm q) 6
π , mp 64B–Sm hm 65
∝ (pp ) 6∼ (wdd) 6, 44ω
q
, 99–121dvd p wh p 103–107 j
j: p 107–113
h (L) 108–109mm (Σ) 109
pv 100, 118,120
pmd vm 19p 99–103
- m 118–121 j
j: pm 118–121m dpd v 101
, p 99–103 q 44A xvhm–m m 65
hm-m–m-m -q 60–66
pp 63–66mp π 64–66mxm 63–64
q d 62m xmp 60p p 61–63m p 61
hm m m mp 62
mp tanx 114
mph p 34
k--h-vp m 78m mp 77mm qd 78pw 10 78
41
B m (
n−2) 76, 113, 116, 121 98
p, h
k h pddv m h mpv
80 mpv
h w 78
p, k k h p
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128
m ffi 96, 107
m d 98
m hm 90, 97 70–73
, CD p 78 kd d 87 d 27 v 57Bkhm, Ed 26
, dm d 31CD-ROM
CDm m CD 77
CD/CD-ROM, p 77–79
h md (p m-d) 44
h m 44hk 78
m m 76
p wh m d 72mp, wh dff
dm 2 - d 35 mp 21
pdm 48j
dd d 105, 119xp 119 100, 103, 104, 105p 105 100, 101, 104, 106, 111, 119
m d 100, 105, 106 pp
S–Bzm 11 pp 5
d 20v, 65, 68vx 104ph k p 82
Cfid, Dvd 105
hh pw 94–97m- ppxm
dvd 86
d 95, 73
d (dff m) 10, 43
d 103dvv 38dvv
ppxm wh z Δx 40 ppxm 38
39mpvd p 39 38v 39
ddm 38
ppxm 38fi-h ppxm
40–41 43Nv–Sk dvv 45 d v 40
v 40d-d mhd 32dff q
hk dm 42z 47, 51–54
m 12pdm 46mp q
43–46p–m m 42–45
x 45pdm q 47
dm dm, mhd ; dm-
pdm
G 10mp hm m 48S–Bzm w 11
dm p 24d 25- pd 24pdm pd 48p–m m 48
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129
dm qdph w 94
h m xp 89hv w p 94
hv w p 81dm
L h 5 5T m 5v 2
dm, mhd 1–12 dm pdv 6hk dff q 42h pfid dm 7,
8–9mpd wh 15 pp 5d 23–26 7–11Kp’ hd w 12
pdm 48–49d- pm 12 v v dff-
q 5S–Bzm w 11
dm 47d (dff) 10dx 10xp 8 9
9
km v ν 22pdm q 47d dvv 38, 43p 43
mm Σ 9d 21–29
dph--w m, ff 93hh Rd m 28w Rd m 30q ff 23
d d
e
h 90h
79
mp 87 13–30
dd dd m 58 - 98 70 d 100hk m 13–17mpd wh dm 15p 16–17p pm 65w 104
w p 103 13–16hh dm 103hh Rd m 27 xp 89w Rd m 30
fi d pd 92, 94
pdm mpd 49–51m mpd 47–48
pm 118
pmd vm 19 tanx = x 114mp 108, 112hz m 17d 21d pmd 18–21
p 17pm 65
p 87
p 87 v 50 mp dv 82–84
ff mm m 83
p xp w-p xp
p mx 81q, kd 6
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130
m dvv dvv, ppxm;dvv, fi-h ppx-m
E 113 B m 98
E–ML mm 112Evolving Brains 57x
v mk 4xmp
dd dd m 58–60hm-m–m-m -
q 60–66
, m 32–33 70–73 d mh 99–103dph w 91–94dvv cosx, m 40–41dvd p wh p 103–107d pp 21–29p 16–17 v m 55 mph pd
m 82–84 36–37
3–6G dm
7–11G
13–16hm 66–70mxmz d 63–64mp 3.15 7.21
h 79–80 w 79
p
h (L) 108–109mm (Σ) 109–113
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