88886220 Street Fighting Mathematics

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SANJOY MAHAJAN

FOREWORD BY CARVER A. MEAD

THE ART OF EDUCATED GUESSING AND

OPPORTUNISTIC PROBLEM SOLVING

STREET-FIGHTING

MATHEMATICS



Street-Fighting Mathematics





Street-Fighting Mathematics

The Art of Educated Guessing andOpportunistic Problem Solving

Sanjoy Mahajan

Fwd Cv A. Md

Th MIT P

Cmd, Mh

Ld, Ed



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

[email protected]

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

http://creativecommons.org/licenses/by-nc-sa/3.0/us/

mailto:[email protected]

mailto:[email protected]

http://creativecommons.org/licenses/by-nc-sa/3.0/us/



For Juliet





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





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


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


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



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


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



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





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:



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.



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 .





1

Dimensions

1.1 Em: Th pw m p 11.2 Nw mh: F 31.3 G 7


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



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!



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 .



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.



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



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- ?



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:



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 dmpd 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)



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

.



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.




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)



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 .



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



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 .



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- ?



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 .



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.



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?



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.



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



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



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.




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.



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




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.



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,




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)



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!




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.


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.



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?



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).



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?



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)



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.



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].



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.



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



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




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’



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




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.



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?



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 ?



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 .



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)



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).



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.)



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 ,




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):



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).




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)



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)




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.



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!


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)




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)





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



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:



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.




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.)



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.




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?




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.




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.




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:




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)



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?




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)



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.




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



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




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?



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.




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:




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

.


F m , v h fid pp .A m hd z p m d qk h d




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.)



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)



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)



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 :




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




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.




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)




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?




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).



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%.




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)




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.




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.




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)




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.



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 ,




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.



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).




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)



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)




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)




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?


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 .)




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 mpmh Mxm.



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 .



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◦.



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



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.



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



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).



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



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.



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.



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)




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



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)




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.



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)



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 .



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)




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)



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 .




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.



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




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



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.



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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[43] Rhd M. Sm d Gd J. Sm. Fwd d dpd-dd kk m mp-dd . AI Mm380, MIT, Afi I L, 1976.

[44] Edw F. T d Jh Ahd Wh. Spacetime Physics: Introduction toSpecial Relativity. W. H. Fm, Nw Yk, 2d d, 1992.

[45] Sv P. Thmp. Calculus Made Easy: Being a Very-Simplest Introduction toThose Beautiful Methods of Reasoning Which are Generally Called by the TerrifyingNames of the Differential Calculus and the Integral Calculus. Mm, Nw Yk,2d d, 1914.

[46] D. J. T. Physical Fluid Dynamics. Oxd Uv P, Nw Yk, 2dd, 1988.

[47] US B h C. Statistical Abstracts of the United States: 1992. Gv-m P Offi, Wh, DC, 112h d, 1992.

[48] Mx Whm. Productive Thinking. Hp, Nw Yk, d d, 1959.

[49] P Zz. The Art and Craft of Problem Solving. W, Hk, Nw J, 2dd, 2007.





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



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



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



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

pdm pd 46–54pw m 1–3pd mp 1/13 84–85 mp fl

86–88p–m dff q

42–45

q 85–86 p CD-ROM CD

77–79mm ln n! 73–75

- m 113–121m 94–97vm d pmd 17–21

xpd, 33 pm 36

xp, dm 8xm

p 36S’ m

E–ML mm 112mp 36–37p 74

mm p 73mm hm 73–75

w m m 78 vd m 78 m mp 78

h 86 83, 84d mp 83h– d 87m wd pw 84

xp −2 86xp 1/4 87 xp 84–90 85, 86dd 79–80

xp 89–90, 95 ppxm 82mp 3.15 7.21 79v d xp

86–88 p v dd 82p 80m h dd 82q 85–86



131

q 82–84- m 117

dm 3–6

dph w 91–94dff q 4mp pd (x) 4wh v 30

d 33 ffi 85

G d m, 14, 16xd m ∞ 96 55

pzd ppxm 14 dm 7–11 13–16 mp 34, 35

GDP, m flw 1m m

hm m; h-m-m–m-m hmdfi 60p 61h m 63

dd 59z 1ph m

p p

hh-p xp w-p xpm qd m 92

How to Solve It xH 48

d p 58m h 81

ppxm mp mp

v dff 109m 14p 109

d 86

pm hm 73

J, Edw Thmp 105 Jff, Hd 26

Kp’ hd w 25km v (ν) 21, 27

Ld I, d m- m 94

L (dm h) 5Ld–J p 41

xp 32 (m d) 10, 43hm

z h 90 dfi 67- ppxm 69

w-p xp fi p 81dm q

81 h 81m v ppxm 93hh-p md p 81dd 80–82

d mx p 81 tanx = x 114

mp 31–551/e h 34mph p 34

md 67dff q 51–54m dvv 37–41d 67 33–37pdm, md mpd 51

pp m 32–33 mh 52

M Cm O, h 3 Mathematics and Plausible Reasoning xmhm, pw 7mxm d mm 41, 70

hm-m–m-m -q 63–64



132

x vm 64

m 64

m dv 33m mp

w w

mhd v k 69mx p 81

Nv–Sk qdffi v 22 m 45m 21v m 46

Nw–Rph mhd 76, 117, 118

m 14

dd m, m 58–60 w

h 79p

dvv (D) 107xp 108

fi dff (Δ) 110 109 h (L) 108–109

h h 109mm (Σ) 109–113

p, wh 76P’ 107

p, k 90pdm

dff q 46 wk v 52pd 46–54

pp 58

p p 57–76dd dd m 58–60 76hm-m–m-m -

q 60–63, 76

70–73mpd d p 58dvd p wh p 107 73–75

hm 66–70Nw–Rph mhd 76

tanx = x 114vm ph 76

p dph w 94

p v w-p xp

P, G 105pp, m 32pw m 1–3pw 78pp 18pmd, d 17

qd m 91hh p 92v v ppxm 93

qd m 80, 82, 84d 85

m 30pd m dv 84–85 69, 101R

Rd md- pm 12w--- k 68, 70, 86Rd m (R) 27

hh 27w 30

xrigor mortis xd

79 w 78

v 40 mp h 86–88 mp fl

v xp 88

ppxm dvv, ppxm

, p 38



133

d dvv dvv, d

Sh–Nq mp hm78

fi-h ppxm dvv, fi-hppxm

m 61, 70mp pm

k h p; mp; ;

, m- ppxmdvd 47d 86

m- ppxm

95 47, 66

-d 86p, dvd wh p 103–107pp 35ph, vm m 76

p–m m 42–45p

dm 43Hk’ w, 42

mh 81

S–Bzm 11, 87S–Bzm w

dv 11q mp Kv 88 mp mp 87

ff p

S’ m : S’ m

v ppxm k h p

dph w 92–94w-p xp 93ph h 93 93v qd m 93

mmppxm 113, 114E–ML 112, 113dfi 110

ppxm 74p 109–113pd dff 112 113–121

74, 113, 115m

v 57m k m 61

mm 72

k h p 77–98dph w 92–94pm xp 106, 107

- m 114, 117–118m 94–97

T d 37 66hm 66, 69

m 68pdm pd 53 118, 120

L (dm h) 5hd, 99The Art and Craft of Problem Solving xhm xp 82

Thmp, Sv 10hh xpm 18, 50

dm, mhd ; ;mp; p p; k h p; ,

mhm 36k 101

pzd ppxm 14k

mp 85w 68, 70, 86v m 36, 101

k v mhd 69 h xv

d- vm?ppxm dph w 92, 93mp q 86



134

mp 54mm ppxm 75

- m 115 w 79

78

M Cm O, h 3p m q 4v dm 2

Whm, Mx 59



Th k w d wh w d . Th x

P, dd Hm Zp d v TX G

P. Th mhm E, dd Hm Zp.

Mxm 5.17.1 d h mpmh Ph dd v .

Th fi w d m v GNU Em d

md h M v- m. Th fi

fi w mpd wh MP 1.208 d Amp 1.88. Th TEX

w mpd PDF CTX 2009.10.27 d PDFTX 1.40.10.Th mp w md wh GNU Mk 3.81 d k 10 m

2006-v pp. A w w D GNU/Lx.

I wm hk h m h w mm.



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