Basic Numerical Mathematics, Vol. 1: Numerical Analysis 9780126924015, 0126924015, 1483247597, 9781483247595

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ISNM ΙΝΤΕΚΝΑΉΟΝΑί SERIES OF NUMERICAL MATHEMATICS INTERNATIONALE SCHRIFTENREIHE ZUR NUMERISCHEN MATHEMATIK SERIE INTERNAΉONALE D'ANALYSE NUMÉRIQUE

Editors: Ch. Blanc, Lausanne; A. Ghizzetti, Roma; P. Henrici, Zürich; A. Ostrowski, Montagnola; J. Todd, Pasadena

VOL. 14

Basic Numerical Mathematics Vol. 1: Numerical Analysis by

John Todd Professor of Mathematics California Institute of Technology

ACADEMIC PRESS

New York

London

Toronto

Sydney

San Francisco

A Subsidiary of Harcourt Brace Jovanovich, Inc.

BIRKHÄUSER VERLAG BASEL UND STUTTGART 1979

1980

C O P Y R I G H T © 1 9 7 9 , BY B I R K H A U S E R V E R L A G B A S E L . A L L RIGHTS R E S E R V E D . NO P A R T O F T H I S P U B L I C A T I O N M A Y BE R E P R O D U C E D O R T R A N S M I T T E D IN A N Y F O R M O R BY A N Y M E A N S , E L E C T R O N I C OR M E C H A N I C A L , INCLUDING PHOTOCOPY, R E C O R D I N G , OR A N Y INFORMATION STORAGE A N D RETRIEVAL SYSTEM, WITHOUT P E R M I S S I O N IN W R I T I N G F R O M T H E P U B L I S H E R .

L I C E N S E D E D I T I O N F O R N O R T H A N D S O U T H A M E R I C A , A C A D E M I C PRESS, INC., NEW Y O R K L O N D O N T O R O N T O S Y D N E Y S A N F R A N C I S C O A SUBSIDIARY O F H A R C O U R T BRACE JOVANOVICH, INC.

ACADEMIC PRESS, I N C . Ill

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ISBN

10003

CARD NUMBER:

0-12-692401-5

P R I N T E D IN T H E U N I T E D S T A T E S O F A M E R I C A

80-28729

**Numerical Analysis" is for DONALD H . SADLER **Numerical Algebra" is for OLGA TAUSSKY T O D D

Notations and Abbreviations

W e use the standard logical symbolism, e.g., € for belongs to, c= for is included in, . φ . for implies. U is the set of real n u m b e r s . If X c : R is hounded above so that there is an MeU such that xeX . φ . χ < Μ , then there is a A G R such that x e X . = > . χ < λ and such that if ε Χ ) there is an Xi^eX such that χ , ) > λ - ε . W e write λ = l u b X ; it is the /east u p p e r b o u n d . If λ € Χ , we write A = m a x X . Conventionally, if X is not b o u n d e d above, e.g., in the case X = { 1 , 2, 3 , . . . } we write lub X = oo. In the same way we define gib X, min X. [α, h] d e n o t e s the set of real n u m b e r s χ such that a < x < / ) . (a, b) denotes the set of real n u m b e r s χ such that

a 200, 2 10 = 1024> 103 • We now give the actual values of a,., bn in this case and we see that our estimates are rather generous. bo =0.0001 ao= 1 b t =0.01 at = 0.50005 b2 = 0.070714214 a2 = 0.25502 5 b 3 = 0.13429 033 a 3 = 0.16286 961 a 4 = 0.14857996 b4 = 0.14789 122 bs = as= b6 = a6= b7 = a7 = bg = ag = 2.

THE ALGORm-IM OF BORCHARDT

We shall now make a small modification of the Gaussian algorithm. This makes a change in the rate of convergence and in this case the limit is readily obtained as a function of ao, 130' Let non-negative a o,13o be given. Define for n = 0, 1,2, ... a n+1 =4(an + 13n), 13n+' = J(~+t13n)' Suppose first that 130> ao. It follows by induction from the definitions that Hence the sequences

{~}, {13n}

13n+t -~+t =

converge. Since

J-;::;J }[13n -~] {2(J ~+t + 13n)

and since the first factor on the right is less than 1. the limits of the sequences coincide. The convergence in this case is ultimately much slower than that in the Gaussian case. The same results obtain when 130 < ao. We shall now determine the limit in closed form. Although the defining relations are still homogeneous they are not symmetrical. Assume first that 130 = 1 > ao. Let 6 be defined by cos 6 = ao. Using the relation 1 + cos x = 2 cos 2 (x/2) repeatedly we find at = cos 2 6/2, 13t = cos 6/2,

a2 =

cos 6/2 cos 2 6/2 2 ,

132 = COS 6/2 COS 6/2

2

•.•

, •••

Chapter 1

16

and establish by induction that (3n

= cos (2- 1 0) cos (2- 2 0) ... cos (2- nO).

If we multiply across by 2n sin (2- n O) and use repeatedly the relation

2 sin x cos x = sin 2x we find 2 n sin (2- nO)(3n = 2n - 1 cos (2- 1 0)

=2

n

-

2

cos (2- n+ 1 0)[2 sin (2- nO) cos (2- nO)]

cos (2- 1 0)

cos (2- n + 2 0) x [2 sin (2- n + 1 0) cos (2- n + 1 0)]

= sin o. Hence (3n

= 0- 1 sin O[O,jsin On]

where On = 2- nO. Clearly, since limx---+o (x/sin x) = 1, we have lim (3n

= 0- 1 sin O.

lf we have ao> (30 we may assume (30 = 1 and a o = cosh t. A similar discussion using hyPerbolic functions instead of trigonometrical ones shows that the common limit of the sequence is

(sinh Oft.

3.

THE CARLSON ALGORITHM

Let non-negative ao, bo be given. Define for n = 0, 1,2, ... ~+1 =.J~(~ +bn)/2,

bn+ 1 =.Jbn(~ +bn )/2.

Suppote that a o > boo Then we find from the definitions that The sequences

{~}, {b n }

therefore converge and since

~+1 - bn + 1 =

+bn)/2} .J;;,. + Jb,. (~ - b {.J(~

n)

where the first factor on the right is always less than v1 < 1 and approaches it follows that the sequences have a common limit, say I. The rate of convergence is, however, ultimately much slower than in the Gaussian case.

!,

The Algorithms of Gauss, Borchardt and Carlson

17

We shall now determine the limit of the Carlson sequences in terms of ao, boo We observe that a~+I-b~+1

(a~-b~)/2

2 log (a,.+I/bn+l)

a~-b~

2 log J a,./bn 2 log (a,./bn )·

Hence, for all n a~-b~

a~-b~

2 log (a,./ bn )

2 log (a o/ bor

By the Mean Value Theorem log (a,./bn) = log a,. -log bn = (a,. where

a~

bn)/a~

is between a,. and bn. Hence, for all n 2 log (ao/b o)·

We now let

n~oo

and then

a,.~l, bn~l, a~~l.

F=

2

This gives

b2

ao- 0 2 log (a o/ bo)

so that

4.

HISTORICAL REMARKS

The arithmetic-geometric mean of Gauss is of the highest importance in the history of mathematics. As a teenager in 1791 Gauss, without computers, made extensive calculations of arithmetic-geometric means. In particular he found that M(J2, 1) = 1.19814 02347 35592 20744.

It seems clear that he was searching for a formula for M(a, b), of the kind given in § 1.2, 1.3 and some of the problems. It was not until 1799 that he

made progress. At that time he computed the definite integral

A=

i J(l1

o

dt

t4 )·

He then recalled his value of M(.J2, 1) given above and observed that the product AM(.J2, 1) coincided to many decimal places with hr. In his diary,

Chapter 1

18

on 30 May 1799, Gauss wrote that if one could prove rigorously that AM(../2, 1) =!1T, then new fields of mathematics would open. In his diary, on 23 December 1799, Gauss noted that he had proved this result, and more; in later years his prophesy was fulfilled. Stirling had actually obtained A to 16D in 1730. In Chapter 9 below we shall show how to evaluate A approximately, using however another method due to Gauss, quite different from that of Stirling. It has been shown that this integral, which can be interpreted geometrically as the quarterperimeter of the lemniscate of Bernoulli, is a transcendental number. There is no obvious way to establish Gauss' result. All known methods have the same character, which is similar to the developments in § 1.3, in Problem 1.8 and in Problem 1.12. The identity AM(../2,l)=!1T is the special case k = 1/../2 of the following theorem. Theorem 1. If 0 $ bo= kao $ ao then

lim a,. = lim bn = ao1T/[2K'(k 2 )] where

a complete elliptic integral.

Change the variables by writing

cos(J=~

t , t+a 2

-adt

d(J=----:::

2(t+a 2 )Jt

and we find 2 b2 ) R ( a,

=!

1T

f= 0

dt . v'[t(t+a 2 )(t+b 2 )]

Change the variable again by writing x(x+bi) t =---'----;:=-:x+ai '

dt

= (x+a1a)(x+a1b) dx (x+ai)2

The Algorithms of Gauss, Borchardt and Carlson

19

where to get R(a 2 , b 2 ) =-1 7T

f"" 0

dx = R(ai, bi). .J[x(x+ai)(x+bm

Applying this result repeatedly we find (compare Problem 1.15) R(a~, b~)

= R(ai, bi) = ... = R(M2 , M 2 )

where M = lim a,. = lim bn • The common value is clearly

~

r.. ./2d6 = MM

7T.lo

1

.

Hence

where

where k = bolao. The algorithm which we have attributed to Borchardt, who discussed it in 1880, was apparently known to Gauss who wrote about it to Pfaff in 1800. For further historical remarks see Problem 1.20, Solution. The Carlson algorithm was discussed in 1971. For an alternative account of Theorem 1, see Problem 1.19. The three algorithms of this chapter are discussed in a geometrical way by I. J. Schoenberg (Delta, 7 (1978), 49-(5). Chapter 1, Problems

1.1. Read again a proof of the fundamental theorem on monotonic sequences: if Xn::S:Xn+1 and if Xn::s:M for all n=O, 1,2, ... then there is an x ::s: M such that lim 1.2. Show that if Xn

~

Yn for all n

=

Xn

0, 1,2, ... and if Xn ~x and Yn ~ Y then

x~Y·

1.3. Show that if

161 < 1 then

lim

=x.

6 = O. n

20

Chapter 1

1.4. Read again the proof of the fact that (1 + n-It~e.

1.5. Show that as x ~oo the sequence (1- x-Iy increases to e- l and the sequence (1- X-l)x-l decreases to the same limit. 1.6. Observe the behavior of the arithmetic-geometric sequences {a,,}, {b n} e.g., when a o = 1, bo = 0.2 and when ao = J2, bo = 1. Specifically, compute enough terms of the sequences to find the arithmetic-geometric means to 8 decimal places. At each step print out the values of an, bn, a" -bn, 2ao[{ao-bo)J(ao+bo)JZ"·

1.7. If M(a o, bo) is the arithmetic-geometric mean of ao, bo, show that for any t~O M(ta o, tbo) = tM(a o, bo),

i.e., M is homogeneous (of degree one). Use the fact to determine, with the help of the result of Problem 1.6, M(6000, 1200). 1.8. For ao, bo given (both different from zero) define

= !(a" + bn ), the arithmetic mean, bn+l = 2J{(1Ja,,) + (1/b n)}, the harmonic mean.

a,,+l

Show that ao~bo implies that {a,,} is monotone decreasing, that {bn} is monotone increasing and that both sequences converge to .Jaob o, the geometric mean of ao, boo Show that a" + I - bn+l

= [a" - bnY/(4a,,+ I)'

Observe the behavior of the sequences {a,,}, {bn} in special cases. 1.9. For ao, bo given (both different from zero) define

Discuss the convergence of the sequences {a,,}, {bn} either directly, or by relating them to the Gaussian sequences for a bOlo

o\

1.10. a) Observe the behavior of the Borchardt sequences in the cases

The Algorithms of Gauss, Borchardt and Carlson

21

when Qo = 0.2, ~o = 1 and when Qo = ./2, ~o = 1. Compare the rates of convergence with those in the Gaussian case. Check the limits you obtain from tables. b) Repeat a) for the Carlson sequences. 1.11. If A o, B o are non-negative and if we define for n = 0, 1, 2, ... B n+1 =.jAnBn

A n + 1 = !(An + B n+ l )

discuss the convergence of the sequences {An}, {B n} either directly, or by relating them to the Borchardt sequences of Q o = A o, ~o = .J AoB o. 1.12. Obtain the common limit of the Borchardt sequences by proving that = 2n arccos (a,j~n) and g(n) = 2n.J(~~-Q~) do not depend on nand then showing that [f(n)/g(n)]-r- I so that the limit l must be g(O)/f(O). f(n)

1.13. Read again proofs of the First Mean Value Theorem. 1.14. Using appropriate tables of elliptic integrals check the value obtained for MO, 0.2). 1.15. Justify the relation R(a~, b~) = R(M2 , M 2 )

in the proof of Theorem 1. 1.16. (B. C. Carlson) a) Discuss the convergence of the algorithm go=JX,

a o =!(l+x),

an+1=!(~+gn),

gn+\=.J~+\gn,

n=0,1, ... ,

where x>O. Calculate 10g./2. Examine the convergence of the sequence H^.

(2)

g(x) = €{l)

as

x^oo.

(3)

h(x) = o{x)

as

(4)

k(x) = o(l)

as

x-^0.

T h e relation (1) m e a n s that for s o m e constant A and s o m e XQ χ>Χοφ|/(χ)|Xo^|g(x)| 0 t h e r e is δ = δ ( ε ) such t h a t h{x) 0Xo4>|h(x)|l.

g ( n ) < 2 n ^ + (3 + 4 + 5)n^ = 2 n ^ + 1 2 n ^

if

n>l

and hence (16)

g(n)12

and (17)

g(n)100.

E a c h of t h e t h r e e relations (15), (16), (17) justifies t h e s t a t e m e n t g ( n ) = 0(n^) and t h e interdependence of t h e HQ a n d A is clear. T h e constant A can b e t a k e n t o b e any n u m b e r bigger t h a n 2 b u t t h e HQ increases as we t a k e A closer t o 2. (b) T o p r o v e that x^^ = oie"") as x->a> w e can p r o c e e d as follows. N o t e that x^^

x^^

12\ x'^

x>0.

W e show that given ε > 0 we can choose Xo = Xoi^) such that χ^^β"^Xo.

Orders of Magnitude and Rates of Convergence

F o r instance, t a k e ε = 1 0 - ^ Since 5 X 1 0 * w e shall have x^^e-'^Xo

27

12! = 4 7 9 0 0 1 6 0 0 = 4 . 7 9 . . . x 1 0 ^
x,; or, one could check at each stage whether Ix~- NI was less than a prescribed tolerance. Whenever a specific algorithm is chosen, it should then be examined thoroughly so that it could be guaranteed that the output 5, corresponding to the input N, would be near to IN either in the sense that IS-JNI E2 are particular (small) numbers. These examinations depend very much on the machine being used, especially on the arithmetic unit, so that the E'S will differ from machine to machine. Returning again to the distinction between theoretical arithmetic and practical computation we note that while in the first we are not concerned

Recurrence Relations for Powers

43

with the size of the numbers which occur in our computation in the second we must be sure that all intermediate results of our computation remains within scale Le., do not cause overflow or underflow on the machine being used. To guarantee this again requires a detailed examination of the algorithm and of the machine. Another example of the distinction we are making is the following: Theoretically the harmonic series 1 +!+l+~+ ...

is divergent. If one tries to sum this on a computer we will have a finite series, for n -1 will become and remain zero from the point of view of the machine, and apparently we have convergence.

Chapter 3, Proble_ 3.1. Demonstrate the relative speeds of the various algorithms for N- 1 and for N 1/ 2 by printing out in parallel columns the corresponding values of x". Observe how the choice of Xo affects the speed of convergence. 3.2. Discuss the behavior of the following sequence wheh O 0, Xl>

°

and if

P =0,1,2, ... , then lim Xl>

= a 1/n •

3.9. What is the order of convergence of the sequence in Problem 3.8? 3.10. Discuss the convergence of rn = pJqn where

= apn-l + Nqn-t> qn = Pn-l +aqn-l' Pn

n = 1,2, ...

where Po, qo are given, in the following cases (i) N>O, a >../N, (ii) N>O, O 2 ^ - "

and

only

2-""^' Τ , ( χ ) = 2 " " ^ ' cos (η arceos χ ) .

if

ρ^{χ) = ±Τ^{χ)

where

Τ^(χ) =

Uniform Convergence and Approximations

61

Proof. T h e proof d e p e n d s o n the equal-ripple behavior of T„(x). First, m a x _ i ^ , ^ i l t x ( x ) | = 2^-". Suppose max

|x"-hbix"-^-f . . . + ί ) „ | = μ < 2 ^ ~ " .

-l=Sx:Sl

Consider 6(X)

= f Jx) -

+ ...+ M.

(X" +

This is a polynomial of d e g r e e n - 1 at most. Its values at t h e points x^ = cos rrnrln, m = 0 , 1 , 2 , . . . , η have t h e s a m e sign as those of T„(x^) i.e., alternately plus a n d minus. H e n c e e(x) has at least η zeros, b u t this m e a n s that it vanishes identically which implies μ = 2 a contradiction. This shows that m a x | x " 4 - a i x " " ' - f . . .-f a j > 2 ' ~ " . T h e proof of t h e equality s t a t e m e n t is left to t h e reader. A n o t h e r case in which t h e polynomials of best approximation can b e given explicitly is that of / ( x ) = ( l + x ) in t h e interval [ 0 , 1 ] , This is discussed in P r o b l e m 5.7. Algorithms which g e n e r a t e a s e q u e n c e of polynomials which converge t o t h e polynomial of best approximation have b e e n constructed. F o r practical purposes it is sufficient to choose a polynomial of " v e r y g o o d " approximation a n d lists of these are available in m a n y places e.g., t h e N B S H a n d b o o k . F o r an informal description of such processes see C. Hastings, Approximations for Digital Computers, Princeton 1 9 5 5 . In m a n y of these algorithms it is desirable t o start off with a " g o o d " approximation. This can often b e d o n e by expanding t h e function in question in a Chebyshev series / U ) = | a o + a i T , ( x ) + a2T2(x) + . . . and taking an a p p r o p r i a t e n u m b e r of terms of this series. T h e coefficients in these Chebyshev series are Fourier coefficients and can b e evaluated as they a r e : since COS

^/T^x^

= 21

mO COS ηθ άθ

TT [COS ( M + η)θ -ΗCOS ( M - η)θ] άθ

Ό

if

m^n

Ítt

if

m = n

62

Chapter 5

we can find t h e a ' s by evaluating the integrals /(cos Θ) COS me de. Tables of Chebyshev coefficients for various functions have b e e n given by C. W, Clenshaw and by C. W . Clenshaw and S. K. Picken in N P L Mathematical Tables 5(1962), 8(1966). F o r examples see Problem 5 . 1 1 .

Chapter 5, Problems 5.1. Let ηο(ε, x) be the least MQ such that (Μ + ΐ 4 - χ ) ~ ^ < ε for η>ηο(ε, x). Tabulate ηο(ε, χ) for say x = 1,0.1,0.01, 0.001,0 8 = 3,1,0.1,0.01,0.001. 5.2. R e p e a t Problem 1 in the case when t h e inequality is (η + 1)-^ + 2(ηχ + 1 ) - ^ < ε . 5.3. Calculate M„ = max |r^(x)| 0 / 2 , θ = arceos ( 2 x - 1 ) a n d 0 < x < l . b) Show that for a certain φ (1 + χ Γ ^ - 7 Γ , ( χ ) = ^ — where

COS

(ηθ-^φ) ^

7rJx) = x / 2 { Ü - c T * ( x ) +

... +(-ir-'c'^-^T*_,(x)] + (-ir-^ Τ ί ( χ ) } 1

c

where T*(x) = cos (n arceos (2x - 1)). c) Evaluate Ei = m a x |(1 + χ Γ * - ί ί ( χ ) | when Í ,(x) = |(7>/2 - 8) - (6V2 - 8)x = ^ί2{^ - cT*(x)} and when d) Write down t h e Bernstein polynomial B , = ß i ( / ( x ) , x) for / ( x ) = ( l + x)~^ a n d evaluate max | ( l + x ) " ' - B i | . e) U s e t h e c o m p u t e r t o find t h e least n u m b e r of terms of t h e series (l+x)-* = l - x + x ^ - x - V . . . which must b e taken t o ensure a r e m a i n d e r less than value when χ = 0.1(0.1)0.9.

10

in absolute

5.8. Write down t h e Bernstein polynomial B^if, x ) in full. [ D o n o t use t h e summation sign.] If this is

B4(/, x) = ao(x)/(0) + a , ( x ) / ( 1/4) + α ^ ί χ ) / ^ ) + α3(χ)/(3/4) + α4(χ)/(1)

64

Chapter 5

by changing t h e variable to y = 2x - 1 get a Bernstein polynomial B * ( / * , y) appropriate for t h e interval [ - 1 , 1 ] . E v a l u a t e B* e*{y) = \y\-B*.

w h e n / * = |y| a n d d r a w a rough graph of fí* a n d of

Find max

-l:Sy:Sl

|e*(y)|.

5.9. T h e F a d e fractions r^,„(x) w h e r e μ, = 0 , 1 , 2 , . . . , corresponding t o a p o w e r series C ( x ) = X c „ x " , a r e t h e u n i q u e rational functions

w h e r e Ν is a polynomial of degree < i / a n d D a polynomial of d e g r e e < μ such that, formally, C{x)D,Jx)-N,Jx) consists of powers of χ greater t h a n μ + v. C o m p u t e t h e first 9 entries in t h e Fade table for e"" i.e., r^,, for μ,ν = 0 , 1 , 2 . 5.10. Is Γ2,ι(χ) of P r o b l e m 5.9 a good approximation t o e"" for small x ? C o m p a r e this approximation with t h e truncated Chebyshev expansion of β^ i.e.,

fix)=t !.

Condition (a) is called "complete monotony" and condition (b) requires that the terms do not decrease faster than a geometric progression with common ratio !. Let 'n' ,~ be the remainders after n + 1 terms in the original series and in its transform. Then

(_l)n'n = -Un+l + Un+2 = aUn+1

Un+3 + Un+4 - ...

+ aUn+3 + ...

Since {aUn} is a monotone null sequence we have

I'nl ~ -![aUn+1 + aUn+2+ aUn+3 + aUn+4 + ... ] = !Un+1 >~uokn'l.

70

Chapter 6

The transfonned series is a positive series and

Hence

1~:lsk(2~r ... , x,. and corresponding values fo, fl, ... , fn are given, find a function f(x) such that f(x;) = fi and use f(p) as the value corresponding to the abscissa p. 1.

LAGRANGIAN INTERPOLATION

The simplest case is when n = 1. Linear interpolation then assigns to f(p) the vallJe

We assert that f(x) is the unique polynomial of degree 1 agreeing with the data at Xo, Xl. Compare Fig. 8.1, where p is taken between xo, Xl only for convenience. This is the simplest example of polynomial interpolation - interpolation by rational functions (the ratio of two polynomials) or by splines (functions which are "piece-wise" polynomials) or by trigonometric or exponential polynomials is often considered, but cannot be discussed here. [See, however, Problem 8.17] In order to discuss polynomial interpolation in general we need the "Fundamental Theorem of Algebra" which is originally due to Gauss. Theorem 1. Every (polynomial) equation

has a root.

Many proofs of this are available. We note that in order that it be true we must admit complex roots even though we restrict the coefficients to be real- the equation z 2 + 1 = 0 shows this. However it is remarkable that if we admit complex numbers as solutions, then the result is true even if we allow complex numbers as coefficients.

Interpolation

85 f,

fof"'--+-------i

o

Xl

Fig. 8.1

If f(a)

= 0 then

f(z)

= f(z)- f(a) = (z -a)[aoz n- I +a~zn-2+ . .. +a~_.]

because It follows from this remark (sometimes called the Remainder Theorem) and the Fundamental Theorem of Algebra that we can factorize any f(z) in the form (2)

where the a j are not necessarily distinct. From this fact we can conclude that if a polynomial vanishes for more than n distinct values of x it vanishes identically. Suppose that f( z) vanishes for z = a I' . . . , an and for z = ao. The relation (2) must hold. If we put z = a o in this we get

None of the factors a o - a j can vanish. Hence an = O. Hence we have a polynomial of degree n - 1, f.(z)

= a1z

n

-

1+ ... +~

which certainly vanishes for more than n -1 distinct values of z. The same argument shows that a\ = O. Proceeding we see that f must be identically zero. It is a corollary to this that we need to establish uniqueness of polynomial interpolation. In fact, if two polynomials of degree n at most agree for more than n distinct values of the variable, they must coincideindeed, their difference, of degree n at most, vanishes for more than n distinct values of the variable.

86

Chapter 8

Suppose that p(x) is a polynomial of degree n at most. Then it has n + 1 unknown coefficients and we may expect it to be determined by n + 1 pieces of information e.g., the data of the problem formulated above. It is possible to write down a polynomial p(x) which interpolates f at the nodes x". To do this we consider (3)

'i(x)

=

:fi (x - Xj)/(x

i -

xJ

j~O

Ni

It is clear that each l;(x) is a polynomial of degree n and that

j1:i,

Hence L(f, x) =

(4)

!

[;li(x)

i=O

is a polynomial of degree n at most and its value at X; is fi' i = 0, 1, ... , n. Thus L(f, x) is a solution to our interpolation problem. It is the unique solution for any other solution L 1 (f, x) would agree with the data and therefore with L(f, x) at n + 1 distinct points - in view of the preceding remark, L 1 == L. This L(f, x) is called the Lagrangian interpolation polynomial. 2.

AITKEN'S ALGORITHM

A method for establishing the existence of such a polynomial is outlined in Problem 8.13. We shall now discuss an elegant algorithm due to A. C. Aitken for the evaluation of L,. at a particular point x. It reduces the general (n + I)-point interpolation to a succession of !n(n + 1) linear interpolations. An important advantage of this scheme is that it does not require the use of auxiliary interpolation tables. Cf. Problem 9.7. We shall describe the 4-point case for simplicity; it will be convenient to change opr notation. Given f(a)=A, f(b)=B, f(c)=C, f(d)=D, we show how to find f(p). Interpolate linearly between (a, A), (b, B) to find (p, B 1 ); then interpolate linearly between (a, A), (c, C) to find (p, C 1 ) and then between (a, A), (d, D) to find (p, D 1 ). The next stage is to interpolate linearly between (b, B 1 ), (c, C 1 ) to find (p, C z) and then between (b, B 1 ), (d, D 1 ) to find (p, D z ). Finally, interpolate linearly between (c, C z ) and (d, D z ) to find (p, P).

The scheme is illustrated graphically in Fig. 8.2. We note that there is no assumption that the a, b, c, d are equally spaced, and p can be anywhere, and so the method can be applied to the case of inverse interpolation.

Interpolation

87

I

I

,fY"f(x)

,, a

b

p

c

d

Fig. 8.2

We give, without comment, two examples which show how the scheme can be carried out and how labor can be saved by dropping common initial figures. Many extensions of the method have been given by Aitken and Neville. Given f(1)= I, f(2) = 125, f(3) =729, f(4) = 2197, find f(2.5). l=a 2=b

125 = B

4=d

729=C 2197=D

3 =c

I=A

187 = B( 547=C t 1099 = D,

367=C2

415 = D 2

343 = P.

Here f(x) = (4x - 3)3 and the interpolation is exact, as it should be. Given f(0)=47.434165, f(1)=47.539457., f(2)=47.644517, f(3)= 47.749346, find f(1.432l).

o 1 2 3

47.434165 539437 644517 749346

0.584954 682 517

837 60

24

-1.4321 -0.4321 +0.5689 + 1.5689

We find f(1.432l)=47.584824, which agrees with the fact that f(x)

= J2250 + 10x.

We shall now show generally that the Aitken algorithm leads to the Lagrangian interpolant. We follow a proof of Feller. We want to evaluate f(p), where f is a polynomial of degree n determined by its values at the

Chapter 8

88

distinct points X o, Xl>

•••

,x". Consider

t ll (x) =

d

p]

[f(xo) Xo et f(x) x - p , X-Xo

xi: Xo'

We observe that fll(x) is a polynomial of degree n -1 and that fll(p) = f(p). Hence our problem is equivalent to that of evaluating fll(p), where fll is determined by its values at Xl> X2' ... , x". Repetition of this process according to the scheme Xo

f(xo)

Xl

[(Xl)

f l l(X1)

X2

f(X2)

fll(X2)

f 2l(X2)

Xn

f(x,,)

fll(x,,)

f 2l(x,,)

'"

f"l(x,,) = f(p)

leads to the determination of f(p). For clarity we write out explicitly the general term r+ll(xj )

(5)

=

fil(xJ +

[Px j

where j 3.

Xj -Xj

][fil(Xj) - fil(Xj)]

= i + 1, i + 2, ... , n.

INVERSE INTERPOLATION

We have noted that the Aitken process does not require the equally spaced. It can be used to solve equations: for to solve

Xi

to be

[(X) = 0

is just the evaluation of

r

1

(0).

We give an example. If [(0) = -342, [(1) = -218, [(2) = 386, [(3) = 1854 find the zero of f between 1,2. We use the Aitken method, recording all the numbers used:

-342 -218 386 1854

0 1 2.7581 2 0.9396 2.1018 3 0.4672 0.5171

124 728 604 1.9926 2196 2072

1468

89

Interpolation

This is a very bad result. Indeed f(x) = (4x + 1)3- 343 and the zero is 1.5 not 1.9926. Inverse interpolation - except linear inverse interpolation, which is the same as direct - is tricky and great care is required. 4.

ERRORS IN INTERPOLATION

We have not yet discussed the error in interpolation:

(6) This discussion cannot begin unless we know something qualitatively about the behavior of f(x) between the nodes Xi. A natural assumption is to assume that the Xj lie in an interval [a, b] and that f(x) is continuously differentiable k times in [a, b] and, say, Irl(x)lsMk

in

[a,b].

It is clear that if f is a polynomial of degree n or less interpolation is exact: f(x)=L,.(f, x). This suggests that we might expect error estimates of the form (7)

for some constant K. This is so and an actual result in the case of n = 1 is (8)

this, and a little more, can be obtained by a slightly sophisticated application of the Mean Value Theorem. We write b - a = h, x = a + ph, 0 s psI. Then Taylor's Theorem gives 2h 2 f(a + ph) = f(a) + phf'(a) + p 2! f'(c), c = c(p), ascsb, f(b)

= f(a) + hf'(a) -

~: f'(d),

We have, for the error in linear interpolation for f(x) between f(a) and f(b): f(a + ph) -[f(a) + p{f(b) - f(a)}] = ~h2{p2f'(C) - pf'(d)}

which can be put in the form (7). We can get the same result with c = d = { by the following trick. Consider F(p) = f(a + ph)-[f(a)+ p{f(b)- f(a)}]- Kp(p -1)

Chapter 8

90

and choose a Po, 0 < Po < 1 and then choose K so that F(po) = O. Then since F(p) also vanishes for p = 0, p = 1. we conclude that F'(p) vanishes twice in [0,1] and F'(p) == h 2 f"(a + ph)-2K

must vanish once in [0, 1], say for p = 8, so that K

= !h 2 f"(a + 8h).

Thus we have, since F(po) = 0, f(a

+ Poh) = f(a) + Po{f(b) - f(a)}+!h 2 (P6 - po)f"(a + 8h).

Clearly 8 depends on Po. We can, however, drop the subscript 0 and conclude that for each P, O:s; P :s; 1, there is a {= {(p), a:S; {:S; b such that f(a

+ ph),= f(a) + p{f(b) - f(a)}+!h 2(p2_ p )f"({).

Thus the error in linear interpolation is and this is bounded by since Ip(p -1)1 ••• , Xn, x. This can be done by repeated applications of Rolle's Theorem to (12)

E(z) = f(z) - H(z) -[f(x) - H(x)]

n

[(z - Xj)2/(X - Xi)],

where X is distinct from the Xj. An extreme case is when we ask for a polynomial T(x) of degree n for which y ... , Xn form a basis for the vector space V n of all polynomials of degree at most n, with real coefficients. 8.14. Interpolate the following table in the neighborhood of the zero of f(x), first at interval 0.1, then at interval 0.01. Is linear interpolation justifiable in the last table? Find an estimate for the zero of f(x). X

0 1 2 3 4 5

f(x)

-855 17443 -513 73755 -172 86901 +167 39929 +507 03555 +846 00809

96

Chapter 8

8.15. (a) If f(x) is a real function of the real variable x, O$x$l, and f(O) = 0, f(l) = 1, what can you say about fm? (b) If, in addition, f is positive and convex, what can you say about fm? (c) If, in addition, f is a quadratic in x, what can you say about f(!)? (d) If, alternatively, f"(x) exists in [0, 1] and is bounded by 1 in absolute value, what can you say about f(!)? (e) If, alternatively, all the derivatives of f at exist, and are bounded by 1 in absolute value and if the Maclaurin series for f converges to f in [0, 1], what can you say about fm?

°

8.16. Discuss the existence of Hermite interpolants, e.g., in the two-point case, along the lines of Problem 8.13. 8.17. Establish the existence of a (cubic) spline in the two panel case. such that, no matter what fo, f±l' f~1> Specifically, show that there is an are, the Hermite interpolants in [X-I' x o] and [x o, Xl] fit smoothly together at X o i.e., that the second derivatives there coincide. Work out the case when

n

X±l

= ±1,

Xo=o,

fo= 1,

f~l

= 0.

Generalize. 8.18. Show how to determine the turning point x and the extreme value f = f(x) of a function f(x) given its values at three points near x, by assuming that f(x) = a +bx +cx 2 • 8.19. Show that if f, then

= f(r), r = 0,1,2, ... , n and if A has its usual meaning

Compute the differences of the function f for which f(O) = 789, f(2) = 2268, f(3) = 3648, f(4) = 5819, f(5) = 9304. Show that f coincides with a quartic at X = 0(1)5. Experiment with interpolation of various orders, to evaluate f(2.5).

f(l)

= 1356,

CHAPTER 9

Quadrature

We are now concerned with the approximate evaluation of definite integrals, i.e., areas under curves, 1=

r

r

f(x)dx.

We shall deal sometimes with weighted integrals 1=

f(x)w(x) dx

where the weight function w(x) is usually nonnegative and continuous in [a, b]. We use a very simple principle: Q, an approximate value of I, is given by the integral of an approximation to f(x). It is natural to approximate f(x) by a polynomial, since these are easy to integrate. If we take f(x) = L,. (f, x) = L l;(x )f(xJ

(1)

rL

r

then we have the so-called Lagrangian quadrature: (2)

1= Q =

a

li(X)f(x;) dx

=

Lf(x;)

a

l;(x) dx = L wJ(x;)

where the coefficients Wi are independent of f, only depending on the nodes Xo, ••. ,x,. and a, b. Clearly this is exact if f(x) is a polynomial of degree :Sn. 1.

TRAPEZOIDAL QUADRATURE

We consider first the very simple trapezoidal case corresponding to linear interpolation when we take n = 1 and a = Xo, XI = b and assume f E C 2 [ a, b]. The area under the curve is approximated by that of the trapezium which is (3)

(b-a)' Gf(a)+!f(b)].

98

Chapter 9

1f"(x)l:5 M 2 in [a, b], using the result about the error in linear interpolation, we can conclude that

If

With a little more care we can improve this to 1-0= --b(b-a)3f"(c),

(4)

for some cE[a,b]. [Compare Problem 9.3.] Supposing that this accuracy is not enough, what can we do? We could approximate f(x) by polynomials of higher degree or, and this is the approach we want to discuss, we can divide the area I into panels and apply the trapezoidal rule (3) to each one. Let us take b = a + nh and take n panels each of width h. Then we have (5)

so that we gain a factor of h 2 in our error estimate at the expense of about !n times as much work: we have to evaluate f at (n + 1) points instead of at 2 points. With a little more trouble (Problem 9.3) we can find 1- Q

(6)

= --b(b -

a)f"(c)h 2

for some c E [a, b]. The corresponding result for an n + I-point Lagrangian quadrature is got by integrating the relation (Chapter 8), f(x)-L,.(f,x)=

tn+l)({(x» n (n+l)! [!(x-x;)

to get n

1

II- 01 3. You should observe that in this case the solution tends to infinity as x ~ x(A) where x(A) depends on the value A of y'(O). In order to estimate x(A) it will be necessary to decrease h as x increases. 10.12. Discuss the differential equation y(O) = 0 = y(l)

y"= -Axy,

in the same way as (31) was handled in the text. 10.13. Discuss the solution of the equation l/J"(x) + (E - V(x»l/J(x)

=0

where E is a parameter and V(x) is given by V(x)

= -5,

Ixl a;t:O, Xo, Xl given. 10.18. Complete the following difference table: f(x)

X

1 771

91

2 3

48

36

5

32

4

-41

5 6 7 10.19. Is it likely that the following sequences are the values of a polynomial at equally spaced arguments? (a)

2,

3,

5,

7, 11, 13, 17, 19, 23, ...

(b)

2,

5,

4, 12, 6,

9, 23, 11, 27, 34, ...

(c) 41, 43, 47, 53, 61, 71, 83, 97, ...

10.20. Evaluate numerically the derivatives of Ai

X

at

X

= 1.4 given the

Chapter 10

138

following set of values: x

Aix

1.1 1.2 1.3 1.4 1.5 1.6 1.7

0.12004943 0.10612576 0.09347467 0.08203805 0.07174950 0.06253691 0.05432479

10.21. Integrate the differential equation y" = xy from x

= 1,

y

= 0.13529 242,

y' = -0.15914744

using the local Taylor series method: y(x ±h) = T(O)± T(l)+ T(2)± T(3) +...

where h2 x h3 T(n +2) = (n + l)(n +2) T(n)+ (n + l)(n +2) T(n -1). 10.22. (D. R. Hartree) Discuss the evaluation of

along the following lines. Express X(O), X'(O) in terms of f-functions. Show that satisfies the differential equation Obtain two solutions Yb Y2 to this for x

= - 2(0.1)8 where

Yl(O) = 1,

Yi(O) = 0,

Y2(0) = 0,

y~(O)

Evaluate X(x) for x

= 1.

= -2(1)8.

10.23. Discuss the solutions of the Emden equation y"+2x- 1 y'+y n =0

Difference Equations, Differentiation and Differential Equations

139

for various values of n, e.g., n = 1(0.5)5, which have y(O) = 1, y'(O) = O. In particular, find the first positive zeros of the solutions. 10.24. Study the solutions of y"+2x- 1 y'-y+yk =0,

which are finite at x

=0

and vanish at x

=

k> 1.

00.

10.25. Use the predictor-corrector method to solve the equation y' = x + y, y(O) = 1, finding in particular y(1).

APPENDIX

Bessel Functions

It is important to be able to carry out controlled computational experiments to evaluate computing procedures. For this purpose it is necessary to have reliable numerical data. The various tables of Bessel Functions, e.g.,

BAAS, Bessel Functions, I (1937), II (1952); The Airy Integral (1946), Cambridge Univ. Press. RSMT, Bessel Functions, III (1960), IV (1964), Cambridge Univ. Press. Harvard, Annals of Computation Lab., 2-14, 1945-1951, Harvard Univ. Press. NBS, Tables of Bessel functions of fractional order, I (1948), II (1949), Columbia Univ. Press. are ideal for this purpose, supplementing the less extensive tables and graphs in E. Jahnke, F. Erode, F. LOsch, Tables of higher functions, McGrawHill, 1960. M. Abramowitz-I. A. Stegun, Handbook of mathematical functions, NBS Applied Math. Series, 55, 1964. We therefore give an account of some aspects of the theory of Bessel Functions which are relevant in the present context. The definitive treatise is G. N. Watson, Bessel Functions, Cambridge Univ. Press, 1922-1941. We shall confine our attention largely to J"(x) - the Bessel function of the first kind of integral order n. Two definitions are: (1)

(2)

1 J"(x) = 277"

1 2

'"

cos (-x sin 0 + nO) dO = 77"1 00

J(X)=L "

.=0

( _

1)'

1'"

(X) "+ -

cos (nO - x sin O)dO 2.

r!(n + r)! 2

Our first objective is to establish the equivalence of (1), (2) and we begin by finding some properties of the J" 's. We indicate by subscripts on the numbers of our formulas the definitions on which they are based. Using the fact that cos A -cos B = 2 sin !(A + B) sin 4(B - A) we obtain

Appendix

142

from (1), with n ~ 1: I n - 1 (X)-J n + 1 (X)=';

f7T sin (-x sin 8+ n8) sin 8d8

= 2J~(x) assuming it permissible to differentiate (1) with respect to x, i.e., to interchange the operations d/dx and S... d8 on the right. We note that when n = 1 we have

r27T cos (8 - x sin 8) d8

1

J 1 (x) = 27T.lo

=-1 7T

i

0

7T

cos 8 cos (x sin 8) d8 +1 7T

1 7T

sin 8 sin (x sin 8) d8.

The first integral vanishes because we can write it as

if we change the variable from 8 to q,=7T-8 in 12 we see that 12 =-11 • Thus J 1 (x)

=1.. 7T

f7T sin

Jo

8 sin (x sin 8) d8.

Assuming that differentiation under the sign of integration is permissible we find

J~(x)

=.; r

-sin (-x sin 8)(-sin 8) d8

= -:

r

sin 8 sin (x sin 8) d8

and so we have

'. THE DIFFERENTIAL EQUATION

We first note that the series (2) is convergent for all x: the ratio of the r+ I-st term to the r-th is -x 2 /(4r(n + r)) and this tends to zero as r ~ 00 for

143

Bessel Functions

any x, n. We can therefore differentiate term by term to find ,

_ ~ (-1Y(n+2r) (~)n+2'-1 r!(n+r)!2 2 '

In(X)-.~o

"( )= f (-1Y(n+2r)(n+2r-l) (~)n+2'-2 Jn x t.. '( )'22 2 .~o r.n+r.

Consider the coefficient of x n+2• in x 2J~(x)+ xJ~(x) + (x 2 - n 2)Jn(x): it is (-l)'/[r!(n + r)!2 n+2.] times (n +2r)(n +2r-1)+(n +2r)-4r(n + r)-n 2

which is zero. This establishes (52)' It is easy to check that for general values of v

L~ (-1)' (X)2' -2 (2X)v .=or!f(r+l+v)

J (x) = -

(2')

v

also satisfies the equation (8)

Indeed, if v is not an integer, J_,,(x) also satisfies this equation and J", Lv form a pair of independent solutions to (8). We note from (2') that, for n=0,1,2, ... J

-n

(x)

= (-x)-n 2

~

(X)2. 2

(-1)'

.~o r!f(r+ 1- n)

and the coefficients of the terms with index 0, I, ... , n - I on the right vanish because of the behavior of the f(t) when t= I-n, 2-n, ... ,0. Hence, putting s = r - n, x)-n ~ (-I)' Ln(x) = ( 2 .~n r!f(r+ 1- n) x)-n ( = 2 =(-I)n

~

(_I)n+s

(X)2. 2

(x)2s+2n

s~of(s+l+n)f(s+l) 2 x)n ~ (-1)S (X)

(2

s~os!f(s+l+n) 2

2s

= (-I)nJn(x).

Therefore, in the case when v is an integer we have to seek elsewhere for another solution of (8). This leads to the introduction of the Bessel functions of the second kind - we shall not be concerned with these here.

144

2.

Appendix MIXED RECURRENCE RELATIONS

These are simple consequences of (2). We deal only with the first relation. The coefficient of x 2,,+2r-l on the right is 1 (-1)' '~'2r+2v=_I_. (-1)' '~'(r+v) 2'" r!f(r+l+v) 22r 2,,-1 r!(r+v)f(r+v) 2 2r which after cancelling the factor (r+ v), is exactly the coefficient of x 2,,+2r-l on the left. If we differentiate out the left hand sides of (9 2 ) and multiply across by x -" and - x" we obtain v

J~(x) +- J,,(x)

x

= J,,_I(X);

From these, by adding and subtracting, we obtain 2v - J,,(x) = J,,_I(X) + J"+I(X);

2J~(x) = J"-1 x) - J"+I(X).

x

We now show how to derive the first part of (11 2 ) on the basis of the original definition (1). [The second part of (11 2 ) is just (3 1),] Elementary trigonometry gives, if x "1= 0, cos «n + 1)8 - x sin 8) + cos «n -1)8 - x sin 8)

= 2 cos 8 cos (n8 -

x sin 8)

= 2n cos (n8 - x sin

x

8)-~ (n x

x cos 8) cos (n8 - x sin 8).

21

If we integrate across with respect to 8 between 0, 21T we find 2n 21TJn+l(X)+ 21TJn _ 1(X) = - 21TJn (X)--

x

x

2n

....

cos q; d8

where q; = n8 - x sin 8. The last integral vanishes and so

We are now able to identify (1), (2). We do this by checking the results first for n = 0, and then using the fact the recurrence relations are satisfied on the basis of (1), (2).

145

Bessel Functions

Since the series for cos t is convergent for all t we can expand the integrand in (1) as a power series and integrate using the fact that for all eve" r

((1/2)'".,

Jo

_(r-l)(r-3) (I).~ r(r-2) (2) 2

sm OdO -

to find

.!. r'" 'fT

Jo

. _.!. r'" (-I)'x z, sin cos (x sm 0) dO Jo ,~o (2r)! 00

z ,0

'fT

= ,~o 00

Z [(-I)'X ' (2r)!

1

.:;;:

dO

r'" sm. z, 0 dO] = ,~o (-1)' (X)Z' (r!f 2: .

Jo

00

This completes the identification for the case "= O. We now note that a special case of the second relation of (9 z) is I6(X) = -I1 (x). Combining this with (4 1) and the identification already established for " = 0, we establish identification for "= 1. To complete our program we use (3 1) and (l1 z) to proceed to the case " + 1 from " and "- 1. 3.

SoLUTION OF TIlE DIFFERENTIAL EQUATION

y" + hy = 0, y(O) = 0

We write I = I 1/3 (jA lIZ X3/Z) and shall show that y = x 1/2 I is the solution. Differentiating and substituting we find y" + Axy

= (AX3/Z_!X-3/Z)I +~A 1/2I' + AX 3/ZJ".

By definition I 1/3 (X) satisfies XZI~/3(x) + xJ~/3(X) +(X Z-~)Il/3(X) = O.

We now replace x by jA l/ZX3/Z in this equation and solve for AX 3/ZJ" getting AX 3/ZJ" = -~A lIZI' - (Ax 3/2 - !x- 3/2)I

which when substituted in the expression for y" + Axy cancels out everything. 4.

SoLUTION OF TIlE DIFFERENTIAL EQUATION

y' = XZ- yZ, yeO) = 1

It has been found convenient to introduce the modified Bessel function. For " integral (12)

satisfies the differential equation (13)

146

Appendix

The properties of the I's can be obtained by translation of those of the ]'s just as we obtain properties of the hyperbolic functions from those of the trigonometric ones, or by a parallel development. Some care has to be taken in the determination of the multiplier on the right in (12) when we deal with cases when n is not an integer. We show that the complete solution of the differential equation u"x 2 u = 0 is (14) u = x'/2(c,11/i!x 2) + c2L1/i!x 2)). Differentiating we find, omitting the argument !x 2 of the I's, u'=!x-1/2(c,1'/4+ ...)+X1/2. x(c,1;/4+ ...)

and u" = -lx-'/2(c,11/4 + ... )+h1/2(c,1~/4+ ...) + ~x '/2(c,1~/4 + ...) + x5/2(c,1~/4 + ...) so that u"- x 2u = -~x-3/2[(1 +4x 4)(c,1'/4 + ...) - 8X2(cd~/4 + ...)-4x 4(c(1 1/ 4 + ...)].

Now 11/ 4 (t) satisfies t 2 y" + ty' -

(-h + t2 )y = 0

and replacing t by !x 2 , this becomes 4

2

4

x x X ) ( 1 41;'/4+21~/4- 16+4 1'/4=0 The same equation is satisfied. by L 1/4 and so we see that u"- x 2 u = O. We have already noted that when v is not an integer, ] ±v are independent solutions of (8) and l±v are consequently independent solutions of (13). This establishes our assertion about (14). The equation y' = x 2- y2 is of the Riccati type and can be transformed to the linear equation u"-x 2u =0 by the substitution y = u'/u. From what we have just seen the solution is 2 C 1(x 1/2/1/4 (I2 X2))' + C 2(x 1/ 1-1/4(I2 X2)' Y= CIX 1/211/4(1'Ix 2) + C2 X1/21-1/4(2) X where the c's will be determined by the initial condition y(O) = 1. We now use the following transforms of (7 2):

147

Bessel Functions

in the case when n = ±! to get y=

c tX3/2 I _3/4(~X2) + C2X3/2 13/4(~x2) CI X 1/2 I t/4(~X2) + C2 X1/2 L1/4(~x2r

For x near zero Iv(x) - (x/2t/f(v + 1) so that y - 2c\f(3/4)/C2f O/4).

Hence (CI/C2) = fO/4)/f(3/4) and the solution is fWI _3/4GX 2) + 2f(~)I 3/4GX 2) Y= x f(!)I I/Ax 2) + 2f(~)I \/4GX 2) . From the NBS tables of Bessel functions of fractional order, II we find y

0) = (3.625609908)(0.9800767696)+2(1.22541 6702)(0.3985850517)

(3.625609908)(0.8196759660)+2(1.225416702)(1.2519701940)

= 1.848449429 = 0.750015703. 2.464547638

5.

RELATIONS OF JACOBI AND HANSEN

We shall now establish the fact that which was mentioned in Chapter 10. We do this on the basis of the series definition (2). For s~m the coefficient of OZ)2s in J 2m (z) is (_1)m-s[((s-m)!) «s+m)!)r\. Hence the coefficient of (~zfS in J o(z)+2I:=t J 2m (z) is 1(-1)' [ s!s!

2 2 ] + .. .+(-1)s_(s-1)!(s+1)! 0!(2s)! (-1)' [(2S) = (2s)! s -

2 (2S) 0 . s _ 1 + ... + (- 1)S 2 (2S)]

The expression [... J can be readily identified with the expansion of (-1)S[1-1fS by replacing, for r = 1,2, ... , s,

2(s-r 2S) b Y

(2S) (2S) s-r + s+r .

Hence all the coefficients in Jo(z) + 2 I:':. ~ \ J 2m (z) vanish, except the first, which is 1. This is what we want, but so far our work has been formal.

148

Appendix

To justify our manipulation, which is essentially the transformation of a sum by rows of a double series into one by columns, it will be enough to establish absolute convergence of the double series. This is easy, for changing signs in the argument of the preceding paragraph shows that the double series is dominated by 1 Iz1 L (!Izlf'(1 +1)2. = L-=cosh Izl b ~ B 2 • The third integral is estimated trivially as

113 ~!x3b-2. 1

153

Bessel Functions

Hence and so the integral I(b) is convergent and indeed uniformly convergent with respect to x in any interval [-X, X]. It follows, from the theorem cited, that Ai'(x) = -7T- 1 i~ t sin Gt 3+ xt) dt. If we differentiate formally again we are led to

I~ t 2 sin (1 t3 + xt) dt: but this integrand does not converge. This follows from the fact stated above (p. 151), or it can be shown directly as follows. If we change the variable from t to Y=1t3+xt then dy=(t 2 +x)dt and since for large t, Y-1t3 the integral becomes approximately

f~ sin Ydy which is divergent. To get the result we require we follow Stolz and Gibson in using a technique of de la Vallee Poussin. We describe it formally in a general case and then deal with the special case rigorously. Suppose that F(x) =

i~ f(x, t) dt

and that fx(x, t) is continuous. Then f(X, t)-f(O, t) =

i

X

fx(x, t) dx.

[iT iX = ;~ [iX iT

If we integrate across with respect to t between 0, T and let T _

(23)

F(X)- F(O) = ;~

dt

fx(x, t) dX]

dx

fx(x, t) dt]

The easy case is when we can interchange lim T F(X)- F(O) =

i

X

dx

i~ fx(x, t) dt

and

00

we get

J... dx to get

154

Appendix

which when differentiated gives F'(x) =

r

fx(x, t) dt.

However we are interested in the case when this inversion is not possible and we proceed as follows. Break up the inner integral in (23) into

IT

fx(x, t) dt = cp(x, T)+

L T

l{J(x, t) dt

where lim

T

while

.lofX cp(t-t, T) dt-t = 0

IT

l{J(x, t) dt

is uniformly convergent with respect to x so that

;~ LX dx L l{J(x, t) dt = LX dx L~l{J(x, t) dt. T

In these circumstances F(X) - F(O) =

LX dx L~ l{J(x, t) dt

and, as before F'(x) = L~ l{J(x, t) dt.

In the special case we have f(x, t) = t sin Gt 3+ xt) and we want to find appropriate cp, l{J so that

IT

t 2 cos Gt 3+ xt) dt = cp(x, T) +

L l{J(x, t) dt. T

We write so that

IT

L + L (~t3 + T

t 2 cos (~t3 + xt) dt =

T

- X

(t 2

cos

x) cos (~t3 + xt) dt xt) dt = sin (~T3 + xT) -

L T

X

cos (~t3 + xt) dt.

155

Bessel Functions

Now as

T~oo,

f°

x

•

(IT 3

SIn:3

+ IL

T) d

3

IL =

cos (!T )-cos (!T

3

+ xT) 0 ~

T

and the uniform convergence of the integral

f

T

r

o

COS (!t 3 +xt) dt

r

has already been established. We are therefore able to conclude that

d~ [-1T-

I

t sin(~t3+xt) dt]

= +1T- I X

3 cos (1t +xt) dt

i.e.

Ai"(x) = xAi(x) as required. REFERENCES G. H. HARDY, On certain definite integrals considered by Airy and Stokes, Quart. J. Math. 41 (1910), 226-240 = Coil. Papers 4 (1969), 460-474. O. STOLZ, Grundzuge der Differential- und Integral-Rechnung, B3 (1899), Leipzig. G. A. GIBSON, Advanced Calculus, Macmillan, London 1931, 439, 452. C. J. de la VALLEE POUSSIN, Etude des integrales limites infinies pour lesquelles la function sous Ie signe est continue, Ann. Soc. Scient. de Bruxelles 168 (1892), 150-180. E. C. TITCHMARSH, Theory of functions, 1939. Oxford.

a

9. REpRESENTATION OF Ai(x) AS A POWER SERIES Let Ai(O)

1 r~ =;:.10 cos (~t3) dt =ao,

Next, differentiating n times the equation y" - xy

=

(24) and then putting x

=0

we get y(n+21(0)_ ny(n-Il(O) = O.

0 we get:

156

Appendix

We therefore find a Maclaurin series for Ai(x) of the form

(1+;,X +...) Ai(x)= (2 4 ) -al x+ 4! x + ... 3

ao

where

ao = T2/3/f(2/3) = 0.35503

al = T

l/3

/f(l/3) = 0.25882'

A real variable derivation of the expressions for a o and a l follows. Write

where n > 0, a> 0, b > O. Put a = r cos 6, b = r sin 6, rx to get u(6)= u = v(6)= v

=

r

ern

= u, srn = v,

fO e-YCOS 9yn-l cos (y sin 6)dy,

r

e-YCOS9yn-l sin (y sin 6) dy.

We differentiate formally to get

~~ =

= y,

e-YCOS 9yn sin 6 cos (y sin 6) dy

-{=

e-YCOS 9yn cos 6 sin (y sin6) dy.

Both these integrals are uniformly convergent with respect to 6 in -Tr/2 < - 60 s 6 s 60 < Tr/2 because they are dominated by

f= Jo e

-cos90y

n _ f(n + 1) y dy - (cos 6 )n+!'

o

Hence the differentiation is legitimate. We can write du = d6

1= yn ~dy {e-YCOS 9sin (y sin 6)} dy

so that, integrating by parts, du = -n f= yn-l . e-YCOS 9 sin(y sin 6) dy d6 Jo

In the same way

dv

-=nu

d6

= -nv.

157

Bessel Functions

and, combining, we get

so that u(O) = A cos nO + B sin nO.

When 0=0 v(O) = 0,

so that v(O) = f(n) sin nO.

= f(n) cos nO,

u(O)

It is clear that c, s are continuous functions of a at a = 0 and we can deduce

from

f

=

o

e

-

ax

x"

-1

cos b d f(n) cos X x=-nO sin r" sin

that =

f

cos x dx x 1 -"

o

=

n'7T

f(n)cosT

_

b"'

b

f(n) sin n '7T

o x 1- "

b"'

=

f 10.

b

.

Sm

REpRESENTATION OF

x dx =

Ai(x)

_z

0 O. Then x" is clearly positive for all nand Xn+l-x"

= 2x,,(N -x~)/(3x~+N)~O

according as x" ~ IN. Also x,,+l -IN= (x" -Jr:tP/(3x 2 + N) ~O

according as x" ~ IN. Consequently if 0 < xoJN we have x" i and x,,:5 IN while if IN < Xo we have x" ~ and x" ~ IN. Convergence takes place for any xo> 0 and the limit satisfies

(31 2 + N)l = x 3 + 3Nl

i.e.,

P = Nl i.e., 1= 0, ±IN

and so I=JN. We have x,,+1-JN=(x,,-,IN'P/(3x~+N) so that convergence is cubic. The behavior of this sequence can be illustrated graphically in the usual way. Observe that if y=(x 3 +3Nx)/(3x 2 +N) then y-3x as x-O and y - xl3 as x~±oo. Also y' = 3[(x 2 - N)/(3x 2 + NW~O for all x and y' = 0 for x = ±N1/2. y 1

Y'x

2

3 x

3.16. Solution This is not entirely trivial. See e.g., J. L. Blue, ACM Trans. Math. Software 4 (1978), 15-23.

181

Chapter 4

Chapter 4 4.1. Solution -0.882027570. 4.2. Solution 0.67434714 ±i 1.1978487. the remaining root is -2. 4.3. Solution

If q(x)=qox n- 2 + ... +qn-3X+qn-2 and r(x)=qn-IX+qn then

i

= 0,1,2, ... , n -2.

4.4. Solution

We take the case of a double root. Then f'(~) = 0 but f'(~)"/; O. We find Xn+I-~=~(Xn -~)+O(Xn -~f

and we have linear convergence, not quadratic. We can, however, restore the quadratic convergence by changing the Newton formula to

in the case of a double root and to in the case of a root of multiplicity r.

4.5. Solution

We show that the conditions (8), (9), (10), (11) are satisfied. First

(8)

H(-I)=28>0

and H(-I-~8)=-6-~83 0 such that for every N (however large) there is an n. > N such that M n• > E. Choose N = no(E). Then I'n(x)! < E for n > N and so M n • :5 E, a contradiction.

184

Solutions to Selected Problems

5.5. Solution The solutions are 1; x; x 2 +[x(1- x)]jn and x 3 +[3x 2 (1-x)]jn + [x(1- x)(2- x)]/n 2 • We shall establish the result in the cases k = 1 and k = 2. In the case k = 1 we have

L (;)x r(1- x)n-r (;) = L

e=:

r r )x (1- xt-

=xL (n -1)x r- 1 (1_ x)(n-l)-(r-l) r-1 = x[x + (1- x)]n-l

=x.

The general term in the case k = 2 which is xr(1- xt-r ~2 ( n) r n can be split into two by cancelling out an r and an n as in the case k = 1 and then writing r = (r - 1) + 1. We find r 1 L (;) x (1- xt-

(;Y =

n: 1 L

e=~) x r(1-xt- r

(n -1)

1 +-L x r(1-xt- r n r-1 n-1 2 L (n-2) x r- 2 (1-xt- r =--x

n

r-2

x (n -1) x

+- L n

r-1

r

-

1

(1- xt- r

n-1 x =--x 2 +n n x(1-x) n

= X +--'---'2

as stated.

5.6. Solution ql(X) is the Bernstein polynomial, adjusted to the interval [-1, 1] and obtained in Problem 5.8 below. q2(X) is (approximately) the polynomial of best approximation, obtained by Remez. Q3(X) is the truncated Legendre expansion. qix) is obtained by truncating the Chebyshev expansion obtained in Problem 5.11 below.

185

Chapter 5

In the evaluation of lie; (x )\1 we can confine our attention to 0 ~ x ~ 1. The value of lIe 1 (x)1I is shown to be 0.375 in Problem 5.8. It is less easy to find the other norms exactly and they can be estimated as max lei(x)l,

x =0(0.01)1

which gives lIe2(x)1I = 0.067621, Remez gives lIe2(x)II+0.06762 1 and this is assumed positively at x = ±1, x=?±0.28 and negatively at 0 and at x=?±0.78 exemplifying the equalripple behavior characteristic of the polynomial of best approximation. 5.7. Solution

a) S = J2!1t{ -! + (1- ceil! + c 2e 2i l! - ...)}

=J2!1t{ -! +(1 + ceil!)-l} =J2H-k 2}{1 +2c cos 0+C 2 t

1

2 3 + cos 0 1 =1+x' b)

en(X)=L~x -7Tn(X)} _ -

2 3 + cos 0

r=

-,,2

{!(1- c 2 ) -

(-I)c"[cos nO + c cos (n -1)0] l+c 2 +2ccosO

no}

+(-1)" c" cos (1- c 2 ) (-I)"J2c" C"-I cos nO 2c(3+cos 0) [cos nO +c cos (n -1)0]-(-1)" 4 (-I)"c"-1

J2

4 2(3+cosO)

[4 cos nO +4c cos (n -1)0 - 3J2 cos nO - J2 cos 0 cos nO]

= (-I)"-lc" X{COS (n + 1)0 + 2c cos nO + c 2 cos (n -1)0}. 4 1 +2c cos 0+c 2

186

Solutions to Selected Problems

The result we need can be established by elementary trigonometry. For it is easy to verify that cos (n + 1)8 + 2c cos nO + C 2 cos (n -1)8 1+2ccosO+c 2 cos (nO+O, =(-l)"2"(2n+l)!'

=(-1)"2" n!(2r-l)!! (2n+2r+l)!!'

These results are easily established using the reduction formulas for l(m, n) =

With this notation we have (1)

1

1/2",

sin m x cos" x dx.

(m + n)l(m, n) = (m -1)I(m - 2, n) = (n -l)I(m, n - 2), l(m, n) = l(n, m),

1(0, 1) =

1

",/2

cos xdx = 1.

r=O.

194

Solutions to Selected Problems

We establish (1) by integrating by parts: we find I(m, n)

=.lor"'/2 (sin

=[

m

x cos x) COS n- 1 X dx

",/2 l",/2 sin - (n -1) cosn-

Sin m + 1 x ] 1 cos n- 1 X + m+ 0

so that (m + l)I(m, n) = (n -1)

m+l

m+1

2

X

sin x dx

["'/2 sin 2x COSn-2X dx m

+

= (n -1) ["'/2 (1-cos 2x) sin

m

x COSn-

2 dx X

=(n-1)I(m, n-2)-(n-1)I(m, n).

Hence (m + n)I(m, n) = (n -l)I(m, n -2).

We therefore obtain, if we start the transform from the beginning,

Observe that this series is much more rapidly convergent than the Gregory series. In fact

We can check that the new series actually has the proper sum, e.g., as follows. Consider the Maclaurin series for f(x)

= (arcsin x)/·J!- x 2 •

This can be found by differentiating to get

so that (1- x 2 )f' - xf -1 = O. If we differentiate this n times by Leibniz' theorem we get

195

Chapter 6

If we put x =0 then

This gives 22 22 • 42 3+ _ _ 5+ f( x ) -- x +3! x 3!5! x .... Putting x =

7r/.J2 we

find

(I 1 [ 1 1.2 ] 4"7r/ ""2= J2 1+ 3+ 3.5 + ... which is the result required.

6.11. Solution Vo- VI

+ V2 +... = (1- E)- lv o = (1/2)(1- m)-I vo = (1/2)[Vo + Mvo + ~vo + ...].

6.12. Solution We take so=O, s"=I:=d-l)"-Ia" for n~1 so that (-1)"-1a,.= S"-S"_I for n~1. By definition S"=I:=12-V(-I)"-IA"-lal. We use induction to prove that S" = 2-" I~=o (~)sr' This is trivial for n = 1 since SI = (1/2)a l , SI = a l . Assume the result established for a particular value, n. Then S"+I = S" + 2-("+\)(-1)" A"a l

"i (-1)"-1 ( v-In ) a" from Problem 6.7 1 = 2-("+\) [2 i (n) s" + 1 S,,_I) x ( : 1)] v ,,=1 l

= S" +2-("+\)

,,=1

,,=0

(s" -

V

by induction hypothesis

= 2-("+\)

1 (n +v 1) s". 1

,,=0

This completes the induction proof. The fact that S" - I implies Sn - I now follows from a general theorem of Toeplitz. [See e.g., K. Knopp, Theory of Infinite Series, p. 72]. We now

196

Solutions to Selected Problems

give a direct proof in the present special case. Because (1 + l)n = 2 n we have

(~) +

m+ ... + (;:) =

and therefore it will be sufficient to establish the result when {sn} is a null sequence. For convenience put a...r = 2-n(~) so that 1 ~ a n •r > 0, L~=o a n •r = 1. Notice that for r fixed a n •r is a polynomial in n (of degree r) divided by an exponential 2n = enlog2 - hence a n •r ~ 0 as n ~ 00. [Compare Problem 2.8]. Given any e > 0, we can find no = no( e) such that Is"I 0, we have [" e-X1t n dt = x- n- 1 Hence G(x)=

roo e-lxtl(xt)n d(xt) = n!/x n+

1

•

2 (2n-2)'} kroo e- xl[l+t 2]-ldt= {IIx'- x;+···+(_l)n-l X2n 1

+(-1)"

1 '

1 00

e-Xlt2n[l+t2]-1 dt

so that

Hence X2n - 1 IG(x)-{... }I = (2n)!/x 2 ~ 0

as

x ~OO.

Also x2nIG(x)-{"'}+0'X-2nl=(2n)!/x~0

The last two relations show that we indeed have

1 2' x x

4' x

G(x)---~+~. ..

according to the strict letter of our definition.

as

x~oo.

206

Solutions to Selected Problems

For x = 5, two terms give G(x) to within 10- 2 ; for x = 10 five terms give G(x) to within 4x 10-5 and for x = 15 seven terms give G(x) to within 2xl0- 7 •

7.9. Solution The only trouble is finding the error in the binomial expansion and from it to show the true asymptotic character of the series derived formally. Integrating the relation d - (1 + tt

dt

from 0 to x we get

= v(1 + tt- 1

r r r r

(l+x)"-I=v

(l+tt-1dt

=v

(l + x - T)"-l dT,

the last line being obtained by changing the variable from t to T Thus we have (l+xt= l+v

Integrating by parts gives

=x -

t.

(l+X-t)"-l dt.

(l+x)"=I+vx+v(v-l)

(l+x-t)"-2 t dt.

Repeating this operation we find (l + x)" ={1 + vx + v(v-l)x 2/2! +... + v(v-l) . .. (v- n + l)x n/n!}+ rn+l(x)

where

Assuming x >0 it is clear that when 0:5 t:5x when 1 + x ~(1 + x - t)~ 1 and so, provided n> v -1, the first factor in the integrand is less than 1 and j

For n = 1, VI =

1

det [ 1

(:x; -Xj).

xo] =

Xl

Xl - Xo·

Suppose we have established the result for n = r. Consider

213

Chapter 8

and expand it in terms of minors of the first row. Clearly V'+l will be a polynomial of degree at most r+ 1 in x. Also V'+l will vanish for x = Xl> X2, •.• ,X,+l so that V'+l

= k(x -

xl)(x - X2) •.. (x - X,+l),

where k is a constant. Now k is the coefficient of X,+l which is also evidently ••• , ~). The induction hypothesis gives

(-I)'+lV(X l , X 2 ,

so that (-1)'+1

= (x -

Xl)(X - X2) ... (x - X'+l)

n (Xj -

Xi)'

r2i>j

and so V(X o, Xl' ... ,X,+l) =

n

r+l2::i>j

(Xj - Xi)'

This completes the proof. It is clear that 1, X, .•• ,x n form a basis for 'Vn and since these are independent (by an application of the Fundamental Theorem of Algebra), the space has dimension n + 1. Since, for r = 0,1, ... ,n, L. = L~=o x;~(x) is a polynomial of degree at most n, coinciding with X' at Xo, Xl> ... , Xn it follows that L. must be identical with X'. In other words 1

1

1

X

n

X

Xo x7

Since the Vandermonde matrix is non-singular, it follows that ,o(x), 'l(x), .. . , In(x) form a basis for 'Vn • 8.14. Solution This is a table of 10 10 Jo(5.495 + 10- 3 X). To the zero

iO.2 = 5.52007 81103 of Jo(x) corresponds a zero of f(x) at X

= 2.50781 103.

214

Solutions to Selected Problems

There is a zero of f(x) between 2 and 3. Using all six points for Lagrangian interpolation to subtabulate indicates a zero between 2.5 and 2.6 since f(2.5) = - 2 65784 f(2.6)

= +3136604.

Estimating the position of the zero by linear interpolation gives 2.5078. We therefore subtabulate again getting f(2.507) = f(2.508) = f(2.509) =

and observe that the second difference is . Linear (inverse) interpolation is permissible and we find the value given.

8.15. Solution For discussions of wide generalizations of parts of this problem see, e.g. P. W. Gaffney, J. Inst. Math. Appl. 21 (1978), 211-226. C. A. Micchelli, T. J. Rivlin, S. Winograd, Numer. Math. 26 (1978), 191-200. For instance, if f(O) = 0, f(l) = 1 and 1f'(x)l:s 2 in [0,'1] then the graph of f(x) must be within the left hand parallelogram. If f(O) = 0, f(1) = 1 and 1f"(x)!:S 2 in [0, 1] the graph of f(x) must lie within the lens shaped region on the right.

y-2x-x 2

215

Chapter 8

8.16. Solution We want to show that there is a cubic H(x) = a + bx + cx 2 + dx 3 such that i =0,1,

H(x;) = fi'

(1)

n,

where Xo ¥- XI and fo, fl' f~ are arbitrary. The relations (1) when written out in full give a set of four linear equations for the unknowns a, b, c, d. The determinant of this system is

1 Xo

h=det

o [

1

o

1

Xl

1

x~

xci

2xI

3xf

2xo 3x~ xf xi

J

.

We evaluate h by row operations as follows. Take row 3 -row 1 and divide the third row through by XI - xo. We find

Taking row 2 -row l and row 3 -row, we find, on dividing through the second and third rows by (x I - xo),

Hence the system is non-singular and a, b, c, d can be found. For information on the use of Hermite interpolation see, e.g., H. E. Salzer, J. Res. Nat. Bur. Standards 52 (1956), 211-216. 8.17. Solution From Problem 8.16, Solution it follows that the second derivatives at X o in the left and right panels are, if h ~ = X I - X lh -It _ = Xl) - X-I'

and

Equating these gives a linear equation for f() with coefficient 4h =..1 + 4h:;: I ,e O. Hence is determined uniquely.

n

216

Solutions to Selected Problems

In the general case, when, fo, fI, ... ,fn and 16, f~ are given, quantities fl, ... ,f~-I can be determined uniquely so that all the abutting cubics fit smoothly, because the system of equations is non-singular, having a dominant diagonal. In the special case f& = 0 and the spline is given by

-2x 3 - 3x 2 + 1 in [-1,0] 2x 3 -3x 2 + 1 in [0,1]. Observe that there is a jump in the third derivative from -12 to 12 at x = O. These results can also be obtained by using the results of Problem 8.12.

8.18. Solution For simplicity suppose the three points are equally spaced and, without loss, take them to be Xl = -1, Xo = 0, Xl = 1. Then we have fo = a, f±I = a±b+c so that b=!(fI-f-I) and c=!(f-I-2fo+fI)' It is clear that i= -b/(2c) and 1= a - b 2 /(4c). For instance, given f-I = -0.3992, fo = -0.4026, fl = -0.4018 we estimate i = (13/42) = 0.3095,

1=0.4028.

For developments of this method see e.g., H. E. Salzer, Formulas for finding the argument for which a function has a given derivative, MfAC 5 (1951), 213-215.

8.19. Solution This is Newton's interpolation formula. It is, by uniqueness, necessarily a rearrangement of the Lagrangian expression. It can be obtained formally by writing f(p) = Ehf(O) = (1 + Atf(O)

and truncating the binomial expansion: if f is a polynomial of degree n, then 0= An+lf = An+lf = .... To establish the formula we can use induction and the basic recurrence relation between binomial coefficients:

The quartic is

so that q(2.5) = 2884.

Chapter 9

217

The difference table for q(x) is: 0

789

1

1356

2

2268

3

3648

4

5819

5 9304

567

345

912

468

1380

791

2171

1314

3485

123 323 523

200 200

The constant term in the quartic is clearly q(O) = 789 and the leading term must be (200/24)x 4 to give a constant fourth difference of 200. If we subtract off these terms and divide through by x we are left with a quadratic whose coefficients are those of the three middle terms of the quartic. The Newton interpolation formula is: {(2.5) = 789 + (2.5 x 567) + (2.5 x 1.5)345/2! +(2.5x 1.5xO.5)123/3! +(2.5 x 1.5 xO.5 x (-0.5))200/4! Truncating this we get successively 789

2206.5,

2853.375,

2891.8125,

2884.

Note that 2206.5 is got by linear extrapolation from q(O), q(l); if we interpolate between q(2), q(3) we get 2958.

Chapter 9 9.1. Solution

Write down the fundamental polynomials and integrate them. For instance l

()_x(x-1)(x-2)_

-1 X

-

(-1)(-2)(-3)

.[

--6

3

2

x -3x +2x],

and

J

2

-1

Ll(X)dx=~,

J 2

-1

'o(X) dx

=~,

J 2

-I

l.(x)dx=t

J 2

-I

'2(x)dx=~.

218

Solutions to Selected Problems

This gives the so called '~ Rule' Q =U(-1)+U(0)+U(1)+U(2).

The fact that the sum of the weights is 3, which is the integral of f(x) == 1 between -1,2, is a check. We shall show that if f E C 4 [-1,2] then

L~

f(x)

dx-[U(-1)+U(0)+U(1)+U(2)]=-for)(~)

where - 1 :::; ~:::; 2. In order that our computations have more symmetry we shall prove, equivalently, E(h)=

L::

f(x) dx-

34h[f(-3h)+3f(-h)+3f(h)+f(3h)]=-~h5r4)(~)

where -3h:::; ~:::;3h and we assume f E C 4 [-3h, 3h]. We differentiate E(h) four times with respect to h obtaining:

i.e.,

fl + f3]+ h[f~3 + f~1 - f~ - f~]; f~1 + f~ - g]- h[3f~3 + f~1 + f~ + 3f~];

~E'(h) = [f-3 - f-l -

~E"(h) = -2[f~-3 -

~E"'(h) = 3[f~3 - f~1 - f~ + f~] + h[9f~~ + f~'1 ~E(4)(h)

= 4[f~ll -

fi' - 9f;]; f~]-·27h[f~j + }\4)]_ h[f~l + fI4)].

Hence, since f4) is continuous, we have ~E(4)( h) =

-8hr)( ~ 1) - 54hr)( ~2) - 2hr)( ~3) = - 64h(4)( ~4)

where -h:::;~I:::;h, -3h:::;~2:::;3h, -h:::;~3:::;h and so -3h:::;~4$3h. We also note that E(O) = E'(O) = E"(O) = E"'(O) = O. Hence, integrating the relation we get E"'(h) = -144

r

hf4)(~4(t)) dt = -72r)(~5)h2;

integrating again E"(h) = -24h3r)(~6)' E'(h) = -6h4r)(~7)'

Chapter 9

219

and, finally, where -3h ~~~3h. 9.2. Solution We may assume the interpolating quadratic to be of the form q(x) = f(0)+ax+bx 2 • Since q(±l)=f(±l) we have two linear equations for a,b. Actually we only need b since Q=

1:

q(x) dx = 4f(0) + (Jf)b =1[2f(-1)-f(0)+2f(1)].

We note that this quadrature is of "open" tyPe; it does not involve the values at the end points and can therefore be used as a "predictor" in the solution of differential equations e.g., in Milne's method. We also note that it is to be expected that the error incurred in open formulas is larger than that in closed ones. It does not seem possible to obtain an error estimate in this case by the method used in the previous problem. We use an essentially general method based on a modification of Steffensen's classical account given by D. R. Hayes and L. Rubin (Amer. Math. Monthly, 77 (1970), 1065-1072). Another general method for this type of problem is due to Peano (d. A. Ghizzetti and A. Ossicini, Quadrature Formulae, Birkhauser and Academic Press, 1970); see also B. Wendroff, Theoretical Numerical Analysis, Academic Press. 1966). Let L(f, x) denote the Lagrangian polynomial based on the nodes -h, 0, h and let '7T(x) = (x + h)x(x - h). Then we put (1)

f(x) - L(f, x)

= '7T(x)R(x)

and get (2)

1- Q =

1 2k

'7T(x)R(x) dx.

-2k

This defines R(x) except at the nodes where we define it by continuity. With this convention it is easy to verify that R(x) is continuously differentiable. If we write l(x) =

fX

1 2k

'7T(t) dt

then l(x)=Hx 2 (x 2 -2h 2 )-Sh 4 ] and 1(±2h)=0. Hence, integrating (2) by

220

Solutions to Selected Problems

parts, we find 1- Q

= [1(t)R(t)]:~k =

_1

2k

-2k

1

2k

-2k

l(t)R'(t) dt

l(t)R'(t) dt.

Now as l(x)sO in (-2h, 2h) we may apply the Mean Value Theorem to get

I-Q=-R'(~)L::

l(t)dt,

-2hs~s2h,

=-R'(~)x-1:~h5. We shall now show that R'(~)=r)(C)/24 where -2hsCs2h, which gives 1- Q

(3)

= Mh5r)(~).

Take a fixed x* in [-2h, 2h] and consider 'l;(x) = f(x)- L(f, x)-1T(x)[R(x*) +(x - x*)R'(x*)].

Since L, 1T are of degree 3 at most we have 'l;(4)(X) = r)(x)-4!R'(x*).

(4)

We shall show that there is a C such that 'l;(4)(C) = O. We observe that (5)

'l;'(x*) = 0

for any x*,

'l;"(x*) = 0

if x* is a node.

The first of these results follows from 'l;'(x) = ['(x) - L'(x) -1T(x)R'(x*) -1T'(x)[R(x*) +(x - x*)R'(x*)]

using the fact that ['(x)- L'(X)-1T(X)R'(x) = 1T'(x)R(x) which comes by differentiating (1). The second of these follows because differentiating (1) again we get 'l;"(x) = rex) -.L"(x) - 21T'(x)R'(x*)-1T"(X)[R(x*) + (x - x*)R'(x*)]

and using f"(x)- L"(x)-21T'(x)R(x) = 1T"(x)R(x) we find ~"(x*)

= 1T(x*)R"(x*)

which vanishes if x* is a node. When x* is in general position, 'l;(x) has (in the special case under discussion) 4 zeros: x* and the three nodes. Hence 'l;'(x) has three zeros by Rolle's Theorem, and an additional distinct one at x* by (5). Hence 'l;"(x) has three zeros.

Chapter 9

221

When x* is a node, ~(x) has three zeros and ~'(x) has two zeros by Rolle's Theorem and an additional distinct one by (5). Hence ~"(x) has two zeros by Rolle's Theorem and an additional distinct one at x* by the second part of (5). Again ~"(x) has three zeros. In both cases, by Rolle's Theorem, ~(4)(X) has a zero, say at ~. It then follows from (4) that

r)(~) = R'(x*) 24

where

~

depends on x*, but x* was arbitrary. This completes the proof of (3).

9.3. Solution Assume that m:5f'(x):5M in [a, b]. By the error estimate for linear interpolation applied to f(x) in the subinterval [a,., a,.+l] where a r = a + rh, h = (b - a)/n, we have f(x) - L(x) = (x - a,.)(x - a,.+1)f'(cr )/2!

where a,. :5cr :5a,.+1' This is true for r = 0,1,2, ... , n -1. Since for a,.:5x:5 a,.+l we have we can integrate these inequalities between a,., a,.+l to get !Mf'+1 (x - a,.)(x - a,.+l) dX:5 f'+1 [f(x) - L(x)] dx a,.

a,.

:5!mf'+1 (x-a,.)(X-a,.+l) dx a,.

which, after a little algebra, reduces to -Mh 3/12:5 f'+' [f(x)- L(x)] dx :5-mh 3/12. a,.

We now sum the last inequalities with respect to r and get m:5 )2 nh3

I [L(x)-f(x)]dx:5M. b

a

Now for a fixed h, the middle term above is a constant between m and M; since f'(x) is continuous in [a, b] and bounded there by m, M, it must assume this value (at least once), say at c. Hence

I a

b

3 [L(x) - f(x)] dx = T- 1= nh f'(c) = (b - a)3f'(c) 12 12n 2 '

222

Solutions to Selected Problems

9.4. Solution

We give two rather similar proofs. (1) We verify by integration by parts, that

f

uv"'dx

= uv"-u'v'+u"v-

f

ulllvdx

r

if u, veach have continuous third derivatives. Taking u.= ~x(l- xf, v = f(x)+f(-x) the above relation gives

fl

f(x) dx

= l[f(-l)+4f(0) +f(l)]-

uv", dx.

Since 1f4)(t)1 $ M4 and since, from the Mean Value Theorem, VIII

= f"'(x)- f"'(- x) = 2xf4l(~),

we have lEI $ 2M4

t:

r

x 2(1- X)2 dx/6 = MJ90.

(2) Let Lz{x) be a quadratic interpolating f(x) at -h, 0, h. Then repeated integration by parts gives for F(x) = f(x) - L 2(x), since F(~h) = F(O) = 0, (x + h)3(3x - h)P4)(X) dx +

f

(x - h)(3x + h)3p4l(X) dx

= 72

f:

F(x) dx.

Now, L 2(x) being a quadratic, p4)(X) = f4l(X) and so IP4l(X)1 $ M 4. Hence 72 IEI$2M4

f

(h-X)3(3x +h) dx =4M4h 5 /5.

This gives the result required. For a.derivation of this result under weaker conditions see Anon, Amer. Math. Monthly 76 (1969), 929-930. 9.5. Solution

We have M=(4/3)[2f(-1)-f(O)+2f(1)] (2/3)[f(-2) +4f(O) +f(2)] so that S- M

and

the

adjusted

S=

= (2/3)[f(-2)-4f(-1) + 6f(O)-4f(l) + f(2)]

= (2/3)f14f(-2)

which is bounded by (2/3)M4. 9.6. Solution It would be enough to do this in the two-point case but the characteris-

tic pattern 1 424 1 first turns up in the 3 point case. We have, when

Chapter 9

223

b-a = 1,

T-ol) = U(O) + ym + U(l) T-02l =U(O)+!fw+um+uw+U(l) so that

=

i2[f(O)+4fW +2fm+4f(~)+f(l)].

9.11. Solution Suppose 1Tn(x) = k"x n + .... Then, by orthogonality,

r

n 1Tn(X)k"x w(x) dx

r

so that

a

=

r

(1Tn(x)fw(x) dx

1Tn(X)X nw(x) dx

=1

= k;;-I.

If there was another orthonormal system {1T~\)(X)} where 1T~I)(X) = k~llxn

+ ... we would have

This implies that

r a

1Tn(X)1T~ll(x)w(x) dx = k.Jk~\) = k~\)/k"

so that k~ = (k~l)f

and, both being positive, this gives k n It follows that (1)

r a

= k~\).

[1Tn(x) -1T~\)(X)]2W(X)dx

= 1- 2 + 1 = o.

The result (1) implies that 1Tn (X)=n::,l\X) if we assume that w(x) is not zero in any subinteroal

of [a, b].

This condition is satisfied in all the classical cases. To establish uniqueness we proceed as follows.

224

Solutions to Selected Problems

If '7Tn(x) ;5 '7T~I)(X) then, these being polynomials and so continuous, there will be an interval (c, d) included in (a, b) such that in it

Hence

0=

l'7Tn (x) - '7T~I)(x)1 ~ 5 > O.

r a

['7Tn(X)-'7T~I)(X)fw(x)dx~Id 5 2w(x)dx>0, c

a contradiction. 9.12. Solution

Let Xl>' •• , Xn be the zeros of 1Tn (X). Let H(x) be the Hermite interpolant introduced in Problem 8.12. If we multiply the error estimate (11), (p. 91), across by w(x) and integrate between [a, b] we get

f a

b

Ibf2n)(~(x))

(f(x)-H(x))w(x)dx=

(2n)!

a

n TI(x-xYw(x)dx.

Since H(x) is of degree 2n -1, and H(xJ = !(X;), we have

f

a

b

H(x)w(x) dx

= L A;H(xJ = L AJ(xJ.

Thus 1- Q = I

b

f2n)(~(x)) (2n)!

a

[0 (x - xY]w(x) dx i=1

TI

f2n)(~) Ib [ n ] = (2n)! a (x - xY w(x) dx

using the Mean Value Theorem, since the last integrand is positive. In the Chebyshev case we have

1+ n(x (2n)! 1

1

-1

2

XJ

dx 1 (1- X 2)l/2 = (2n)!

n(X)}2 J+l {T2n-l dx (1- X2)1/2 -1

1 = (2n)!22n- 2

1..

2

cos nO dO

'7T (2n)l2 2n - 1·

To deal with the Legendre case we note that it can be verified (e.g., by integration by parts, Apostol II, 179) that if Pn(x) = [1!(2n(n !))]D n {(x 2-lt}

Chapter 9

225

then

Also

Pn (x) = [(2n)!/(2 n(n !)2)][X n - ... ]. Hence

f\

(i'n(X»2 dx = [2 2n (n !)4/(2n)!2](N +!)-I

and so the coefficient of

f2n)(~)

is

9.13. Solution (a) The indefinite integral is

1 x 2 -x+ 1 1 x13 - log +- arctan - 6 (x+l)2 13 2-x when -1 < x < 2. Hence the definite integral is -log 4 + arctan 13 = _log 2 +~=0.60459 9788-0.23014 9060 6 13 3 313 = 0.374450728. (b) The indefinite integrals are, respectively:

+ !log (l + x) +

I

x

dt

13

- -6 = -log

o l+t

12

v'(l0+2J5) 10

4x+v'S-1 4x-v'S-1 arctan v' arctan v' ; (l0+2J5) (l0-2J5)

x 2+ x13+ 1 x 13 + arctan - -2+ ~ arctan x. 2 x -x 3+1 l-x

t

226

Solutions to Selected Problems

The definite integrals, between 0 and 1, are respectively:

.j2

8

[log (3 +.j2) + 7T] = 0.86697299;

./5 3+./5 1 2 "/10+2../5 3+./5 - 1og - r ; +:5 log + arctan -;r===:::::: 10 3-v5 10 v 10+2.)5 +

../10-2./5

10

3-./5 arctan ---,==.==:::::: ../10 - 2.)5

./5 7T./5 =Slog!(./5 + 1)+! log 2+ ../(10+2.)5) =0.88831357;

50

The general results of which these are special cases are due to A. F. Timofeyev (1933) - see, e.g., formula 2.142 in I. S. Gradshteyn and I. M. Ryzhik, Table of integrals, series and products, Academic Press, 1965. See also I. J. Schwatt, An introduction to the operations with series. Philadelphia, 1924.

9.14. Solution A short table of this integral in which the argument of y is in degrees is given in NBS Handbook p. 1001. When y = 0.5 in radians we find S(0.5, 0.5) = 0.296657503.

The differences between the given estimates are 9312, 253, 1 in units of the tenth decimal. The ratio 9312/253 is about the fourth power of the ratio of the intervals. We may suspect that Simpson's Rule, or some other of the same accuracy was used. We can use Richardson extrapolation to get a correction of 16 (625 -16) x 9312 = 245 leading to an estimate of 0.2966575002.

9.15. Solution These integrals were evaluated by E. Brixy (ZAMM 20 (1940), 236238); unfortunately some of his results seem wrong. It is desirable to draw rough graphs of the integrands and to check that the integrals all exist. Note

227

Chapter 9

that, e.g., the integrand in h is of the form % at x = 0 and so some minor adjustment in the program may be necessary. Since JQ(x)=-J1(x) we have Is = -

r

[J~(x)/Jo(x)] dx = [log [Jo(xn- j]6 = log [1/Jo(1)] = 0.267621.

Since xJ2(x)=J j(x)-xJ;(x) we have

=

[lOg

Lj;xJI

= log [1/J j(1)]-log 2 = 0.127718. Since xJo(x)=2J j(x)-xJ 2(x) and since, as we have just seen in the discussion of 16 , [J2(X)/J j(x)] = [log {x/Jj(x)}]' we have, on integration by parts,

14 =

r

r

(2-x[log{x/J j(x)}]')dx

=2-[x log {x/J j(x)}]6+

log{x/Jj(x)}dx

= 2+ logJ 1(1)+ log 2+ 17 This result can be used as a check on the values of 14 , 17 ,

12 = 2.08773,

II = 1.09407,

13 = 0.086734,

14 = 1.914485, 17 = 0.04220 4,

II! = 0.02795 4.

9.16. Solution If 0:50:5!7T then 0:5 cos 0:51 and so which implies 12n - j ~ 12n ~ 12n +j where

cos2n-jO~cos2nO~cos2n+jo

l k= 1',./2 cos k0 dO. lt is easy to find a reduction formula for this integral: in fact

nln = (n -1)In-2'

n ~ 2,

and 10 = !7T, I j = 1. These give I

_(2 n (n!))2 2n+1 - (2n + 1)1'

228

Solutions to Selected Problems

Inserting these values in the inequalities we get

which gives 1+~~(2n)!(2n+l)! ~~1 2n 2 4n (n!)4 2

which is the result required.

9.17. Solution This is established in a way similar to that used to get Wallis' Formula. We write

r

In = .lo

1

t n dt (1- t 4 )1/2

and observe that In decreases as n increases. It is easy to show that 13 =! for [-!(l- t 4 )1/2], = t 3 (1- t 4 )-1/2. We next obtain the reduction formula

I n + 3 = [n/(n +2)]In -

1.

We then use this, and the facts that

to conclude that and hence that

Use of the reduction formula gives I 4r

=1.5 3.7

(4r-3)L (4r-l) 0,

I

- 3.7 5.9

4r+2 -

and I

_!.. 4r+3 -

2.4 2 3.5

(2r) (2r+ 1)"

(4r-l) I (4r + 1) 2

229

Chapter 9

We now observe that 10 12 = (4r + 1)14)4r+2 = (4r + 1)(14r+3)2 X(14 jI4r + 3 ) X(14r+2/14r+3)'

We let r_ oo and the last two factors tend to unity and the first, because of Wallis' Formula, tends to !1T.

r

9.18. Solution

1=

x 4 dx = (l/5)(b S-as) = (l/5)(b - a)[a 4+a 3b +a 2b 2+ab 3+b 4].

r+3e

a b 4 Q = 3~:3a) [a +3e 3+

~2br +b 4]

=(b -a) x[27a4+27b4+(16a4+ 32a3b +26a 2b 2+8ab 3+b 4) 8x27

+(a 4+8a 3b +24a 2b 2+32ab 3+ 16b 4)]

= ~:2~ [lla 4+ 10a 3b+ 12a 2b 2+ 10ab 3+ llb 4 ]. Hence the actual error is: 1-0= -(b-a)[ 4-4 3b+6 2b2-4 b 3+b 4]= -(b-a)5 270 a a a a 270 . The estimate is

In the case of N panels, the estimate will be

~x (b-a))S xM = (b-a)S . M4

NX 5

6N

4

5x64

N4

where M 4 is maxasxS;b If(x)l. If we take M 4 to be 1 then we have to choose N to make i.e., N>(6480 £)-1/4. E.g., with £ = 10-8 we must take N> (6480)-114 X 102 * 102/9* 11.

and so about 34 evaluations of the integrand. Note that if we change M 4 to 10 or to 0.1 we only change N by a factor of 10",1/4* 1.78 or 0.60.

230

Solutions to Selected Problems

9.19. Solution We have to show that

r r

J =-

a

~(x)(l-li(x»w(x) dx = O.

Since ~(.~;) = 1 it follows that (x - x;) divides 1-l;(x), the quotient being a polynomial of qn-ix) of degree at most n - 2. Thus J=-

a

((x-Xj)~(X)}qn_2(X)W(X) dx

and, as the quantity in braces is a multiple of 1Tn (X), it follows by orthogonality that J = O.

9.20. Solution Since 1f4)(x)l:s; 1, the absolute values of the errors in Simpson's Rule and in the ~- Rule are bounded by h 5 /90

and

3h 5 /80,

h being the interval. Suppose we use N panels. Then the total error in the Simpson case is

and that in the ~-case is

b -a)5 3 N· ( 3N . 80' If the error is to be less than E(b - a)5 we must have N=~1/2880E

and N

= ~1/6480E.

The cost, in terms of evaluations of the function 2· ~1/2880E=~1/180E and

t, is

3· ~1/6480E=~1/80E.

The basic error estimate in the Gauss-Legendre case is (Problem 9.12)

when we use the interval (-1,1). For a general interval

(a,~)

we have to

Chapter 9

r

231

write

f(x)dx=

[lfG(~-a)t+!(a+~)).!(~-a)dt=!(~-a)[lF(t)dt

where t=(2x-a-~)/(~-a). As Fr)(t)=W~-a)Yfr)(x) our error estimate should be mulitplied by W~ - a )]2n + 1. For n = 2 we find

Returning to our main problem, if we again use N panels, the total error will be

-a)5 '4320 1

b N· ( N

and the relevant value of N is N

= ~1/4320e.

There are two evaluations per subinterval (at "awkward" abscissas) and so the corresponding cost is 2N = ~1/270e.

The relative efficiencies are therefore about 1.36 G-L.;

1.22 S;

1(~).

9.21. Solution See NBS Handbook, p. 492. In particular f(3)

= 1.3875672520-0.9197304101 = 0.46783 68419.

9.22. Solution We find 'Y1O = 0.62638 316; 'Y30 = 0.59378 975;

At the first stage At the second stage

= 'Y1O - (1/20) = 0.5763832; 'Y~o = 'Y30 - (1/60) = 0.57712 30. 'Y~O

Solutions to Selected Problems

232

9.23. Solution This is a table of Pix) = (35x 4 -30x 2+3)/8; the endings 4,6 are exactly 375,625. The exact value of the integral is [(7x 5 -lOx 3 + 3x)/8]~·2 = 0.46728. Simpson: (Milne)3: (3/8)4:

0.467352 0.466304 0.467439

We shall examine these results more closely. The errors in the first three methods are estimated as multiples of a mean of the fourth derivative of the integrand i.e., 105. The multiples are obtained in Problems 9.4, 9.2, 9.1. We have to be careful in scaling. The observed errors are respectively -72 x 10-6 ,

976x 10-6 ,

-159x 10-6 •

The theoretical errors are 6 x -ifo X 10-5 x 105,

3 xHx 10-5 x 105,

-3 4x-xlO- 5 x105 80

i.e., -7x10- 5 ,

98x 10-5 ,

-15.75 X 10-5

in close agreement. 9.24. Solution The relation given is ascribed to C. Hermite (E. W. Hobson, Plane Trigonometry (1925), 378). Write for r = 1, 2, ... , n, 2 sin ra cos (nx - rx) = sin (nx + r(a - x» + sin (-nx + r(a + x». We then use the formula sin6+sin(6+cp)+ ... +sin(8+(n-1)cp)=

cos (6 _!cp) - cos (6 +!(2n -l)cp) 2 2. 1 2 SlD'2CP

to get { }=cos(nx+!(a-x»-cos(na+!(a-x» ... 2sin!(a-x) +

cos (nx -!(a + x»-cos (na +!(a + x» 2 sin !(a + x)

sin na.

Using the relation 2 cos A sin B = sin (A + B)-sin (A - B) four times and the fact that 4 sin !(a - x) sin !(a + x) = 2(cos x -cos a)

233

Chapter 9

we find

2(cos x -cos a){. .. } = sin (nx +a)-sin (n -I)x -sin (n + I)a +sin (na - x) + sin (n -I)x -sin (nx -a) -sin (n + I)a +sin (na + x) -2 sin na cos x +2 sin na cos a

= sin (nx +a)-sin (nx -a)-2 sin (n + I)a +sin (na + x)

+sin (na - x)

-2 sin na cos x +2 sin na cos a

= 2 cos nx sin a - 2 sin na cos a - 2 sin a cos na

+ 2 sin na cos x

-2sin na cos x +2 sin na cos a

= 2 sin a(cos nx -cos na), which gives the result required. If we take a = 8m = ~(2m -1)17'/n, x

= 8 we

get

1 {(_I)m-t + 2 sm . (m - l)6m cos 8 - -cosn8 - - - =-.-cos 8-cos 8m sm 8m

+2sin (m -2)8m cos 28 + ... +2 sin 8m cos (m -1)6} so that

a,=

2 sin (m - r)8m sin 8m

r

= I, 2, ... , m-1.

9.25. Solution (a) For a popular method using double integrals see T. M. Apostol, Calculus, II, §11.28, Exercise 16. (b) The following method is given by A. M. Ostrowski, Aufgaben sammlung ... III, p. 51,257. We prove that (1)

[f

e-,2 dtr =!17' -

r

e- x2 (\+l2) dt/(1 + t 2 )

by observing first that both sides vanish for x = 0 since J~ dt/(1 + t 2 ) =!17' and then noting that the derivatives of the two sides are identical: the derivative of the right hand side is -( -2x)

r

e- x2(1+r") dt = 2

r

e- x2 - u2 du,

= 2e- X2 LX e- u2 duo

if

u = xt

Solutions to Selected Problems

234

If we now let x~

(n + 1)L,. = «2n + 1)- x)L,. - nL,.-t>

(c)

(d)

H n + l =2xHn -2nHn -

l •

10.3. Solution

The relation is equivalent to

Vn

Vn+l - Vn = (v n - k)(vn -(1- k)), = aUn +b, k(1- k) = ac-b 2 +b.

Convergence can only take place to k or 1- k. We may assume that k::::! when it is real. The following gives a complete answer: (1) k not real, V n ~ 00 monotonically (2) IVol> k, V n ~ 00 monotonically (3) IVol> k, !:5 k:5t V n ~ 1- k and monotonically if !:5 k:5 1 (4) IVol < k, ~< k :52, V n oscillates finitely (5) IVol < k, k > 2, V n ~ 00 except when Vo belongs to a certain set (cardinal c, measure zero) in which case it oscillates finitely.

237

Chapter 10

For further details see: T. W. Chaundy and Eric Phillips, Quarterly Journal (Oxford), 7 (1936), 74-80.

lOA. Solution. For further details see E. Hansen, Signum Newsletter 3 (1968), #3. 19 =0.0916123,

110 = 0.0838771.

10.7. Solution The exact solution to this equation is Y= (1- x)-\ with a pole at x = 1. Consider integrating at an interval h = N- 1 and write X. = rh, Yr = y(rh) where Yr+1 = Yr + hy;. We have h 1 h 2 Yr 1 1 --=------=--h+---Yr+1 Yr 1+hYr Yr 1+hy:

r

= 0,1, ... , n-1.

Summing we get 1 1 r-1 Y -=--rh+h 2 s -= 1-rh+hR, say. Yr ~o s=0 1 + hys

L

We shall now estimate the R, which obviously increase steadily from R o = 0 to RN • Oearly hYr 1 1 R+1-R=1+hYr =h- 1 -r+Rr+1 N+1+Rr-r We use the method of the "Integral Test". First, neglecting the positive quantities Rr we have RN


N-1

r~o

iN

1

L N + 1- r
N + 1+ log (N + 1) - r

f

N

1

-

1

N

dx 1- =log(N+1).

+

x

dx N + 1+ log (N + 1) - x

=10 N+log(N+1) g log(N+1)+2'

We have therefore shown that R N = log N +O'(log log N)

= log h- 1 + O'(log log (h- 1»

238

Solutions to Selected Problems

and hence

1

=hlogh-1

{

-I))}.

"'log log (h l+v\ log(h 1)

For further details see M. L. J. Hautus and G. W. Veltkamp, Nieuw Arch. Wisk. {3}, 17 (1969), 79-80.

10.8. Solution In practice there will be round-off so that, e.g., 100 times 0.001 will not be 0.1 exactly. A refined program will allow for this and take a small step (backwards or forwards) before printing so as to get the argument 0.1 (or, as near this as the machine allows!). Library programs will often have an automatic choice of an appropriate interval incorporated. The solution of y' = x 2 - y2, y(O) = 1 has the value 0.75001 5703 at x = 1. See Appendix for details. The results from a Univac computation for the equation y' = ~y'~+~y'y,

y(O) = 0

are given below. x

h =0.00025

h = 0.00125

h =0.0025

h =0.0125

0 0.0125 0.0250 0.0375 0.0500

00000000000 00054793700 00159912174 00299888889 00469064935

00000000000 00054084766 00159108463 00299020417 00468145066

00000000000 00052730731 00157549032 00297323965 00466341119

00000000000 00034938562 00135795112 00273099005 00440237794

0.0625 0.0750 0.0875 0.1000 0.1125

00664121481 00882794453 01123406696 01384648727 01665459906

00663158054 00881792703 01122370425 01383580841 01664362708

00661263636 00879819092 01120325726 01381471217 01662193028

00633609674 00850825638 01090141626 01350207833 01629937288

0.1250 0.1375 0.1500 0.1625 0.1750

01964957501 02282391333 02617113183 02968555308 03336214844

01963832876 02281240853 02615938188 02967356957 03334994148

01961607102 02278962286 02613609625 02964980810 03332572516

01928429044 02244919744 02578751306 02929348394 03296202153

Chapter 10

239

x

h =0.00025

h=0.00125

h = 0.0025

h =0.0125

0.1875 0.2000 0.2125 0.2250 0.2375

03719642145 04118431874 04532216083 04960658706 05403451t39

03718399993 04117169062 04530933327 04959356649 05402130364

03715934729 04114661815 04528385574 04956769724 05399505479

03678858123 04076907044 04489977709 04917731319 05359856951

0.2500

05860308623

05858969672

05856307932

05816067873

This is a classical problem studied by Moigno in 1844. Awkwardnesses arise because the expansion of the solution around the origin is of the form:

10.9. Solution The solution is y = e E 1(x) and the following values are obtained from NBS Handbook, p. 243: x = 1 0.596347361 2 0.36132 8617 3 0.262083740 4 0.206345650 5 0.177297535 10 0.0915633339 20 0.0477185455 X

10. 11. Solution Writing z = y' we replace our scalar differential equation by a vector differential equation where

[Y] z

is given for x = 0:

[Y,(O)]. y (0)

We have y*(l)= y(O)+hz(O) { z*(l) = z(O)+ hf(O, y(O), z(O))

and {

Y**(l) = y(O) + h[z(O) + hf(O, y(O), z(O))] z**(l) = z(O) + hf(h, y(O) + hz(O), z(O) + hf(O, y(O), z(O)))

giving {

y(l) = y(O) + hz(O) + ~h2f(0, y(O), z(O)) z(l) = z(O) +~hf(O, y(O), z(O)) +~hf(h, y(O) + hz(O), z(O) + hf(O, y(O), z(O)))

Solutions to Selected Problems

240

Clearly y(1), z(1) only depend on y(O), z(O), hand f. We have to evaluate f twice, once with arguments 0, y(O), z(O) (which we use three times), and once with arguments y(O) + hz(O),

h,

z(O)+ hf(O, y(O), z(O».

We are free to change h whenever it is desirable without change in the formulas. Equation (14) has been studied by P. Painleve - the solution becomes singular, as in the simple case of y(O) = A

which has a solution y = (A - X 2)-1. Thus the problem is one of "polar exploration". The Painleve equation in the case y'(O) = 1 and on the real axis has a series of poles of order 2 of which the first two are at 2.97 ...

1.70 ... ,

10.12. Solution Discretizing using a mesh size h = t we get a system of three homogeneous linear equations for y(l), yG), y(~):

-k' !Ay(!) y(l)-2y(!)+y(~) = -k' !AyG) y(!)-2y(~)+(y(1» = -k' ~Ay(~)

(y(0»-2y(!)+ y(!) =

Note that y(O) = y(1) = O. In order that we should have a nontrivial solution the determinant of this system must vanish, i.e., (A -128)

det

[

32

o

64 (A -64) 64

3

o ] =0. 32 (A _1~8)

This reduces to X

X

X

A3_ A2[11 ;64]+A[128\64 5]_ [64 6: 128] =0.

Writing A = 64A we get i.e.,

241

Chapter 10

The last equation has an obvious root A = 1 and the residual equation is 3A2 -8A+2=0 which has roots A = (4±JiO)/3. Thus the roots are A. = 64(4-JiO)/3 = 17.8688,

A3 = 64(4+ JiO)/3 = 152.794.

A2 =64,

The corresponding values of y(~), y(~), y(~) are:

2+~J'

[ 2+JiO -2+JiO

[:], -1

[2-~]. -2+JiO 2+JiO

These have sign patterns (+ + +), (+ + -), (+ - + ). Integrating the differential equation with A = 17, 18, 19 enables us to get a better estimate for the smallest eigenvalue. In order to check our integration subroutines we proceed as follows. Integrate the differential equation with A = 18.6624,

18.9225,

19.1844,

The corresponding values of the solution at x 0.01066,

0.00122,

19.4481

= 1 should be

-0.00816,

-0.01747

and we interpolate inversely by the Aitken method 1066 122 -816 -1747

6624 9225 9561 11844 581 14481 601

564 564

giving A = 18.9564 which is the coyrect result. The motivation of our choice of the four A's is the following: more details are given in the Appendix. The solution to y" - Axy =0,

is a multiple of which has the value

y(O)

= 0,

y'(O) = 1

242

Solutions to Selected Problems

when x = 1. We want to choose A so that this vanishes. The zeros of J 1/3 have been tabulated in the NBS volume cited in the Appendix and are, to 4D, 2.9026,

6.0327,

9.1705, ...

These correspond to the following values of A: 18.9564,

81.8844,

189.219, ...

(which we can compare with the roots of the cubic). It is Al = 18.9564 with which we are concerned. The assigned values of A correspond to the following values of ~A ~ ~(18.6624)1/2 = ~(4.32) =

2.88,

2.90,

2.92,

2.94

The values of J 1/3 (X), x = 2.88(0.02)2.94 are 0.01066, 0.00122, -0.00816, -0.01747.

10.13. Solution [For a more theoretical discussion of this problem see, e.g., p. 279 of G. Birkhoff and G. C. Rota, Ordinary differential equations, Ginn, Boston, 1962.] This is a simple form of the Schrodinger equation. Consider the case when V is even, vanishing outside [-1, 1]. We ask whether there are values of E for which there are nontrivial solutions (i.e., l/J¢: 0) which are even, or odd and which are such that l/J(x) ~ 0 as x ~ ±oo. Since l/J is required to be even or odd we can restrict our attention to x ~ O. When x ~ 1 the equation reduces to l/J"(x) + El/J(x)

=0

with general solution l/J(x) = A exp (.J=E(x -1»+ B exp (J=E(I- x».

In order that cp(x) ~ 0 as x ~ 00 we must have E < 0, A = O. When Os;xs;l, the solution depends on the form of V(x). However it will involve two arbitrary constants, say l/J(O), l/J'(O). Now if l/J is even, l/J'(O) = 0 and if l/J is odd, l/J(O) = O. Hence one constant is determined and the solution is of the form l/J(x) = Ccp(x)

where C is arbitrary and cp involves E. To match the solutions at x = 1 we must have Ccp(l) = l/J(l) = B,

Ccp'(I) = l/J'(I) = -J=EB

Chapter 10

243

which can be satisfied if and only if

T(E);: J=Ecp(l) + cp'(l) = O. This is the equation which determines the characteristic values (if any) of E. We now return to the special Vex) of the problem and note that an odd solution of

I/J"(x) + (E + 5)I/J(x) = 0 IS

cp(x) = sin.JE + 5x for which cp(l) = sin.JE + 5, cp'(l) =.JE + 5 cos.JE + 5. The characteristic values are determined by J-Esin JE+5 +JE+5 cos JE+5 =0 i.e., tan.JE+5= -

~E+5 ~.

Remembering that E is negative we consider the behavior of the two sides of the last equation in the range - 5 s; E :s 0 where they are real. The left side increases from 0 to 00 in (-5,-5+~7T2) and from -00 to tan.J5 in (-5+~7T2,0). The right side decreases from 0 at E=-5 to -00 at E=O. There is exactly one intersection near E = - 1. This can be estimated by Newton's method. If

feE) = tan.JE + 5 +.J(E + 5)/E then f(-l)=tan 2+2, and the next approximation is -1- 4(2+tan2). _1+0.74016._ 5 +sec 2 2· 10.7744· 0.931. 10. 14. Solution In general it is not possible to obtain I/J(x) explicitly as in Problem 10.13. We must proceed as follows. According as to whether we want an even or odd solution we integrate the equation

I/J"(x) = (V(x) - E)I/J(x)

Solutions to Selected Problems

244

numerically from 0 to 1, beginning with If/(O) = 0, If/'(O) = 1 or 1/1(0) = 1, If/'(O) = 0, for some value of E. We then check whether T(E) = J="Elf/(l) + If/'(1)

is or is not zero. If it is not zero we try to get a better guess for E and repeat the process until we get T(E) close enough to zero. Various artifices can be used to determine E, e.g., inverse interpolation as on p. 133. The result is that there is an even solution corresponding to E ~ 0.337.

10.15. Solution The table is meant to be one of (3x+912)3_38299 99877. There were errors in the entries for arguments 217, 218, 219: the corrected values are respectivelY-116 39330; 10389619; 325 03132 10.16. Solution Compare Problem 7.7. y(1.5) = 0.7127926,

y(2.0) = 0.6429039.

10.17. Solution a ).

_anf(n+(b/a)) Xn f(b/a)

b). Xn

Xo·

= (Ain + Bin+l)/C

where C = ci~+(a + b)joil - ii

A ={cio+(a+b)jl}xo-iIXb

and in

= (_c)1/2n J n+(bla)(2a- l (_C)1/2).

See D. H. Lehmer, Proc. Conference Numerical Mathematics, Winnipeg (1971), 15-30.

10.18. Solution f(7) =2983

10.19. Solution a) These are the odd prime numbers. b) If Vn is the nth prime number of the form 4r+ 1 so that VI = 5, V2 = 13, V3 = 17, . .. then the sequence is that of the least integer N for

Chapter 10

245

which Pn is a factor of W+1. Thus 2 2 +1=5,5 2 +1=2.13,42 +1=17, 122 + 1 = 5.29, ... c) These are the values of n 2 - n +41 for n = 0, 1,2, ... and are all prime numbers until n = 41.

10.20. Solution Ai'(1.4) = -0.10850 959 Ai"(1.4) = 1.4Ai(1.4) = 0.11485327 10.21. Solution See preceding SOlution, or British Assoc. Math. Tables, Part Vol. A. 10.22. Solution x(O) =

X'(O) =

L"" e-,4 dt = ~ L"" eL"" 2t e2

14

T

•

T- 3 / 4 dT = ~rw = 0.906402479.

dt =! L"" e- T r

1 4 /

dT = !f(~) = 0.61270 8351.

Differentiating we find y"= x 2 y + L"" t(4t 3 -3xt) exp (-!x 2 +2xt 2 - t4 ) dt

- L"" exp ( - !x

2

+ 2xt 2 - t4 ) dt.

Integrating by parts the second term on the right we find

1""... dt = [-t exp (!x

2

+ 2xt 2 - t 4 >:t + L"" exp ( - !x 2 + 2xt 2 - t 4 ) dt

The first term on the right vanishes at both limits and the second cancels the third term on the right in the previous display. Hence

The solution is

and for x = -2(1) 8 this has values 0.0078, 0.2059, 0.9064, 0.9869, 0.6702, 0.5241, 0.4487, 0.3637, 0.3363, 0.3143.

246

Solutions to Selected Problems

10.23. Solution See, British Assoc. Math. Tables, Vol. 2. For recent theoretical work see papers by E. Hille. 10.24. Solution See, J. L. Synge, On a certain non-linear equation, Proc. Royal Irish Academy 62A (1961), 17-41. Z. Nehari, On a non-linear differential equation arising in nuclear physics, Proc. Royal Irish Academy 62A (1963), 117-135. 10.25. Solution We use an interval h = 0.1 and work to 4D. From the power-series y(x) = 1 + x + x 2 + (x 3 /3)+ (x 4 /12) + ... we compute y(O.l), y(0.2), y(O.3) and [(0.1), [(0.2), [(0.3) and we then predict yp(O.4) = 1 +(4/30)[2.4206-1.4428+ 3.3994] = 1.5836 We then compute f(O.4) = 1.9836 and check by Yc(O.4) = 1.2428 + (1.30)[1.4428+6.7988 + 8.9836] = 1.5836. We can accept this value and proceed, or alternatively, try a larger h = 0.2. x

yp

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

1.0000 1.1103 1.2428 1.3997 1.5836 1.7974

Yc

y'=[(x,y)=x+y

1.5836 1.7974

1.0000 1.2103 1.4428 1.6997 1.9836 2.2974

The correct value at x = 1 is 2e - 2 = 3.43656 ...

Bibliographical Remarks

RECOMMENDED LITERATURE

T . M . APÓSTOL, Calculus,

I, II (Wiley, 1 9 6 7 - 9 ) .

A . M . OSTROWSKI, Vorlesungen vols., Aufgabensammlung (Birkhäuser,

über Differentialand Integralrechnung, 3 zur Infinitesimalrechnung, 3 vols.

1965-72).

TESTS AND MONOGRAPHS

F. S. A c T O N , Numerical methods that work ( H a φ e r a n d R o w , 1 9 7 0 ) . E . K . BLUM, Numerical analysis and computation; theory and practice (Addison-Wesley, 1 9 7 2 ) C . BREZINSKI, Acceleration de la convergence en analyse numérique (Springer, 1 9 7 1 ) .

S. D .

CoNTE

a n d C . DE B O O R , Elementary

numerical

analysis

(McGraw-Hill,

1973).

G.

DAHLOUIST

and

Ä . BJÖRCK,

tr.

N . ANDERSON,

Numerical

methods

(Prentice-Hall, 1 9 7 6 ) .

J. W . DANIEL a n d R. E . M O O R E , Computation and theory in ordinary differential equations (Freeman, 1 9 7 0 ) . P. J. DAVIS and P. RABINOWITZ, Methods of numerical integration (Academic Press, 1 9 7 5 ) .

C . - E . FRÖBERG, Introduction to numerical analysis (Addison-Wesley, 1 9 6 5 ) . R. W . HAMMING, Introduction to applied numerical analysis (McGraw-Hill, 1971).

H . M . NAUTICAL ALMANAC OFFICE, Interpolation

and

allied

tables

Stationery Office, 1 9 5 6 ) . P. HENRICI, Elements of numerical analysis (Wiley, 1 9 6 4 ) . P. HENRICI, Discrete variable methods in ordinary differential

(H. M .

equations

(Wiley, 1 9 6 2 ) .

P. HENRICI, Computational 1977).

analysis

with H P 2 5 pocket

calculator

(Wiley,

248

Bibliographical Remarks

F. Η . HILDEBRAND, Introduction to numerical analysis, 2 n d e d . (McGrawHiU, 1976). A . S. HOUSEHOLDER, Principles of numerical analysis (McGraw-Hill, 1 9 5 3 ; Dover, 1974). A . S. HOUSEHOLDER, The numerical treatment of a single non-linear equation (McGraw-Hill, 1970). E . ISAACSON a n d H . B . KELLER, Analysis of numerical methods (Wiley, 1966). L. M . MILNE-THOMSON, The calculus of finite differences (Macmillan, 1933). Modem Computing Methods ( H . M . Stationery Office, 1961). B. R . MORTON, Numerical approximation (Routledge a n d Kegan Paul, 1961). B. NOBLE, Numerical methods, 2 vols. (Oliver a n d Boyd, 1964). A . M . OSTROWSKI, Solution of equations, 3 r d e d . (Academic Press, 1973). A . RALSTON a n d P . RABINOWFTZ, A first course in numerical analysis, 2 n d e d . (McGraw-HiU, 1978). T . J . R i v u N , The Chebyshev polynomials (Wiley, 1976). H . R u T i s H A U S E R , Vorlcsungcn über numerische Mathematik, 2 vols. (Birkhäuser, 1976). Η . R u T i s H A u s E R , Numcrischc Prozeduren - ed. W . G a n d e r , L . Molinari a n d H . Svecová (Birkhäuser, 1977). F. SCHEID, Theory and problems of numerical analysis, (Schaum-McGrawHiU, 1968). L. F . SHAMPINE a n d R . C. ALLEN, Numerical computing: an introduction (Saunders, 1973). L. F . SHAMPINE a n d M . K. G O R D O N , Computer solution of ordinary differential equations. The initial value problem (Freeman, 1975). E . STIEFEL, tr. W . C. a n d C. J . RHEINBOLDT, A n introduction

mathematics (Academic Press, 1963). J . STOER, Einführung in die numerische Mathematik, 1976).

to

numerical

I 2 n d e d . (Springer,

J . STOER and R. BULIRSCH, Einführung in die numerische Mathematik, II 2nd e d . (Springer, 1978). JOHN T O D D , e d . . Survey of numerical analysis (McGraw-Hill, 1962). JOHN T O D D , Numerical analysis. C h a p t e r 7 , Part 1 of E . U . C o n d o n - H . Odishaw, H a n d b o o k of Physics, 2 n d e d . ( M c G r a w - H ü l , 1967). JOHN T O D D , Introduction to the constructive theory of functions (Birkhäuser, 1963). J . F . T R A U B , Iterative 1966).

methods

for the solution

of equations

(Prentice-Hall,

Bibliographical Remarks

249

Β . WENDROFF, Theoretical numerical analysis (Academic Press, 1966). E . T . W H m A K E R a n d G . ROBINSON, The calculus of observations, 4 t h e d . (Blackie, 1966). J . H . WILKINSON, Rounding errors in algebraic processes (Prentice-Hall, 1963). D.

M . YOUNG and R . T. GREGORY, A

survey

of numerical

mathematics,

2

vols. (Addison-Wesley, 1 9 7 2 - 7 3 ) .

TABULAR MATERIAL M . ABRAMOWFFZ a n d

I. A . STEGUN, e d s . Handbook

of mathematical

func­

tions. National B u r e a u of Standards, Applied M a t h . Series, 55 ( U . S . G o v e r n m e n t Printing Office, 1964). B A R L O W S T A B L E S , e d . L . J . C o m r i e (Spon, 1961).

L . J . COMRIE, Chamber's 1950).

Shorter six-figure

mathematical

tables, ( C h a m b e r s ,

A . FLETCHER, J . C. P . M I L L E R , L . R O S E N H E A D a n d L . J . C O M R I E , An

mathematical

index

of

tables, 2 n d e d . , 2 vols. (Addison-Wesley, 1962).

In addition t o t h e literature m e n t i o n e d above, t h e r e a r e m a n y useful expository articles available, a n d for u p - t o - d a t e surveys of special topics, reference can b e m a d e t o Symposia Proceedings, such as those which a p p e a r in t h e I S N M series. T h e r e is a developing interest in t h e history of numerical mathematics and computing machines. F o r t h e classical material s e e , respectively. H . H . G o L D S Ή N E , A history of numerical analysis from the 16th through the 19th century (Springer, 1978), B. RANDELL, The origins of digital computers, 2 n d e d . (Springer, 1975). For m o r e recent history see t h e obituaries of t h e founders a n d t h e periodical Annals of t h e history of computing, 1 9 7 9 - .

Contents Vol. 2, Numerical Algebra

Notations and Abbreviations

7

Preface .

9

Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Manipulation of Vectors and Matrices Norms of Vectors and Matrices . . . Induced Norms . The Inversion Problem I: Theoretical Arithmetic The Inversion Problem II: Practical Computation The Characteristic Value Problem - Generalities. The Power Method, Deflation, Inverse Iteration. Characteristic Values. . . . . . . . . . . . . Iterative Methods for the Solution of Systems Ax = b Application: Solution of a Boundary Value Problem. Application: Least Squares Curve Fitting. . . . . Singular Value Decomposition and Pseudo-Inverses

Solutions to Selected Problems

.13 .16 .19 .29 .44 .53 .65 .71 .83 .99 105 110 117

Bibliographical Remarks

.212

Index

.214

.

INDEX

Μ. Abramowitz 141, 249 Acceleration methods 66, 68 A. C. Aitken algorithm 86, 87, 241 transform 67 G. B. Airy integral 124, 129, 137, 150, 155, 157, 245 Algorithms Borchardt 15, 166, 170 Carlson 16 Gauss 13 E. Ε. Allen 190 Arithmetic-geometric mean 13, 161 Asymptotic series 75 Bad examples 47, 88, 119, 130 F. L. Bauer 102 S. N. Bernstein 58, 211 polynomials 59, 63, 184, 187 F. W. Bessel 149 functions 64, 65, 115, 116, 119, 189, 227, 241 modified functions 145 Garrett BirkhofT 242 G. D. Birkhoff 91 J. L. Blue 180 R. P. Boas, Jr. 172 C. W. Borchardt 15, 163, 166 A. J. Bosch 234 Ε. Brixy 226 G. H. Brown, Jr. 182 B. C. Carlson 16, 18, 21, 22, 164, 167, 168 T. W. Chaundy 237 P. L. Chebyshev 186 coefficients 61 polynomials 60, 64, 186 Ε. B. Christo fTel-J. G. Darboux formula 109 C. W. Clenshaw 62 L. Collatz 54, 181 Complementary error function 77 L. J. Comrie 81, 249 M. Davies 182

B. Dawson 182 R. Dedekind 37, 40 Deferred approach 100 Difference equation 118 Differencing 121 Elliptic integral 18, 21, 165 F. Emde 141 R. Emden equation 138 O. Emersleben 178 Error analysis backward 48 forward 48 interpolation 89, 91 quadrature 98, 99, 105, 109, 113, 218, 229 Error function 77, 117 L. Euler 116 constant 31, 116, 231 method 125 transform 68, 69, 71, 72, 192, 193 Exponential integral 73, 81 R. M. Federova 82 W. Feller 87 Fibonacci 133, 179, 235 Fixed point 36, 38 A. Fletcher 81 FMRC 81, 202 G. E. Forsythe 46, 47 J. B. J. Fourier coefficients 61 method 50 series 64 L. Fox 83, 202 A. J. Fresnel integrals 76, 202 Fundamental theorem of algebra 84, 213 P. W. Gaffney 214 Gamma function 79 K. F. Gauss arithmetic-geometric mean 161 K. F. Gauss-P. L. Chebyshev quadrature 105, 224 K. F. Gauss-A. M. Legendre quadrature 116, 224, 225,230

252 Walter Gautschi 121 Α. Ghizzetti 219 G. Α. Gibson 153, 155 Ε. Τ. Goodwin 82, 199, 235 I. S. Gradshteyn 226 J. P. Gram-Ε. Schmidt process 103 J. A. Greenwood 81 E. Halley 54, 182 E. Hansen 134, 237 P. A. Hansen 148 G. H. Hardy 151, 155, 172 H. O. Hartley 81 D. R. Hartree 138, 235, 245 C. Hastings 61 M. L. J. Hautus 135, 237 D. R. Hayes 219 P. Henrici 50 C. Hermite 232 interpolation 90, 224 K. Heun method 125, 135, 238, 239 E. Hille 246 Κ. Ε. Hirst 180 W. G. Horner's method 52 W. G. Hwang 43, 176 Instability 119, 130 Interpolation Aitken algorithm 86 errors 89, 91 Hermite 90, 224 inverse 88, 241 Lagrange 84, 210 Newton 96, 216 spline 96, 215 Inverse inteφolation 88, 241 Iteration 38 C. G. J. Jacobi relation 147 E. Jahnke 141 E. Kamke 124 A. N. Khovanski 180 K. Knopp 195 J. L. Lagrange ¡nteφolation 84 C. A. Laisant 235 E. Landau 24 J. Landen 22, 168 A. V. Lebedev 82 H. Lebesgue constants 91 A. M, Legendre

Index expansion 184 polynomial 184, 236 D. H. Lehmer 244 G. W. Leibniz' Theorem 129, 207 Lemniscate constants 17, 109, 114 J. Liang 191 L. Lichtenstein-S. A. Gerschgorin equa­ tion 191 Local Taylor series 129 F. Lösch 141 Ε. C. J. von Lommel 149 Y. L. Luke 188 Mean value theorems first 51, 89, 98, 99, 167, 224 second 151, 152 C. A. MicchelU 214 J. C. P. Miller 81 W. E. Milne quadrature 113 I. P. Natanson 211 NBS Handbook 76, 81, 92, 93, 94, 163, 164, 165, 178, 190, 198, 199, 200, 202, 231, 236, 239 Z. Nehari 246 E. H. Neville 87 1. Newton process 39, 48 inteφolation formula 216 J. W. Nicholson 151, 157 F. W. J. Olver 121 Order of convergence 29 Order symbols 24 Orthogonal polynomials 103 Chebyshev 104, 134, 236 Hermite 134, 236 Laguerre 104, 134, 236 Legendre 104, 134, 236 A. Ossicini 219 A. M. Ostrowski 47, 233 Η. Padé fractions 64, 188 P. Painlevé 240 J. F. Pfaff 19 Eric Phillips 237 É. Picard method 125 S. K. Picken 62 H. Poincaré 75 Practical computation 42 Principal value integral 106

253

Index Quadrature Gaussian 103, 115 Lagrangian 98 Milne 113, 219, 232 Romberg 100 I Rule 99, 113, 218, 229, 232 Simpson 99, 113, 222, 232 Trapezoidal 97, 221 Predictor - corrector 127 Recurrence relations 33 for reciprocals 33 for square root 37 Remainder Theorem 85 E. Ja. Remez 184 L. F. Richardson 66, 100, 126 T. J. Rivlin 214 G. Robinson 180 M. Rolle's Theorem 90, 212, 220 W. Romberg quadrature 100 L. Rosenhead 81 G. C. Rota 242 L. Rubin 219 H. Rutishauser 102 I. M. Ryzhik 226 D. H. Sadler 3, 83, 202 H. E. Salzer 215, 216 I. J. Schoenberg 19, 169 E. Schrödinger 136, 242 J. Schwab 22, 169 I. J. Schwatt 226 Second mean value theorem 151, 152 R. M. Sievert integral 115 T. Simpson's rule 102, 113, 222, 230 Spline 96 J. Staton 82, 199 I. A. Stegun 141, 269

E. Stiefel 102 J. Stirling 18 formula 79, 111, 174, 211 G. G. Stokes 56 O. Stolz 153, 155 Stopping rule 42 J. C. F. Sturm-J. Liouville problem 124, 131 Subtabulation 95, 214 J. L. Synge 246 Synthetic division 53 Olga Taussky Todd 3 H. C. Thacher 121 Theoretical arithmetic 42 i quadrature 99, 116, 229, 230 A. F. Timofeyev 226 E. C. Titchmarsh 152, 155 O. Toeplitz 195 UNIVAC 238 C. J. de la Vallée Poussin 153, 155 A. Vandermonde 95, 148, 213 G. W. Veltkamp 135, 237 H. S. Wall 44, 179 J. Wallis' formula 116, 229 S. E. Warschawski 191 G. N. Watson 141, 198, 203 B. Wendroff 219 E. T. Whittaker 180 K. T. W. Weierstrass' Theorem 58 A. van Wijngaarden 72 J. H. Wilkinson 47 S. Winograd 214 W. Wirtinger 151, 157 J. W. Wrench, Jr. 172