# b
a
ω(x)f(x)dx−n
i=1
wif(xi) =
# b
a
ω(x)(f(x)−h(x))dx.
By Theorem (2.1.5.9), and since thexi are the roots ofpn(x))∈Π¯n, f(x)−h(x) = f(2n)(ζ)
(2n)! (x−x1)2· · ·(x−xn)2=f(2n)(ζ) (2n)! p2n(x) for someζ=ζ(x) in the intervalI[x, x1, . . . , xn] spanned byxandx1,. . ., xn. Next,
f(2n)(ζ(x))
(2n)! = f(x)−h(x) p2n(x)
is continuous on [a, b] so that the mean-value theorem of integral calculus applies:
# b
a
ω(x)(f(x)−h(x))dx= 1 (2n)!
# b
a
ω(x)f(2n)(ζ(x))p2n(x)dx
= f(2n)(ξ)
(2n)! (pn, pn)
for someξ∈(a, b).
Comparing the various integration rules (Newton-Cotes formulas, ex- trapolation methods, Gaussian integration), we find that, computational efforts being equal, Gaussian integration yields the most accurate results.
If only one knew ahead of time how to chosen so as to achieve specified accuracy for any given integral, then Gaussian integration would be clearly superior to other methods. Unfortunately, it is frequently not possible to use the error formula (3.6.24) for this purpose, because the 2nth derivative is difficult to estimate. For these reasons, one will usually apply Gaussian integration for increasing values of n until successive approximate values agree within the specified accuracy. Since the function values which had been calculated forncannot be used forn+ 1 (at least not in the classical caseω(x)≡1), the apparent advantages of Gauss integration as compared with extrapolation methods are soon lost. There have been attempts to remedy this situation [e.g. Kronrod (1965)]. A collection offortran pro- grams is given in Piessens et al. (1983).
# b
a
f(x)dx, a,b finite,
was justified provided the integrand f(x) was sufficiently often differen- tiable in [a, b]. For many practical problems, however, the function f(x) turns out to be not differentiable at the end points of [a, b], or at some isolated point in its interior. In what follows, we suggest several ways of dealing with this and related situations.
1)f(x) is sufficiently often differentiable on the closed subintervals of a partitiona=a1< a2<· · ·< am+1=b. Puttingfi(x) :≡f(x) on [ai,ai+1] and defining the derivatives offi(x) atai as the one-sided right derivative and atai+1 as the one-sided left derivative, we find that standard methods can be applied to integrate the functionsfi(x) separately. Finally
# b
a
f(x)dx= m i=1
# ai+1
ai
fi(x)dx.
2) Suppose there is a point ˜x∈[a, b] for which not even one-sided derivatives off(x) exist. For instance, the functionf(x) =√
xsinxis such thatf(x) will not be continuous for any choice of the valuef(0). Nevertheless, the variable transformationt:=√
xyields
# b
0
√xsinx dx=
# √b
0
2t2sint2dt
and leads to an integral with an integrand which is now arbitrarily often differentiable in [0,√
b].
3) Another way to deal with the previously discussed difficulty is to split the integral:
# b
0
√xsinx dx=
# ε
0
√xsinx dx+
# b
ε
√xsinx dx, ε >0.
The second integrand is arbitrarily often differentiable in [ε, b]. The first integrand can be developed into an uniformly convergent series on [0, ε] so that integration and summation can be interchanged:
# ε
0
√xsinx dx=
# ε
0
√x
x−x3 3!±· · ·
dx=
∞ ν=0
(−1)ν ε2ν+5/2 (2ν+ 1)!(2ν+ 5/2). For sufficiently small ε, only few terms of the series need be considered.
The difficulty lies in the choice of ε: if ε is selected too small, then the proximity of the singularity at x = 0 causes the speed of convergence to deteriorate when we calculate the remaining integral.
4) Sometimes it is possible to subtract from the integrandf(x) a func- tion whose indefinite integral is known, and which has the same singularities asf(x). For the above example,x√
xis such a function:
# b
0
√xsinx dx=
# b
0
(√
xsinx−x√ x)dx+
# b
0
x√ x dx
=
# b
0
√x(sinx−x)dx+25b5/2.
The new integrand has a continuous third derivative and is therefore better amenable to standard integration methods. In order to avoid cancellation when calculating the difference sinx−xfor smallx, it is recommended to evaluate the power series
sinx−x=−x3 1
3!− 1
5!x2± · · ·
=−x3 ∞ ν=0
(−1)ν (2ν+ 3)!x2ν. 5) For certain types of singularities, as in the case of
I=
# b
0
xαf(x)dx, 0< α <1,
withf(x) sufficiently often differentiable on [0, b], the trapezoidal sumT(h) does not have an asymptotic expansion of the form (3.4.1), but rather of the more general form (3.5.1):
T(h)∼τ0+τ1hγ1+τ2hγ2+· · ·, where
{γi}={1 +α,2,2 +α,4,4 +α,6,6 +α, . . .}
[see Bulirsch (1964)]. Suitable step-length sequences for extrapolation methods in this case are discussed in Bulirsch and Stoer (1964).
6) Often the following scheme works surprisingly well: if the integrand of
I=
# b
a
f(x)dx
is not, or not sufficiently often, differentiable forx=a, put aj:=a+b−a
j , j= 1, 2, . . . ,
in effect partitioning the half-open interval (a, b] into infinitely many subin- tervals over which to integrate separately:
Ij:=
# aj
aj+1
f(x)dx,
using standard methods. Then
I=I1+I2+I3+· · ·.
The convergence of this sequence can often be accelerated using for in- stance, Aitken’s∆2 method [see Section 5.10]. Obviously, this scheme can be adapted to calculatingimproperintegrals
# ∞
a
f(x)dx.
7) The range of improper integrals can be made finite by suitable trans- formations. Forx= 1/twe have, for instance,
# ∞
1
f(x)dx=
# 1
0
1 t2f
1 t
dt.
If the new integrand is singular at 0, then one of the above approaches may be tried. Note that the Gaussian integration rules based on Laguerre and Hermite polynomials [see Section 3.6] apply directly to improper integrals
of the forms # ∞
0
f(x)dx,
# +∞
−∞
f(x)dx, respectively.
Exercises for Chapter 3
1. Leta≤x0 < x1< x2 <· · ·< xn≤bbe an arbitrary fixed partition of the interval [a, b]. Show that there exist uniqueγ0,γ1,. . .,γnwith
n i=0
γiP(xi) =
# b a
P(x)dx for all polynomialsP with degree (P)≤n.
Hint:P(x) = 1,x,. . .,xn. Compare the resulting system of linear equations with that representing the polynomial interpolation problem with support abscissasxi,i= 0,. . .,n.
2. By construction, thenth Newton-Cotes formula yields the exact value of the integral for integrands which are polynomials of degree at mostn. Show that for even values ofn, polynomials of degreen+ 1 are also integrated exactly.
Hint: Consider the integrandxn+1in the interval [−k,+k],n= 2k+ 1.
3. If f ∈ C2[a, b] then there exists an ˜x ∈ (a, b) such that the error of the trapezoidal rule is expressed as follows:
12(b−a)(f(a) +f(b))−
# b a
f(x)dx=121(b−a)3f(˜x).
Derive this result from the error formula in (2.1.4.1) by showing thatf(ξ(x)) is continuous inx.
4. Derive the error formula (3.1.6) using Theorem (2.1.5.9).Hint: See Exercise 3.
5. Letf∈C6[−1,+1], and letP ∈Π5be the Hermite interpolation polynomial withP(xi) =f(xi),P(xi) =f(xi),xi=−1, 0, +1.
(a) Show that
# +1
−1
P(t)dt=157f(−1) +1615f(0) + 157f(+1) +151f(−1)−151f(+1).
(b) By construction, the above formula represents an integration rule which is exact for all polynomials of degree 5 or less. Show that it need not be exact for polynomials of degree 6.
(c) Use Theorem (2.1.5.9) to derive an error formula for the integration rule in (a).
(d) Given a uniform partitionxi=a+ih,i= 0,. . ., 2n,h= (b−a)/2n, of the interval [a, b], what composite integration rule can be based on the integration rule in (a) ?
6. Consider an arbitrary partition ∆:={a=x0 <· · ·< xn =b}of a given interval [a, b]. In order to approximate
# b a
f(t)dt
using the function valuesf(xi),i= 0,. . .,n, spline interpolation [see Section 2.4] may be considered. Derive an integration rule in terms off(xi) and the moments (2.4.2.1) of the “natural” spline (2.4.1.2a).
7. Determine the Peano kernel for Simpson’s rule andn = 2 instead ofn= 3 in [−1,+1]. Does it change sign in the interval of integration?
8. Consider the integration rule of Exercise 5.
(a) Show that its Peano kernel does not change its sign in [−1,+1].
(b) Use (3.2.8) to derive an error term.
9. Prove n
k=0
k3=
n(n+ 1) 2
2
using the Euler-Macluarin summation formula.
10. Integration over the interval [0,1] by Romberg’s method using Neville’s al- gorithm leads to the tableau
h20 = 1 T00=T(h0) T11 h21 = 14 T10=T(h1) T22
T21 T33
h22=161 T20=T(h2) T32 ... T31 ... h23=641 T30=T(h3) ...
... ...
In Section 3.4, it is shown thatT11is Simpson’s rule.
(a) Show thatT22is Milne’s rule.
(b) Show thatT33is not the Newton-Cotes formula forn= 8.
11. Leth0:=b−a,h1:=h0/3. Show that extrapolatingT(h0) andT(h1) linearly toh= 0 gives the 3/8-rule.
12. One wishes to approximate the numbereby an extrapolation method.
(a) Show that T(h) := (1 +h)1/h,h= 0,|h|<1, has an expansion of the form
T(h) =e+ ∞
i=1
τihi which converges if|h|<1.
(b) ModifyT(h) in such a way that extrapolation toh= 0 yields, for a fixed valuex, an approximation toex.
13. Consider integration by a polynomial extrapolation method based on a ge- ometric step-size sequence hj =h0bj,j = 0, 1, . . ., 0 < b <1. Show that small errors∆Tjin the computation of the trapezoidal sumsT(hj),j= 0, 1, . . .,m, will cause an error∆Tmmin the extrapolated valueTmmsatisfying
|∆Tmm| ≤Cm(b2) max
0≤j≤m|∆Tj|,
whereCm(θ) is the constant given in (3.5.5). Note thatCm(θ)→ ∞asθ→1, so that the stability of the extrapolation method deteriorates sharply as b approaches 1.
14. Consider a weight function ω(x) ≥ 0 which satisfies (3.6.1a) and (3.6.1b).
Show that (3.6.1c) is equivalent to
# a b
ω(x)dx >0.
Hint: The mean-value theorem of integral calculus applied to suitable subin- tervals of [a, b].
15. The integral
(f, g) :=
# +1
−1
f(x)g(x)dx
defines a scalar product for functionsf,g∈C[−1,+1]. Show that iff andg are polynomials of degree less thann, ifxi,i= 1, 2,. . .,n, are the roots of thenth Legendre polynomial (3.6.18), and if
γi:=
# +1
−1
Li(x)dx with
Li(x) :=
!n k=ik=1
x−xk
xi−xk, i= 1,2, . . . , n, then
(f, g) = n
i=1
γif(xi)g(xi).
16. Consider the Legendre polynomialspj(x) in (3.6.18).
(a) Show that the leading coefficient ofpj(x) has value 1.
(b) Verify the orthogonality of these polynomials: (pi, pj) = 0 ifi < j.
Hint: Integration by parts, noting that d2i+1
dx2i+1(x2−1)i≡0 and that the polynomial
dl
dxl(x2−1)k is divisible byx2−1 ifl < k.
17. Consider Gaussian integration, [a, b] = [−1,1],ω(x)≡1.
(a) Show thatδi= 0 fori >0 in the recursion (3.6.5) for the corresponding orthogonal polynomialspj(x) (3.6.18).Hint:pj(x)≡(−1)j+1pj(−x).
(b) Verify # +1
−1
(x2−1)jdx= (−1)j22j+1 2j
j
(2j+ 1).
Hint: Repeated integration by parts of the integrand (x2−1)j≡(x+ 1)j(x−1)j.
(c) Calculate (pj, pj) using integration by parts [see Exercise 16] and the result (b) of this exercise. Show that
γj2= j2 (2j+ 1)(2j−1) forj >0 in the recursion (3.6.5).
18. Consider Gaussian integration in the interval [−1,+1] with the weight func- tion
ω(x)≡ 1
√1−x2.
In this case, the orthogonal polynomials pj(x) are the classical Chebyshev polynomials T0(x)≡1,T1(x)≡x,T2(x)≡2x2−1,T3(x)≡4x3−3x,. . ., Tj+1(x)≡2xTj(x)−Tj−1(x), up to scalar factors.
(a) Prove that pj(x) ≡(1/2j−1)Tj(x) forj ≥ 1. What is the form of the tridiagonal matrix (3.6.19) in this case ?
(b) For n = 3, determine the equation system (3.6.13). Verify thatw1 = w2 =w3 =π/3. (In the Chebyshev case, the weightswi are equal for generaln).
19. Denote byT(f;h) the trapezoidal sum of step lengthhfor the integral
# 1
0
f(x)dx.
Forα >1,T(xα;h) has the asymptotic expansion T(xα;h)∼
# 1
0
xαdx+a1h1+α+a2h2+a4h4+a6h6+· · ·.
Show that, as a consequence, every functionf(x) which is analytic on a disk
|z| ≤r in the complex plane withr >1 admits an asymptotic expansion of the form
T(xαf(x);h)∼
# 1
0
xαf(x)dx+b1h1+α+b2h2+α+b3h3+α+· · · +c2h2+c4h4+c6h6+· · ·.
Hint: Expandf(x) into a power series and apply T(ϕ+ψ;h) =T(ϕ;h) + T(ψ;h).