Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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10.1 Taylor Series 283

Indeed, this is obvious for n = 0, where q

= 1. We proceed by induction. If it is

true for n, then by the product rule

(n+1)

(x) = f

(n)0

(x) = e

−1/x

(x)

−

3nq

(x)

3n+1

(x)

−2

(x) − 3nx

(x) − 2q

(x)

3n+3

−1/x

To check the degree of

n+1

(x) = x

(x) − 3nx

(x) − 2q

(x),

note that

deg(x

) ≤ 3 + degq

≤ 2 + degq

≤ 2n + 2

and

deg(3nx

) ≤ 2 + degq

≤ 2n + 2.

Clearly this nice algebraic formula for the derivatives is continuous on both

(−∞, 0) and (0, ∞). So f is C

∞

everywhere except possibly at x = 0. Also,

lim

x→0

−1/x

= lim

t→±∞

where we substitute t = 1/x. Since

∞

n=0

≥

we see that

lim

t→±∞

≤ lim

t→±∞

|t|

= 0.

Therefore, if q

j=0

, we obtain

lim

x→0

(n)

(x) = lim

x→0

j=0

−1/x

j=0

lim

x→0

−1/x

3n−j

= 0.

We use the same fact to show that f

(n)

(0) = 0 for n ≥ 1. Indeed,

(n+1)

(0) = lim

h→0

(n)

(h) − f

(n)

(0)

= lim

h→0

(h)e

−1/h

3n+1

= 0.

So f

(n)

is deﬁned on the whole line and is continuous for each n. Therefore, f is

∞

Because all of the derivatives of f vanish at x = 0, all of the Taylor polynomi-

als are P

(x) = 0. While this certainly converges rapidly, it converges to the wrong

function! The Taylor polynomials completely fail to approximate f anywhere ex-

cept at the one point x = 0.

284 Approximation by Polynomials

Exercises for Section 10.1

A. Find the Taylor polynomials of order 3 for each of the following functions at the given

point a, and estimate the error at the point b.

(a) f(x) = tanx about a =

and b = .75

(b) g(x) =

√

1 + x

about a = 0 and b = .1

about a = 1 and b = 0.99

(d) k(x) = sinhx about a = 0 and b = .003

B. Let a ∈ [A, B], f ∈ C

[A, B], and let P

(x) = f(a) + f

(a)(x − a) be the ﬁrst-order

Taylor polynomial. Fix a point x

in [A, B].

(a) Deﬁne h(t) = f(t) + f

(t)(x

− t) + A(x

− t)

. Find the constant A that makes

h(a) = h(x

(b) Apply Rolle’s Theorem to h to obtain a point c between a and x

so that

f(x

) − P

) = f

(c)

− a)

for all x ∈ [A, B].

C. Let f satisfy the hypotheses of Taylor’s Theorem at x = a.

(a) Show that lim

x→a

f(x) − P

(x)

(x − a)

= 0.

(b) If Q(x) ∈ P

and lim

x→a

f(x) − Q(x)

(x − a)

= 0, prove that Q = P

D. (a) Find the Taylor series for sinx about x = 0, and prove that it converges to sin x

uniformly on any bounded interval [−N, N].

(b) Find the Taylor expansion of sin x about x = π/6. Hence show howto approximate

sin(31

◦

) to 10 decimal places. Do careful estimates.

E. (a) Verify that 4 tan

−1

(

) − tan

−1

(

239

) =

. HINT: Take the tan of both sides.

(b) Using the estimates for tan

−1

(x) derived in Example 10.1.6, compute how many

terms are needed to approximate π to 1000 decimals accuracy using this formula.

F. Let f(x) = logx.

(a) Find the Taylor series of f about x = 1.

(b) What is the radius of convergence of this series?

series converging to log2.

(d) By observing that log2 = log 4/3−log 2/3, ﬁnd another series converging tolog2.

Why is this series more useful?

(e) Show that log3 = 3 log.96 + 5 log

− 11log.9. Find a ﬁnite expression that

estimates log3 to 50 decimal places.

G. Compute the following limits using Taylor polynomials.

(a) lim

x→0

− 3

. HINT: a

= e

x loga

(b) lim

x→0

x cos x − sinx

x sin

x→0

tanx

1/x

. HINT: Take the log, and use Taylor polynomials for

both tan(x) and log(1 + x) about 0.

10.2 How Not to Approximate a Function 285

H. Suppose that f and g have n + 1 continuous derivatives on [a − δ, a + δ] and that

(k)

(a) = g

(k)

(a) = 0 for 0 ≤ k < n and g

(n)

(a) 6= 0. Use Taylor polynomials to

show that lim

x→a

f(x)

g(x)

(n)

(a)

(n)

(a)

I. Show that

+∞

−∞

log

1 + x

1 − x

= π

as follows:

(a) Reduce the integral to 4

log

1 + x

1 − x

(b) Use the Taylor series for log x about x = 1 found in Exercise 10.1.F. Use your

knowledge of convergence and integration to evaluate

log

x + 1

x − 1

as a se-

ries when r < 1.

(d) Use the famous identity

∞

n=1

to complete the argument.

J. Let f(x) = (1 + x)

−1/2

(a) Find a formula for f

(k)

(x). Hence show that

(k)

(0) =

–

(

–

− 1) ···(

–

+ 1 −k)

(−1)(−3) ···(1 − 2k)

(−1)

(2k)!

(k!)

−1

(b) Show that the Taylor series for f about x = 0 is

∞

k=0

−x

, and compute

the radius of convergence.

√

2 = 1.4f(−.02). Hence compute

√

2 to 8 decimal places.

(d) Express

√

2 = 1.415f (ε) where ε is expressed as a fraction in lowest terms. Use

this to obtain an alternating series for

√

2. How many terms are needed to estimate

√

2 to 100 decimal places?

K. Let a be the decimal number with 198 ones (i.e., a = 11. . . 11

{z }

198 ones

). Find

√

a to 500

decimal places. HINT: a =

(1 −10

−198

). Your decimal expansion should end

in 97916.

10.2. How Not to Approximate a Function

Given a continuous function f : [a, b] → R, can we approximate f by a poly-

nomial? For example, suppose you need to write a computer program to evaluate

f. Since some round-off errors are inevitable, why not replace f with a polynomial

that is close to f, as the polynomial will be easy to evaluate?

286 Approximation by Polynomials

What precisely do we mean by close? This depends on the context, but in the

context of the preceding programming example, we mean a polynomial p so that

kf − pk

∞

= max

x∈[a,b]

|f(x) − p(x)|

is small. This is known as uniform approximation. Such approximations are im-

portant both in practical work and in theory. Later in this chapter, there are several

methods for computing such approximations, including methods well adapted to

programming.

We start by looking at several plausible ways to answer this problem that do

not work.

It might seem that the Taylor polynomials of the previous section are a good

answer to this problem. However, there are several serious ﬂaws. First, and most

important, the function f may not be differentiable at all and f must have at least n

derivatives in order to compute P

. Moreover, the bound on the (n+1)st derivative

is the crucial factor in the error estimate, so f must have n + 1 derivatives. The

bounds may be so large that this estimate is useless. Second, even for very nice

functions, the Taylor series may converge only on a small interval about the point

a. Recall Example 10.1.6, which shows that the Taylor series for tan

−1

(x) only

converges on the small interval [−1, 1] even though the function is C

∞

on all of

R. Moreover, the convergence is very slow unless we further restrict the interval to

something like [−.5, .5]. Worse yet was f(x) = e

−1/x

of Example 10.1.9, which

is C

∞

and for which the Taylor series about x = 0 converges everywhere, but to

the wrong function. Because the Taylor series only uses information at one point,

it cannot be expected always to do a good job over an entire interval.

Differentiation is a very unstable process when the function is known only ap-

proximately in the uniform norm—a small error in evaluating the function can re-

sult in a huge error in the derivative (see Example 10.2.1). So even when the Taylor

series does converge, it can be difﬁcult to compute the coefﬁcients numerically.

10.2.1. EXAMPLE. One reason that Taylor polynomials fail is that they only

use information available at one point. Another failing is that they try too hard to

approximate derivatives at the same time. Here is an example where convergence

for the function is good, but not for the derivative. Let

(x) = x +

√

sinnx for − π ≤ x ≤ π.

It is easy to verify that f

converges uniformly on [−π, π] to the function

f(x) = x. Indeed,

kf − f

∞

= max

−π≤x≤π

√

|sinnx| =

√

for n ≥ 1.

However f

(x) = 1 everywhere, while f

(x) = 1 +

√

n cosnx. Therefore,

− f

∞

= max

−π≤x≤π

√

n |cosnx| =

√

n for n ≥ 1.

As the approximants f

get closer to f, they oscillate more and more dramatically.

So the derivatives are very far from the derivative of f.

10.2 How Not to Approximate a Function 287

Another possible method for ﬁnding good approximants is to use polynomial

interpolation. Pick n + 1 points distributed over [a, b], for example,

= a +

i(b − a)

for i = 0, 1, . . . , n.

There is a unique polynomial p

of degree at most n that goes through the n + 1

points (x

, f(x

)), i = 0, 1, . . . , n. One way to ﬁnd this polynomial is to solve the

system of linear equations in n + 1 variables a

, a

, . . . , a

+ x

+ ··· + x

= f (x

)

+ x

+ ··· + x

= f (x

)

+ x

+ ··· + x

= f (x

This determines the van der Monde matrix X and vectors a and f :

X =







1 x

. . . x

1 x

. . . x

1 x

. . . x







a =













and f =







f(x

)

f(x

)

f(x

)







The system becomes Xa = f . This always has a unique solution because the

matrix X is invertible (see Exercise 10.2.N). This determines the polynomial p

which agrees with f at the n + 1 points x

, . . . , x

It seems reasonable to suspect that as n increases, p

will converge uniformly

to f. However, this is not true. In 1901, Runge showed that the polynomial inter-

polants on [−5, 5] to the function

f(x) =

1 + x

do not converge to f. In fact, if p

is polynomial interpolant of degree at most n,

then

lim

n→∞

kf − p

∞

= ∞.

Proving this is more than a little tricky. If instead of choosing n + 1 equally spaced

points, we choose the points x

= 5 cos(iπ/n), i = 0, 1, . . . , n, then the interpolat-

ing polynomials p

(x) will converge uniformly to this particular function.

However, if you specify the points in advance, then no matter which points you

choose at each stage, there is some continuous function f ∈ C[a, b] so that the

interpolating polynomials of degree n do not converge uniformly to f. This was

proved by Bernstein and Faber independently in 1914.

There are algorithms using interpolation that involvevaryingthe points of inter-

polation strategically. There are also ways of making interpolation into a practical

method by using splines instead of polynomials. We’ll return to the latter idea in

the last two sections of this chapter.

After all of these negative results, we might wonder if it is possible to approx-

imate an arbitrary continuous function by a polynomial. It is a remarkable and

important theorem, proved by Weierstrass in 1885, that this is possible.

288 Approximation by Polynomials

10.2.2. WEIERSTRASS APPROXIMATION THEOREM.

Let f be any continuous real-valued function on [a, b]. Then there is a sequence of

polynomials p

that converges uniformly to f on [a, b].

In the language of normed vector spaces, this theorem says that the polynomials

are dense in C[a, b] in the max norm.

In fact, this theorem is sufﬁciently important that many different proofs have

been found. The proof we give was found in 1912 by Bernstein, a Russian mathe-

matician. It explicitly constructs the approximating polynomial. This algorithm is

not the most efﬁcient, but the problem of ﬁnding efﬁcient algorithms can wait until

we have proved that the theorem is true.

Exercises for Section 10.2

A. Assumethe Weierstrass Theorem is true for C[0, 1], and then proveit is true for C[a, b],

for an arbitrary interval [a, b].

HINT: For f ∈ C[a, b], consider g(t) := f

a + (b − a)t

in C[0, 1].

B. Let α > 0. Using the Weierstrass Theorem, prove that every continuous function

f : [0, +∞] → R with lim

x→∞

f(x) = 0 can be uniformly approximated as closely as

we like by a function of the form q(x) =

n=1

−nαx

HINT: Consider g(y) = f(−log(y)/α) on (0, 1].

C. (a) Show that every continuous function f on [a, b] is the uniform limit of polynomials

of the form p

(b) Describe the subspace of C[−1, 1] functions which are uniform limits of polyno-

mials of the form p

D. Suppose that f is a continuous function on [0, 1] such that

f(x)x

dx = 0 for all

n ≥ 0. Prove that f = 0.

HINT: Use the Weierstrass Theorem to show that

|f(x)|

dx = 0.

E. Let X be a compact subset of [−N, N].

(a) Show that every continuous function f on X may be extended to a continuous

function g deﬁned on [−N, N ] with kgk

∞

= kfk

∞

(b) Show that every continuous function on X is the uniform limit of polynomials.

F. Show that if f is continuously differentiable on [0, 1], then there is a sequence of

polynomials p

converging uniformly to f such that p

converge uniformly to f

well. HINT: Approximate f

ﬁrst.

G. Show that if f is in C

∞

[0, 1], then there is a sequence p

such that the kth derivatives

(k)

converge uniformly to f

(k)

for every k ≥ 0.

HINT: Adapt Exercise 10.2.F to ﬁnd p

with kf

(k)

− p

(k)

∞

for 0 ≤ k ≤ n.

H. Prove that e

is not a polynomial. HINT: Consider behaviour at ±∞.

I. (a) If 0 6∈ [a, b], show that every continuous function f on [a, b] is the uniform limit of

a sequence of polynomials (q

), where q

(x) = x

(x) for polynomials p

10.2 How Not to Approximate a Function 289

(b) If 0 ∈ [a, b], show that a continuous function f on [a, b] is the uniform limit of a

sequence of polynomials (q

), where q

(x) = x

(x) for polynomials p

, if and

only if f(0) = 0.

J. (a) If x

, . . . , x

are points in [a, b] and a = (a

, . . . , a

) ∈ R

n+1

, show that there is a

unique polynomial p

in P

such that p(x

) = a

for 0 ≤ i ≤ n.

HINT: Find polynomials q

such that q

) is 1 if i = j and is 0 if 0 ≤ i 6= j ≤ n.

(b) Show that there is a constant M so that kp

∞

≤ Mkak

K. Suppose that f ∈ C[a, b], ε > 0 and x

, . . . , x

are points in [a, b]. Prove that there is

a polynomial p such that p(x

) = f(x

) for 1 ≤ i ≤ n and kf − pk

∞

< ε.

HINT: First approximate f closely by some polynomial. Then use the previous exer-

cise to adjust the difference.

L. Show that every continuous function h(x, y) on [a, b] × [c, d] is the uniform limit of

polynomials in two variables as follows:

(a) Show that every function of the form f (x)g(y), where f ∈ C[a, b] and g ∈ C[c, d]

is the uniform limit of polynomials.

(b) Let k(x) = max{1 − |x|, 0}. Set k

(n)

(x, y) = k(2

x − i)k(2

y − j) for i, j ∈ Z

and n ≥ 0. Show that

(n)

(x, y) = 1.

(x, y) =

h(2

−n

i, 2

−n

j)k

(n)

(x, y). Show that h

is the uniform

limit of polynomials.

(d) Use the uniform continuity of h to prove that h

converges to h uniformly.

M. Let α > 0 and ε > 0. Using the following outline, show by induction that for each

positive integer n, there is a polynomial p so that |e

−nαx

− e

−αx

p(x)| < ε for all

x ∈ [0, +∞).

(a) Show that for each n, it sufﬁces to prove the result for α = 1.

(b) For n = 2, ﬁrst show that for f(x) = e

−2x

− e

−x

N−1

k=0

(−x)

/k! satisﬁes

kfk

∞

≤

sup{e

−x

: x ≥ 0}. HINT: Example 10.1.4.

−x

, show that g has a unique maximum at x = N and

hence |f(x)| ≤ K/

√

N for some constant K. HINT: Stirling’s formula.

(d) Deduce that the result holds for n = 2.

(e) For the inductive step, ﬁnda polynomial q with |e

−(n+1)x

−e

−nx/2−x/2

q(x)| ≤ ε/2

for all x ∈ [0, +∞]. HINT: Use part (d) for e

−2αx

with α = (n + 1)/2.

(f) Next, ﬁnd a polynomial r so that |e

−nx/2−x/2

q(x) − e

−x

q(x)r(x)| ≤ ε/2 for all

x ∈ [0, +∞). HINT: Note that h(x) = e

−x/2

|q(x)| is bounded on [0, +∞) and

use the inductive result for e

−nαx

with α = 1/2.

(g) Deduce that the result holds for e

−(n+1)x

(h) Finally, use this result and Exercise 10.2.B to show that every continuous function

f : [0, +∞) → R with lim

n→∞

f(x) = 0 can be uniformly approximated as closely

as we like by a function of the form e

−αx

p(x) for some polynomial p.

N. (a) Show that the determinant of the van der Monde matrix X is a polynomial of degree

n(n + 1)/2 in the variables x

, . . . , x

(b) Show that the determinant is 0 if x

= x

for some 0 ≤ i < j ≤ n. Hence show

that it is a scalar multiple of

0≤i<j≤n

− x

). By looking at the coefﬁcient of

. . . x

in both the determinant and the product, show that the scalar is 1.

, . . . , x

are distinct.

290 Approximation by Polynomials

10.3. Bernstein’s Proof of the Weierstrass Theorem

Recall the binomial formula, (x + y)

k=0

n−k

. If we set y = 1 − x,

then we obtain

1 =

k=0

(1 − x)

n−k

Bernstein started by considering the functions

(x) =

(1 − x)

n−k

for k = 0, 1, . . . , n,

now called Bernstein polynomials. They have several virtues. They are all poly-

nomials of degree n. They take only nonnegative values on [0, 1]. And they add

up to 1. Moreover, P

is a “bump” function with a maximum at k/n, as a routine

calculus calculation shows. For example, the four functions P

for 0 ≤ k ≤ 3 are

given in Figure 10.3.

Given a continuous function f on [0, 1], deﬁne a polynomial B

f by

f)(x) =

k=0

(x) =

k=0

(1 − x)

n−k

This is a linear combination of the polynomials P

; and so B

f is a polynomial of

degree at most n. We think of B

as a function from the vector space C[0, 1] into

itself. This map has several easy but important properties. If f, g ∈ C[0, 1], we say

that f ≥ g if f(x) ≥ g(x) for all 0 ≤ x ≤ 1.

10.3.1. PROPOSITION. The map B

is linear and monotone. That is, for all

f, g ∈ C[0, 1] and α ∈ R,

(1) B

(f + g) = B

f + B

(2) B

(αf) = αB

(3) B

f ≥ 0 if f ≥ 0

(4) B

f ≥ B

g if f ≥ g

(5) |B

f| ≤ B

g if |f| ≤ g.

The only part that requires any cleverness is the monotonicity. However, since

each P

≥ 0, it follows that when f ≥ 0, then B

f is also positive. In particular,

|f| ≤ g means that −g ≤ f ≤ g; and hence −B

g ≤ B

f ≤ B

g. The details are

left to the reader.

Next let us compute B

f for three basic polynomials: 1, x, and x

10.3.2. LEMMA. B

1 = 1, B

x = x, and

n − 1

x = x

x − x

10.3 Bernstein’s Proof of the Weierstrass Theorem 291

FIGURE 10.3. The Bernstein polynomials of degree 3.

PROOF. For the ﬁrst equation, we use the Binomial Theorem to get

1 =

k=0

(1 − x)

n−k

= 1.

Next, notice that

k!(n − k)!

(n − 1)!

(k − 1)!(n − k)!

n − 1

k − 1

Using this result and the Binomial Theorem, we have

x =

k=0

(1 − x)

n−k

= x

k=1

n − 1

k − 1

k−1

(1 − x)

n−k

= x

x + (1 −x)

n−1

= x.

Finally, notice that

k!(n − k)!

(k − 1) + 1

(n − 1)!

(k − 1)!(n − k)!

n − 1

k − 1

n − 1

(n − 1)!

(k − 1)!(n − k)!

(n − 1)!

(k − 1)!(n − k)!

n − 1

n − 2

k − 2

n − 1

k − 1

292 Approximation by Polynomials

Note that

−2

−1

= 0. Hence

k=0

(1 − x)

n−k

n − 1

k=2

n − 2

k − 2

(1 − x)

n−k

k=1

n − 1

k − 1

(1 − x)

n−k

n−1

k=2

n−2

k−2

(1 − x)

n−k

k=1

n−1

k−1

(1 − x)

n−k

n − 1

x + (1 −x)

n−2

x + (1 −x)

n−1

n − 1

x = x

x − x

PROOF OF WEIERSTRASS’S THEOREM. By Exercise10.2.A, it sufﬁces to prove

the theorem for the interval [0, 1]. Fix a continuous function f in C[0, 1]. We will

prove that for each ε > 0, there is some N > 0 so that

kf(x) − B

f(x)k < ε for all n ≥ N.

Since [0, 1] is compact, f is uniformly continuous on [0, 1] by Theorem 5.5.9.

Thus for our given ε > 0, there is some δ > 0 so that

|f(x) − f(y)| ≤

for all |x − y| ≤ δ, x, y ∈ [0, 1].

Also, f is bounded on [0, 1] by the Extreme Value Theorem (Theorem 5.4.4). So

let

M = kf k

∞

= sup

x∈[0,1]

|f(x)|.

Fix any point a ∈ [0, 1]. We claim that

|f(x) − f(a)| ≤

(x − a)

Indeed, if |x − a| ≤ δ, then

|f(x) − f(a)| ≤

≤

(x − a)

by our estimate of uniform continuity. And if |x − a| ≥ δ, then

|f(x) − f(a)| ≤ 2M ≤ 2M

x − a

≤

(x − a)

Notice that by linearity and the fact that B

1 = 1, we obtain

(f − f(a))(x) = B

f(x) − f(a).