TAYLOR’S FORMULA FOR SINGLE-VARIABLE FUNCTIONS - Multivariable Differential Calculus - Advanced Calculus of Several Variables

Advanced Calculus of Several Variables (1973)

Part II. Multivariable Differential Calculus

Chapter 6. TAYLOR'S FORMULA FOR SINGLE-VARIABLE FUNCTIONS

In order to generalize the results of Section 4, and in particular to apply the Lagrange multiplier method to classify critical points for functions of n variables, we will need Taylor‘s formula for functions on Imagen. As preparation for the treatment in Section 7 of the multivariable Taylor’s formula, this section is devoted to the single-variable Taylor's formula.

Taylor's formula provides polynomial approximations to general functions. We will give examples to illustrate both the practical utility and the theoretical applications of such approximations.

If f: Image is differentiable at a, and R(h) is defined by

Image

then it follows immediately from the definition of f′(a) that

Image

With x = a + h, (1) and (2) become

Image

where

Image

The linear function P(x − a) = f(a) + f′(a)(x − a) is simply that first degree polynomial in (x − a) whose value and first derivative at a agree with those of f at a. The kth degree polynomial in (x − a), such that the values of it and of its first k derivatives at a agree with those of f and its first k derivatives f′, f′, f(3), . . . , f(k) at a, is

Image

This fact may be easily checked by repeated differentiation of Pk(x − a). The polynomial Pk(x − a) is called the kth degree Taylor polynomial of f at a.

The remainder f(x) − Pk(x − a) is denoted by Rk(x − a), so

Image

With x − a = h, this becomes

Image

where

Image

In order to make effective use of Taylor polynomials, we need an explicit formula for Rk(x − a) which will provide information as to how closely Pk(x − a) approximates f(x) near a. For example, whenever we can show that

Image

this will mean that f is arbitrarily closely approximated by its Taylor polynomials; they can then be used to calculate f(x) as closely as desired. Equation (4), or (4′), together with such an explicit expression for the remainder Rk, is referred to as Taylor's formula. The formula for Rk given in Theorem 6.1 below is known as the Lagrange form of the remainder.

Theorem 6.1 Suppose that the (k + 1)th derivative f(k+1) of f: Image exists at each point of the closed interval I with endpoints a and x. Then there exists a point ζ between a and x such that

Image

Hence

Image

or

Image

with h = x − a.

REMARK This is a generalization of the mean value theorem; in particular, P0(x − a) = f(a), so the case k = 0 of the theorem is simply the mean value theorem

Image

for the function f on the interval I. Moreover the proof which we shall give for Taylor's formula is a direct generalization of the proof of the mean value theorem. So for motivation we review the proof of the mean value theorem (slightly rephrased).

First we define R0(t) for Image (for convenience we assume h > 0) by

Image

and note that

Image

while

Image

Then we define φ: [0, h] Image by

Image

where the constant K is chosen so that Rolle's theorem [the familiar fact that, if f is a differentiable function on [a, b] with f(a) = f(b) = 0, then there exists a point Image will apply to φ on [0, h], that is,

Image

so it follows that φ(h) = 0. Hence Rolle's theorem gives a Image such that

Image

Hence K = f′(ζ) where ζ = a + Image, so from (9) we obtain R0(h) = f′(ζ)h as desired.

PROOF OF THEOREM 6.1 We generalize the above proof, labeling the formulas with the same numbers (primed) to facilitate comparison.

First we define Rk(t) for Image by

Image

and note that

Image

while

Image

The reason for (6′) is that the first k derivatives of Pk(x − a) at a, and hence the first k derivatives of Pk(t) at 0, agree with those of f at a, while (7′) follows from the fact that Image because Pk(t) is a polynomial of degree k.

Now we define φ: [0, h] Image by

Image

where the constant K is chosen so that Rolle's theorem will apply to φ on [0, h], that is,

Image

so it follows that φ(h) = 0. Hence Rolle's theorem gives a point Image such that φ′(Image1) = 0.

It follows from (6′) and (7′) that

Image

while

Image

Therefore we can apply Rolle's theorem to φ′ on the interval [0, Image1] to obtain a point Image such that φ′(Image2) = 0.

Image

By (10), φ′ satisfies the hypotheses of Rolle‘s theorem on [0, Image2], so we can continue in this way. After k + 1 applications of Rolle’s theorem, we finally obtain a point Image such that φ(k+1)(tk+1) = 0. From the second equation in (10) we then obtain

Image

with ζ = a + Imagek+1. Finally (9′) gives

Image

as desired.

Image

Corollary 6.2 If, in addition to the hypotheses of Theorem 6.1, Image

Image

It follows that

Image

In particular, (12) holds if f(k+1) is continuous at a, because it will then necessarily be bounded (by some M) on some open interval containing a.

Example 1 As a standard first example, we take f(x) = ex, a = 0. Then f(k)(x) = ex, so f(k)(0) = 1 for all k. Then

Image

and

Image

for some ζ between 0 and x. Therefore

Image

In either case the elementary fact that

Image

implies that limk→∞ Rk(x) = 0 for all x, so

Image

[To verify the elementary fact used above, choose a fixed integer m such that Image. If k > m, then

Image

as k → ∞.]

In order to calculate the value of ex with preassigned accuracy by simply calculating Pk(x), we must be able to estimate the error Rk(x). For this we need the preliminary estimate e < 4. Since log e = 1 and log x is a strictly increasing function, to verify that e < 4 it suffices to show that log 4 > 1. But

Image

From (see 13) we now see that Rk(x) < 4/(k + 1)! if Image this can be used to compute e to any desired accuracy (Exercise 6.1).

Example 2 To calculate Image, and consider the first degree Taylor formula

Image

Image.

Image

Since Image

Image

so we conclude that

Image

(actually Image to five places).

The next two examples give a good indication of the wide range of application of Taylor's formula.

Example 3 We show that the number e is irrational. To the contrary, suppose that e = p/q where p and q are positive integers. Since e = 2.718 to three decimal places (see Exercise 6.1), it is clear that e is not an integral multiple of 1, Image, By Example 1, we can write

Image

where

Image

since 0 < ζ < 1 and e < 3. Upon multiplication of both sides of the above equation by q!, we obtain

Image

But this is a contradiction, because the left-hand side (q − 1)! p is an integer, but the right-hand side is not, because

Image

since q > 3.

Example 4 We use Taylor's formula to prove that, if f′ − f = 0 on R and f(0) = f′(0) = 0, then f = 0 on Image.

Since f′ = f, we see by repeated differentiation that f(k) exists for all k; in particular,

Image

Since f(0) = f′(0) = 0, it follows that f(k)(0) = 0 for all k. Consequently Theorem 4.1 gives, for each k, a point Image such that

Image

Since there are really only two different derivatives involved, and each is continuous because it is differentiable, there exists a constant M such that

Image

Hence Image so we conclude that f(x) = 0.

Now we apply Taylor's formula to give sufficient conditions for local maxima and minima of real-valued single-variable functions.

Theorem 6.3 Suppose that f(k+1) exists in a neighborhood of a and is continuous at a. Suppose also that

Image

but f(k)(a) ≠ 0. Then

(a)f has a local minimum at a if k is even, and f(k)(a) > 0;

(b)f has a local maximum at a if k is even and f(k)(a) < 0;

(c)f has neither a maximum nor a minimum at a if k is odd.

This is a generalization of the familiar “second derivative test” which asserts that, if f′(a) = 0, then f has a local minimum at a if f′(a) > 0, and a local maximum at a if f′(a) < 0. The three cases can be remembered by thinking of the three graphs in Fig. 2.35.

If f(k)(a) = 0 for all k, then Theorem 6.3 provides no information as to the behavior of f in a neighborhood of a. For instance, if

Image

then it turns out that f(k)(0) = 0 for all k, so Theorem 6.3 does not apply. However it is obvious that f has a local minimum at 0, since f(x) > 0 for x ≠ 0 (Fig. 2.36).

Image

Figure 2.35

As motivation for the proof of Theorem 6.3, let us consider first the “second-derivative test.” If f′(a) = 0, then Taylor's formula with k = 2 is

Image

where limx→a R2(x − a)/(x − a)2 = 0 by Corollary 6.2 (assuming that f(3) is continuous at a). By transposing f(a) and dividing by (x − a)2, we obtain

Image

so it follows that

Image

If f′(a) > 0, this implies that f(x) − f(a) > 0 if x is sufficiently close to a, since (x − a)2 > 0 for all x ≠ a. Thus f(a) is a local minimum. Similarly f(a) is a local maximum if f′(a) < 0.

Image

Figure 2.36

In similar fashion we can show that, if f′(a) = f′(a) = 0 while f(3)(a) ≠ 0, then f has neither a maximum nor a minimum at a (this fact might be called the “third-derivative test”). To see this, we look at Taylor's formula with k = 3,

Image

where limx→a R3(x − a)/(x − a)3 = 0. Transposing f(a) and then dividing by (x − a)3, we obtain

Image

so it follows that

Image

If, for instance, f(3)(a) > 0, we see that [f(x) − f(a)]/(x − a)3 > 0 if x is sufficiently close to a. Since (x − a)3 > 0 if x > a and (x − a)3 < 0 if x < a, it follows that, for x sufficiently close to a, f(x) − f(a) < 0 if x < a, and f(x) − f(a) < 0 if x < a. These inequalities are reversed if f(3)(a) < 0. Consequently f(a) is neither a local maximum nor a local minimum.

The proof of Theorem 6.3 simply consists of replacing 2 and 3 in the above discussion by k, the order of the first nonzero derivative of f at the critical point a. If k is even the argument is the same as when k = 2, while if k is odd it is the same as when k = 3.

PROOF OF THEOREM 6.3 Because of the hypotheses, Taylor's formula takes the form

Image

where limx→a Rk(x − a)/(x − a)k = 0 by Corollary 6.2. If we transpose f(a), divide by (x − a)k, and then take limits as x → a, we therefore obtain

Image

In case (a), limx→a[f(x) − f(a)]/(x − a)k > 0 by (14), so it follows that there exists a Δ > 0 such that

Image

Since k is even in this case, (x − a)k > 0 whether x > a or x < a, so

Image

Therefore f(a) is a local minimum.

The proof in case (b) is the same except for reversal of the inequalities.

In case (c), supposing f(k)(a) > 0, there exists (just as above) a Δ > 0 such that

Image

But now, since k is odd, the sign of (x − a)k depends upon whether x < a or x > a. The same is then true of f(x) − f(a), so f(x) < f(a) if x > a, and f(x) > f(a) if x > a; the situation is reversed if f(k)(a) < 0. In either event it is clear that f(a) is neither a local maximum nor a local minimum.

Image

Let us look at the case k = 2 of Theorem 6.3 in a bit more detail. We have

Image

where limx→a R2(x − a)/(x − a)2 = 0. Therefore, given Image, there exists a Δ > 0 such that

Image

which implies that

Image

Substituting (16) into (15), we obtain

Image

If f′(a) > 0, then Image are both positive because Image. It follows that the graphs of the equations

Image

are then parabolas opening upwards with vertex (a, f(a)) (see Fig. 2.37). The fact that inequality (17) holds if 0 < Imagex − aImage < Δ means that the part of the graph of y = f(x), over the interval (a − Δ, a + Δ), lies between these two parabolas. This makes it clear that f has a local minimum at a if f′(a) > 0.

The situation is similar if f′(a) < 0, except that the parabolas open downward, so f has a local maximum at a.

In the case k = 3, these parabolas are replaced by the cubic curves

Image

which look like Fig. 2.35c above, so f has neither a maximum nor a minimum at a.

Image

Figure 2.37

Exercises

6.1Show that e = 2.718, accurate to three decimal places. Hint: Refer to the error estimate at the end of Example 1; choose k such that 4/(k + 1)! < 10−4.

6.2Prove that, if we compute ex by the approximation

Image

then the error will not exceed 0.001 if Image. Then compute Image accurate to two decimal places.

6.3If Image.Conclude that

Image

is the only kth degree polynomial in (x − a) such that the values of it and its first k derivatives at a agree with those of f at a.

6.4(a) Show that the values of the sine function for angles between 40° and 50° can be computed by means of the approximation

Image

with 4-place accuracy. Hint: With f(x) = sin x, a = π/4, k = 3, show that the error is less than 10−5, since 5° = π/36 < 1/10 rad.

(b) Compute sin 50°, accurate to four decimal places.

6.5Show that

Image

for all x.

6.6Show that the kth degree Taylor polynomial of f(x) = log x at a = 1 is

Image

and that limk→∞ Rk(x − 1) = 0 if xImage(1,2). Then compute Image with error < 10−3

Hint: Show by induction that f(k)(x) = (−1)k−1(k − 1)!/xk.

6.7If f′(x) = f(x) for all x, show that there exist constants a and b so that

Image

Hint: Let g(x) = f(x) − a exb ex, show how to choose a and b so that g(0) = g′(0) = 0. Then apply Example 4.

6.8If α is a fixed real number and n is a positive integer, show that the nth degree Taylor polynomial at a = 0 for

Image

is Image, where the “binomial coefficient” Image is defined by

Image

(remember that 0! = 1). If α = n, then

Image

so it follows that

Image

since Rn(x) ≡ 0, because f(n+1)(x) ≡ 0.

If α is not an integer, then Image ≠ 0 for all j, so the series Image is infinite. The binomial theorem asserts that this infinite series converges to f(x) = (1 + x)α if ImagexImage < 1, and can be proved by showing that limn→∞ Rn(x) = 0 for ImagexImage < 1.

6.9Locate the critical points of

Image

and apply Theorem 6.3 to determine the character of each. Hint: Do not expand before differentiating.

6.10Let f(x) = x tan−1 x − sin2 x. Assuming the fact that the sixth degree Taylor polynomials at a = 0 of tan−1 x and sin2 x are

Image

respectively, prove that

Image

where limx→0 R(x) = 0. Deduce by the proof of Theorem 6.3 that f has a local minimum at 0.

Contemplate the tedium of computing the first six derivatives of f. If one could endure it, he would find that

Image

but f(6)(0) = 112 > 0, so the statement of Theorem 6.3(a) would then give the above result.

6.11 (a) This problem gives a form of “l‘Hospital’s rule.” Suppose that f and g have k + 1 continuous derivatives in a neighborhood of a, and that both f and g and their first k − 1 derivatives vanish at a. If g(k)(a) ≠ 0, prove that

Image

Hint: Substitute the kth degree Taylor expansions of f(x) and g(x), then divide numerator and denominator by (x − a)k before taking the limit as x → a.

(b)Apply (a) with k = 2 to evaluate

Image

6.12In order to determine the character of f(x) = (ex − 1)(tan−1(x) − x) at the critical point 0, substitute the fourth degree Taylor expansions of ex and tan−1 x to show that

Image

where limx→0 R4(x)/x4 = 0. What is your conclusion?