Taylor's Theorem

We have so far explored the Mean Value theorem, which can be rewritten as

f (a + h) = f (a) + hf'(c)

where c is some point between a and a + h. [By writing the definition of c in this way, we have a statement that works whether h > 0 or h < 0.] We have already met the approximation

f (a + h) $$\sim$$ f (a) + hf'(a)

when we studied the Newton - Raphson method for solving an equation, and have already observed that the Mean Value Theorem provides a more accurate version of this. Now consider what happens when f is a polynomial of degree n

f (x) = a₀ + a₁x + a₂x² +...+ a_n-1x^n-1 + a_nxⁿ.

Note that f (0) = a₀. Differentiating gives

f'(x) = a₁ + 2a₂x + 3a₃x²...+ (n - 1)a_n-1x^n-2 + na_nx^n-1,

and so f'(0) = a₁. Again, we have

f''(x) = 2a₂ + 3.2a₃x...+ (n - 1)(n - 2)a_n-1x^n-3 + n(n - 1)a_nx^n-2,

and f''(0) = 2a₂. After the next differentiation, we get f'''(0) = 3!a₃, while after k differentiations, we get, f^(k)(0) = k!a_k, provided k$\le$n. Thus we can rewrite the polynomial, using its value, and the value of its derivatives at 0, as

f (x) = f (0) + f'(0)x + $${\frac{{f''(0)}}{{2!}}}$$x² + $${\frac{{f'''(0)}}{{3!}}}$$x³ +...+ $${\frac{{f^{(n-1)}(0)}}{{(n-1)!}}}$$x^n-1 + $${\frac{{f^{(n)}(0)}}{{n!}}}$$xⁿ.

This opens up the possibility of representing more general functions than polynomials in this way, and so getting a generalisation of the Mean Value Theorem.

Theorem 5.29 (Taylors Theorem - Lagrange form of Remainder)   Let f be continuous on [a, x], and assume that each of f', f'',..., f⁽ⁿ⁺¹⁾ is defined on [a, x]. Then we can write

f (x) = P_n(x) + R_n(x),

where P_n(x), the Taylor polynomial of degree n about a, and R_n(x), the corresponding remainder, are given by

$${\frac{{f''(a)}}{{2!}}} {\frac{{f^{(n)}(a)}}{{n!}}}$$ P_n(x) =

$${\frac{{f^{(n+1)}(c)}}{{(n+1)!}}}$$ R_n(x) =

where c is some point between a and x.

We make no attempt to prove this, although the proof can be done with the tools we have at our disposal. Some quick comments:

• the theorem is also true for x < a; just restate it for the interval [x, a] etc;
• if n = 0, we have f (x) = f (a) + (x - a)f'(c) for some c between a and x; this is a restatement of the Mean Value Theorem;
• if n = 1, we have
  f (x) = f (a) + (x - a)f'(x) + $${\frac{{f''(c)}}{{2!}}}$$(x - a)²
  for some c between a and x; this often called the Second Mean Value Theorem;
• in general we can restate Taylor's Theorem as
  f (x) = f (a) + (x - a)f'(x) +...+ $${\frac{{f^{(n)}(a)}}{{2!}}}$$(x - a)ⁿ + $${\frac{{f^{(n+1)}(c)}}{{(n+1)!}}}$$(x - a)ⁿ⁺¹,
  for some c between a and x;
• the special case in which a = 0 has a special name; it is called Maclaurin's Theorem;
• just as with Rolle, or the Mean Value Theorem, there is no useful information about the point c.

We now explore the meaning and content of the theorem with a number of examples.

Example 5.30   Find the Taylor polynomial of order n about 0 for f (x) = e$\nolimits^{x}_{}$, and write down the corresponding remainder term.

Solution. There is no difficulty here in calculating derivatives -- clearly f^(k)(x) = e$\nolimits^{x}_{}$ for all k, and so f^(k)(0) = 1. Thus by Taylor's theorem,

e$$\nolimits^{x}_{}$$ = 1 + x + $${\frac{{x^2}}{{2!}}}$$ + $${\frac{{x^2}}{{2!}}}$$ +...$${\frac{{x^n}}{{n!}}}$$ + $${\frac{{x^{n+1}}}{{(n+1)!}}}$$ e$$\nolimits^{c}_{}$$

for some point c between 0 and x. In particular,

P_n(x) = 1 + x + $${\frac{{x^2}}{{2!}}}$$ + $${\frac{{x^2}}{{2!}}}$$ +...$${\frac{{x^n}}{{n!}}}$$    and    R_n(x) = $${\frac{{x^{n+1}}}{{(n+1)!}}}$$ e$$\nolimits^{c}_{}$$.

We can actually say a little more about this example if we recall that x is fixed. We have

e^x = P_n(x) + R_n(x) = P_n(x) + $${\frac{{x^{n+1}}}{{(n+1)!}}}$$ e$$\nolimits^{c}_{}$$

We show that R_n(x)$\to$ 0 as n$\to$$\infty$, so that (again for fixed x), the sequence P_n(x)$\to$e^x as n$\to$$\infty$. If x < 0, e$\nolimits^{c}_{}$ < 1, while if x$\ge$1, then since c < x, we have e$\nolimits^{c}_{}$ < e$\nolimits^{x}_{}$. thus

$$\left\vert\vphantom{R_n(x)}\right.$$R_n(x)$$\left.\vphantom{R_n(x)}\right\vert$$ = $$\left\vert\vphantom{\frac{x^{n+1}}{(n+1)!}\mathop\mathrm{e}\nolimits ^c}\right.$$$${\frac{{x^{n+1}}}{{(n+1)!}}}$$ e$$\nolimits^{c}_{}$$$$\left.\vphantom{\frac{x^{n+1}}{(n+1)!}\mathop\mathrm{e}\nolimits ^c}\right\vert$$$$\le$$$${\frac{{\vert x\vert^{n+1}}}{{(n+1)!}}}$$max( e$$\nolimits^{x}_{}$$, 1)$$\to$$0    as n$\to$$\infty$.    

We think of the limit of the polynomial as forming a series, the Taylor series for e$\nolimits^{x}_{}$. We study series (and then Taylor series) in Section 7.

Example 5.31   Find the Taylor polynomial of order 1 about a for f (x) = e$\nolimits^{x}_{}$, and write down the corresponding remainder term.

Solution. Using the derivatives computed above, by Taylor's theorem,

e$$\nolimits^{x}_{}$$ = e$$\nolimits^{a}_{}$$ + (x - a) e$$\nolimits^{a}_{}$$ + $${\frac{{(x-a)^2}}{{2!}}}$$ e$$\nolimits^{c}_{}$$

for some point c between a and x. In particular,

P₁(x) = e$$\nolimits^{a}_{}$$ + (x - a) e$$\nolimits^{a}_{}$$    and    R₁(x) = $${\frac{{(x-a)^2}}{{2!}}}$$ e$$\nolimits^{c}_{}$$.

Example 5.32   Find the Maclaurin polynomial of order n > 3 about 0 for f (x) = (1 + x)³, and write down the corresponding remainder term.

Solution. We have

        $$\begin{array}{l} f(0) = 1\\ f'(x) = 3\\ f''(x) = 6\\ f'''(x) = 6\\ \end{array}$$

and so, by Taylor's theorem

(1 + x)³ = 1 + 3x + $${\frac{{6}}{{2!}}}$$x² + $${\frac{{6}}{{3!}}}$$x³,

a result we could have got directly, but which is at least reassuring.

Example 5.33   Find the Taylor polynomial of order n about 0 for f (x) = sin x, and write down the corresponding remainder term.

Solution. There is no difficulty here in calculating derivatives -- we have

        $$\begin{array}{l} f(0) = 0\\ f'(x) = 1\\ f''(x) = 0\\ f'''(x) = -1.\\ \end{array}$$

Thus by Taylor's theorem,

sin x = x - $${\frac{{x^3}}{{3!}}}$$ + $${\frac{{x^5}}{{5!}}}$$ +...+ (- 1)ⁿ⁺¹$${\frac{{x^{2n+1}}}{{(2n+1)!}}}$$ +...

Writing down the remainder term isn't particularly useful, but the important point is that

| R_2n+1(x)|$$\le$$$$\left\vert\vphantom{ \frac{x^{2n+3}}{(2n+3)!}}\right.$$$${\frac{{x^{2n+3}}}{{(2n+3)!}}}$$$$\left.\vphantom{ \frac{x^{2n+3}}{(2n+3)!}}\right\vert$$$$\to$$0    as n$\to$$\infty$.    

Exercise 5.34   Recall that cosh x = $${\frac{{\mathop\mathrm{e}\nolimits ^x + \mathop\mathrm{e}\nolimits ^{-x}}}{{2}}}$$, and that sinh x = $${\frac{{\mathop\mathrm{e}\nolimits ^x - \mathop\mathrm{e}\nolimits ^{-x}}}{{2}}}$$. Now check the shape of the following Taylor polynomials:

$${\frac{{x^2}}{{2!}}} {\frac{{x^4}}{{4!}}} {\frac{{x^{2n}}}{{2n!}}}$$ cos x =

$${\frac{{x^3}}{{3!}}} {\frac{{x^5}}{{5!}}} {\frac{{x^{2n+1}}}{{(2n+1)!}}}$$ sinh x =

$${\frac{{x^2}}{{2!}}} {\frac{{x^4}}{{4!}}} {\frac{{x^{2n}}}{{2n!}}}$$ cosh x =

Example 5.35   Find the maximum error in the approximation

sin(x) $$\sim$$ x - $${\frac{{x^3}}{{3!}}}$$

given that | x| < 1/2.

Solution. We use the Taylor polynomial for sin x of order 4 about 0, together with the corresponding remainder. Thus

sin x = x - $${\frac{{x^3}}{{3!}}}$$ + $${\frac{{x^5}}{{5!}}}$$cos c

for some c with 0 < c < 1/2 or -1/2 < c < 0. In any case, since | x| < 1/2,

$$\left\vert\vphantom{ \frac{x^5}{5!}\cos c }\right.$$$${\frac{{x^5}}{{5!}}}$$cos c$$\left.\vphantom{ \frac{x^5}{5!}\cos c }\right\vert$$$$\le$$$$\left\vert\vphantom{ \frac{x^5}{5!}}\right.$$$${\frac{{x^5}}{{5!}}}$$$$\left.\vphantom{ \frac{x^5}{5!}}\right\vert$$$$\le$$$$\left\vert\vphantom{ \frac{1}{2^5.5!}}\right.$$$${\frac{{1}}{{2^5.5!}}}$$$$\left.\vphantom{ \frac{1}{2^5.5!}}\right\vert$$$$\le$$$${\frac{{1}}{{120.32}}}$$.

Warning:The Taylor polynomial always exists, providing f is suitably differentiable. But it need not be useful. Consider the example

f (x) = $$\left\{\vphantom{\begin{array}{ll} \exp(-1/x^2)&\mbox{if $x > 0$;}\\ 0&\mbox{if $x \le 0$.} \end{array}}\right.$$$$\begin{array}{ll} \exp(-1/x^2)&\mbox{if $x > 0$;}\\ 0&\mbox{if $x \le 0$.} \end{array}$$

The interest in f is at 0; it is well behaved everywhere else. It turns out that

f (0) = f'(0) = f''(0) =...= f⁽ⁿ⁾(0) =...= 0.

So the Taylor polynomial of degree n for f about 0 is P_n(x) = 0 + 0x + 0x² +...+ 0.xⁿ = 0, and so for every n, R_n(x) = f (x). Clearly in this case, P_n tells us nothing useful about the function.

Example 5.36   Find the Taylor polynomial of order n about 0 for f (x) = (1 + x)^$\alpha$, and note that this gives a derivation of the binomial theorem. In fact, the remainder | R_n(x)|$\to$ 0 as n$\to$$\infty$, provided | x| < 1.

Solution. There is again no difficulty here in calculating derivatives -- we have

$$\alpha$$

$$\alpha \alpha$$

$$\alpha \alpha \alpha$$

$$\alpha \alpha \alpha \alpha$$     and in general    

$$\alpha \alpha \alpha \alpha$$

        $$\begin{array}{l} f(0) = 1\\ f'(x) = \alpha\\ f''(x) = \alp... ...ad\text{}\quad \\ f^{(n)} = \alpha(\alpha-1)\ldots(\alpha-n+1). \end{array}$$

Thus by Taylor's theorem,

$$\alpha \alpha {\frac{{\alpha(\alpha-1)}}{{2!}}} {\frac{{\alpha(\alpha-1)(\alpha-2)}}{{3!}}}$$

$${\frac{{\alpha(\alpha-1)\ldots(\alpha-n+1)}}{{n!}}}$$

The remainder is not hard to deal with, but we omit the proof; in fact | R_n(x)|$\to$ 0 when n$\to$$\infty$.

Note also that if $\alpha$ > 0 is an integer, say $\alpha$ = n then | R_n(x)| = 0 and f (x) = P_n(x). This is another way to get the Binomial theorem described in Section 1.8.