
# 11.11: Taylor's Theorem

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554
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One of the most important uses of infinite series is the potential for using an initial portion of the series for $$f$$ to approximate $$f$$. We have seen, for example, that when we add up the first $$n$$ terms of an alternating series with decreasing terms that the difference between this and the true value is at most the size of the next term. A similar result is true of many Taylor series.

Theorem 11.11.1: Taylor's Theorem

Suppose that $$f$$ is defined on some open interval $$I$$ around $$a$$ and suppose $$f^{(N+1)}(x)$$ exists on this interval. Then for each $$x\not=a$$ in $$I$$ there is a value $$z$$ between $$x$$ and $$a$$ so that $$f(x) = \sum_{n=0}^N {f^{(n)}(a)\over n!}\,(x-a)^n + {f^{(N+1)}(z)\over (N+1)!}(x-a)^{N+1}.$$

Proof

The proof requires some cleverness to set up, but then the details are quite elementary. We want to define a function $$F(t)$$. Start with the equation $$F(t)=\sum_{n=0}^N{f^{(n)}(t)\over n!}\,(x-t)^n + B(x-t)^{N+1}.$$ Here we have replaced $$a$$ by $$t$$ in the first $$N+1$$ terms of the Taylor series, and added a carefully chosen term on the end, with $$B$$ to be determined. Note that we are temporarily keeping $$x$$ fixed, so the only variable in this equation is $$t$$, and we will be interested only in $$t$$ between $$a$$ and $$x$$. Now substitute $$t=a$$: $$F(a)=\sum_{n=0}^N{f^{(n)}(a)\over n!}\,(x-a)^n + B(x-a)^{N+1}.$$ Set this equal to $$f(x)$$: $$f(x)=\sum_{n=0}^N{f^{(n)}(a)\over n!}\,(x-a)^n + B(x-a)^{N+1}.$$ Since $$x\not=a$$, we can solve this for $$B$$, which is a "constant''---it depends on $$x$$ and $$a$$ but those are temporarily fixed.

Now we have defined a function $$F(t)$$ with the property that $$F(a)=f(x)$$. Consider also $$F(x)$$: all terms with a positive power of $$(x-t)$$ become zero when we substitute $$x$$ for $$t$$, so we are left with $$F(x)=f^{(0)}(x)/0!=f(x).$$ So $$F(t)$$ is a function with the same value on the endpoints of the interval $$[a,x]$$. By Rolle's theorem (6.5.1), we know that there is a value $$z\in(a,x)$$ such that $$F'(z)=0$$. Let's look at $$F'(t)$$. Each term in $$F(t)$$, except the first term and the extra term involving $$B$$, is a product, so to take the derivative we use the product rule on each of these terms.

It will help to write out the first few terms of the definition: \eqalign{ F(t)=f(t)&+{f^{(1)}(t)\over 1!}(x-t)^1+{f^{(2)}(t)\over 2!}(x-t)^2+ {f^{(3)}(t)\over 3!}(x-t)^3+\cdots\cr &+{f^{(N)}(t)\over N!}(x-t)^N+ B(x-t)^{N+1}.\cr} Now take the derivative: \eqalign{ F'(t) = f'(t) &+ \left({f^{(1)}(t)\over 1!}(x-t)^0(-1)+{f^{(2)}(t)\over 1!}(x-t)^1\right)\cr &+\left({f^{(2)}(t)\over 1!}(x-t)^1(-1)+{f^{(3)}(t)\over 2!}(x-t)^2\right)\cr &+\left({f^{(3)}(t)\over 2!}(x-t)^2(-1)+{f^{(4)}(t)\over 3!}(x-t)^3\right)+…+\cr &+\left({f^{(N)}(t)\over (N-1)!}(x-t)^{N-1}(-1)+{f^{(N+1)}(t)\over N!}(x-t)^N\right)\cr &+B(N+1)(x-t)^N(-1).\cr} Now most of the terms in this expression cancel out, leaving just $$F'(t) = {f^{(N+1)}(t)\over N!}(x-t)^N+B(N+1)(x-t)^N(-1).$$ At some $$z$$, $$F'(z)=0$$ so \eqalign{ 0&={f^{(N+1)}(z)\over N!}(x-z)^N+B(N+1)(x-z)^N(-1)\cr B(N+1)(x-z)^N&={f^{(N+1)}(z)\over N!}(x-z)^N\cr B&={f^{(N+1)}(z)\over (N+1)!}.\cr } Now we can write $$F(t)=\sum_{n=0}^N{f^{(n)}(t)\over n!}\,(x-t)^n + {f^{(N+1)}(z)\over (N+1)!}(x-t)^{N+1}.$$ Recalling that $$F(a)=f(x)$$ we get $$f(x)=\sum_{n=0}^N{f^{(n)}(a)\over n!}\,(x-a)^n + {f^{(N+1)}(z)\over (N+1)!}(x-a)^{N+1},$$ which is what we wanted to show.

$$\square$$

It may not be immediately obvious that this is particularly useful; let's look at some examples.

Example 11.11.1

Find a polynomial approximation for $$\sin x$$ accurate to $$\pm 0.005$$.

Solution

From Taylor's theorem: $$\sin x= \sum_{n=0}^N{f^{(n)}(a)\over n!}\,(x-a)^n + {f^{(N+1)}(z)\over (N+1)!}(x-a)^{N+1}.$$ What can we say about the size of the term $${f^{(N+1)}(z)\over (N+1)!}(x-a)^{N+1}?$$ Every derivative of $$\sin x$$ is $$\pm\sin x$$ or $$\pm\cos x$$, so $$|f^{(N+1)}(z)|\le 1$$. The factor $$(x-a)^{N+1}$$ is a bit more difficult, since $$x-a$$ could be quite large. Let's pick $$a=0$$ and $$|x|\le\pi/2$$; if we can compute $$\sin x$$ for $$x\in[-\pi/2,\pi/2]$$, we can of course compute $$\sin x$$ for all $$x$$.

We need to pick $$N$$ so that $$\left|{x^{N+1}\over (N+1)!}\right| < 0.005.$$ Since we have limited $$x$$ to $$[-\pi/2,\pi/2]$$, $$\left|{x^{N+1}\over (N+1)!}\right| < {2^{N+1}\over (N+1)!}.$$ The quantity on the right decreases with increasing $$N$$, so all we need to do is find an $$N$$ so that $${2^{N+1}\over (N+1)!} < 0.005.$$ A little trial and error shows that $$N=8$$ works, and in fact $$2^{9}/9! < 0.0015$$, so

\eqalign{ \sin x &=\sum_{n=0}^8{f^{(n)}(0)\over n!}\,x^n \pm 0.0015\cr &=x-{x^3\over 6}+{x^5\over 120}-{x^7\over 5040}\pm 0.0015.\cr }

Figure 11.11.1 shows the graphs of $$\sin x$$ and and the approximation on $$[0,3\pi/2]$$. As $$x$$ gets larger, the approximation heads to negative infinity very quickly, since it is essentially acting like $$-x^7$$.

Example 11.11.2

Figure 11.11.1. $$\sin x$$ and a polynomial approximation.

Solution

We can extract a bit more information from this example. If we do not limit the value of $$x$$, we still have $$\left|{f^{(N+1)}(z)\over (N+1)!}x^{N+1}\right|\le \left|{x^{N+1}\over (N+1)!}\right|$$ so that $$\sin x$$ is represented by

$$\sum_{n=0}^N{f^{(n)}(0)\over n!}\,x^n \pm \left|{x^{N+1}\over (N+1)!}\right|.$$

If we can show that $$\lim_{N\to\infty} \left|{x^{N+1}\over (N+1)!}\right|=0$$ for each x then

$$\sin x=\sum_{n=0}^\infty{f^{(n)}(0)\over n!}\,x^n = \sum_{n=0}^\infty (-1)^n{x^{2n+1}\over (2n+1)!},$$

that is, the sine function is actually equal to its Maclaurin series for all x. How can we prove that the limit is zero? Suppose that N is larger than $$|x|$$, and let M be the largest integer less than $$|x|$$ (if $$M=0$$ the following is even easier). Then

\eqalign{ {|x^{N+1}|\over (N+1)!} &= {|x|\over N+1}{|x|\over N}{|x|\over N-1}\cdots {|x|\over M+1}{|x|\over M}{|x|\over M-1}\cdots {|x|\over 2}{|x|\over 1}\cr &\le {|x|\over N+1}\cdot 1\cdot 1\cdots 1\cdot {|x|\over M}{|x|\over M-1}\cdots {|x|\over 2}{|x|\over 1}\cr &={|x|\over N+1}{|x|^M\over M!}. }

The quantity $$|x|^M/ M!$$ is a constant, so $$\lim_{N\to\infty} {|x|\over N+1}{|x|^M\over M!} = 0$$ and by the Squeeze Theorem (11.1.3)

$$\lim_{N\to\infty} \left|{x^{N+1}\over (N+1)!}\right|=0$$ as desired. Essentially the same argument works for $$\cos x$$ and $$e^x$$; unfortunately, it is more difficult to show that most functions are equal to their Maclaurin series.

Example 11.11.3

Find a polynomial approximation for $$e^x$$ near $$x=2$$ accurate to $$\pm 0.005$$.

Solution

From Taylor's theorem: $$e^x= \sum_{n=0}^N{e^2\over n!}\,(x-2)^n + {e^z\over (N+1)!}(x-2)^{N+1},$$ since $$f^{(n)}(x)=e^x$$ for all n. We are interested in x near 2, and we need to keep $$|(x-2)^{N+1}|$$ in check, so we may as well specify that $$|x-2|\le 1$$, so $$x\in[1,3]$$. Also $$\left|{e^z\over (N+1)!}\right|\le {e^3\over (N+1)!},$$ so we need to find an N that makes $$e^3/(N+1)!\le 0.005$$. This time $$N=5$$ makes $$e^3/(N+1)! < 0.0015$$, so the approximating polynomial is $$e^x=e^2+e^2(x-2)+{e^2\over2}(x-2)^2+{e^2\over6}(x-2)^3+ {e^2\over24}(x-2)^4+{e^2\over120}(x-2)^5 \pm 0.0015.$$ This presents an additional problem for approximation, since we also need to approximate $$e^2$$, and any approximation we use will increase the error, but we will not pursue this complication.

Note well that in these examples we found polynomials of a certain accuracy only on a small interval, even though the series for $$\sin x$$ and $$e^x$$ converge for all $$x$$; this is typical. To get the same accuracy on a larger interval would require more terms.

### Contributors

• Integrated by Justin Marshall.