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 13.11.1 Suppose that $f$ is defined on some open interval $I$ around $a$ and suppose $\ds f^{(N+1)}(x)$ exists on this interval. Then for each $x\not=a$ in $I$ there is a value $z$ between $x$ and $a$ so that $$ f(x) = \sum_{n=0}^N {f^{(n)}(a)\over n!}\,(xa)^n + {f^{(N+1)}(z)\over (N+1)!}(xa)^{N+1}. $$
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
$$F(t)=\sum_{n=0}^N{f^{(n)}(t)\over n!}\,(xt)^n + B(xt)^{N+1}.$$
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 $t=a$:
$$F(a)=\sum_{n=0}^N{f^{(n)}(a)\over n!}\,(xa)^n + B(xa)^{N+1}.$$
Set this equal to $f(x)$:
$$f(x)=\sum_{n=0}^N{f^{(n)}(a)\over n!}\,(xa)^n + B(xa)^{N+1}.$$
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
$(xt)$ become zero when we substitute $x$ for $t$, so we are left
with $\ds F(x)=f^{(0)}(x)/0!=f(x)$. 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:
$$\eqalign{
F(t)=f(t)&+{f^{(1)}(t)\over 1!}(xt)^1+{f^{(2)}(t)\over 2!}(xt)^2+
{f^{(3)}(t)\over 3!}(xt)^3+\cdots\cr
&+{f^{(N)}(t)\over N!}(xt)^N+
B(xt)^{N+1}.\cr}
$$
Now take the derivative:
$$\eqalign{
F'(t) = f'(t) &+
\left({f^{(1)}(t)\over 1!}(xt)^0(1)+{f^{(2)}(t)\over
1!}(xt)^1\right)\cr
&+\left({f^{(2)}(t)\over 1!}(xt)^1(1)+{f^{(3)}(t)\over
2!}(xt)^2\right)\cr
&+\left({f^{(3)}(t)\over 2!}(xt)^2(1)+{f^{(4)}(t)\over
3!}(xt)^3\right)+…+\cr
&+\left({f^{(N)}(t)\over (N1)!}(xt)^{N1}(1)+{f^{(N+1)}(t)\over
N!}(xt)^N\right)\cr
&+B(N+1)(xt)^N(1).\cr}
$$
Now most of the terms in this expression cancel out,
leaving just
$$F'(t) = {f^{(N+1)}(t)\over N!}(xt)^N+B(N+1)(xt)^N(1).$$
At some $z$, $F'(z)=0$ so
$$\eqalign{
0&={f^{(N+1)}(z)\over N!}(xz)^N+B(N+1)(xz)^N(1)\cr
B(N+1)(xz)^N&={f^{(N+1)}(z)\over N!}(xz)^N\cr
B&={f^{(N+1)}(z)\over (N+1)!}.\cr
}$$
Now we can write
$$
F(t)=\sum_{n=0}^N{f^{(n)}(t)\over n!}\,(xt)^n +
{f^{(N+1)}(z)\over (N+1)!}(xt)^{N+1}.
$$
Recalling that $F(a)=f(x)$ we get
$$
f(x)=\sum_{n=0}^N{f^{(n)}(a)\over n!}\,(xa)^n +
{f^{(N+1)}(z)\over (N+1)!}(xa)^{N+1},
$$
which is what we wanted to show.
It may not be immediately obvious that this is particularly useful; let's look at some examples.
Example 13.11.2 Find a polynomial approximation for $\sin x$ accurate to $\pm 0.005$.
From Taylor's theorem: $$ \sin x= \sum_{n=0}^N{f^{(n)}(a)\over n!}\,(xa)^n + {f^{(N+1)}(z)\over (N+1)!}(xa)^{N+1}. $$ What can we say about the size of the term $${f^{(N+1)}(z)\over (N+1)!}(xa)^{N+1}?$$ Every derivative of $\sin x$ is $\pm\sin x$ or $\pm\cos x$, so $\ds f^{(N+1)}(z)\le 1$. The factor $\ds (xa)^{N+1}$ is a bit more difficult, since $xa$ 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 $\ds 2^{9}/9!< 0.0015$, 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 13.11.1 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 $\ds x^7$.

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 $$ \eqalign{ {x^{N+1}\over (N+1)!} &= {x\over N+1}{x\over N}{x\over N1}\cdots {x\over M+1}{x\over M}{x\over M1}\cdots {x\over 2}{x\over 1}\cr &\le {x\over N+1}\cdot 1\cdot 1\cdots 1\cdot {x\over M}{x\over M1}\cdots {x\over 2}{x\over 1}\cr &={x\over N+1}{x^M\over M!}. } $$ 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 (13.1.3) $$ \lim_{N\to\infty} \left{x^{N+1}\over (N+1)!}\right=0 $$ as desired. Essentially the same argument works for $\cos x$ and $\ds e^x$; unfortunately, it is more difficult to show that most functions are equal to their Maclaurin series.
Example 13.11.3 Find a polynomial approximation for $\ds e^x$ near $x=2$ accurate to $\pm 0.005$.
From Taylor's theorem: $$ e^x= \sum_{n=0}^N{e^2\over n!}\,(x2)^n + {e^z\over (N+1)!}(x2)^{N+1}, $$ since $\ds f^{(n)}(x)=e^x$ for all $n$. We are interested in $x$ near 2, and we need to keep $\ds (x2)^{N+1}$ in check, so we may as well specify that $x2\le 1$, 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 $\ds e^3/(N+1)!\le 0.005$. This time $N=5$ makes $\ds e^3/(N+1)!< 0.0015$, so the approximating polynomial is $$ e^x=e^2+e^2(x2)+{e^2\over2}(x2)^2+{e^2\over6}(x2)^3+ {e^2\over24}(x2)^4+{e^2\over120}(x2)^5 \pm 0.0015. $$ This presents an additional problem for approximation, since we also need to approximate $\ds e^2$, 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 $\ds e^x$ converge for all $x$; this is typical. To get the same accuracy on a larger interval would require more terms.
Exercises 13.11
Ex 13.11.1 Find a polynomial approximation for $\cos x$ on $[0,\pi]$, accurate to $\ds \pm 10^{3}$ (answer)
Ex 13.11.2 How many terms of the series for $\ln x$ centered at 1 are required so that the guaranteed error on $[1/2,3/2]$ is at most $\ds 10^{3}$? What if the interval is instead $[1,3/2]$? (answer)
Ex 13.11.3 Find the first three nonzero terms in the Taylor series for $\tan x$ on $[\pi/4,\pi/4]$, and compute the guaranteed error term as given by Taylor's theorem. (You may want to use Sage or a similar aid.) (answer)
Ex 13.11.4 Show that $\cos x$ is equal to its Taylor series for all $x$ by showing that the limit of the error term is zero as $N$ approaches infinity.
Ex 13.11.5 Show that $\ds e^x$ is equal to its Taylor series for all $x$ by showing that the limit of the error term is zero as $N$ approaches infinity.