Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/255

This page needs to be proofread.

a special case of Taylor's Series that is very useful. The statements made concerning the remainder and the convergence of Taylor's Series apply with equal force to Maclaurin's Series, the latter being merely a special case of the former.

The student should not fail to note the importance of such an expansion as (65). In all practical computations results correct to a certain number of decimal places are sought, and since the process in question replaces a function perhaps difficult to calculate by an ordinary polynomial with constant coefficients, it is very useful in simplifying such computations. Of course we must use terms enough to give the desired degree of accuracy.

In the case of an alternating series ( §139) the error made by stopping at any term is numerically less than that term, since the sum of the series after that term is numerically less than that term.

Illustrative Example 1. Expand $\cos x$ into an infinite power series and determine for what values of $x$ it converges.

Solution. Differentiating first and then placing

x=0

, we get

{\begin{array}{rclrcl}f(x)&=&\cos x,&f(0)&=&1,\\f'(x)&=&-\sin x,&f'(0)&=&0,\\f''(x)&=&-\cos x,&f''(0)&=&-1,\\f'''(x)&=&\sin x,&f'''(0)&=&0,\\f^{iv}(x)&=&\cos x,&f^{iv}(0)&=&1,\\f^{v}(x)&=&-\sin x,&f^{v}(0)&=&0,\\f^{vi}(x)&=&-\cos x,&f^{vi}(0)&=&-1,\\&etc.&,&&etc.&\end{array}}

Substituting in (65),

(A)	$\cos x=1-{\tfrac {x^{2}}{2!}}+{\tfrac {x^{4}}{4!}}-{\tfrac {x^{6}}{6!}}+\cdots .$

Comparing with Ex. 20, §20, we see that the series converges for all values of $x$ . In the same way for $\sin x$ .

(B)	$\sin x=1-{\tfrac {x^{3}}{3!}}+{\tfrac {x^{5}}{5!}}-{\tfrac {x^{7}}{7!}}+\cdots ,$

which converges for all values of $x$ (Ex. 21, §142). ^[1]

↑ Since here

f^{n}(x)=\sin \left(x+{\tfrac {n\pi }{2}}\right)

and

f^{n}(x_{1})=\sin \left(x_{i}+{\tfrac {n\pi }{2}}\right),

we have, by substituting in the last term of (65),

remainder

={\tfrac {x^{n}}{n!}}\sin \left(x_{i}+{\tfrac {n\pi }{x}}\right).

0<x_{1}<x

But $\sin \left(x_{1}+{\tfrac {n\pi }{x}}\right)$ can never exceed unity, and from (Ex. 19, §142), $\lim _{n\to \infty }{\tfrac {x^{n}}{n!}}=0$ for all values of $x$ ; that is, in this case the limit of the remainder is for all values of $x$ for which the series converges. This is also the case for all the functions considered in this book.

[1] Since here $f^{n}(x)=\sin \left(x+{\tfrac {n\pi }{2}}\right)$ and $f^{n}(x_{1})=\sin \left(x_{i}+{\tfrac {n\pi }{2}}\right),$ we have, by substituting in the last term of (65),

remainder $={\tfrac {x^{n}}{n!}}\sin \left(x_{i}+{\tfrac {n\pi }{x}}\right).$ $0<x_{1}<x$

But $\sin \left(x_{1}+{\tfrac {n\pi }{x}}\right)$ can never exceed unity, and from (Ex. 19, §142), $\lim _{n\to \infty }{\tfrac {x^{n}}{n!}}=0$ for all values of $x$ ; that is, in this case the limit of the remainder is for all values of $x$ for which the series converges. This is also the case for all the functions considered in this book.

[1]