Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/251

EXPANSION OF FUNCTIONS

143. Introduction. The student is already familiar with some methods of expanding certain functions into series. Thus, by the Binomial Theorem,

(A)	$\left(a+x\right)^{4}=a^{4}+4a^{3}x+6a^{2}x^{2}+4ax^{3}+x^{4},$

giving a finite power series from which the exact value of $(a+x)^{4}$ for any value of $x$ may be calculated. Also by actual division,

(B)	${\frac {1}{1-x}}=1+x+x^{2}+x^{3}+\cdots +x^{n-1}+\left({\frac {1}{1-x}}\right)x^{n},$

we get an equivalent series, all of whose coefficients except that of $x^{n}$ are constants, $n$ being a positive integer.

Suppose we wish to calculate the value of this function when $x=.5$, not by substituting directly in

${\frac {1}{1-x}},$

but by substituting $x=.5$ in the equivalent series

(C)	$\left(1+x+x^{2}+x^{3}+\cdots +x^{x-1}\right)+\left(frac{1}{1-x}\right)x^{n}.$

Assuming $n=8$, (C) gives for $x=.5$

(D)	${\frac {1}{1-x}}=1.9921875+.0078125.$

If we then assume the value of the function to be the sum of the first eight terms of series (C), the error we make is .0078125. However, in case we need the value of the function correct to two decimal places only, the number 1.99 is as close an approximation to the true value as we care for, since the error is less than .01. It is evident that if a greater degree of accuracy is desired, all we need to do is to use more terms of the power series

(E)	$1+x+x^{2}+x^{3}+\cdots .$