Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/253

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If the series converges for values of $x$ for which the remainder, does not approach zero as $n$ increases without limit, then the limit of the sum of the series is not equal to the function $f(x)$.

The infinite series (62) represents the function for those values of $x$, and those only, for which the remainder approaches zero as the number of terms increases without limit.

It is usually easier to determine the interval of convergence of the series than that for which the remainder approaches zero; but in simple cases the two intervals are identical.

When the values of a function and its successive derivatives are known for some value of the variable, as $x = a$, then (62) is used for finding the value of the function for values of $x$ near $a$, and (62) is also called the expansion off $f(x)$ in the vicinity of $x = a$.

Illustrative Example 1. Expand $log x$ in powers of $(x - 1)$.

Solution

$\begin{array}{rclrcl} f(x) & = & \log x, & f(1) & = & 0; \\ f'(x) & = & \tfrac{1}{x}, & f'(1) & = & 1; \\ f''(x) & = & \tfrac{1}{x^2}, & f''(1) & = & -1; \\ f'''(x) & = & \tfrac{2}{x^3}, & f'''(1) & = & 2. \end{array}$

Substituting in (62), $\log x = x - \tfrac{1}{2}(x-1)^2 + \tfrac{1}{3}(x-1)^3 - \cdots.$ Ans.

This converges for values of $x$ between and 2 and is the expansion of $\log x$ in the vicinity of $x = 1$, the remainder converging to zero.

When a function of the sum of two numbers $a$ and $x$ is given, say $f(a + x)$, it is frequently desirable to expand the function into a power series in one of them, say $x$. For this purpose we use another form of Taylor's Series, got by replacing $x$ by $a + x$ in (62), namely,

(63)	$f(a + x) = f(a) + \frac{x}{1!}f'\left(a\right) + \frac{x^2}{2!}f''\left(a\right) + \frac{x^3}{3!}f'''\left(a\right) + \cdots.$

Illustrative Example 2. Expand $\sin(a+x)$ in power of $x$

Solution. Here $f(a+x) = \sin(a+x).$

Hence, placing

$\begin{array}{rcl} x &=& 0, \\ f(a) &=& \sin a, \\ f'(a) &=& \cos a, \\ f''(a) &=& -\sin a, \\ f'''(a) &=& -\cos a, \\ \end{array}$

Substituting in (61),

$\sin(a+x) = \sin a +\tfrac{x}{1}\cos a - \tfrac{x^2}{2!}\sin a - \tfrac{x^3}{3!}\cos a + \cdots.$ Ans.