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## Functions as Power Series A power series $\displaystyle\sum_{n=0}^\infty c_n x^n$ can be
thought of as a function of $x$ whose domain is the interval of
convergence.
most basic
examples, found by manipulating
the geometric series.Consider $\displaystyle\sum_{n=0}^\infty r^n$. This is geometric series converges when
$|r|<1$ and diverges otherwise. When it
converges, its value is $\frac{1}{1-r}$. We take that
series, and replace $r$ with $x$, getting a power series. We
can express this as a function, as you see below below, as long as
$|x|<1$. $$f(x)=\frac{1}{1-x} = \sum_{n=0}^\infty x^n = 1+x+x^2+x^3+x^4+\cdots\qquad\text{ as long as }\lvert x\rvert<1$$ We get a different function by replacing $x$ with $-x$: $$\displaystyle{g(x)=\frac{1}{1+x} = \sum_{n=0}^\infty (-x)^n = 1-x+x^2+\cdots}\quad\text{ as long as } \lvert -x\rvert=\lvert x\rvert<1.$$ The video will look at these and other examples arising from the geometric series.
DO: Work on the following
two examples before reading ahead.Example 1: Find a power series
representation of the function $\displaystyle\frac{x}{1+x^2}$,
and determine for which $x$ it would be defined.Example 2: Find a power series representation of
the function $\displaystyle\frac{1}{7+2x}$, and determine for
which $x$ it would be defined.Solution 1: Replace $x$ (in our original $f(x)$
before the video) by $-x^2$, and multiply the expression by
$x$. $$\displaystyle{\frac{x}{1+x^2} = x\sum_{n=0}^\infty
(-x^2)^n =\sum_{n=0}^\infty x(-x^2)^n=\sum_{n=0}^\infty (-1)^n
x^{2n+1}= x-x^3+x^5-x^7+\cdots}.$$For convergence, we need
$|x^2|<1$, which simplifies to $\lvert x\rvert<1$. -----------------------------------------------------------------------------------------Solution 2: Divide out a $7$ in the denominator, in
order to have the constant equal to 1:
$\displaystyle\frac{1}{7+2x}=\frac{1}{7}\frac{1}{1+\left(\frac{2x}7\right)}.$
Now we can see that we replace our original $x$ by
$-\frac{2x}{7}$ and multiply the expression by $\frac17$.$$\displaystyle \frac{7x}{1+2x}=\left(\frac17\right)\sum_{n=0}^\infty\left(-\frac{2x}{7}\right)^n=\left(\frac17\right)\sum_{n=0}^\infty(-1)^n\left(\frac{2x}7 \right)^n=\sum_{n=0}^\infty(-1)^n\frac{(2x)^n}{7^{n+1}}=\frac17-\frac{2}{7^2}x+\frac{2^2}{7^3}x^2+\frac{2^3}{7^4}x^3+\cdots .$$For convergence, we need $\lvert\frac{2x}{7}\rvert<1$. This simplifies to $|x|<\frac72$. |