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 $|x|<\frac72$.