## 175 – Quiz 8

Here is quiz 8.

Problem 1 asks to find the power series ${a_0+a_1x+a_2x^2+a_3x^3+\cdots}$ for ${(2-x)^3}$.

Method 1: Using basic algebra, expand the cube:

$\displaystyle \begin{array}{rcl} (2-x)^3&=&2^3-3\cdot2^2\cdot x+3\cdot 2\cdot x^2-x^3\\ &=&8-12x+6x^2-x^3. \end{array}$

This gives ${(2-x)^3}$ as a sum of powers of ${x}$. Since there is only one way of doing this, this is the series we were looking for (and we have ${a_4=a_5=a_6=\dots=0}$).

Method 2: Let ${f(x)=(2-x)^3}$. Then we have:

$\displaystyle \begin{array}{ccc} f(x)=(2-x)^3&\mbox{ and }&f(0)=8\\ f'(x)=-3(2-x)^2&\mbox{ and }&f'(0)=-12\\ f''(x)=6(2-x)&\mbox{ and }&f''(0)=12\\ f'''(x)=-6&\mbox{ and }&f'''(0)=-6\\ f^{(4)}(x)=0&\mbox{ and }&f^{(4)}(0)=f^{(5)}(0)=\dots=0. \end{array}$

Using that ${f(x)=}$

$\displaystyle f(0)+f'(0)x+\frac{f''(0)}2 x^2+\frac{f'''(0)}6 x^3+\frac{f^{(4)}(0)}{4!}x^4+\dots,$

we have

$\displaystyle f(x)=8-12x+6x^2-x^3+0+0+0+\dots\,.$

Problem 2 asks to find the power series ${a_0+a_1x+a_2x^2+a_3x^3+\cdots}$ for ${\displaystyle \frac{x}{1+x^2}}$.

Perhaps the easiest way to proceed is to note that

$\displaystyle \begin{array}{rcl} \displaystyle \frac1{1+x^2}&=&\displaystyle \frac1{1-(-x^2)}\\ &=&\displaystyle 1+(-x^2)+(-x^2)^2+(-x^2)^3+\dots\\ &=&\displaystyle 1-x^2+x^4-x^6+x^8-\dots \end{array}$

and therefore

$\displaystyle \frac x{1+x^2}=x-x^3+x^5-x^7+x^9-\dots$

Problem 3 asks to find the radius ${R}$ of convergence of the series

$\displaystyle \sum_{n=0}^\infty \frac{(2x)^{2n+1}}{n^2+1}=2x+4x^3+\frac{32}5 x^5+\frac{64}{5}x^7+\cdots\,.$

As usual, we look at ${\displaystyle b_n=\left|\frac{(2x)^{2n+1}}{n^2+1}\right|}$ and ${b_{n+1}}$, which we find by replacing ${n}$ with ${n+1}$ in the previous formula, so we have

$\displaystyle b_{n+1}=\left|\frac{(2x)^{2(n+1)+1}}{(n+1)^2+1}\right|.$

We then take the quotient ${b_{n+1}/b_n}$ and simplify:

$\displaystyle \begin{array}{rcl} \displaystyle \frac{b_{n+1}}{b_n}&=&\displaystyle \left|\frac{(2x)^{2(n+1)+1}}{(n+1)^2+1}\cdot\frac{n^2+1}{(2x)^{2n+1}}\right|\\ &=&\displaystyle |(2x)^2|\frac{n^2+1}{(n+1)^2+1}, \end{array}$

and this last expression converges to ${|(2x)^2|}$ as ${n\rightarrow\infty}$. To see this, note that

$\displaystyle \begin{array}{rcl} \displaystyle \frac{n^2+1}{(n+1)^2+1}&=&\displaystyle \frac{n^2+1}{n^2+2n+2}\\ &=&\displaystyle \frac{1+\displaystyle \frac1{n^2}}{\displaystyle 1+\frac2n+\frac2{n^2}}\\ &\rightarrow& 1. \end{array}$

By the ratio test, the series converges if ${|(2x)^2|<1}$ and diverges if ${|(2x)^2|>1}$. Note that ${|(2x)^2|<1}$ is the same as ${|2x|<1}$ or ${|x|<1/2,}$ i.e., ${R=1/2}$.

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