Solutions - 0230-03

(a)

$$\begin{array}{c}\frac{x^2}{x^4+81}\gg \gg \frac{x^2}{81}\cdot \frac{1}{1-(-\frac{x^4}{81})}\\ \\ \gg \gg \frac{x^2}{81}\sum _{n=0}^{\infty }{(-\frac{x^4}{81})}^n\gg \gg f(x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{3^{4n+4}}{x}^{4n+2}\end{array}$$

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Another approach:

$$\begin{array}{ccc}& \frac{x^2}{x^4+81}\gg \gg \frac{x^2}{a}\frac{a}{{\left(x^2\right)}^2+a^2},a=9& \text{(A)}\end{array}$$

We know that:

$$\begin{array}{c}\frac{a}{u^2+a^2}=\frac{d}{du}{\tan }^{-1}\left(\frac{u}{a}\right)\gg \gg \frac{d}{du}\sum _{n=0}^{\infty }{(-1)}^n\frac{{(u/a)}^{2n+1}}{2n+1}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n{\left(\frac{u}{a}\right)}^{2n}\frac{1}{a}\end{array}$$

Plug in $u=x^2$:

$$\gg \gg \sum _{n=0}^{\infty }{(-1)}^n{\left(\frac{x^2}{a}\right)}^{2n}\frac{1}{a}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{a^{2n+1}}{x}^{4n}$$

Complete:

$$\begin{array}{c}\text{A:}\frac{x^2}{a}\frac{a}{{\left(x^2\right)}^2+a^2}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{a^{2n+2}}{x}^{4n+2}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{9^{2n+2}}{x}^{4n+2}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{3^{4n+4}}{x}^{4n+2}\end{array}$$

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(b)

Notice:

$$\frac{d}{dx}\ln (1+x)=\frac{1}{1+x}\gg \gg \sum _{n=0}^{\infty }{(-x)}^n$$

Integrate:

$$\ln (1+x)=\int \sum _{n=0}^{\infty }{(-x)}^{n}dx\gg \gg C+\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n+1}{x}^{n+1}$$

Plug in $x=0$ to solve and find $C=0$.

Now then:

$$\begin{array}{c}{x}^2\ln (1+x)\gg \gg x^2\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n+1}{x}^{n+1}\\ \\ \gg \gg g(x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n+1}{x}^{n+3}\end{array}$$