Question

Now, lets evaluate the same integral using power series. first, find the power series for the function f(x = \frac{32}{x^2+4}. then, integrate it from 0 to 2, and call it s. s should be an infinite series. what are the first few terms of s ?

Answer 1

No idea what the previous part of the problem is, but you have


`f(x)=(32)/(x^2+4)=\frac8{1-\left(-\frac{x^2}4\right)}=\displaystyle8\sum_(n\ge0)\left(-\frac{x^2}4\right)^n`

`f(x)=\displaystyle8\sum_(n\ge0)\left(-\frac14\right)^nx^(2n)`

which is valid for
`\left|-\frac{x^2}4\right|<1`, or
`|x|<2`. So the integral from 0 to 2 is


`\displaystyle\int_0^2f(x)\,\mathrm dx=\int_0^28\sum_(n\ge0)\left(-\frac14\right)^nx^(2n)\,\mathrm dx`

`=\displaystyle8\sum_(n\ge0)\left(-\frac14\right)^n\int_0^2x^(2n)\,\mathrm dx`

Note that since the power series only converges on the interval if
`x` is strictly less than 2, which means we have to treat this as an improper integral.


`=\displaystyle8\sum_(n\ge0)\left(-\frac14\right)^n\lim_(t\to2^-)\int_0^tx^(2n)\,\mathrm dx`[/tex]

`=\displaystyle8\sum_(n\ge0)\left(-\frac14\right)^n\lim_(t\to2^-)(x^(2n+1))/(2n+1)\bigg|_(x=0)^(x=t)`

`=\displaystyle8\sum_(n\ge0)((-1)^n)/(2^(2n)(2n+1))\lim_(t\to2^-)t^(2n+1)`

`=\displaystyle16\sum_(n\ge0)((-1)^n)/(2n+1)`

`=16-\frac{16}3+\frac{16}5-\frac{16}7+\frac{16}9+\cdots`