Question

write this function as a power series

Answer 1

f(x)=arcsin(x^2)

Answer 2

Well I would say first find a power series representation for arcsine, so the derivative is \[\frac{ 1 }{ \sqrt{1-x^2} }\]
\[\implies (1-x^2)^{-1/2} \implies (1+(-x^2))^{-1/2}\] now use a binomial series

Answer 3

\[\sum_{n=0}^{\infty} \left(\begin{matrix}k \\ n\end{matrix}\right)x^n = (1+x)^k\]

Answer 4

Can you find the pattern?

Answer 5

\[(1+(-x^2))^{-1/2} = 1+ \left( -\frac{ 1 }{ 2 } \right)(-x^2)+\frac{ (-1/2)(-3/2) }{ 2! }(-x^2)^2+...\] something like that

Answer 6

\[\sum_{n=0}^{\infty}\ \left(\begin{matrix}-1/2 \\ n\end{matrix}\right)(-x^2)^n\]

Answer 7

\[= \sum_{n=1}^{\infty} \frac{ 1*3...(2n-1)^{2n} }{ 2^nn! }x^2n\] well I was specifically looking for this

Answer 8

mhm

Answer 9

\[= 1+ \sum_{n=1}^{\infty} \frac{ 1*3...(2n-1)^{2n} }{ 2^nn! }x^{2n}\] there, I think that looks right, now lets integrate haha...

Answer 10

\[\arcsin(x) = \int\limits \frac{ 1 }{ \sqrt{1-x^2} }dx \implies \int\limits 1+ \sum_{n=1}^{\infty} \frac{ 1*3...(2n-1)^{2n} }{ 2^nn! }x^2n dx\]

Answer 11

Well we could've done a little u sub for the x^2, I'll let you do that :P, it's just one extra step. But the integration shouldn't be too bad, most of it can just be treated as a constant.

Answer 12

the true power series http://i.imgur.com/1VTOp.jpg?1