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the 36 kinda threw me off

did you forget the 'x' in x*y' ?

no

because before the x it was
\[y'=\sum_{n=1}^{\infty}a_nnx^{n-1}\]

I can write it all out of you want me to

it's ok...

let me catch up

ok

I don't know anymore

@Outkast3r09 watcha think?

we've got something there!

and a sign changing term...

could also do
\[\frac{ 1 }{ 2^{k}k! }\]

k=1,2,3...

but would the k! work?

I think I've done enough math for tonight...g'night!

definitely an alternating series interp

\[y=\sum_{n=0}^{\infty}a_nx^n=a_0\sum_{n=0}^{\infty}(-1)^{2k}\frac{x^{2k}}{2^kk!}\]

for the even numbers

so your x corresponds to \[x^{2k-1}\]

would you agree?

the negatives are constant with each n so you can write that as
\[(-1)^k\]

I would agree, but how do we put the denominator in terms of k's?
(2k+1)!

\[\sum_{n=0} ^\infty \frac{(-1)^kx^{2k-1}}{1*3*5*7*.....(2k+1)}\]

i believe this is your series =]

This is right lol i overlookd my own self haha

however i'm thinking that your zero term is outside not sure =/

\[\frac{-x^5}{7!}\]

write it as 1*3*5*7