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I know I'm a disappointment :(

i would have written denominator of the form
2^m m!

num = x^(2m)

\(\large y=a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\)
doesn't this work ??

yeah, that's it @hart

#9

how and why?

got that ^ ?

still thinking

I didn't quite get that part...

which part ?
x^(2m) or 2^m or m! ??

ok I'm back. I can make sense of the answer but how do I come up with this on my own....
=(

so as m increases ...
that makes sense
Problem is will I be able to do this on my own?

ok

ok I'll try writing the sum a_1 let me know how I did

*odds I mean

product of odds is not difficult...u'll get it turing....

hint : 1/(1.3.5) = 2.4 /(5!)

ohhhhhhhhhhhhhh

I understand the hint.... :(

I don't understand I meant

1/(1.3.5) = 2.4 /(5!) = 2^2 . (1.2) / (5!) = 2^2 (2!)/(5!)
got that ?

so how ....Ohhh by having 2x4 in the numerator your eliminating the evens in the denominator!!!!

i try to bring factorial form wherever possible.

and yes this part makes sense

why m=3 ?
i would say m=2
5 = 2m+1
so, the term will be
x^(2m+1) 2^m m!/ (2m+1)!
got that ?

nevermind

ohh..

start from m=0

m=0 <-1st term
m=1 <-2nd term
m=2 <-3rd term

*my not me

soo many typos sorry

2m+1

LOL

<---frustrated koala :S

oh so 2m+1 would be right!

yes.

sorry koalas are fairly slow...I'm still thinking

I'm just very visual

finally it will be
\(\large a_1[\sum\limits _0^\infty\frac{x^{(2m+1)}2^mm!}{(2m+1)!}]\)

do either of you have a similar problem somewhere? LOL I wanna try this again

welcome ^_^

no I think this is enough. Ok thanks

does this make sense to anyone?
\[\sum_{n=0}^{\infty}\frac{x^{2n}}{2^nn!}\]

for both a_0 and a_1

thats for a0 only...

It's the solution in the back of my book for y''-xy'-y=0

u mean in the last image ?
he just wrote solution of initial value problem....
with a1=0

ohhhhhhh....durrrr
sorry. Ok I got now haha