how do I write this in sum form?
\[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\]

- anonymous

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- anonymous

I know I'm a disappointment :(

- TuringTest

no you're not, I don't have it either\[y=a_0\sum_{n=0}^\infty\frac{x^{2n}}{}\]darn I know there is a way to do the denom, but I can't think of it...
I have seen final answers for power series DE's written as you have them though, so the summation notation is not necesarry

- hartnn

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

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## More answers

- hartnn

num = x^(2m)

- anonymous

this is the whole (series?)
\[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]

- hartnn

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

- TuringTest

yeah, that's it @hart

- anonymous

##### 1 Attachment

- anonymous

#9

- anonymous

how and why?

- hartnn

2 = 2 (1) = 2^1 (1!)
2.4 = 2.2(1.2) = 2^2 (2!)
2.4.6 = 2.2.2 (1.2.3) = 2^3 (3!)
in general
denominator = (2^n n!)

- hartnn

got that ^ ?

- anonymous

still thinking

- anonymous

ok from the top.
so you look at this and you think..........? What is your thought process?
\[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]

- hartnn

my 1st thought : did u change the Q ?
where does these a1's jump in..... ?
anyways, i would first factor out a0 and a1....

- anonymous

that's the whole sum (series?)
what I wrote is just the a_1, then I added the a_0
It's from a previous problem
http://openstudy.com/users/jennifersmart1#/updates/50d63d0fe4b0d6c1d541ef3a

- hartnn

ok, so did u get how , after factoring a0, u get
\(\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\)
now your next step is to find sum for a1, right ?

- anonymous

I didn't quite get that part...

- hartnn

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

- anonymous

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

- anonymous

\[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\]
\[\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\]
Let's look at this

- anonymous

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

- TuringTest

Well as you saw, I had forgotten about the 2^m for the product of only the even numbers in the factorial, but this is kind of a tricky pattern to write if you ask me. They are not all this bad.
You can just reason to yourself about needing a 2 for each of the m elements in the factorial to make each factor double, and therefor even.

- anonymous

ok

- anonymous

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

- TuringTest

I'm not sure I know that one... I actually remember trying to come up with this before, a summation representing the product of evens and could not. Maybe I'll be more lucky this time.

- TuringTest

*odds I mean

- hartnn

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

- hartnn

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

- TuringTest

ohhhhhhhhhhhhhh

- anonymous

=\[a_1x^1+\frac{a_1}{3}x^3+\frac{a_1}{3\cdot5}x^5+\frac{a_1}{3\cdot5\cdot7}x^7+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9\]
=\[a_1 \left(x^1+\frac{x^3}{3}+\frac{x^5}{3\cdot5}+\frac{x^7}{3\cdot5\cdot7}+\frac{x^9}{3\cdot5\cdot7\cdot9}+\frac{x^{10}}{3\cdot5\cdot7\cdot9\cdot10}\right)\]
=
ok let's start with the exponents....or where would I look first?

- anonymous

I understand the hint.... :(

- anonymous

I don't understand I meant

- hartnn

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

- anonymous

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

- hartnn

i try to bring factorial form wherever possible.

- hartnn

for 1.3.5 i needed 2.4 to get 5!
so i multiplied and divided by 2.4
now from 2.4, i get 1.2=2! by factoring out 2, twice.

- anonymous

\[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\]
doesn't m=3 for this?

- anonymous

and yes this part makes sense

- hartnn

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 ?

- anonymous

no I mean this is the 3rd ....whatchamacallit?
"m" is from 0 to infinity ...and this would be the third term?

- anonymous

nevermind

- hartnn

ohh..

- hartnn

start from m=0

- hartnn

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

- anonymous

\[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\]
my point is....where does me thought process continue from here...how do I change these numbers to variables.
ok so were working on the third term now. where m=2

- anonymous

*my not me

- anonymous

soo many typos sorry

- anonymous

2m+1

- anonymous

LOL

- anonymous

<---frustrated koala :S

- hartnn

what i do is always take 3rd or 4th term., bring it in factorial form, take lowest number =m
all other numbers in terms of m....

- anonymous

oh so 2m+1 would be right!

- hartnn

yes.

- anonymous

where would it go?
\[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\]
shall we put it somewhere in the numerator?

- anonymous

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

- hartnn

all this exercise was to find the general term , which we found as
x^(2m+1) 2^m m!/ (2m+1)!
right ?
now just put a summation sign b4 this, with limits from 0 to infinity.

- anonymous

\[\frac{x^5}{3\cdot5}\]
m=2
\[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\]
\[\frac{x^{(2m+1)}2^mm!}{(2m+1)!}\]
ok I can see it now

- anonymous

I'm just very visual

- hartnn

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

- anonymous

thanks @hartnn

- anonymous

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

- hartnn

welcome ^_^

- anonymous

no I think this is enough. Ok thanks

- anonymous

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

- anonymous

for both a_0 and a_1

- hartnn

thats for a0 only...

- anonymous

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

- anonymous

I think I wrote it out correctly, and the the sums we wrote seem right too...but somehow he took those two and made one compact solution

- hartnn

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

- anonymous

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

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