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JenniferSmart1
how do I write this in sum form? \[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\]
I know I'm a disappointment :(
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
i would have written denominator of the form 2^m m!
this is the whole (series?) \[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]
\(\large y=a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\) doesn't this work ??
yeah, that's it @hart
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!)
ok from the top. so you look at this and you think..........? What is your thought process? \[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]
my 1st thought : did u change the Q ? where does these a1's jump in..... ? anyways, i would first factor out a0 and a1....
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
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 ?
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.... =(
\[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\] \[\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\] Let's look at this
so as m increases ... that makes sense Problem is will I be able to do this on my own?
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.
ok I'll try writing the sum a_1 let me know how I did
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.
product of odds is not difficult...u'll get it turing....
hint : 1/(1.3.5) = 2.4 /(5!)
=\[a_1x^1+\frac{a_1}{3}x^3+\frac{a_1}{3\cdot5}x^5+\frac{a_1}{3\cdot5\cdot7}x^7+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9\] =\[a_1 \left(x^1+\frac{x^3}{3}+\frac{x^5}{3\cdot5}+\frac{x^7}{3\cdot5\cdot7}+\frac{x^9}{3\cdot5\cdot7\cdot9}+\frac{x^{10}}{3\cdot5\cdot7\cdot9\cdot10}\right)\] = ok let's start with the exponents....or where would I look first?
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.
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.
\[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\] doesn't m=3 for this?
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 ?
no I mean this is the 3rd ....whatchamacallit? "m" is from 0 to infinity ...and this would be the third term?
m=0 <-1st term m=1 <-2nd term m=2 <-3rd term
\[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\] 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
soo many typos sorry
<---frustrated koala :S
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....
oh so 2m+1 would be right!
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?
sorry koalas are fairly slow...I'm still thinking
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.
\[\frac{x^5}{3\cdot5}\] 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
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
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
It's the solution in the back of my book for y''-xy'-y=0
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
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