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

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JenniferSmart1Best ResponseYou've already chosen the best response.0
I know I'm a disappointment :(
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
i would have written denominator of the form 2^m m!
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
this is the whole (series?) \[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
\(\large y=a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\) doesn't this work ??
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yeah, that's it @hart
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
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!)
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
ok from the top. so you look at this and you think..........? What is your thought process? \[a_0+a_1x^1+\frac{a_0}{2}x^2+\frac{a_1}{3}x^3+\frac{a_0}{2\cdot4}x^4+\frac{a_1}{3\cdot5}x^5+\frac{a_0}{2\cdot4\cdot6}x^6+\frac{a_1}{3\cdot5\cdot7}x^7\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
my 1st thought : did u change the Q ? where does these a1's jump in..... ? anyways, i would first factor out a0 and a1....
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
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 ?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I didn't quite get that part...
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
which part ? x^(2m) or 2^m or m! ??
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
ok I'm back. I can make sense of the answer but how do I come up with this on my own.... =(
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\] \[\large a_0\sum \limits_{m=0}^\infty\frac{x^{2m}}{2^m m!}\] Let's look at this
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
so as m increases ... that makes sense Problem is will I be able to do this on my own?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
ok I'll try writing the sum a_1 let me know how I did
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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.
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
product of odds is not difficult...u'll get it turing....
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
hint : 1/(1.3.5) = 2.4 /(5!)
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
=\[a_1x^1+\frac{a_1}{3}x^3+\frac{a_1}{3\cdot5}x^5+\frac{a_1}{3\cdot5\cdot7}x^7+\frac{a_1}{3\cdot5\cdot7\cdot9}x^9\] =\[a_1 \left(x^1+\frac{x^3}{3}+\frac{x^5}{3\cdot5}+\frac{x^7}{3\cdot5\cdot7}+\frac{x^9}{3\cdot5\cdot7\cdot9}+\frac{x^{10}}{3\cdot5\cdot7\cdot9\cdot10}\right)\] = ok let's start with the exponents....or where would I look first?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I understand the hint.... :(
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I don't understand I meant
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
1/(1.3.5) = 2.4 /(5!) = 2^2 . (1.2) / (5!) = 2^2 (2!)/(5!) got that ?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
so how ....Ohhh by having 2x4 in the numerator your eliminating the evens in the denominator!!!!
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
i try to bring factorial form wherever possible.
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
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.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\] doesn't m=3 for this?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
and yes this part makes sense
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
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 ?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
no I mean this is the 3rd ....whatchamacallit? "m" is from 0 to infinity ...and this would be the third term?
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
m=0 <1st term m=1 <2nd term m=2 <3rd term
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\frac{1}{1\cdot3\cdot 5}=\frac{2\cdot4}{5!}=\frac{2^2\cdot2\cdot1}{5!}=\frac{2^22!}{5!}\] 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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
soo many typos sorry
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
<frustrated koala :S
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
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....
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
oh so 2m+1 would be right!
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
sorry koalas are fairly slow...I'm still thinking
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
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.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\frac{x^5}{3\cdot5}\] 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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I'm just very visual
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
finally it will be \(\large a_1[\sum\limits _0^\infty\frac{x^{(2m+1)}2^mm!}{(2m+1)!}]\)
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
do either of you have a similar problem somewhere? LOL I wanna try this again
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
no I think this is enough. Ok thanks
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
does this make sense to anyone? \[\sum_{n=0}^{\infty}\frac{x^{2n}}{2^nn!}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
for both a_0 and a_1
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
It's the solution in the back of my book for y''xy'y=0
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.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
 one year ago

hartnnBest ResponseYou've already chosen the best response.4
u mean in the last image ? he just wrote solution of initial value problem.... with a1=0
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
ohhhhhhh....durrrr sorry. Ok I got now haha
 one year ago
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