A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
how do I write this in sum form?
\[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\]
anonymous
 4 years ago
how do I write this in sum form? \[y=a_0\left(1+\frac{x^2}{2}+\frac{x^4}{2\cdot4}+\frac{x^6}{2\cdot4\cdot6}+...\right)\]

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know I'm a disappointment :(

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no 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
 4 years ago
Best ResponseYou've already chosen the best response.4i would have written denominator of the form 2^m m!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this 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
 4 years ago
Best 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 ??

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, that's it @hart

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.42 = 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!)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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
 4 years ago
Best ResponseYou've already chosen the best response.4my 1st thought : did u change the Q ? where does these a1's jump in..... ? anyways, i would first factor out a0 and a1....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that'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
 4 years ago
Best ResponseYou've already chosen the best response.4ok, 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
 4 years ago
Best ResponseYou've already chosen the best response.0I didn't quite get that part...

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.4which part ? x^(2m) or 2^m or m! ??

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

anonymous
 4 years ago
Best 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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Well 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok I'll try writing the sum a_1 let me know how I did

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I'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.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.4product of odds is not difficult...u'll get it turing....

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.4hint : 1/(1.3.5) = 2.4 /(5!)

anonymous
 4 years ago
Best 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I understand the hint.... :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't understand I meant

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.41/(1.3.5) = 2.4 /(5!) = 2^2 . (1.2) / (5!) = 2^2 (2!)/(5!) got that ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how ....Ohhh by having 2x4 in the numerator your eliminating the evens in the denominator!!!!

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.4i try to bring factorial form wherever possible.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.4for 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
 4 years ago
Best 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and yes this part makes sense

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.4why 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
 4 years ago
Best ResponseYou've already chosen the best response.0no I mean this is the 3rd ....whatchamacallit? "m" is from 0 to infinity ...and this would be the third term?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.4m=0 <1st term m=1 <2nd term m=2 <3rd term

anonymous
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0<frustrated koala :S

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.4what 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
 4 years ago
Best ResponseYou've already chosen the best response.0oh so 2m+1 would be right!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where 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
 4 years ago
Best ResponseYou've already chosen the best response.0sorry koalas are fairly slow...I'm still thinking

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.4all 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
 4 years ago
Best 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do either of you have a similar problem somewhere? LOL I wanna try this again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no I think this is enough. Ok thanks

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's the solution in the back of my book for y''xy'y=0

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhhhhhh....durrrr sorry. Ok I got now haha
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.