Question

solve DE y=1/2(xy'+y") with y(0)=1 and y'(0)=0 with power series and what do you expect of the convergence behaviour of the power series? What is the convergence radius?
With the power series I found that y=1+x²... but not sure how to evaluate its convergence behaviour or get the radius from hereon

Answer 1

Please give a few more terms so that we can get the general form. Form there then we can work out the convergence.

Answer 2

recursive relationship
\[c _{n+2}=\frac{ -(n-2)c _{n} }{ (n+2)(n+1) }\]

since y(0)=1 and y'(0)=0
=> c0=1
=> c1= 0
=>c2=1 (the constants that I put out of the summations to equalize to one are 2(c2-c0)=0)
and every further coefficient =0 after that

Answer 3

It looks like your power series is: (x^2)^0 + (x^2)^1+(x^2)^2+... which is a geometric series. Which means it can be rewritten as 1/(1-x^2) so that means we need the denominator to be not zero or smaller because if it's larger then it really diverges.

Answer 4

I can explain anything that you don't understand about my reasoning if it seems weird to help you fill in the gaps and understand. =)

Answer 5

Wait, though are you saying the whole answer is just

y=1+x² and that's it? Now that I've seen your recursion it sounds like you're saying all coefficients are 0 except those two.

Answer 6

yes, I had thought of the geometric series, but wasn't sure what to put for r, but the 1+x² and that's it threw me for a loop.

But it's 0 because of the conditions given along of y(0)=1 and y'(0)=0

Answer 7

Ok I think I am discovering a problem.

What is the coefficient on the first 4 even terms?

Answer 8

I'm not sure I see how you reasoned out how all the even terms past C0 and C2 are zero. I see how the odd ones will be all 0 because your recursion relation and y'(0)=0.

Answer 9

Ahh wait no I'm wrong, sorry you're right I just did some algebra wrong.

Well that's fairly straight forward then.

y=1+x^2. There are no more terms. It's not 1+x^2+... since all the higher terms are zero.

So then it's fairly simple, it is not limited because you don't have an infinite series anymore.

Answer 10

Here's DE I have before applying the extra conditions and finding the recursive relationship
\[2(c _{2}-c _{0})+\sum_{n=1}^{\infty}\left[ \left( n+2 \right)\left( n+1 \right) c_{n+2}+\left( n-2 \right) c _{n}\right]x ^{n}=0\]

So I know that c2=c0

If i ignore the extra conditions I get

n=1 -> c3=c1/3!
n=2 -> c4=0
n=3 -> c5=-c3/5!
n=4 -> c6=0

Answer 11

So it converges for all values of x, its radius of convergence is infinite. It's not an infinite series anymore, so it's just like any other differential equation you solved without power series.

Answer 12

One half of the solution is not infinite while the other half is infinite

Answer 13

@DominicNg There is no infinite half.

Answer 14

Ah ok. Sorry, I'm used to sometimes writing ... in my sentences, and in that one that would have been confusing. But at least I have an answer.

I was stumped for finding a limit of something that wasn't a series in this case and had no n's. And I wasn't sure what to do regarding convergence and radius if I didn't have a limit. You've given me the understanding now.

Answer 15

Awesome, yeah. In some way you can kind of pretend that in the limit of n it converges for all n since... n isn't even there! You could do anything to n and it will converge since it's nowhere to be found. =)

Answer 16

well x can be any real number, and y will always be a positive real number. It won't converge to negative infinity. Is that what you mean @DominicNg?

Answer 17

Let me clarify the DE. Is it:\[y=\frac{ 1 }{ 2 }(xy'+y'')\]or\[y=\frac{ 1 }{2(xy'+y'') }\]

Answer 18

Let me now and i will go away and work in detail rather than making statement off the cuff.

Answer 19

It's the first

Answer 20

OK I will revert

Answer 21

So you get 2y=(xy'+y")

Answer 22

Thanks for wanting to check on my calculations.