ODE - Power Series Convergence

thejinx0r
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Homework Statement



Solve
(1-4x^2)y''+34x\cdot y'-70y=0

Homework Equations



Basically, I found the recurrence relationship to be:
a_{n+2}=\frac{2 (-7 + n) (-5 + 2 n)}{(n+1)(n+2)}a_n}



Now, I solve for y1 where y1 had a_0=0 and a_1 = 1. It is a simple polynomial of degree 7.
That was the first part.


The second part wants me to find the radius of convergence of y2, where y2 has a_0=1 and a_1 = 0.
In this case, all the odd coefficients are equal to 0. But I am having a hard time trying to find the radius of convergence. I actually don't have a clue here.
I tried doing the ratio test directly on the recurrence term, but led me to 1/x^2, but that didn't work, and I think I did wrong to.

I tried finding the pattern and trying to find what the series actually was, but couldn't. :S
So, now I'm stuck.

(P.S. For the patternn, I got a_n = \frac{2*(-7)(-5)(..)(2n-9)(-5)(-1)(3)(7)(4n-9)}{(2n)!}
But it appears to be invalid when I do the ratio test on it according to webworks.
 
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thejinx0r said:

Homework Statement



Solve
(1-4x^2)y''+34x\cdot y'-70y=0

Homework Equations



Basically, I found the recurrence relationship to be:
a_{n+2}=\frac{2 (-7 + n) (-5 + 2 n)}{(n+1)(n+2)}a_n}



Now, I solve for y1 where y1 had a_0=0 and a_1 = 1. It is a simple polynomial of degree 7.
That was the first part.


The second part wants me to find the radius of convergence of y2, where y2 has a_0=1 and a_1 = 0.
In this case, all the odd coefficients are equal to 0. But I am having a hard time trying to find the radius of convergence. I actually don't have a clue here.
I tried doing the ratio test directly on the recurrence term, but led me to 1/x^2, but that didn't work, and I think I did wrong to.
No, the ratio test will work fine. If it led you to 1/x^2, then you have it upside down! Look at \frac{a_{2(n+1)}x^{2(n+1)}}{a_{2n}x^{2n}}= \frac{a_{2n+2}}{a_{2n}}x^2[/itex]. For what values of x is that less than 1?<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I tried finding the pattern and trying to find what the series actually was, but couldn&#039;t. :S<br /> So, now I&#039;m stuck.<br /> <br /> (P.S. For the patternn, I got a_n = \frac{2*(-7)(-5)(..)(2n-9)(-5)(-1)(3)(7)(4n-9)}{(2n)!}<br /> But it appears to be invalid when I do the ratio test on it according to webworks. </div> </div> </blockquote>
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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