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Radius of convergence and Frobenius method

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    I am solving a differential equation using the Frobenius method, however halfway through, I have introduced a new power series. I have been told that I need to show the radius of convergence of this series. Why would I have to do that?
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    (Original post by claret_n_blue)
    I am solving a differential equation using the Frobenius method, however halfway through, I have introduced a new power series. I have been told that I need to show the radius of convergence of this series. Why would I have to do that?
    So that you know when it converges :p: For instance, if you introduced a power series whose radius of convergence is zero, your solution isn't going to be of much use.

    Without knowing more details I can't give you much more of a helpful answer than this!
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    (Original post by nuodai)
    So that you know when it converges :p: For instance, if you introduced a power series whose radius of convergence is zero, your solution isn't going to be of much use.

    Without knowing more details I can't give you much more of a helpful answer than this!
    It's that same differential equation I've been working on for the past, what, 2 weeks? 3 weeks?

    I still can't seem to solve it.

    This was just because I got told that I needed to make a point about the radius of convergence.

    Seeing as your online, please could you help me wit it.I am really stuck on one bit about solving DE's using Frobenius. I have a recurrance relation as:

      \frac{h^2}{2m}(\lambda)(\lambda + 1)b_{\lambda+2}  -A b_{\lambda + 1} + (E - Aa)b_{\lambda } = 0

    Where h , A < 0 and a and E are constants. How do I solve this from here? I can solve DE's using power series if there are two "unkowns" but how do I do it using three?

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