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If I had a power series given by $\sum_{n=0}^{\infty}a_{n}x^{n}$ and it was the solution to a differential equation, how would I go about finding the radius of convergence of said power series? TIA

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  • $\begingroup$ The same way you would find the radius of convergence of any power series. $\endgroup$
    – David Reed
    Apr 24 '20 at 11:33
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Assuming this is a linear DE and it can be written

$$y^{(n)} + p_{n-1}(x)y^{(n-1)} + \cdot +p_0(x)y = g(x)$$,

then you can find all the singularities in all the coefficient functions $p_k(t)$. The distance from $0$ to the nearest singularity is the radius of convergence.

Notes: If you center the series at $x_0$, then it's the distance from $x_0$ to the nearest singularity.

You have to account for the complex singularities. If one of your coefficients is $1/(1+x^2)$ then the radius of convergence is no more than the distance from $0$ to $i$.

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