Does the series of $\frac{x^n}{(n!)^2}$ converge to a function?

Asked Nov 27 '19 at 21:19

Active Nov 27 '19 at 21:54

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So, I was resolving the equation $xy''+y'-y=0$ by calculating Frobenius series, and I got that the first linear independent solution is $y_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{(n!)^2}$, which kind of looks like the power series for $e^x$, but I don't know what to do to the other $\frac{1}{n!}$.

Does this series converges to a function or should I just leave it indicated as the series?

edited Nov 27 '19 at 21:45

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asked Nov 27 '19 at 21:19

Armando Rosas

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Looks like a Bessel function to me. – Angina Seng Nov 27 '19 at 21:21
@Lord so my ODE is a bessel equation? but what about the $-y$? If a multiply by x again, i have $x^2y''+xy'-xy=0$, which looks like a bessel equation, but $-x$ doesn't look like $x^2-c^2$ – Armando Rosas Nov 27 '19 at 21:29

Does the series of $\frac{x^n}{(n!)^2}$ converge to a function?

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