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Without actually solving the differential equation $$(\cos x)y'' + y' + 5y = 0,$$ find lower bounds for the radii of convergence of the power series solutions about $x=0$ and $x=1$.

Any idea guys?

I thinks singular points are required to answer this. But the problem is I don't understand singular points really well. Can you help me?

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  • $\begingroup$ For the series about $x=0$, the radius of convergence should be $\pi/2$. Try to find the one for $x=1$. $\endgroup$ – Mhenni Benghorbal Jul 16 '13 at 9:24
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    $\begingroup$ @MhenniBenghorbal, how did you get $\pi/2$ ? $\endgroup$ – user84275 Jul 16 '13 at 9:34
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First, note that your ODE can be written as $y'' + p(x)y' + q(x)y = 0$. Moreover, there ought to be a theorem in the textbook you are using which states that the radius of convergence of the power series solution is at least as large as the minimum of the radii of convergence of the power series of $p(x)$ and $q(x)$. The radius of convergence of $f(x)$, about $x_0$, is the distance between $x_0$ and the nearest singularity of $f(x)$. In the complex plane, $0$ is $(0, 0)$ and $1$ is $(1, 0)$, and the nearest singularity of $1/\cos x$ (as well as $5/\cos x$) is $\pi/2$, or $(\pi/2, 0)$. Now you only need to use the Pythagorean Theorem to compute the distance between $(0, 0)$ and $(\pi/2, 0)$, and the distance between $(1, 0)$ and $(\pi/2, 0)$.

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  • $\begingroup$ Do we really need the pythagorean theorem to compute distances on a line? $\endgroup$ – Emily May 7 '14 at 18:17
  • $\begingroup$ @Arkamis No, but in general we do. $\endgroup$ – glebovg May 8 '14 at 0:15

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