MHB Are the Singular Points of the Chebyshev Equation Regular?

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The singular points of the Chebyshev equation have been identified as x = 1 and x = -1. To determine if these points are regular singular points, one must evaluate the limits of p(x) and q(x) as x approaches these singularities. If both limits exist and are finite, the points qualify as regular singular points. The discussion confirms that both x = 1 and x = -1 meet these criteria, thus establishing them as regular singular points. This conclusion is supported by the definitions and calculations provided.
Poirot1
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I have computed the singular points of Chebyshev equation to be x= 1, -1. What is the best way to find whether they are regular? Thanks.
 
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The Chebisheff DE is...

$\displaystyle y^{\ ''} - \frac{x}{1-x^{2}}\ y^{\ '} + \frac{\alpha^{2}}{1-x^{2}}\ y= y^{\ ''} + p(x)\ y^{\ '} + q(x)\ y=0$ (1)

If $x_{0}$ is a singularity of p(x) and q(x) and both the limits...

$\displaystyle \lim_{x \rightarrow x_{0}} (x-x_{0})\ p(x)$

$\displaystyle \lim_{x \rightarrow x_{0}} (x-x_{0})^{2}\ q(x)$ (2)

... exist finite, then $x_{0}$ is a regular singular point. You can verify that $x_{0}=1$ and $x_{0}=-1$ are both regular singular points...

Kind regards

$\chi$ $\sigma$
 
Thanks
 

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