Are the Singular Points of the Chebyshev Equation Regular?

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SUMMARY

The singular points of the Chebyshev equation, specifically at x = 1 and x = -1, have been determined to be regular singular points. This conclusion is reached by evaluating the limits of the functions p(x) and q(x) as defined in the Chebyshev differential equation. The conditions for regular singular points are satisfied, confirming that both x = 1 and x = -1 are indeed regular singular points.

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  • Understanding of Chebyshev differential equations
  • Familiarity with singular points in differential equations
  • Knowledge of limit evaluation in calculus
  • Basic concepts of regular and irregular singular points
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  • Study the properties of Chebyshev polynomials
  • Learn about the Frobenius method for solving differential equations
  • Explore the implications of regular singular points in differential equations
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Mathematicians, students of differential equations, and researchers interested in the properties of Chebyshev equations and their applications in mathematical modeling.

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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