Frobenius Solution To The Hypergeometric Equation - Solution Around x = 1

Solution Around x = 1

Let us now study the singular point x = 1. To see if it is regular,

$\begin{align} &\lim_{x \to a} \frac{(x - a) P_1(x)}{P_2(x)} = \lim_{x \to 1} \frac{(x - 1) (\gamma - (1 + \alpha + \beta)x)}{x(1 - x)} \\ &\quad = \lim_{x \to 1} \frac{-(\gamma - (1 + \alpha + \beta)x)}{x} = 1 + \alpha + \beta - \gamma \\ &\lim_{x \to a} \frac{(x - a)^2 P_0(x)}{P_2(x)} = \lim_{x \to 1} \frac{(x - 1)^2 (-\alpha\beta)}{x(1 - x)} = \lim_{x \to 1} \frac{(x - 1) \alpha \beta}{x} = 0 \end{align}$

Hence, both limits exist and x = 1 is a regular singular point. Now, instead of assuming a solution on the form

we will try to express the solutions of this case in terms of the solutions for the point x = 0. We proceed as follows: we had the hypergeometric equation

Let z = 1 - x. Then

$\begin{align} &\frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} = -\frac{dy}{dz} = -y' \\ &\frac{d^2 y}{dx^2} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{d}{dx}\left( -\frac{dy}{dz} \right) = \frac{d}{dz}\left( -\frac{dy}{dz} \right) \times \frac{dz}{dx} =\frac{d^{2}y}{dz^{2}} = y'' \end{align}$

Hence, the equation takes the form

Since z = 1 - x, the solution of the hypergeometric equation at x = 1 is the same as the solution for this equation at z = 0. But the solution at z = 0 is identical to the solution we obtained for the point x = 0, if we replace each γ by α + β - γ + 1. Hence, to get the solutions, we just make this substitution in the previous results. Note also that for x = 0, c₁ = 0 and c₂ = 1 - γ. Hence, in our case, c₁ = 0 while c₂ = γ - α - β. Let us now write the solutions. In the following we replaced each z by 1 - x.

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