- #1

Kirjava

- 27

- 1

Apologies in advance for the TeX in this post, I'm new and having difficulty with the formatting.

I'm trying to understand the logic my professor uses to derive a second linearly independent solution to the hypergeometric DE:

[tex]z(1-z)\frac{d^2 w}{dz^2} + (\gamma - z(1+\alpha + \beta)\frac{dw}{dz} -\alpha\beta w = 0 [/tex].

It is given that the solutions have exponents at each of the singularities (0,1,[tex]\infty[/tex]) which are listed in the Riemann P-function:

[tex]

P\left(

\begin{array}{ccc}

0 & 1 & \infty\\

0 & 0 & \alpha\\

1-\gamma & \gamma -\alpha -\beta & \beta

\end{array} \right) [/tex].

The hypergeometric series [tex]$ _2F_1 \left( \alpha , \beta ; \gamma ; z\right)$[/tex] is the first solution with exponents given in the second row. To "derive" an expression for the second solution whose exponents appear in the last row, the following manipulations are made:

[tex]

$ y = P\left(

\begin{array}{ccc}

0 & 1 & \infty\\

0 & 0 & \alpha\\

1-\gamma & \gamma -\alpha -\beta & \beta

\end{array} \right) = z^{1-\gamma} P\left(

\begin{array}{ccc}

0 & 1 & \infty\\

\gamma -1 & 0 & \alpha+1-\gamma\\

0 & \gamma -\alpha -\beta & \beta+1-\gamma

\end{array} \right)$ [/tex]

and therefore (apparently) [tex] z^{1-\gamma} \, _2F_1 \left(1+ \alpha-\gamma , 1+\beta-\gamma ; 2-\gamma ; z\right)[/tex] is the second linearly independent solution.

All above

These manipulations seem dishonest to me (i.e. I don't understand them). If the Riemann P-function is just a table for summarizing each solution's exponents at the singular points (0,1,[tex]\infty[/tex]), then I'm not even sure what the statement y = P(...) could mean. I can certainly see that if we have a solution w with exponents given in a particular row, then taking out a factor [tex]z^{1-\gamma}[/tex] leaves us with [tex]wz^{\gamma-1}[/tex], which will have exponents given in the respective row of the second P function above. Now the "derivation" seems to assume that [tex]wz^{\gamma-1}[/tex] is the solution to a hypergeometric equation with parameters adjusted to match its exponents at each of the singular points. I don't see how this is warranted.

I'm aware that there are other more direct ways of deriving the second solution (if the DE is tackled directly using a Frobenius series the two solutions pop right out). However I'd rather like to understand these manipulations since they seem rather clever, and we're expected to understand and use many others of a similar kind.

## Homework Statement

I'm trying to understand the logic my professor uses to derive a second linearly independent solution to the hypergeometric DE:

[tex]z(1-z)\frac{d^2 w}{dz^2} + (\gamma - z(1+\alpha + \beta)\frac{dw}{dz} -\alpha\beta w = 0 [/tex].

It is given that the solutions have exponents at each of the singularities (0,1,[tex]\infty[/tex]) which are listed in the Riemann P-function:

[tex]

P\left(

\begin{array}{ccc}

0 & 1 & \infty\\

0 & 0 & \alpha\\

1-\gamma & \gamma -\alpha -\beta & \beta

\end{array} \right) [/tex].

The hypergeometric series [tex]$ _2F_1 \left( \alpha , \beta ; \gamma ; z\right)$[/tex] is the first solution with exponents given in the second row. To "derive" an expression for the second solution whose exponents appear in the last row, the following manipulations are made:

[tex]

$ y = P\left(

\begin{array}{ccc}

0 & 1 & \infty\\

0 & 0 & \alpha\\

1-\gamma & \gamma -\alpha -\beta & \beta

\end{array} \right) = z^{1-\gamma} P\left(

\begin{array}{ccc}

0 & 1 & \infty\\

\gamma -1 & 0 & \alpha+1-\gamma\\

0 & \gamma -\alpha -\beta & \beta+1-\gamma

\end{array} \right)$ [/tex]

and therefore (apparently) [tex] z^{1-\gamma} \, _2F_1 \left(1+ \alpha-\gamma , 1+\beta-\gamma ; 2-\gamma ; z\right)[/tex] is the second linearly independent solution.

## Homework Equations

All above

## The Attempt at a Solution

These manipulations seem dishonest to me (i.e. I don't understand them). If the Riemann P-function is just a table for summarizing each solution's exponents at the singular points (0,1,[tex]\infty[/tex]), then I'm not even sure what the statement y = P(...) could mean. I can certainly see that if we have a solution w with exponents given in a particular row, then taking out a factor [tex]z^{1-\gamma}[/tex] leaves us with [tex]wz^{\gamma-1}[/tex], which will have exponents given in the respective row of the second P function above. Now the "derivation" seems to assume that [tex]wz^{\gamma-1}[/tex] is the solution to a hypergeometric equation with parameters adjusted to match its exponents at each of the singular points. I don't see how this is warranted.

I'm aware that there are other more direct ways of deriving the second solution (if the DE is tackled directly using a Frobenius series the two solutions pop right out). However I'd rather like to understand these manipulations since they seem rather clever, and we're expected to understand and use many others of a similar kind.

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