Hypergeometric equation solution

[Anonymous No.15674132]

The solutions to the hypergeometric equation may be given in the form of a Frobenius series. One of the solutions of the indicial equation has σ=1-c. From this we get the recurrence relation for the coefficients:
a_n+1 = [(a-c+1+n)*(b-c+1+n)/[(n+1)*(2-c+n)]]*a_n
Can you show that we require c, a-b, and c-a-b to all be non-integers? I assume it has something to do with requiring the series doesn't terminate. See the sentence before equation (18.143)

[Anonymous No.15674141]

>homework thread

[Anonymous No.15674147]

>>15674141
It ain't, honestly. The textbook just drops this claim, but doesn't justify it.

[Anonymous No.15674189]

Clearly when c=1 the two solutions in (18.143) are identical. When c is an integer <1, then F(a,b,c;x) is undefined. You can see that from the gamma function representation or from the recursion relation. Similarly when c is an integer >1 then F(..., ..., 2-c;x) is undefined. That covers all integers.

F(a,b,c;x) is undefined at c being non-positive integers. You can see that in the gamma function representation. So you only have the other solution in that case.

Similarly F(a-c+1,b-c+1,2-c;x) is undefined for 2-c being non-positive integers, or in other words if c is an integer strictly greater than 1.

[Anonymous No.15674191]

Clearly when c=1 the two solutions in (18.143) are identical. When c is an integer <1, then F(a,b,c;x) is undefined. You can see that from the gamma function representation or from the recursion relation. Similarly when c is an integer >1 then F(..., ..., 2-c;x) is undefined. That covers all integers.

[Anonymous No.15674194]

>>15674189
I don't see how a negative integer c ruins it? Say, c=-3. Gamma functions are defined for negative integers aren't they?

[Anonymous No.15674196]

>>15674194
No. There is a pole at the negative integers and zero

[Anonymous No.15674202]

>>15674196
Ok, thanks, that makes sense. What about the requirements for a-b and c-a-b to be non-integers?

[Anonymous No.15674207]

>>15674202
No clue for that statement, sorry

[Jealous No.15674219]

>>15674132
That's the good stuff.

[Anonymous No.15674307]

Bump

[Anonymous No.15674395]

>>15674191
Hang on, F(..., ..., 2-c;x) is surely well defined for c as a negative integer. The gamma would be of 2-c, not c like in the case for σ=0.

[Anonymous No.15674412]

>>15674395
No, because the condition is for the full solution which requires both series.

[Anonymous No.15674576]

So, does anyone see why a-b and c-a-b must be non-integers? The only thing I can see is that c-a-1 and c-b-1 are not positive integers (otherwise the recurrence relation terminates).

[Anonymous No.15674762]

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[Anonymous No.15675602]

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[Anonymous No.15675755]

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[Anonymous No.15675977]

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[Anonymous No.15676231]

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[Anonymous No.15676489]

Bump

[Anonymous No.15676522]

>>15676489
Just ask on math stack exchange or dig around on this site:
https://dlmf.nist.gov/15

Stop bumping the thread