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Is this a typo? An exercise in Riemann Surfaces by Donaldson
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I'm trying to solve an exercise in the book "Riemann Surfaces" by Donaldson. But I think there might be a typo in it. Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu,$$ where $a,b,cinmathbb{C}$ are some fixed constants. This exercise asks us to, among other things, show that the indicial equation of the hypergeometric equation at $z=0$ has roots $0$ and $c$. But according to my calculations, and also according to Wikipedia, the roots of the indicial equation should be $0$ and $1-c$ (Wikipedia uses $gamma$ while I use $c$, but whatever). Is this a typo in the book?
ordinary-differential-equations riemann-surfaces
asked Jan 26 at 17:54
user638526user638526
213
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add a comment |
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I'm trying to solve an exercise in the book "Riemann Surfaces" by Donaldson. But I think there might be a typo in it. Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu,$$ where $a,b,cinmathbb{C}$ are some fixed constants. This exercise asks us to, among other things, show that the indicial equation of the hypergeometric equation at $z=0$ has roots $0$ and $c$. But according to my calculations, and also according to Wikipedia, the roots of the indicial equation should be $0$ and $1-c$ (Wikipedia uses $gamma$ while I use $c$, but whatever). Is this a typo in the book?
ordinary-differential-equations riemann-surfaces
asked Jan 26 at 17:54
user638526user638526
213
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Yes, clearly a typo. To find the indicial equation, we substitute $u = z^alpha$ into the differential equation and equate the coefficient at the lowest power of $z$ to zero. The lowest power will be $z^{alpha - 1}$: $$[z^{alpha - 1}] ,z (1 - z) (z^alpha)'' = alpha (alpha - 1), \ [z^{alpha - 1}] ,(c - (a + b + 1) z) (z^alpha)' = c alpha, \ alpha (alpha - 1) + c alpha = 0.$$
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– Maxim
Feb 1 at 12:49
add a comment |
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I'm trying to solve an exercise in the book "Riemann Surfaces" by Donaldson. But I think there might be a typo in it. Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu,$$ where $a,b,cinmathbb{C}$ are some fixed constants. This exercise asks us to, among other things, show that the indicial equation of the hypergeometric equation at $z=0$ has roots $0$ and $c$. But according to my calculations, and also according to Wikipedia, the roots of the indicial equation should be $0$ and $1-c$ (Wikipedia uses $gamma$ while I use $c$, but whatever). Is this a typo in the book?
ordinary-differential-equations riemann-surfaces
asked Jan 26 at 17:54
user638526user638526
213
$endgroup$
I'm trying to solve an exercise in the book "Riemann Surfaces" by Donaldson. But I think there might be a typo in it. Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu,$$ where $a,b,cinmathbb{C}$ are some fixed constants. This exercise asks us to, among other things, show that the indicial equation of the hypergeometric equation at $z=0$ has roots $0$ and $c$. But according to my calculations, and also according to Wikipedia, the roots of the indicial equation should be $0$ and $1-c$ (Wikipedia uses $gamma$ while I use $c$, but whatever). Is this a typo in the book?
ordinary-differential-equations riemann-surfaces
ordinary-differential-equations riemann-surfaces
asked Jan 26 at 17:54
user638526user638526
213
asked Jan 26 at 17:54
user638526user638526
213
asked Jan 26 at 17:54
user638526user638526
213
asked Jan 26 at 17:54
user638526user638526
213
asked Jan 26 at 17:54
user638526user638526
213
213
$begingroup$
Yes, clearly a typo. To find the indicial equation, we substitute $u = z^alpha$ into the differential equation and equate the coefficient at the lowest power of $z$ to zero. The lowest power will be $z^{alpha - 1}$: $$[z^{alpha - 1}] ,z (1 - z) (z^alpha)'' = alpha (alpha - 1), \ [z^{alpha - 1}] ,(c - (a + b + 1) z) (z^alpha)' = c alpha, \ alpha (alpha - 1) + c alpha = 0.$$
$endgroup$
– Maxim
Feb 1 at 12:49
add a comment |
$begingroup$
Yes, clearly a typo. To find the indicial equation, we substitute $u = z^alpha$ into the differential equation and equate the coefficient at the lowest power of $z$ to zero. The lowest power will be $z^{alpha - 1}$: $$[z^{alpha - 1}] ,z (1 - z) (z^alpha)'' = alpha (alpha - 1), \ [z^{alpha - 1}] ,(c - (a + b + 1) z) (z^alpha)' = c alpha, \ alpha (alpha - 1) + c alpha = 0.$$
$endgroup$
– Maxim
Feb 1 at 12:49
$begingroup$
Yes, clearly a typo. To find the indicial equation, we substitute $u = z^alpha$ into the differential equation and equate the coefficient at the lowest power of $z$ to zero. The lowest power will be $z^{alpha - 1}$: $$[z^{alpha - 1}] ,z (1 - z) (z^alpha)'' = alpha (alpha - 1), \ [z^{alpha - 1}] ,(c - (a + b + 1) z) (z^alpha)' = c alpha, \ alpha (alpha - 1) + c alpha = 0.$$
$endgroup$
– Maxim
Feb 1 at 12:49
$begingroup$
Yes, clearly a typo. To find the indicial equation, we substitute $u = z^alpha$ into the differential equation and equate the coefficient at the lowest power of $z$ to zero. The lowest power will be $z^{alpha - 1}$: $$[z^{alpha - 1}] ,z (1 - z) (z^alpha)'' = alpha (alpha - 1), \ [z^{alpha - 1}] ,(c - (a + b + 1) z) (z^alpha)' = c alpha, \ alpha (alpha - 1) + c alpha = 0.$$
$endgroup$
– Maxim
Feb 1 at 12:49
add a comment |
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Yes, clearly a typo. To find the indicial equation, we substitute $u = z^alpha$ into the differential equation and equate the coefficient at the lowest power of $z$ to zero. The lowest power will be $z^{alpha - 1}$: $$[z^{alpha - 1}] ,z (1 - z) (z^alpha)'' = alpha (alpha - 1), \ [z^{alpha - 1}] ,(c - (a + b + 1) z) (z^alpha)' = c alpha, \ alpha (alpha - 1) + c alpha = 0.$$
$endgroup$
– Maxim
Feb 1 at 12:49