Polygamma reflection formula
$begingroup$
How does one prove the polygamma reflection formula:
$$psi^{(n)}(1-z)+(-1)^{n+1}psi^{(n)}(z)=(-1)^n pi frac{d^n}{d z^n} cot pi z $$
Do we have to invoke the power of contour integration and kernels? I searched the web but the proof was to be found nowhere.
real-analysis analysis special-functions
asked Apr 3 '15 at 15:29
TolasoTolaso
3,4151231
$endgroup$
add a comment |
$begingroup$
How does one prove the polygamma reflection formula:
$$psi^{(n)}(1-z)+(-1)^{n+1}psi^{(n)}(z)=(-1)^n pi frac{d^n}{d z^n} cot pi z $$
Do we have to invoke the power of contour integration and kernels? I searched the web but the proof was to be found nowhere.
real-analysis analysis special-functions
asked Apr 3 '15 at 15:29
TolasoTolaso
3,4151231
$endgroup$
add a comment |
$begingroup$
How does one prove the polygamma reflection formula:
$$psi^{(n)}(1-z)+(-1)^{n+1}psi^{(n)}(z)=(-1)^n pi frac{d^n}{d z^n} cot pi z $$
Do we have to invoke the power of contour integration and kernels? I searched the web but the proof was to be found nowhere.
real-analysis analysis special-functions
asked Apr 3 '15 at 15:29
TolasoTolaso
3,4151231
$endgroup$
How does one prove the polygamma reflection formula:
$$psi^{(n)}(1-z)+(-1)^{n+1}psi^{(n)}(z)=(-1)^n pi frac{d^n}{d z^n} cot pi z $$
Do we have to invoke the power of contour integration and kernels? I searched the web but the proof was to be found nowhere.
real-analysis analysis special-functions
real-analysis analysis special-functions
asked Apr 3 '15 at 15:29
TolasoTolaso
3,4151231
asked Apr 3 '15 at 15:29
TolasoTolaso
3,4151231
asked Apr 3 '15 at 15:29
TolasoTolaso
3,4151231
asked Apr 3 '15 at 15:29
TolasoTolaso
3,4151231
asked Apr 3 '15 at 15:29
TolasoTolaso
3,4151231
3,4151231
add a comment |
add a comment |
1 Answer
1
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$begingroup$
You just need to prove the reflection formula:
$$ psi(1-z)-psi(z)=picot(pi z)tag{1}$$
then differentiate it multiple times. In order to prove $(1)$, let's start from the Weierstrass product for the $Gamma$ function:
$$Gamma(t+1) = e^{-gamma t}prod_{n=1}^{+infty}left(1+frac{t}{n}right)^{-1}e^{frac{t}{n}}tag{2}$$
leading to:
$$ Gamma(z),Gamma(1-z) = frac{pi}{sin(pi z)}tag{3} $$
(the reflection formula for the $Gamma$ function), then consider the logarithmic derivative of $(3)$:
$$psi(z)-psi(1-z) = frac{d}{dz},logsin(pi z)tag{4} $$
and $(1)$ is proved.
answered Apr 3 '15 at 16:26
Jack D'AurizioJack D'Aurizio
289k33280660
$endgroup$
1
$begingroup$
Wow... Jack... super! I did not think that this formula is SOOOO easy to prove... I thought we would invoke contour integration or apply many acrobatics in order to prove that.. Thank you.
$endgroup$
– Tolaso
Apr 3 '15 at 16:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You just need to prove the reflection formula:
$$ psi(1-z)-psi(z)=picot(pi z)tag{1}$$
then differentiate it multiple times. In order to prove $(1)$, let's start from the Weierstrass product for the $Gamma$ function:
$$Gamma(t+1) = e^{-gamma t}prod_{n=1}^{+infty}left(1+frac{t}{n}right)^{-1}e^{frac{t}{n}}tag{2}$$
leading to:
$$ Gamma(z),Gamma(1-z) = frac{pi}{sin(pi z)}tag{3} $$
(the reflection formula for the $Gamma$ function), then consider the logarithmic derivative of $(3)$:
$$psi(z)-psi(1-z) = frac{d}{dz},logsin(pi z)tag{4} $$
and $(1)$ is proved.
answered Apr 3 '15 at 16:26
Jack D'AurizioJack D'Aurizio
289k33280660
$endgroup$
1
$begingroup$
Wow... Jack... super! I did not think that this formula is SOOOO easy to prove... I thought we would invoke contour integration or apply many acrobatics in order to prove that.. Thank you.
$endgroup$
– Tolaso
Apr 3 '15 at 16:32
add a comment |
$begingroup$
You just need to prove the reflection formula:
$$ psi(1-z)-psi(z)=picot(pi z)tag{1}$$
then differentiate it multiple times. In order to prove $(1)$, let's start from the Weierstrass product for the $Gamma$ function:
$$Gamma(t+1) = e^{-gamma t}prod_{n=1}^{+infty}left(1+frac{t}{n}right)^{-1}e^{frac{t}{n}}tag{2}$$
leading to:
$$ Gamma(z),Gamma(1-z) = frac{pi}{sin(pi z)}tag{3} $$
(the reflection formula for the $Gamma$ function), then consider the logarithmic derivative of $(3)$:
$$psi(z)-psi(1-z) = frac{d}{dz},logsin(pi z)tag{4} $$
and $(1)$ is proved.
answered Apr 3 '15 at 16:26
Jack D'AurizioJack D'Aurizio
289k33280660
$endgroup$
1
$begingroup$
Wow... Jack... super! I did not think that this formula is SOOOO easy to prove... I thought we would invoke contour integration or apply many acrobatics in order to prove that.. Thank you.
$endgroup$
– Tolaso
Apr 3 '15 at 16:32
add a comment |
$begingroup$
You just need to prove the reflection formula:
$$ psi(1-z)-psi(z)=picot(pi z)tag{1}$$
then differentiate it multiple times. In order to prove $(1)$, let's start from the Weierstrass product for the $Gamma$ function:
$$Gamma(t+1) = e^{-gamma t}prod_{n=1}^{+infty}left(1+frac{t}{n}right)^{-1}e^{frac{t}{n}}tag{2}$$
leading to:
$$ Gamma(z),Gamma(1-z) = frac{pi}{sin(pi z)}tag{3} $$
(the reflection formula for the $Gamma$ function), then consider the logarithmic derivative of $(3)$:
$$psi(z)-psi(1-z) = frac{d}{dz},logsin(pi z)tag{4} $$
and $(1)$ is proved.
answered Apr 3 '15 at 16:26
Jack D'AurizioJack D'Aurizio
289k33280660
$endgroup$
You just need to prove the reflection formula:
$$ psi(1-z)-psi(z)=picot(pi z)tag{1}$$
then differentiate it multiple times. In order to prove $(1)$, let's start from the Weierstrass product for the $Gamma$ function:
$$Gamma(t+1) = e^{-gamma t}prod_{n=1}^{+infty}left(1+frac{t}{n}right)^{-1}e^{frac{t}{n}}tag{2}$$
leading to:
$$ Gamma(z),Gamma(1-z) = frac{pi}{sin(pi z)}tag{3} $$
(the reflection formula for the $Gamma$ function), then consider the logarithmic derivative of $(3)$:
$$psi(z)-psi(1-z) = frac{d}{dz},logsin(pi z)tag{4} $$
and $(1)$ is proved.
answered Apr 3 '15 at 16:26
Jack D'AurizioJack D'Aurizio
289k33280660
answered Apr 3 '15 at 16:26
Jack D'AurizioJack D'Aurizio
289k33280660
answered Apr 3 '15 at 16:26
Jack D'AurizioJack D'Aurizio
289k33280660
answered Apr 3 '15 at 16:26
Jack D'AurizioJack D'Aurizio
289k33280660
289k33280660
1
$begingroup$
Wow... Jack... super! I did not think that this formula is SOOOO easy to prove... I thought we would invoke contour integration or apply many acrobatics in order to prove that.. Thank you.
$endgroup$
– Tolaso
Apr 3 '15 at 16:32
add a comment |
1
$begingroup$
Wow... Jack... super! I did not think that this formula is SOOOO easy to prove... I thought we would invoke contour integration or apply many acrobatics in order to prove that.. Thank you.
$endgroup$
– Tolaso
Apr 3 '15 at 16:32
1
1
$begingroup$
Wow... Jack... super! I did not think that this formula is SOOOO easy to prove... I thought we would invoke contour integration or apply many acrobatics in order to prove that.. Thank you.
$endgroup$
– Tolaso
Apr 3 '15 at 16:32
$begingroup$
Wow... Jack... super! I did not think that this formula is SOOOO easy to prove... I thought we would invoke contour integration or apply many acrobatics in order to prove that.. Thank you.
$endgroup$
– Tolaso
Apr 3 '15 at 16:32
add a comment |
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