Mixing
How do I prove $zDelta(z+1) = Delta(z)$, where $Gamma(z)=1/Delta(z)$?
$begingroup$
EDIT notes: Deleted irrelevant information
I'm asked to prove that $zDelta(z+1)=Delta(z)$, where $Delta(z)equiv ze^{gamma z}prodlimits_{m=1}^infty(1+z/m)e^{-z/m}$ and $gamma =lim_{ntoinfty} 1+ 1/2 + 1/3 + ... + 1/n-log n$.
My idea was to look at finite n both in $gamma$ and in the product. First I assumed the theorem to be true, thus getting $Delta(z+1)=Delta(z)/z$. I divided both sides by $e^{gamma z}$ and then took the logarithm. From here, managed to prove equivalence. Is there another, "cleaner" proof?
complex-analysis gamma-function
asked Nov 30 '18 at 14:37
Idea FluxIdea Flux
86
$endgroup$
add a comment |
$begingroup$
EDIT notes: Deleted irrelevant information
I'm asked to prove that $zDelta(z+1)=Delta(z)$, where $Delta(z)equiv ze^{gamma z}prodlimits_{m=1}^infty(1+z/m)e^{-z/m}$ and $gamma =lim_{ntoinfty} 1+ 1/2 + 1/3 + ... + 1/n-log n$.
My idea was to look at finite n both in $gamma$ and in the product. First I assumed the theorem to be true, thus getting $Delta(z+1)=Delta(z)/z$. I divided both sides by $e^{gamma z}$ and then took the logarithm. From here, managed to prove equivalence. Is there another, "cleaner" proof?
complex-analysis gamma-function
asked Nov 30 '18 at 14:37
Idea FluxIdea Flux
86
$endgroup$
add a comment |
$begingroup$
EDIT notes: Deleted irrelevant information
I'm asked to prove that $zDelta(z+1)=Delta(z)$, where $Delta(z)equiv ze^{gamma z}prodlimits_{m=1}^infty(1+z/m)e^{-z/m}$ and $gamma =lim_{ntoinfty} 1+ 1/2 + 1/3 + ... + 1/n-log n$.
My idea was to look at finite n both in $gamma$ and in the product. First I assumed the theorem to be true, thus getting $Delta(z+1)=Delta(z)/z$. I divided both sides by $e^{gamma z}$ and then took the logarithm. From here, managed to prove equivalence. Is there another, "cleaner" proof?
complex-analysis gamma-function
asked Nov 30 '18 at 14:37
Idea FluxIdea Flux
86
$endgroup$
EDIT notes: Deleted irrelevant information
I'm asked to prove that $zDelta(z+1)=Delta(z)$, where $Delta(z)equiv ze^{gamma z}prodlimits_{m=1}^infty(1+z/m)e^{-z/m}$ and $gamma =lim_{ntoinfty} 1+ 1/2 + 1/3 + ... + 1/n-log n$.
My idea was to look at finite n both in $gamma$ and in the product. First I assumed the theorem to be true, thus getting $Delta(z+1)=Delta(z)/z$. I divided both sides by $e^{gamma z}$ and then took the logarithm. From here, managed to prove equivalence. Is there another, "cleaner" proof?
complex-analysis gamma-function
complex-analysis gamma-function
asked Nov 30 '18 at 14:37
Idea FluxIdea Flux
86
asked Nov 30 '18 at 14:37
Idea FluxIdea Flux
86
edited Jan 12 at 17:01
Idea Flux
asked Nov 30 '18 at 14:37
Idea FluxIdea Flux
86
asked Nov 30 '18 at 14:37
Idea FluxIdea Flux
86
asked Nov 30 '18 at 14:37
Idea FluxIdea Flux
86
86
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Consider the telescoping sum
$$sum_{m=1}^n left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) = logleft(1+frac{z}{n+1}right) - log(1+z)$$
Then
$$- log(1+z) =
sum_{m=1}^infty left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-logleft(frac{m+1}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-frac{1}{m}+frac{1}{m}-logleft(frac{m+1}{m}right)right)
=
sum_{m=1}^infty left(logleft(frac{1+(z+1)/m}{1+z/m}right)-frac{1}{m}right)+gamma$$
Apply exponential to get
$$frac{1}{1+z} = e^gamma prod_{m=1}^inftyfrac{1+(z+1)/m}{1+z/m}e^{-1/m}=frac{e^{gamma(z+1)}}{e^{gamma z}}prod_{m=1}^infty frac{(1+(z+1)/m)e^{-(z+1)/m}}{(1+z/m)e^{-z/m}}$$
Then
$$z e^{gamma z} prod_{m=1}^infty (1+z/m)e^{-z/m} = z(z+1) e^{gamma(z+1)}prod_{m=1}^infty (1+(z+1)/m)e^{-(z+1)/m}$$
and we are done
answered Nov 30 '18 at 15:41
jjagmathjjagmath
2247
$endgroup$
$begingroup$
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
$endgroup$
– Idea Flux
Nov 30 '18 at 15:49
$begingroup$
Thank you very much. It is a clean/neat proof!
$endgroup$
– Idea Flux
Dec 3 '18 at 10:51
$begingroup$
How does $1/m -log[(m+1)/m]$ give $gamma$? I get $1+1/2+...+1/n - (log(2/1)+log(3/2)+...+log[(n+1)/n])=1+1/2+...+1/n-log(n+1)$ which is almost equal to $gamma$. I can only think of something like $log(n+1)/log(n)to 0$ as $ntoinfty$
$endgroup$
– Idea Flux
Dec 4 '18 at 21:05
$begingroup$
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
$endgroup$
– jjagmath
Dec 4 '18 at 23:58
$begingroup$
Thank you * character filling *
$endgroup$
– Idea Flux
Dec 5 '18 at 11:29
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Consider the telescoping sum
$$sum_{m=1}^n left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) = logleft(1+frac{z}{n+1}right) - log(1+z)$$
Then
$$- log(1+z) =
sum_{m=1}^infty left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-logleft(frac{m+1}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-frac{1}{m}+frac{1}{m}-logleft(frac{m+1}{m}right)right)
=
sum_{m=1}^infty left(logleft(frac{1+(z+1)/m}{1+z/m}right)-frac{1}{m}right)+gamma$$
Apply exponential to get
$$frac{1}{1+z} = e^gamma prod_{m=1}^inftyfrac{1+(z+1)/m}{1+z/m}e^{-1/m}=frac{e^{gamma(z+1)}}{e^{gamma z}}prod_{m=1}^infty frac{(1+(z+1)/m)e^{-(z+1)/m}}{(1+z/m)e^{-z/m}}$$
Then
$$z e^{gamma z} prod_{m=1}^infty (1+z/m)e^{-z/m} = z(z+1) e^{gamma(z+1)}prod_{m=1}^infty (1+(z+1)/m)e^{-(z+1)/m}$$
and we are done
answered Nov 30 '18 at 15:41
jjagmathjjagmath
2247
$endgroup$
$begingroup$
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
$endgroup$
– Idea Flux
Nov 30 '18 at 15:49
$begingroup$
Thank you very much. It is a clean/neat proof!
$endgroup$
– Idea Flux
Dec 3 '18 at 10:51
$begingroup$
How does $1/m -log[(m+1)/m]$ give $gamma$? I get $1+1/2+...+1/n - (log(2/1)+log(3/2)+...+log[(n+1)/n])=1+1/2+...+1/n-log(n+1)$ which is almost equal to $gamma$. I can only think of something like $log(n+1)/log(n)to 0$ as $ntoinfty$
$endgroup$
– Idea Flux
Dec 4 '18 at 21:05
$begingroup$
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
$endgroup$
– jjagmath
Dec 4 '18 at 23:58
$begingroup$
Thank you * character filling *
$endgroup$
– Idea Flux
Dec 5 '18 at 11:29
|
show 1 more comment
$begingroup$
Consider the telescoping sum
$$sum_{m=1}^n left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) = logleft(1+frac{z}{n+1}right) - log(1+z)$$
Then
$$- log(1+z) =
sum_{m=1}^infty left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-logleft(frac{m+1}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-frac{1}{m}+frac{1}{m}-logleft(frac{m+1}{m}right)right)
=
sum_{m=1}^infty left(logleft(frac{1+(z+1)/m}{1+z/m}right)-frac{1}{m}right)+gamma$$
Apply exponential to get
$$frac{1}{1+z} = e^gamma prod_{m=1}^inftyfrac{1+(z+1)/m}{1+z/m}e^{-1/m}=frac{e^{gamma(z+1)}}{e^{gamma z}}prod_{m=1}^infty frac{(1+(z+1)/m)e^{-(z+1)/m}}{(1+z/m)e^{-z/m}}$$
Then
$$z e^{gamma z} prod_{m=1}^infty (1+z/m)e^{-z/m} = z(z+1) e^{gamma(z+1)}prod_{m=1}^infty (1+(z+1)/m)e^{-(z+1)/m}$$
and we are done
answered Nov 30 '18 at 15:41
jjagmathjjagmath
2247
$endgroup$
$begingroup$
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
$endgroup$
– Idea Flux
Nov 30 '18 at 15:49
$begingroup$
Thank you very much. It is a clean/neat proof!
$endgroup$
– Idea Flux
Dec 3 '18 at 10:51
$begingroup$
How does $1/m -log[(m+1)/m]$ give $gamma$? I get $1+1/2+...+1/n - (log(2/1)+log(3/2)+...+log[(n+1)/n])=1+1/2+...+1/n-log(n+1)$ which is almost equal to $gamma$. I can only think of something like $log(n+1)/log(n)to 0$ as $ntoinfty$
$endgroup$
– Idea Flux
Dec 4 '18 at 21:05
$begingroup$
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
$endgroup$
– jjagmath
Dec 4 '18 at 23:58
$begingroup$
Thank you * character filling *
$endgroup$
– Idea Flux
Dec 5 '18 at 11:29
|
show 1 more comment
$begingroup$
Consider the telescoping sum
$$sum_{m=1}^n left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) = logleft(1+frac{z}{n+1}right) - log(1+z)$$
Then
$$- log(1+z) =
sum_{m=1}^infty left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-logleft(frac{m+1}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-frac{1}{m}+frac{1}{m}-logleft(frac{m+1}{m}right)right)
=
sum_{m=1}^infty left(logleft(frac{1+(z+1)/m}{1+z/m}right)-frac{1}{m}right)+gamma$$
Apply exponential to get
$$frac{1}{1+z} = e^gamma prod_{m=1}^inftyfrac{1+(z+1)/m}{1+z/m}e^{-1/m}=frac{e^{gamma(z+1)}}{e^{gamma z}}prod_{m=1}^infty frac{(1+(z+1)/m)e^{-(z+1)/m}}{(1+z/m)e^{-z/m}}$$
Then
$$z e^{gamma z} prod_{m=1}^infty (1+z/m)e^{-z/m} = z(z+1) e^{gamma(z+1)}prod_{m=1}^infty (1+(z+1)/m)e^{-(z+1)/m}$$
and we are done
answered Nov 30 '18 at 15:41
jjagmathjjagmath
2247
$endgroup$
Consider the telescoping sum
$$sum_{m=1}^n left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) = logleft(1+frac{z}{n+1}right) - log(1+z)$$
Then
$$- log(1+z) =
sum_{m=1}^infty left(logleft(1+frac{z}{m+1}right)-logleft(1+frac{z}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-logleft(frac{m+1}{m}right)right) =
sum_{m=1}^infty left(logleft(frac{z+m+1}{z+m}right)-frac{1}{m}+frac{1}{m}-logleft(frac{m+1}{m}right)right)
=
sum_{m=1}^infty left(logleft(frac{1+(z+1)/m}{1+z/m}right)-frac{1}{m}right)+gamma$$
Apply exponential to get
$$frac{1}{1+z} = e^gamma prod_{m=1}^inftyfrac{1+(z+1)/m}{1+z/m}e^{-1/m}=frac{e^{gamma(z+1)}}{e^{gamma z}}prod_{m=1}^infty frac{(1+(z+1)/m)e^{-(z+1)/m}}{(1+z/m)e^{-z/m}}$$
Then
$$z e^{gamma z} prod_{m=1}^infty (1+z/m)e^{-z/m} = z(z+1) e^{gamma(z+1)}prod_{m=1}^infty (1+(z+1)/m)e^{-(z+1)/m}$$
and we are done
answered Nov 30 '18 at 15:41
jjagmathjjagmath
2247
answered Nov 30 '18 at 15:41
jjagmathjjagmath
2247
answered Nov 30 '18 at 15:41
jjagmathjjagmath
2247
answered Nov 30 '18 at 15:41
jjagmathjjagmath
2247
2247
$begingroup$
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
$endgroup$
– Idea Flux
Nov 30 '18 at 15:49
$begingroup$
Thank you very much. It is a clean/neat proof!
$endgroup$
– Idea Flux
Dec 3 '18 at 10:51
$begingroup$
How does $1/m -log[(m+1)/m]$ give $gamma$? I get $1+1/2+...+1/n - (log(2/1)+log(3/2)+...+log[(n+1)/n])=1+1/2+...+1/n-log(n+1)$ which is almost equal to $gamma$. I can only think of something like $log(n+1)/log(n)to 0$ as $ntoinfty$
$endgroup$
– Idea Flux
Dec 4 '18 at 21:05
$begingroup$
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
$endgroup$
– jjagmath
Dec 4 '18 at 23:58
$begingroup$
Thank you * character filling *
$endgroup$
– Idea Flux
Dec 5 '18 at 11:29
|
show 1 more comment
$begingroup$
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
$endgroup$
– Idea Flux
Nov 30 '18 at 15:49
$begingroup$
Thank you very much. It is a clean/neat proof!
$endgroup$
– Idea Flux
Dec 3 '18 at 10:51
$begingroup$
How does $1/m -log[(m+1)/m]$ give $gamma$? I get $1+1/2+...+1/n - (log(2/1)+log(3/2)+...+log[(n+1)/n])=1+1/2+...+1/n-log(n+1)$ which is almost equal to $gamma$. I can only think of something like $log(n+1)/log(n)to 0$ as $ntoinfty$
$endgroup$
– Idea Flux
Dec 4 '18 at 21:05
$begingroup$
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
$endgroup$
– jjagmath
Dec 4 '18 at 23:58
$begingroup$
Thank you * character filling *
$endgroup$
– Idea Flux
Dec 5 '18 at 11:29
$begingroup$
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
$endgroup$
– Idea Flux
Nov 30 '18 at 15:49
$begingroup$
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
$endgroup$
– Idea Flux
Nov 30 '18 at 15:49
$begingroup$
Thank you very much. It is a clean/neat proof!
$endgroup$
– Idea Flux
Dec 3 '18 at 10:51
$begingroup$
Thank you very much. It is a clean/neat proof!
$endgroup$
– Idea Flux
Dec 3 '18 at 10:51
$begingroup$
How does $1/m -log[(m+1)/m]$ give $gamma$? I get $1+1/2+...+1/n - (log(2/1)+log(3/2)+...+log[(n+1)/n])=1+1/2+...+1/n-log(n+1)$ which is almost equal to $gamma$. I can only think of something like $log(n+1)/log(n)to 0$ as $ntoinfty$
$endgroup$
– Idea Flux
Dec 4 '18 at 21:05
$begingroup$
How does $1/m -log[(m+1)/m]$ give $gamma$? I get $1+1/2+...+1/n - (log(2/1)+log(3/2)+...+log[(n+1)/n])=1+1/2+...+1/n-log(n+1)$ which is almost equal to $gamma$. I can only think of something like $log(n+1)/log(n)to 0$ as $ntoinfty$
$endgroup$
– Idea Flux
Dec 4 '18 at 21:05
$begingroup$
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
$endgroup$
– jjagmath
Dec 4 '18 at 23:58
$begingroup$
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
$endgroup$
– jjagmath
Dec 4 '18 at 23:58
$begingroup$
Thank you * character filling *
$endgroup$
– Idea Flux
Dec 5 '18 at 11:29
$begingroup$
Thank you * character filling *
$endgroup$
– Idea Flux
Dec 5 '18 at 11:29
|
show 1 more comment
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