How do I prove $zDelta(z+1) = Delta(z)$, where $Gamma(z)=1/Delta(z)$?
EDIT notes: I found a mistake in my calculations. * something * means that this term should be omitted (I'm not going to delete those terms, since my question would become meaningless)
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 $gamma =lim_{ntoinfty} 1+ 1/2 + 1/3 + ... + 1/n-log n$, but this far I have not succeeded.
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, getting
$$-log n + log (z+1)+left[sum_{m=1}^nlog(1+cfrac{z+1}{m})+ (-z/m)right] =*log z* + left(sum_{m=1}^nlog (1+z/m)+(-z/m)right)$$
Here $1+1/2+...+n$ (from $gamma$) cancelled against $-1/m$ in the sum on the LHS. This is how I got the above equation.
I then substracted the RHS from the LHS, getting LHS - RHS $= 0$
Obviously, the "$-z/m$" terms cancel and upon taking the e-power, I got
$$(z+1)*(1/z)*(1/n)left[prod_{m=1}^n(k+z+1)/k cdot k/(k+z)right]=0$$
This can be simplified a bit, but I don't see why this equality holds, not even in $ntoinfty$. I have asked a friend of mine and we both think all the steps in between are correct, thus perhaps I should try a completely different approach?
I can elaborate on my steps, but I thought it better to keep it concise.
complex-analysis gamma-function
asked Nov 30 at 14:37
Idea Flux
86
add a comment |
EDIT notes: I found a mistake in my calculations. * something * means that this term should be omitted (I'm not going to delete those terms, since my question would become meaningless)
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 $gamma =lim_{ntoinfty} 1+ 1/2 + 1/3 + ... + 1/n-log n$, but this far I have not succeeded.
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, getting
$$-log n + log (z+1)+left[sum_{m=1}^nlog(1+cfrac{z+1}{m})+ (-z/m)right] =*log z* + left(sum_{m=1}^nlog (1+z/m)+(-z/m)right)$$
Here $1+1/2+...+n$ (from $gamma$) cancelled against $-1/m$ in the sum on the LHS. This is how I got the above equation.
I then substracted the RHS from the LHS, getting LHS - RHS $= 0$
Obviously, the "$-z/m$" terms cancel and upon taking the e-power, I got
$$(z+1)*(1/z)*(1/n)left[prod_{m=1}^n(k+z+1)/k cdot k/(k+z)right]=0$$
This can be simplified a bit, but I don't see why this equality holds, not even in $ntoinfty$. I have asked a friend of mine and we both think all the steps in between are correct, thus perhaps I should try a completely different approach?
I can elaborate on my steps, but I thought it better to keep it concise.
complex-analysis gamma-function
asked Nov 30 at 14:37
Idea Flux
86
add a comment |
EDIT notes: I found a mistake in my calculations. * something * means that this term should be omitted (I'm not going to delete those terms, since my question would become meaningless)
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 $gamma =lim_{ntoinfty} 1+ 1/2 + 1/3 + ... + 1/n-log n$, but this far I have not succeeded.
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, getting
$$-log n + log (z+1)+left[sum_{m=1}^nlog(1+cfrac{z+1}{m})+ (-z/m)right] =*log z* + left(sum_{m=1}^nlog (1+z/m)+(-z/m)right)$$
Here $1+1/2+...+n$ (from $gamma$) cancelled against $-1/m$ in the sum on the LHS. This is how I got the above equation.
I then substracted the RHS from the LHS, getting LHS - RHS $= 0$
Obviously, the "$-z/m$" terms cancel and upon taking the e-power, I got
$$(z+1)*(1/z)*(1/n)left[prod_{m=1}^n(k+z+1)/k cdot k/(k+z)right]=0$$
This can be simplified a bit, but I don't see why this equality holds, not even in $ntoinfty$. I have asked a friend of mine and we both think all the steps in between are correct, thus perhaps I should try a completely different approach?
I can elaborate on my steps, but I thought it better to keep it concise.
complex-analysis gamma-function
asked Nov 30 at 14:37
Idea Flux
86
EDIT notes: I found a mistake in my calculations. * something * means that this term should be omitted (I'm not going to delete those terms, since my question would become meaningless)
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 $gamma =lim_{ntoinfty} 1+ 1/2 + 1/3 + ... + 1/n-log n$, but this far I have not succeeded.
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, getting
$$-log n + log (z+1)+left[sum_{m=1}^nlog(1+cfrac{z+1}{m})+ (-z/m)right] =*log z* + left(sum_{m=1}^nlog (1+z/m)+(-z/m)right)$$
Here $1+1/2+...+n$ (from $gamma$) cancelled against $-1/m$ in the sum on the LHS. This is how I got the above equation.
I then substracted the RHS from the LHS, getting LHS - RHS $= 0$
Obviously, the "$-z/m$" terms cancel and upon taking the e-power, I got
$$(z+1)*(1/z)*(1/n)left[prod_{m=1}^n(k+z+1)/k cdot k/(k+z)right]=0$$
This can be simplified a bit, but I don't see why this equality holds, not even in $ntoinfty$. I have asked a friend of mine and we both think all the steps in between are correct, thus perhaps I should try a completely different approach?
I can elaborate on my steps, but I thought it better to keep it concise.
complex-analysis gamma-function
complex-analysis gamma-function
asked Nov 30 at 14:37
Idea Flux
86
asked Nov 30 at 14:37
Idea Flux
86
edited Dec 3 at 10:59
asked Nov 30 at 14:37
Idea Flux
86
asked Nov 30 at 14:37
Idea Flux
86
asked Nov 30 at 14:37
Idea Flux
86
86
add a comment |
add a comment |
1 Answer
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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 at 15:41
jjagmath
2057
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
– Idea Flux
Nov 30 at 15:49
Thank you very much. It is a clean/neat proof!
– Idea Flux
Dec 3 at 10:51
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 $ntoinfty$
– Idea Flux
Dec 4 at 21:05
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
– jjagmath
Dec 4 at 23:58
Thank you * character filling *
– Idea Flux
Dec 5 at 11:29
add a comment |
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1 Answer
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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 at 15:41
jjagmath
2057
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
– Idea Flux
Nov 30 at 15:49
Thank you very much. It is a clean/neat proof!
– Idea Flux
Dec 3 at 10:51
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 $ntoinfty$
– Idea Flux
Dec 4 at 21:05
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
– jjagmath
Dec 4 at 23:58
Thank you * character filling *
– Idea Flux
Dec 5 at 11:29
add a comment |
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 at 15:41
jjagmath
2057
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
– Idea Flux
Nov 30 at 15:49
Thank you very much. It is a clean/neat proof!
– Idea Flux
Dec 3 at 10:51
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 $ntoinfty$
– Idea Flux
Dec 4 at 21:05
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
– jjagmath
Dec 4 at 23:58
Thank you * character filling *
– Idea Flux
Dec 5 at 11:29
add a comment |
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 at 15:41
jjagmath
2057
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 at 15:41
jjagmath
2057
answered Nov 30 at 15:41
jjagmath
2057
answered Nov 30 at 15:41
jjagmath
2057
answered Nov 30 at 15:41
jjagmath
2057
2057
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
– Idea Flux
Nov 30 at 15:49
Thank you very much. It is a clean/neat proof!
– Idea Flux
Dec 3 at 10:51
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 $ntoinfty$
– Idea Flux
Dec 4 at 21:05
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
– jjagmath
Dec 4 at 23:58
Thank you * character filling *
– Idea Flux
Dec 5 at 11:29
add a comment |
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
– Idea Flux
Nov 30 at 15:49
Thank you very much. It is a clean/neat proof!
– Idea Flux
Dec 3 at 10:51
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 $ntoinfty$
– Idea Flux
Dec 4 at 21:05
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
– jjagmath
Dec 4 at 23:58
Thank you * character filling *
– Idea Flux
Dec 5 at 11:29
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
– Idea Flux
Nov 30 at 15:49
Thank you for the answer :) Unfortunately, monday will be the earliest I can take a close look at it.
– Idea Flux
Nov 30 at 15:49
Thank you very much. It is a clean/neat proof!
– Idea Flux
Dec 3 at 10:51
Thank you very much. It is a clean/neat proof!
– Idea Flux
Dec 3 at 10:51
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 $ntoinfty$
– Idea Flux
Dec 4 at 21:05
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 $ntoinfty$
– Idea Flux
Dec 4 at 21:05
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
– jjagmath
Dec 4 at 23:58
$1+1/2+...+1/n - log(n+1) = 1+1/2+...+1/n -log(n) + log(n) - log(n+1)$ But $log(n)-log(n+1) = log(n/(n+1)) to 0$ as $n to infty$
– jjagmath
Dec 4 at 23:58
Thank you * character filling *
– Idea Flux
Dec 5 at 11:29
Thank you * character filling *
– Idea Flux
Dec 5 at 11:29
add a comment |
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