Mixing
How to show that $ prodlimits_{r=1}^{n}Gamma{ left({frac {r}{n+1}}right)}={sqrt {frac {(2pi )^{n}}{n+1}}}$?
$begingroup$
I saw the following equation on Wikipedia, but I am not sure how to approach it.
$$ prod_{r=1}^{n}Gamma{ left({frac {r}{n+1}}right)}={sqrt {frac {(2pi )^{n}}{n+1}}}$$
Here are some other values listed on Wikipedia
$$ prod _{r=1}^{2}Gamma left({tfrac {r}{3}}right)={frac {2pi }{sqrt {3}}}$$
$$ prod _{r=1}^{3}Gamma left({tfrac {r}{4}}right)={sqrt {2pi ^{3}}}$$
$$prod _{r=1}^{4}Gamma left({tfrac {r}{5}}right)={frac {4pi ^{2}}{sqrt {5}}}$$
Should I use induction for this problem? Any hints on where I should start on?
sequences-and-series gamma-function
edited Jan 9 at 9:19
mrtaurho
4,36421234
asked Dec 20 '18 at 22:08
LarryLarry
2,39131129
$endgroup$
add a comment |
$begingroup$
I saw the following equation on Wikipedia, but I am not sure how to approach it.
$$ prod_{r=1}^{n}Gamma{ left({frac {r}{n+1}}right)}={sqrt {frac {(2pi )^{n}}{n+1}}}$$
Here are some other values listed on Wikipedia
$$ prod _{r=1}^{2}Gamma left({tfrac {r}{3}}right)={frac {2pi }{sqrt {3}}}$$
$$ prod _{r=1}^{3}Gamma left({tfrac {r}{4}}right)={sqrt {2pi ^{3}}}$$
$$prod _{r=1}^{4}Gamma left({tfrac {r}{5}}right)={frac {4pi ^{2}}{sqrt {5}}}$$
Should I use induction for this problem? Any hints on where I should start on?
sequences-and-series gamma-function
edited Jan 9 at 9:19
mrtaurho
4,36421234
asked Dec 20 '18 at 22:08
LarryLarry
2,39131129
$endgroup$
2
$begingroup$
It might be of help to take a look at Gauss Multiplication Formula for which a proof can be found within the given link.
$endgroup$
– mrtaurho
Dec 20 '18 at 22:12
2
$begingroup$
The formula $Gamma(1-z)Gamma(z) = frac{pi}{sin(picdot z)}$ should be useful. Just consider when $n$ is odd and when it's even, and use the formula.
$endgroup$
– Jakobian
Dec 20 '18 at 22:13
3
$begingroup$
If you square both sides you see that $$prod_{r=1}^{n}Gamma^2bigg(frac{r}{1+n}bigg)=frac{(2pi)^n}{1+n}$$ Each $Gamma^2$ term can be put into the form $$Gammabigg(frac{2r}{n+1}bigg)Bbigg(frac{r}{1+n},frac{r}{1+n}bigg)$$ After that I have no idea
$endgroup$
– clathratus
Dec 20 '18 at 22:15
add a comment |
$begingroup$
I saw the following equation on Wikipedia, but I am not sure how to approach it.
$$ prod_{r=1}^{n}Gamma{ left({frac {r}{n+1}}right)}={sqrt {frac {(2pi )^{n}}{n+1}}}$$
Here are some other values listed on Wikipedia
$$ prod _{r=1}^{2}Gamma left({tfrac {r}{3}}right)={frac {2pi }{sqrt {3}}}$$
$$ prod _{r=1}^{3}Gamma left({tfrac {r}{4}}right)={sqrt {2pi ^{3}}}$$
$$prod _{r=1}^{4}Gamma left({tfrac {r}{5}}right)={frac {4pi ^{2}}{sqrt {5}}}$$
Should I use induction for this problem? Any hints on where I should start on?
sequences-and-series gamma-function
edited Jan 9 at 9:19
mrtaurho
4,36421234
asked Dec 20 '18 at 22:08
LarryLarry
2,39131129
$endgroup$
I saw the following equation on Wikipedia, but I am not sure how to approach it.
$$ prod_{r=1}^{n}Gamma{ left({frac {r}{n+1}}right)}={sqrt {frac {(2pi )^{n}}{n+1}}}$$
Here are some other values listed on Wikipedia
$$ prod _{r=1}^{2}Gamma left({tfrac {r}{3}}right)={frac {2pi }{sqrt {3}}}$$
$$ prod _{r=1}^{3}Gamma left({tfrac {r}{4}}right)={sqrt {2pi ^{3}}}$$
$$prod _{r=1}^{4}Gamma left({tfrac {r}{5}}right)={frac {4pi ^{2}}{sqrt {5}}}$$
Should I use induction for this problem? Any hints on where I should start on?
sequences-and-series gamma-function
sequences-and-series gamma-function
edited Jan 9 at 9:19
mrtaurho
4,36421234
asked Dec 20 '18 at 22:08
LarryLarry
2,39131129
edited Jan 9 at 9:19
mrtaurho
4,36421234
asked Dec 20 '18 at 22:08
LarryLarry
2,39131129
edited Jan 9 at 9:19
mrtaurho
4,36421234
edited Jan 9 at 9:19
mrtaurho
4,36421234
edited Jan 9 at 9:19
mrtaurho
4,36421234
4,36421234
asked Dec 20 '18 at 22:08
LarryLarry
2,39131129
asked Dec 20 '18 at 22:08
LarryLarry
2,39131129
asked Dec 20 '18 at 22:08
LarryLarry
2,39131129
2,39131129
2
$begingroup$
It might be of help to take a look at Gauss Multiplication Formula for which a proof can be found within the given link.
$endgroup$
– mrtaurho
Dec 20 '18 at 22:12
2
$begingroup$
The formula $Gamma(1-z)Gamma(z) = frac{pi}{sin(picdot z)}$ should be useful. Just consider when $n$ is odd and when it's even, and use the formula.
$endgroup$
– Jakobian
Dec 20 '18 at 22:13
3
$begingroup$
If you square both sides you see that $$prod_{r=1}^{n}Gamma^2bigg(frac{r}{1+n}bigg)=frac{(2pi)^n}{1+n}$$ Each $Gamma^2$ term can be put into the form $$Gammabigg(frac{2r}{n+1}bigg)Bbigg(frac{r}{1+n},frac{r}{1+n}bigg)$$ After that I have no idea
$endgroup$
– clathratus
Dec 20 '18 at 22:15
add a comment |
2
$begingroup$
It might be of help to take a look at Gauss Multiplication Formula for which a proof can be found within the given link.
$endgroup$
– mrtaurho
Dec 20 '18 at 22:12
2
$begingroup$
The formula $Gamma(1-z)Gamma(z) = frac{pi}{sin(picdot z)}$ should be useful. Just consider when $n$ is odd and when it's even, and use the formula.
$endgroup$
– Jakobian
Dec 20 '18 at 22:13
3
$begingroup$
If you square both sides you see that $$prod_{r=1}^{n}Gamma^2bigg(frac{r}{1+n}bigg)=frac{(2pi)^n}{1+n}$$ Each $Gamma^2$ term can be put into the form $$Gammabigg(frac{2r}{n+1}bigg)Bbigg(frac{r}{1+n},frac{r}{1+n}bigg)$$ After that I have no idea
$endgroup$
– clathratus
Dec 20 '18 at 22:15
2
2
$begingroup$
It might be of help to take a look at Gauss Multiplication Formula for which a proof can be found within the given link.
$endgroup$
– mrtaurho
Dec 20 '18 at 22:12
$begingroup$
It might be of help to take a look at Gauss Multiplication Formula for which a proof can be found within the given link.
$endgroup$
– mrtaurho
Dec 20 '18 at 22:12
2
2
$begingroup$
The formula $Gamma(1-z)Gamma(z) = frac{pi}{sin(picdot z)}$ should be useful. Just consider when $n$ is odd and when it's even, and use the formula.
$endgroup$
– Jakobian
Dec 20 '18 at 22:13
$begingroup$
The formula $Gamma(1-z)Gamma(z) = frac{pi}{sin(picdot z)}$ should be useful. Just consider when $n$ is odd and when it's even, and use the formula.
$endgroup$
– Jakobian
Dec 20 '18 at 22:13
3
3
$begingroup$
If you square both sides you see that $$prod_{r=1}^{n}Gamma^2bigg(frac{r}{1+n}bigg)=frac{(2pi)^n}{1+n}$$ Each $Gamma^2$ term can be put into the form $$Gammabigg(frac{2r}{n+1}bigg)Bbigg(frac{r}{1+n},frac{r}{1+n}bigg)$$ After that I have no idea
$endgroup$
– clathratus
Dec 20 '18 at 22:15
$begingroup$
If you square both sides you see that $$prod_{r=1}^{n}Gamma^2bigg(frac{r}{1+n}bigg)=frac{(2pi)^n}{1+n}$$ Each $Gamma^2$ term can be put into the form $$Gammabigg(frac{2r}{n+1}bigg)Bbigg(frac{r}{1+n},frac{r}{1+n}bigg)$$ After that I have no idea
$endgroup$
– clathratus
Dec 20 '18 at 22:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider Gauss Multiplication Formula in its general form
$$forall z notin left{{-frac m n: m in mathbb N}right}: prod_{k = 0}^{n - 1} Gamma left(z + frac k nright) = (2 pi)^{(n - 1) / 2 }n^{1/2 - n z} Gamma(n z)tag1$$
Now by plugging in $z=1$ and $n=n+1$ we further get
$$begin{align}
prod_{k = 0}^{(n+1)-1} Gamma left(1 + frac k{(n+1)}right) &= (2 pi)^{((n+1) - 1) / 2} (n+1)^{1/2 - (n+1)cdot 1} Gamma((n+1)cdot1)\
prod_{k = 1}^{n} frac k{n+1}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
prod_{k = 1}^{n} frac k{n+1}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
frac{n!}{(n+1)^n}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}
end{align}$$
$$therefore~prod_{k = 1}^{n}Gamma left(frac k{n+1}right) = sqrt{frac{(2 pi)^{n}}{n+1}}$$
We started by utilizing the fundamental functional property of the Gamma Function $($note that the index $k$ has been uplifted to starting by $k=1$ due the fact that within the original product the first term $Gamma(1)$ equals $1$ afterall and is so of minor matter$)$, followed by splitting up the product, rewriting the first subproduct in finite terms, and finally making use of the fact that $n!=Gamma(n+1)$ for all $ninmathbb N$ .
answered Dec 20 '18 at 22:39
mrtaurhomrtaurho
4,36421234
$endgroup$
1
$begingroup$
+1. It was really fine.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:16
$begingroup$
@FelixMarin Thank you kindly.
$endgroup$
– mrtaurho
Dec 22 '18 at 18:02
add a comment |
$begingroup$
Given the reflection formula for the $Gamma$ function, $Gamma(s)Gamma(1-s)=frac{pi}{sin(pi s)}$, the problem immediately boils down to showing that
$$ prod_{k=1}^{n}sinleft(frac{pi k}{n+1}right) = frac{2(n+1)}{2^{n+1}}, $$
which is well-known. One may use roots of unity or Chebyshev polynomials of the second kind, together with Vieta's formulas.
answered Dec 21 '18 at 16:26
Jack D'AurizioJack D'Aurizio
1
$endgroup$
$begingroup$
I guess the RHS is inversed since a product of sines is bounded by $1$.
$endgroup$
– yultan
Jan 9 at 9:30
$begingroup$
@yultan: of course, now fixed.
$endgroup$
– Jack D'Aurizio
Jan 9 at 22:32
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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votes
active
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$begingroup$
Consider Gauss Multiplication Formula in its general form
$$forall z notin left{{-frac m n: m in mathbb N}right}: prod_{k = 0}^{n - 1} Gamma left(z + frac k nright) = (2 pi)^{(n - 1) / 2 }n^{1/2 - n z} Gamma(n z)tag1$$
Now by plugging in $z=1$ and $n=n+1$ we further get
$$begin{align}
prod_{k = 0}^{(n+1)-1} Gamma left(1 + frac k{(n+1)}right) &= (2 pi)^{((n+1) - 1) / 2} (n+1)^{1/2 - (n+1)cdot 1} Gamma((n+1)cdot1)\
prod_{k = 1}^{n} frac k{n+1}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
prod_{k = 1}^{n} frac k{n+1}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
frac{n!}{(n+1)^n}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}
end{align}$$
$$therefore~prod_{k = 1}^{n}Gamma left(frac k{n+1}right) = sqrt{frac{(2 pi)^{n}}{n+1}}$$
We started by utilizing the fundamental functional property of the Gamma Function $($note that the index $k$ has been uplifted to starting by $k=1$ due the fact that within the original product the first term $Gamma(1)$ equals $1$ afterall and is so of minor matter$)$, followed by splitting up the product, rewriting the first subproduct in finite terms, and finally making use of the fact that $n!=Gamma(n+1)$ for all $ninmathbb N$ .
answered Dec 20 '18 at 22:39
mrtaurhomrtaurho
4,36421234
$endgroup$
1
$begingroup$
+1. It was really fine.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:16
$begingroup$
@FelixMarin Thank you kindly.
$endgroup$
– mrtaurho
Dec 22 '18 at 18:02
add a comment |
$begingroup$
Consider Gauss Multiplication Formula in its general form
$$forall z notin left{{-frac m n: m in mathbb N}right}: prod_{k = 0}^{n - 1} Gamma left(z + frac k nright) = (2 pi)^{(n - 1) / 2 }n^{1/2 - n z} Gamma(n z)tag1$$
Now by plugging in $z=1$ and $n=n+1$ we further get
$$begin{align}
prod_{k = 0}^{(n+1)-1} Gamma left(1 + frac k{(n+1)}right) &= (2 pi)^{((n+1) - 1) / 2} (n+1)^{1/2 - (n+1)cdot 1} Gamma((n+1)cdot1)\
prod_{k = 1}^{n} frac k{n+1}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
prod_{k = 1}^{n} frac k{n+1}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
frac{n!}{(n+1)^n}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}
end{align}$$
$$therefore~prod_{k = 1}^{n}Gamma left(frac k{n+1}right) = sqrt{frac{(2 pi)^{n}}{n+1}}$$
We started by utilizing the fundamental functional property of the Gamma Function $($note that the index $k$ has been uplifted to starting by $k=1$ due the fact that within the original product the first term $Gamma(1)$ equals $1$ afterall and is so of minor matter$)$, followed by splitting up the product, rewriting the first subproduct in finite terms, and finally making use of the fact that $n!=Gamma(n+1)$ for all $ninmathbb N$ .
answered Dec 20 '18 at 22:39
mrtaurhomrtaurho
4,36421234
$endgroup$
1
$begingroup$
+1. It was really fine.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:16
$begingroup$
@FelixMarin Thank you kindly.
$endgroup$
– mrtaurho
Dec 22 '18 at 18:02
add a comment |
$begingroup$
Consider Gauss Multiplication Formula in its general form
$$forall z notin left{{-frac m n: m in mathbb N}right}: prod_{k = 0}^{n - 1} Gamma left(z + frac k nright) = (2 pi)^{(n - 1) / 2 }n^{1/2 - n z} Gamma(n z)tag1$$
Now by plugging in $z=1$ and $n=n+1$ we further get
$$begin{align}
prod_{k = 0}^{(n+1)-1} Gamma left(1 + frac k{(n+1)}right) &= (2 pi)^{((n+1) - 1) / 2} (n+1)^{1/2 - (n+1)cdot 1} Gamma((n+1)cdot1)\
prod_{k = 1}^{n} frac k{n+1}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
prod_{k = 1}^{n} frac k{n+1}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
frac{n!}{(n+1)^n}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}
end{align}$$
$$therefore~prod_{k = 1}^{n}Gamma left(frac k{n+1}right) = sqrt{frac{(2 pi)^{n}}{n+1}}$$
We started by utilizing the fundamental functional property of the Gamma Function $($note that the index $k$ has been uplifted to starting by $k=1$ due the fact that within the original product the first term $Gamma(1)$ equals $1$ afterall and is so of minor matter$)$, followed by splitting up the product, rewriting the first subproduct in finite terms, and finally making use of the fact that $n!=Gamma(n+1)$ for all $ninmathbb N$ .
answered Dec 20 '18 at 22:39
mrtaurhomrtaurho
4,36421234
$endgroup$
Consider Gauss Multiplication Formula in its general form
$$forall z notin left{{-frac m n: m in mathbb N}right}: prod_{k = 0}^{n - 1} Gamma left(z + frac k nright) = (2 pi)^{(n - 1) / 2 }n^{1/2 - n z} Gamma(n z)tag1$$
Now by plugging in $z=1$ and $n=n+1$ we further get
$$begin{align}
prod_{k = 0}^{(n+1)-1} Gamma left(1 + frac k{(n+1)}right) &= (2 pi)^{((n+1) - 1) / 2} (n+1)^{1/2 - (n+1)cdot 1} Gamma((n+1)cdot1)\
prod_{k = 1}^{n} frac k{n+1}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
prod_{k = 1}^{n} frac k{n+1}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}\
frac{n!}{(n+1)^n}prod_{k = 1}^{n}Gamma left(frac k{n+1}right) &= sqrt{frac{(2 pi)^{n}}{n+1}} frac{Gamma(n+1)}{(n+1)^{n}}
end{align}$$
$$therefore~prod_{k = 1}^{n}Gamma left(frac k{n+1}right) = sqrt{frac{(2 pi)^{n}}{n+1}}$$
We started by utilizing the fundamental functional property of the Gamma Function $($note that the index $k$ has been uplifted to starting by $k=1$ due the fact that within the original product the first term $Gamma(1)$ equals $1$ afterall and is so of minor matter$)$, followed by splitting up the product, rewriting the first subproduct in finite terms, and finally making use of the fact that $n!=Gamma(n+1)$ for all $ninmathbb N$ .
answered Dec 20 '18 at 22:39
mrtaurhomrtaurho
4,36421234
edited Dec 24 '18 at 10:58
answered Dec 20 '18 at 22:39
mrtaurhomrtaurho
4,36421234
answered Dec 20 '18 at 22:39
mrtaurhomrtaurho
4,36421234
answered Dec 20 '18 at 22:39
mrtaurhomrtaurho
4,36421234
4,36421234
1
$begingroup$
+1. It was really fine.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:16
$begingroup$
@FelixMarin Thank you kindly.
$endgroup$
– mrtaurho
Dec 22 '18 at 18:02
add a comment |
1
$begingroup$
+1. It was really fine.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:16
$begingroup$
@FelixMarin Thank you kindly.
$endgroup$
– mrtaurho
Dec 22 '18 at 18:02
1
1
$begingroup$
+1. It was really fine.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:16
$begingroup$
+1. It was really fine.
$endgroup$
– Felix Marin
Dec 22 '18 at 16:16
$begingroup$
@FelixMarin Thank you kindly.
$endgroup$
– mrtaurho
Dec 22 '18 at 18:02
$begingroup$
@FelixMarin Thank you kindly.
$endgroup$
– mrtaurho
Dec 22 '18 at 18:02
add a comment |
$begingroup$
Given the reflection formula for the $Gamma$ function, $Gamma(s)Gamma(1-s)=frac{pi}{sin(pi s)}$, the problem immediately boils down to showing that
$$ prod_{k=1}^{n}sinleft(frac{pi k}{n+1}right) = frac{2(n+1)}{2^{n+1}}, $$
which is well-known. One may use roots of unity or Chebyshev polynomials of the second kind, together with Vieta's formulas.
answered Dec 21 '18 at 16:26
Jack D'AurizioJack D'Aurizio
1
$endgroup$
$begingroup$
I guess the RHS is inversed since a product of sines is bounded by $1$.
$endgroup$
– yultan
Jan 9 at 9:30
$begingroup$
@yultan: of course, now fixed.
$endgroup$
– Jack D'Aurizio
Jan 9 at 22:32
add a comment |
$begingroup$
Given the reflection formula for the $Gamma$ function, $Gamma(s)Gamma(1-s)=frac{pi}{sin(pi s)}$, the problem immediately boils down to showing that
$$ prod_{k=1}^{n}sinleft(frac{pi k}{n+1}right) = frac{2(n+1)}{2^{n+1}}, $$
which is well-known. One may use roots of unity or Chebyshev polynomials of the second kind, together with Vieta's formulas.
answered Dec 21 '18 at 16:26
Jack D'AurizioJack D'Aurizio
1
$endgroup$
$begingroup$
I guess the RHS is inversed since a product of sines is bounded by $1$.
$endgroup$
– yultan
Jan 9 at 9:30
$begingroup$
@yultan: of course, now fixed.
$endgroup$
– Jack D'Aurizio
Jan 9 at 22:32
add a comment |
$begingroup$
Given the reflection formula for the $Gamma$ function, $Gamma(s)Gamma(1-s)=frac{pi}{sin(pi s)}$, the problem immediately boils down to showing that
$$ prod_{k=1}^{n}sinleft(frac{pi k}{n+1}right) = frac{2(n+1)}{2^{n+1}}, $$
which is well-known. One may use roots of unity or Chebyshev polynomials of the second kind, together with Vieta's formulas.
answered Dec 21 '18 at 16:26
Jack D'AurizioJack D'Aurizio
1
$endgroup$
Given the reflection formula for the $Gamma$ function, $Gamma(s)Gamma(1-s)=frac{pi}{sin(pi s)}$, the problem immediately boils down to showing that
$$ prod_{k=1}^{n}sinleft(frac{pi k}{n+1}right) = frac{2(n+1)}{2^{n+1}}, $$
which is well-known. One may use roots of unity or Chebyshev polynomials of the second kind, together with Vieta's formulas.
answered Dec 21 '18 at 16:26
Jack D'AurizioJack D'Aurizio
1
edited Jan 9 at 22:32
answered Dec 21 '18 at 16:26
Jack D'AurizioJack D'Aurizio
1
answered Dec 21 '18 at 16:26
Jack D'AurizioJack D'Aurizio
1
answered Dec 21 '18 at 16:26
Jack D'AurizioJack D'Aurizio
1
1
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I guess the RHS is inversed since a product of sines is bounded by $1$.
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– yultan
Jan 9 at 9:30
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@yultan: of course, now fixed.
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– Jack D'Aurizio
Jan 9 at 22:32
add a comment |
$begingroup$
I guess the RHS is inversed since a product of sines is bounded by $1$.
$endgroup$
– yultan
Jan 9 at 9:30
$begingroup$
@yultan: of course, now fixed.
$endgroup$
– Jack D'Aurizio
Jan 9 at 22:32
$begingroup$
I guess the RHS is inversed since a product of sines is bounded by $1$.
$endgroup$
– yultan
Jan 9 at 9:30
$begingroup$
I guess the RHS is inversed since a product of sines is bounded by $1$.
$endgroup$
– yultan
Jan 9 at 9:30
$begingroup$
@yultan: of course, now fixed.
$endgroup$
– Jack D'Aurizio
Jan 9 at 22:32
$begingroup$
@yultan: of course, now fixed.
$endgroup$
– Jack D'Aurizio
Jan 9 at 22:32
add a comment |
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2
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It might be of help to take a look at Gauss Multiplication Formula for which a proof can be found within the given link.
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– mrtaurho
Dec 20 '18 at 22:12
2
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The formula $Gamma(1-z)Gamma(z) = frac{pi}{sin(picdot z)}$ should be useful. Just consider when $n$ is odd and when it's even, and use the formula.
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– Jakobian
Dec 20 '18 at 22:13
3
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If you square both sides you see that $$prod_{r=1}^{n}Gamma^2bigg(frac{r}{1+n}bigg)=frac{(2pi)^n}{1+n}$$ Each $Gamma^2$ term can be put into the form $$Gammabigg(frac{2r}{n+1}bigg)Bbigg(frac{r}{1+n},frac{r}{1+n}bigg)$$ After that I have no idea
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– clathratus
Dec 20 '18 at 22:15