Formula for the ratio $frac{Gammaleft(n + frac{1}{2}right)}{Gamma(n + 1)}$ of two values of the Gamma...
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Show that $$frac{Gammaleft(n + frac{1}{2}right)}{Gamma(n + 1)} = frac{1 cdot 3 cdot cdots (2 n - 3) (2 n - 1)}{2 cdot 4 cdot cdots (2 n - 2) cdot 2n} .$$
I have proved that $$Gammaleft(n + frac{1}{2}right) = frac{1 cdot 3 cdot cdots (2 n - 3) (2 n - 1) sqrt{pi}}{2^n} qquad textrm{and} qquadGamma(n+1) = n! ,$$
but when I divide, I am not able to proceed. Could you please help me?
gamma-function beta-function
edited Nov 16 at 6:06
Travis
58.7k765142
asked Nov 16 at 5:47
Renuka
73
|
show 1 more comment
up vote
-1
down vote
favorite
Show that $$frac{Gammaleft(n + frac{1}{2}right)}{Gamma(n + 1)} = frac{1 cdot 3 cdot cdots (2 n - 3) (2 n - 1)}{2 cdot 4 cdot cdots (2 n - 2) cdot 2n} .$$
I have proved that $$Gammaleft(n + frac{1}{2}right) = frac{1 cdot 3 cdot cdots (2 n - 3) (2 n - 1) sqrt{pi}}{2^n} qquad textrm{and} qquadGamma(n+1) = n! ,$$
but when I divide, I am not able to proceed. Could you please help me?
gamma-function beta-function
edited Nov 16 at 6:06
Travis
58.7k765142
asked Nov 16 at 5:47
Renuka
73
Please see math.meta.stackexchange.com/questions/5020/…
– Lord Shark the Unknown
Nov 16 at 5:49
$Gamma(n+1/2)/Gamma(n)$ really is a rational multiple of $sqrtpi$ for $ninBbb N$.
– Lord Shark the Unknown
Nov 16 at 5:51
1
I've taken the liberty of formatting the post using MathJax. Please make sure it preserves the intended meaning. In particular, (1) the identity seems to be missing a factor of $sqrtpi$, and (2) the last equation in the original post was nonsensical; you might like to reinstate a corrected version.
– Travis
Nov 16 at 6:08
1
The statement is false, as @Travis points out. Check it for $n=1$: $$frac{Gamma(3/2)}{Gamma(2)} = frac{sqrt{pi}}{2} neq frac{1}{2}$$ Missing a $sqrt{pi}$ somewhere?
– Clement C.
Nov 16 at 6:15
1
(Once you put that $sqrt{pi}$ back, what you have proven gives the result. Simply note that $$2^n n! = 2^n prod_{k=1}^n k = prod_{k=1}^n (2k) = 2cdot 4cdot 6cdots (2n-2)cdot 2n$$ to conclude)
– Clement C.
Nov 16 at 6:17
|
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Show that $$frac{Gammaleft(n + frac{1}{2}right)}{Gamma(n + 1)} = frac{1 cdot 3 cdot cdots (2 n - 3) (2 n - 1)}{2 cdot 4 cdot cdots (2 n - 2) cdot 2n} .$$
I have proved that $$Gammaleft(n + frac{1}{2}right) = frac{1 cdot 3 cdot cdots (2 n - 3) (2 n - 1) sqrt{pi}}{2^n} qquad textrm{and} qquadGamma(n+1) = n! ,$$
but when I divide, I am not able to proceed. Could you please help me?
gamma-function beta-function
edited Nov 16 at 6:06
Travis
58.7k765142
asked Nov 16 at 5:47
Renuka
73
Show that $$frac{Gammaleft(n + frac{1}{2}right)}{Gamma(n + 1)} = frac{1 cdot 3 cdot cdots (2 n - 3) (2 n - 1)}{2 cdot 4 cdot cdots (2 n - 2) cdot 2n} .$$
I have proved that $$Gammaleft(n + frac{1}{2}right) = frac{1 cdot 3 cdot cdots (2 n - 3) (2 n - 1) sqrt{pi}}{2^n} qquad textrm{and} qquadGamma(n+1) = n! ,$$
but when I divide, I am not able to proceed. Could you please help me?
gamma-function beta-function
gamma-function beta-function
edited Nov 16 at 6:06
Travis
58.7k765142
asked Nov 16 at 5:47
Renuka
73
edited Nov 16 at 6:06
Travis
58.7k765142
asked Nov 16 at 5:47
Renuka
73
edited Nov 16 at 6:06
Travis
58.7k765142
edited Nov 16 at 6:06
Travis
58.7k765142
edited Nov 16 at 6:06
Travis
58.7k765142
58.7k765142
asked Nov 16 at 5:47
Renuka
73
asked Nov 16 at 5:47
Renuka
73
asked Nov 16 at 5:47
Renuka
73
73
Please see math.meta.stackexchange.com/questions/5020/…
– Lord Shark the Unknown
Nov 16 at 5:49
$Gamma(n+1/2)/Gamma(n)$ really is a rational multiple of $sqrtpi$ for $ninBbb N$.
– Lord Shark the Unknown
Nov 16 at 5:51
1
I've taken the liberty of formatting the post using MathJax. Please make sure it preserves the intended meaning. In particular, (1) the identity seems to be missing a factor of $sqrtpi$, and (2) the last equation in the original post was nonsensical; you might like to reinstate a corrected version.
– Travis
Nov 16 at 6:08
1
The statement is false, as @Travis points out. Check it for $n=1$: $$frac{Gamma(3/2)}{Gamma(2)} = frac{sqrt{pi}}{2} neq frac{1}{2}$$ Missing a $sqrt{pi}$ somewhere?
– Clement C.
Nov 16 at 6:15
1
(Once you put that $sqrt{pi}$ back, what you have proven gives the result. Simply note that $$2^n n! = 2^n prod_{k=1}^n k = prod_{k=1}^n (2k) = 2cdot 4cdot 6cdots (2n-2)cdot 2n$$ to conclude)
– Clement C.
Nov 16 at 6:17
|
show 1 more comment
Please see math.meta.stackexchange.com/questions/5020/…
– Lord Shark the Unknown
Nov 16 at 5:49
$Gamma(n+1/2)/Gamma(n)$ really is a rational multiple of $sqrtpi$ for $ninBbb N$.
– Lord Shark the Unknown
Nov 16 at 5:51
1
I've taken the liberty of formatting the post using MathJax. Please make sure it preserves the intended meaning. In particular, (1) the identity seems to be missing a factor of $sqrtpi$, and (2) the last equation in the original post was nonsensical; you might like to reinstate a corrected version.
– Travis
Nov 16 at 6:08
1
The statement is false, as @Travis points out. Check it for $n=1$: $$frac{Gamma(3/2)}{Gamma(2)} = frac{sqrt{pi}}{2} neq frac{1}{2}$$ Missing a $sqrt{pi}$ somewhere?
– Clement C.
Nov 16 at 6:15
1
(Once you put that $sqrt{pi}$ back, what you have proven gives the result. Simply note that $$2^n n! = 2^n prod_{k=1}^n k = prod_{k=1}^n (2k) = 2cdot 4cdot 6cdots (2n-2)cdot 2n$$ to conclude)
– Clement C.
Nov 16 at 6:17
Please see math.meta.stackexchange.com/questions/5020/…
– Lord Shark the Unknown
Nov 16 at 5:49
Please see math.meta.stackexchange.com/questions/5020/…
– Lord Shark the Unknown
Nov 16 at 5:49
$Gamma(n+1/2)/Gamma(n)$ really is a rational multiple of $sqrtpi$ for $ninBbb N$.
– Lord Shark the Unknown
Nov 16 at 5:51
$Gamma(n+1/2)/Gamma(n)$ really is a rational multiple of $sqrtpi$ for $ninBbb N$.
– Lord Shark the Unknown
Nov 16 at 5:51
1
1
I've taken the liberty of formatting the post using MathJax. Please make sure it preserves the intended meaning. In particular, (1) the identity seems to be missing a factor of $sqrtpi$, and (2) the last equation in the original post was nonsensical; you might like to reinstate a corrected version.
– Travis
Nov 16 at 6:08
I've taken the liberty of formatting the post using MathJax. Please make sure it preserves the intended meaning. In particular, (1) the identity seems to be missing a factor of $sqrtpi$, and (2) the last equation in the original post was nonsensical; you might like to reinstate a corrected version.
– Travis
Nov 16 at 6:08
1
1
The statement is false, as @Travis points out. Check it for $n=1$: $$frac{Gamma(3/2)}{Gamma(2)} = frac{sqrt{pi}}{2} neq frac{1}{2}$$ Missing a $sqrt{pi}$ somewhere?
– Clement C.
Nov 16 at 6:15
The statement is false, as @Travis points out. Check it for $n=1$: $$frac{Gamma(3/2)}{Gamma(2)} = frac{sqrt{pi}}{2} neq frac{1}{2}$$ Missing a $sqrt{pi}$ somewhere?
– Clement C.
Nov 16 at 6:15
1
1
(Once you put that $sqrt{pi}$ back, what you have proven gives the result. Simply note that $$2^n n! = 2^n prod_{k=1}^n k = prod_{k=1}^n (2k) = 2cdot 4cdot 6cdots (2n-2)cdot 2n$$ to conclude)
– Clement C.
Nov 16 at 6:17
(Once you put that $sqrt{pi}$ back, what you have proven gives the result. Simply note that $$2^n n! = 2^n prod_{k=1}^n k = prod_{k=1}^n (2k) = 2cdot 4cdot 6cdots (2n-2)cdot 2n$$ to conclude)
– Clement C.
Nov 16 at 6:17
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
You may
- Define $f(n)$ as $frac{Gamma(n+1/2)}{Gamma(n+1)}$ and $g(n)$ as $frac{(2n-1)!!}{(2n)!!}$
- Compute $frac{f(n+1)}{f(n)}$ (via $Gamma(z+1)=z,Gamma(z)$) and $frac{g(n+1)}{g(n)}$, then check they are equal
- Compare $f(1)$ with $g(1)$ to derive that $f(n)$ is a constant multiple of $g(n)$ for any $ninmathbb{N}^+$.
answered Nov 16 at 18:59
Jack D'Aurizio
283k33275653
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You may
- Define $f(n)$ as $frac{Gamma(n+1/2)}{Gamma(n+1)}$ and $g(n)$ as $frac{(2n-1)!!}{(2n)!!}$
- Compute $frac{f(n+1)}{f(n)}$ (via $Gamma(z+1)=z,Gamma(z)$) and $frac{g(n+1)}{g(n)}$, then check they are equal
- Compare $f(1)$ with $g(1)$ to derive that $f(n)$ is a constant multiple of $g(n)$ for any $ninmathbb{N}^+$.
answered Nov 16 at 18:59
Jack D'Aurizio
283k33275653
add a comment |
up vote
0
down vote
You may
- Define $f(n)$ as $frac{Gamma(n+1/2)}{Gamma(n+1)}$ and $g(n)$ as $frac{(2n-1)!!}{(2n)!!}$
- Compute $frac{f(n+1)}{f(n)}$ (via $Gamma(z+1)=z,Gamma(z)$) and $frac{g(n+1)}{g(n)}$, then check they are equal
- Compare $f(1)$ with $g(1)$ to derive that $f(n)$ is a constant multiple of $g(n)$ for any $ninmathbb{N}^+$.
answered Nov 16 at 18:59
Jack D'Aurizio
283k33275653
add a comment |
up vote
0
down vote
up vote
0
down vote
You may
- Define $f(n)$ as $frac{Gamma(n+1/2)}{Gamma(n+1)}$ and $g(n)$ as $frac{(2n-1)!!}{(2n)!!}$
- Compute $frac{f(n+1)}{f(n)}$ (via $Gamma(z+1)=z,Gamma(z)$) and $frac{g(n+1)}{g(n)}$, then check they are equal
- Compare $f(1)$ with $g(1)$ to derive that $f(n)$ is a constant multiple of $g(n)$ for any $ninmathbb{N}^+$.
answered Nov 16 at 18:59
Jack D'Aurizio
283k33275653
You may
- Define $f(n)$ as $frac{Gamma(n+1/2)}{Gamma(n+1)}$ and $g(n)$ as $frac{(2n-1)!!}{(2n)!!}$
- Compute $frac{f(n+1)}{f(n)}$ (via $Gamma(z+1)=z,Gamma(z)$) and $frac{g(n+1)}{g(n)}$, then check they are equal
- Compare $f(1)$ with $g(1)$ to derive that $f(n)$ is a constant multiple of $g(n)$ for any $ninmathbb{N}^+$.
answered Nov 16 at 18:59
Jack D'Aurizio
283k33275653
answered Nov 16 at 18:59
Jack D'Aurizio
283k33275653
answered Nov 16 at 18:59
Jack D'Aurizio
283k33275653
answered Nov 16 at 18:59
Jack D'Aurizio
283k33275653
283k33275653
add a comment |
add a comment |
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Please see math.meta.stackexchange.com/questions/5020/…
– Lord Shark the Unknown
Nov 16 at 5:49
$Gamma(n+1/2)/Gamma(n)$ really is a rational multiple of $sqrtpi$ for $ninBbb N$.
– Lord Shark the Unknown
Nov 16 at 5:51
1
I've taken the liberty of formatting the post using MathJax. Please make sure it preserves the intended meaning. In particular, (1) the identity seems to be missing a factor of $sqrtpi$, and (2) the last equation in the original post was nonsensical; you might like to reinstate a corrected version.
– Travis
Nov 16 at 6:08
1
The statement is false, as @Travis points out. Check it for $n=1$: $$frac{Gamma(3/2)}{Gamma(2)} = frac{sqrt{pi}}{2} neq frac{1}{2}$$ Missing a $sqrt{pi}$ somewhere?
– Clement C.
Nov 16 at 6:15
1
(Once you put that $sqrt{pi}$ back, what you have proven gives the result. Simply note that $$2^n n! = 2^n prod_{k=1}^n k = prod_{k=1}^n (2k) = 2cdot 4cdot 6cdots (2n-2)cdot 2n$$ to conclude)
– Clement C.
Nov 16 at 6:17