Formula for the ratio $frac{Gammaleft(n + frac{1}{2}right)}{Gamma(n + 1)}$ of two values of the Gamma...
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
|
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
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
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
58.9k765143
58.9k765143
asked Nov 16 at 5:47
Renuka
74
74
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 $ninmathbb{N}^+$.
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 $ninmathbb{N}^+$.
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 $ninmathbb{N}^+$.
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 $ninmathbb{N}^+$.
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
283k33275653
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000769%2fformula-for-the-ratio-frac-gamma-leftn-frac12-right-gamman-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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