Mixing
Where did I go wrong in proving $mathbb E[X^{2n}] = prod_{1 leq k leq 2n, k operatorname{odd}}k$
$begingroup$
Let $X$ ~ $mathcal{N}(0,1)$
Show that:
$mathbb E[X^{2n}] = prod_{1 leq k leq 2n, k operatorname{odd}}k$
Idea:
$mathbb E[X^{2n}]=frac{1}{sqrt{2pi}}int_{mathbb R}x^{2n}e^{-frac{x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx$, then set $t=frac{x^2}{2}Rightarrow dt=xdx$ and
$frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx=frac{2^{n+1}}{sqrt{pi}}int_{0}^{infty}t^{n-frac{1}{2}}e^{-t}dt=frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
I know that in order to get the desired result, I need to show that:
$frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx=frac{2}{sqrt{pi}}Gamma(n+frac{1}{2})$
But I am far off it, where did I go wrong?
probability probability-theory probability-distributions expected-value
asked Jan 24 at 0:10
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
Let $X$ ~ $mathcal{N}(0,1)$
Show that:
$mathbb E[X^{2n}] = prod_{1 leq k leq 2n, k operatorname{odd}}k$
Idea:
$mathbb E[X^{2n}]=frac{1}{sqrt{2pi}}int_{mathbb R}x^{2n}e^{-frac{x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx$, then set $t=frac{x^2}{2}Rightarrow dt=xdx$ and
$frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx=frac{2^{n+1}}{sqrt{pi}}int_{0}^{infty}t^{n-frac{1}{2}}e^{-t}dt=frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
I know that in order to get the desired result, I need to show that:
$frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx=frac{2}{sqrt{pi}}Gamma(n+frac{1}{2})$
But I am far off it, where did I go wrong?
probability probability-theory probability-distributions expected-value
asked Jan 24 at 0:10
SABOYSABOY
656311
$endgroup$
$begingroup$
I didn't check constants for certain, but you definitely should get something of the sort of $frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
$endgroup$
– Jakobian
Jan 24 at 0:17
1
$begingroup$
Apart from apparent typos in your final formula, I can't see anything wrong with what you've done so far. Is the problem you're having difficulty with that of showing $$frac{2}{sqrt{pi}}Gamma(n+frac{1}{2}) =prod_{1 leq k leq 2n, k operatorname{odd}}k ?$$
$endgroup$
– lonza leggiera
Jan 24 at 0:59
add a comment |
$begingroup$
Let $X$ ~ $mathcal{N}(0,1)$
Show that:
$mathbb E[X^{2n}] = prod_{1 leq k leq 2n, k operatorname{odd}}k$
Idea:
$mathbb E[X^{2n}]=frac{1}{sqrt{2pi}}int_{mathbb R}x^{2n}e^{-frac{x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx$, then set $t=frac{x^2}{2}Rightarrow dt=xdx$ and
$frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx=frac{2^{n+1}}{sqrt{pi}}int_{0}^{infty}t^{n-frac{1}{2}}e^{-t}dt=frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
I know that in order to get the desired result, I need to show that:
$frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx=frac{2}{sqrt{pi}}Gamma(n+frac{1}{2})$
But I am far off it, where did I go wrong?
probability probability-theory probability-distributions expected-value
asked Jan 24 at 0:10
SABOYSABOY
656311
$endgroup$
Let $X$ ~ $mathcal{N}(0,1)$
Show that:
$mathbb E[X^{2n}] = prod_{1 leq k leq 2n, k operatorname{odd}}k$
Idea:
$mathbb E[X^{2n}]=frac{1}{sqrt{2pi}}int_{mathbb R}x^{2n}e^{-frac{x^2}{2}}dx=frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx$, then set $t=frac{x^2}{2}Rightarrow dt=xdx$ and
$frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx=frac{2^{n+1}}{sqrt{pi}}int_{0}^{infty}t^{n-frac{1}{2}}e^{-t}dt=frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
I know that in order to get the desired result, I need to show that:
$frac{2}{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}{2}}dx=frac{2}{sqrt{pi}}Gamma(n+frac{1}{2})$
But I am far off it, where did I go wrong?
probability probability-theory probability-distributions expected-value
probability probability-theory probability-distributions expected-value
asked Jan 24 at 0:10
SABOYSABOY
656311
asked Jan 24 at 0:10
SABOYSABOY
656311
edited Jan 24 at 0:16
SABOY
asked Jan 24 at 0:10
SABOYSABOY
656311
asked Jan 24 at 0:10
SABOYSABOY
656311
asked Jan 24 at 0:10
SABOYSABOY
656311
656311
$begingroup$
I didn't check constants for certain, but you definitely should get something of the sort of $frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
$endgroup$
– Jakobian
Jan 24 at 0:17
1
$begingroup$
Apart from apparent typos in your final formula, I can't see anything wrong with what you've done so far. Is the problem you're having difficulty with that of showing $$frac{2}{sqrt{pi}}Gamma(n+frac{1}{2}) =prod_{1 leq k leq 2n, k operatorname{odd}}k ?$$
$endgroup$
– lonza leggiera
Jan 24 at 0:59
add a comment |
$begingroup$
I didn't check constants for certain, but you definitely should get something of the sort of $frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
$endgroup$
– Jakobian
Jan 24 at 0:17
1
$begingroup$
Apart from apparent typos in your final formula, I can't see anything wrong with what you've done so far. Is the problem you're having difficulty with that of showing $$frac{2}{sqrt{pi}}Gamma(n+frac{1}{2}) =prod_{1 leq k leq 2n, k operatorname{odd}}k ?$$
$endgroup$
– lonza leggiera
Jan 24 at 0:59
$begingroup$
I didn't check constants for certain, but you definitely should get something of the sort of $frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
$endgroup$
– Jakobian
Jan 24 at 0:17
$begingroup$
I didn't check constants for certain, but you definitely should get something of the sort of $frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
$endgroup$
– Jakobian
Jan 24 at 0:17
1
1
$begingroup$
Apart from apparent typos in your final formula, I can't see anything wrong with what you've done so far. Is the problem you're having difficulty with that of showing $$frac{2}{sqrt{pi}}Gamma(n+frac{1}{2}) =prod_{1 leq k leq 2n, k operatorname{odd}}k ?$$
$endgroup$
– lonza leggiera
Jan 24 at 0:59
$begingroup$
Apart from apparent typos in your final formula, I can't see anything wrong with what you've done so far. Is the problem you're having difficulty with that of showing $$frac{2}{sqrt{pi}}Gamma(n+frac{1}{2}) =prod_{1 leq k leq 2n, k operatorname{odd}}k ?$$
$endgroup$
– lonza leggiera
Jan 24 at 0:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You were actually off only by a factor of 2. With your choice of change of variable,
begin{align}
frac2{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}2 }dx &= frac2{sqrt{2pi}}int_{0}^{infty}x^{2(n - frac12)} e^{-frac{x^2}2 } cdot xdx \
&=frac{ 2^{1/2} }{sqrt{pi}}int_{0}^{infty}(2t)^{n - frac12} e^{-t^2 } cdot dt \
&= frac{2^n}{sqrt{pi}}int_{0}^{infty}t^{n-frac12}e^{-t}dt=frac{2^n}{sqrt{pi}}Gamma(n+frac12)
end{align}
Invoking the definition of Gamma function that $Gamma(x) = (x-1)Gamma(x-1)$, for $n in mathbb{N}$ we have
begin{align}
&phantom{ {}={} }frac{2^n}{sqrt{pi}}Gammabigl( n + frac12 bigr) \
&= frac{ 2^{n-1} }{sqrt{pi}} cdot 2cdot bigl( n - frac12 bigr)cdot Gammabigl( n - frac12 bigr) \
&= frac{ 2^{n-2} }{sqrt{pi}} cdot (2n - 1) cdot 2bigl( n - frac32 bigr) cdot Gammabigl( n - frac32 bigr) \
& hspace{36pt}vdots \
&=frac1{sqrt{pi}} cdot (2n-1)underbrace{(2n-3)}_{text{from}~n-frac12-(1)~text{so 2nd}} overbrace{(2n-5)}^{text{from}~n-frac12-(2)~text{so 3rd}} ldots 5cdot 3cdot underbrace{bigl(2 cdot frac12 bigr)}_{text{from}~n-frac12-(n-1)~text{so $n$ th}} cdotGammabigl( frac12 bigr) \
&= (2n-1)(2n-3)(2n-5)cdots 5cdot 3 cdot 1
end{align}
as desired, where it is well-known that $Gammabigl( frac12 bigr) = sqrt{pi}$.
answered Jan 24 at 1:20
Lee David Chung LinLee David Chung Lin
4,39541242
$endgroup$
add a comment |
$begingroup$
I think your computation is actually correct. Note that
$$Gamma(n+frac{1}{2})=frac{1cdot 3cdot 5cdotcdotscdot (2n-1)}{2^n}sqrt{pi}$$
see here, for instance. Thus, the expression you derived is actually exactly what you want.
answered Jan 24 at 0:56
user293794user293794
1,748613
$endgroup$
add a comment |
$begingroup$
You can also prove this by induction using integration by parts:
begin{align}
int x^{2n}e^{-x^2/2},dx
&=int x^{2n-1}cdot xe^{-x^2/2},dx
\&=-int x^{2n-1}dbig(e^{-x^2/2}big)
\&=-[x^{2n-1}e^{-x^2/2}]Big|^{infty}_{-infty}+int e^{-x^2/2}dbig(x^{2n-1}big)
\&=0+(2n-1)int x^{2n-2}e^{-x^2/2},dx
end{align}
Now apply the induction hypothesis.
answered Jan 24 at 2:49
Mike EarnestMike Earnest
24.7k22151
$endgroup$
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
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active
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$begingroup$
You were actually off only by a factor of 2. With your choice of change of variable,
begin{align}
frac2{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}2 }dx &= frac2{sqrt{2pi}}int_{0}^{infty}x^{2(n - frac12)} e^{-frac{x^2}2 } cdot xdx \
&=frac{ 2^{1/2} }{sqrt{pi}}int_{0}^{infty}(2t)^{n - frac12} e^{-t^2 } cdot dt \
&= frac{2^n}{sqrt{pi}}int_{0}^{infty}t^{n-frac12}e^{-t}dt=frac{2^n}{sqrt{pi}}Gamma(n+frac12)
end{align}
Invoking the definition of Gamma function that $Gamma(x) = (x-1)Gamma(x-1)$, for $n in mathbb{N}$ we have
begin{align}
&phantom{ {}={} }frac{2^n}{sqrt{pi}}Gammabigl( n + frac12 bigr) \
&= frac{ 2^{n-1} }{sqrt{pi}} cdot 2cdot bigl( n - frac12 bigr)cdot Gammabigl( n - frac12 bigr) \
&= frac{ 2^{n-2} }{sqrt{pi}} cdot (2n - 1) cdot 2bigl( n - frac32 bigr) cdot Gammabigl( n - frac32 bigr) \
& hspace{36pt}vdots \
&=frac1{sqrt{pi}} cdot (2n-1)underbrace{(2n-3)}_{text{from}~n-frac12-(1)~text{so 2nd}} overbrace{(2n-5)}^{text{from}~n-frac12-(2)~text{so 3rd}} ldots 5cdot 3cdot underbrace{bigl(2 cdot frac12 bigr)}_{text{from}~n-frac12-(n-1)~text{so $n$ th}} cdotGammabigl( frac12 bigr) \
&= (2n-1)(2n-3)(2n-5)cdots 5cdot 3 cdot 1
end{align}
as desired, where it is well-known that $Gammabigl( frac12 bigr) = sqrt{pi}$.
answered Jan 24 at 1:20
Lee David Chung LinLee David Chung Lin
4,39541242
$endgroup$
add a comment |
$begingroup$
You were actually off only by a factor of 2. With your choice of change of variable,
begin{align}
frac2{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}2 }dx &= frac2{sqrt{2pi}}int_{0}^{infty}x^{2(n - frac12)} e^{-frac{x^2}2 } cdot xdx \
&=frac{ 2^{1/2} }{sqrt{pi}}int_{0}^{infty}(2t)^{n - frac12} e^{-t^2 } cdot dt \
&= frac{2^n}{sqrt{pi}}int_{0}^{infty}t^{n-frac12}e^{-t}dt=frac{2^n}{sqrt{pi}}Gamma(n+frac12)
end{align}
Invoking the definition of Gamma function that $Gamma(x) = (x-1)Gamma(x-1)$, for $n in mathbb{N}$ we have
begin{align}
&phantom{ {}={} }frac{2^n}{sqrt{pi}}Gammabigl( n + frac12 bigr) \
&= frac{ 2^{n-1} }{sqrt{pi}} cdot 2cdot bigl( n - frac12 bigr)cdot Gammabigl( n - frac12 bigr) \
&= frac{ 2^{n-2} }{sqrt{pi}} cdot (2n - 1) cdot 2bigl( n - frac32 bigr) cdot Gammabigl( n - frac32 bigr) \
& hspace{36pt}vdots \
&=frac1{sqrt{pi}} cdot (2n-1)underbrace{(2n-3)}_{text{from}~n-frac12-(1)~text{so 2nd}} overbrace{(2n-5)}^{text{from}~n-frac12-(2)~text{so 3rd}} ldots 5cdot 3cdot underbrace{bigl(2 cdot frac12 bigr)}_{text{from}~n-frac12-(n-1)~text{so $n$ th}} cdotGammabigl( frac12 bigr) \
&= (2n-1)(2n-3)(2n-5)cdots 5cdot 3 cdot 1
end{align}
as desired, where it is well-known that $Gammabigl( frac12 bigr) = sqrt{pi}$.
answered Jan 24 at 1:20
Lee David Chung LinLee David Chung Lin
4,39541242
$endgroup$
add a comment |
$begingroup$
You were actually off only by a factor of 2. With your choice of change of variable,
begin{align}
frac2{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}2 }dx &= frac2{sqrt{2pi}}int_{0}^{infty}x^{2(n - frac12)} e^{-frac{x^2}2 } cdot xdx \
&=frac{ 2^{1/2} }{sqrt{pi}}int_{0}^{infty}(2t)^{n - frac12} e^{-t^2 } cdot dt \
&= frac{2^n}{sqrt{pi}}int_{0}^{infty}t^{n-frac12}e^{-t}dt=frac{2^n}{sqrt{pi}}Gamma(n+frac12)
end{align}
Invoking the definition of Gamma function that $Gamma(x) = (x-1)Gamma(x-1)$, for $n in mathbb{N}$ we have
begin{align}
&phantom{ {}={} }frac{2^n}{sqrt{pi}}Gammabigl( n + frac12 bigr) \
&= frac{ 2^{n-1} }{sqrt{pi}} cdot 2cdot bigl( n - frac12 bigr)cdot Gammabigl( n - frac12 bigr) \
&= frac{ 2^{n-2} }{sqrt{pi}} cdot (2n - 1) cdot 2bigl( n - frac32 bigr) cdot Gammabigl( n - frac32 bigr) \
& hspace{36pt}vdots \
&=frac1{sqrt{pi}} cdot (2n-1)underbrace{(2n-3)}_{text{from}~n-frac12-(1)~text{so 2nd}} overbrace{(2n-5)}^{text{from}~n-frac12-(2)~text{so 3rd}} ldots 5cdot 3cdot underbrace{bigl(2 cdot frac12 bigr)}_{text{from}~n-frac12-(n-1)~text{so $n$ th}} cdotGammabigl( frac12 bigr) \
&= (2n-1)(2n-3)(2n-5)cdots 5cdot 3 cdot 1
end{align}
as desired, where it is well-known that $Gammabigl( frac12 bigr) = sqrt{pi}$.
answered Jan 24 at 1:20
Lee David Chung LinLee David Chung Lin
4,39541242
$endgroup$
You were actually off only by a factor of 2. With your choice of change of variable,
begin{align}
frac2{sqrt{2pi}}int_{0}^{infty}x^{2n}e^{-frac{x^2}2 }dx &= frac2{sqrt{2pi}}int_{0}^{infty}x^{2(n - frac12)} e^{-frac{x^2}2 } cdot xdx \
&=frac{ 2^{1/2} }{sqrt{pi}}int_{0}^{infty}(2t)^{n - frac12} e^{-t^2 } cdot dt \
&= frac{2^n}{sqrt{pi}}int_{0}^{infty}t^{n-frac12}e^{-t}dt=frac{2^n}{sqrt{pi}}Gamma(n+frac12)
end{align}
Invoking the definition of Gamma function that $Gamma(x) = (x-1)Gamma(x-1)$, for $n in mathbb{N}$ we have
begin{align}
&phantom{ {}={} }frac{2^n}{sqrt{pi}}Gammabigl( n + frac12 bigr) \
&= frac{ 2^{n-1} }{sqrt{pi}} cdot 2cdot bigl( n - frac12 bigr)cdot Gammabigl( n - frac12 bigr) \
&= frac{ 2^{n-2} }{sqrt{pi}} cdot (2n - 1) cdot 2bigl( n - frac32 bigr) cdot Gammabigl( n - frac32 bigr) \
& hspace{36pt}vdots \
&=frac1{sqrt{pi}} cdot (2n-1)underbrace{(2n-3)}_{text{from}~n-frac12-(1)~text{so 2nd}} overbrace{(2n-5)}^{text{from}~n-frac12-(2)~text{so 3rd}} ldots 5cdot 3cdot underbrace{bigl(2 cdot frac12 bigr)}_{text{from}~n-frac12-(n-1)~text{so $n$ th}} cdotGammabigl( frac12 bigr) \
&= (2n-1)(2n-3)(2n-5)cdots 5cdot 3 cdot 1
end{align}
as desired, where it is well-known that $Gammabigl( frac12 bigr) = sqrt{pi}$.
answered Jan 24 at 1:20
Lee David Chung LinLee David Chung Lin
4,39541242
edited Jan 24 at 1:25
answered Jan 24 at 1:20
Lee David Chung LinLee David Chung Lin
4,39541242
answered Jan 24 at 1:20
Lee David Chung LinLee David Chung Lin
4,39541242
answered Jan 24 at 1:20
Lee David Chung LinLee David Chung Lin
4,39541242
4,39541242
add a comment |
add a comment |
$begingroup$
I think your computation is actually correct. Note that
$$Gamma(n+frac{1}{2})=frac{1cdot 3cdot 5cdotcdotscdot (2n-1)}{2^n}sqrt{pi}$$
see here, for instance. Thus, the expression you derived is actually exactly what you want.
answered Jan 24 at 0:56
user293794user293794
1,748613
$endgroup$
add a comment |
$begingroup$
I think your computation is actually correct. Note that
$$Gamma(n+frac{1}{2})=frac{1cdot 3cdot 5cdotcdotscdot (2n-1)}{2^n}sqrt{pi}$$
see here, for instance. Thus, the expression you derived is actually exactly what you want.
answered Jan 24 at 0:56
user293794user293794
1,748613
$endgroup$
add a comment |
$begingroup$
I think your computation is actually correct. Note that
$$Gamma(n+frac{1}{2})=frac{1cdot 3cdot 5cdotcdotscdot (2n-1)}{2^n}sqrt{pi}$$
see here, for instance. Thus, the expression you derived is actually exactly what you want.
answered Jan 24 at 0:56
user293794user293794
1,748613
$endgroup$
I think your computation is actually correct. Note that
$$Gamma(n+frac{1}{2})=frac{1cdot 3cdot 5cdotcdotscdot (2n-1)}{2^n}sqrt{pi}$$
see here, for instance. Thus, the expression you derived is actually exactly what you want.
answered Jan 24 at 0:56
user293794user293794
1,748613
answered Jan 24 at 0:56
user293794user293794
1,748613
answered Jan 24 at 0:56
user293794user293794
1,748613
answered Jan 24 at 0:56
user293794user293794
1,748613
1,748613
add a comment |
add a comment |
$begingroup$
You can also prove this by induction using integration by parts:
begin{align}
int x^{2n}e^{-x^2/2},dx
&=int x^{2n-1}cdot xe^{-x^2/2},dx
\&=-int x^{2n-1}dbig(e^{-x^2/2}big)
\&=-[x^{2n-1}e^{-x^2/2}]Big|^{infty}_{-infty}+int e^{-x^2/2}dbig(x^{2n-1}big)
\&=0+(2n-1)int x^{2n-2}e^{-x^2/2},dx
end{align}
Now apply the induction hypothesis.
answered Jan 24 at 2:49
Mike EarnestMike Earnest
24.7k22151
$endgroup$
add a comment |
$begingroup$
You can also prove this by induction using integration by parts:
begin{align}
int x^{2n}e^{-x^2/2},dx
&=int x^{2n-1}cdot xe^{-x^2/2},dx
\&=-int x^{2n-1}dbig(e^{-x^2/2}big)
\&=-[x^{2n-1}e^{-x^2/2}]Big|^{infty}_{-infty}+int e^{-x^2/2}dbig(x^{2n-1}big)
\&=0+(2n-1)int x^{2n-2}e^{-x^2/2},dx
end{align}
Now apply the induction hypothesis.
answered Jan 24 at 2:49
Mike EarnestMike Earnest
24.7k22151
$endgroup$
add a comment |
$begingroup$
You can also prove this by induction using integration by parts:
begin{align}
int x^{2n}e^{-x^2/2},dx
&=int x^{2n-1}cdot xe^{-x^2/2},dx
\&=-int x^{2n-1}dbig(e^{-x^2/2}big)
\&=-[x^{2n-1}e^{-x^2/2}]Big|^{infty}_{-infty}+int e^{-x^2/2}dbig(x^{2n-1}big)
\&=0+(2n-1)int x^{2n-2}e^{-x^2/2},dx
end{align}
Now apply the induction hypothesis.
answered Jan 24 at 2:49
Mike EarnestMike Earnest
24.7k22151
$endgroup$
You can also prove this by induction using integration by parts:
begin{align}
int x^{2n}e^{-x^2/2},dx
&=int x^{2n-1}cdot xe^{-x^2/2},dx
\&=-int x^{2n-1}dbig(e^{-x^2/2}big)
\&=-[x^{2n-1}e^{-x^2/2}]Big|^{infty}_{-infty}+int e^{-x^2/2}dbig(x^{2n-1}big)
\&=0+(2n-1)int x^{2n-2}e^{-x^2/2},dx
end{align}
Now apply the induction hypothesis.
answered Jan 24 at 2:49
Mike EarnestMike Earnest
24.7k22151
answered Jan 24 at 2:49
Mike EarnestMike Earnest
24.7k22151
answered Jan 24 at 2:49
Mike EarnestMike Earnest
24.7k22151
answered Jan 24 at 2:49
Mike EarnestMike Earnest
24.7k22151
24.7k22151
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$begingroup$
I didn't check constants for certain, but you definitely should get something of the sort of $frac{2^{n+1}}{sqrt{pi}}Gamma(n+frac{1}{2})$
$endgroup$
– Jakobian
Jan 24 at 0:17
1
$begingroup$
Apart from apparent typos in your final formula, I can't see anything wrong with what you've done so far. Is the problem you're having difficulty with that of showing $$frac{2}{sqrt{pi}}Gamma(n+frac{1}{2}) =prod_{1 leq k leq 2n, k operatorname{odd}}k ?$$
$endgroup$
– lonza leggiera
Jan 24 at 0:59