Showing $binom{2n}{n} = (-4)^n binom{-1/2}{n}$
$begingroup$
Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?
Let $n$ be a non-negative integer.
$$binom{2n}{n} = (-4)^n binom{-frac{1}{2}}{n}$$
binomial-coefficients factorial
$endgroup$
add a comment |
$begingroup$
Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?
Let $n$ be a non-negative integer.
$$binom{2n}{n} = (-4)^n binom{-frac{1}{2}}{n}$$
binomial-coefficients factorial
$endgroup$
1
$begingroup$
Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
$endgroup$
– KCd
Jan 13 '15 at 2:17
add a comment |
$begingroup$
Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?
Let $n$ be a non-negative integer.
$$binom{2n}{n} = (-4)^n binom{-frac{1}{2}}{n}$$
binomial-coefficients factorial
$endgroup$
Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?
Let $n$ be a non-negative integer.
$$binom{2n}{n} = (-4)^n binom{-frac{1}{2}}{n}$$
binomial-coefficients factorial
binomial-coefficients factorial
asked Jan 13 '15 at 1:47
MarkusMarkus
234
234
1
$begingroup$
Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
$endgroup$
– KCd
Jan 13 '15 at 2:17
add a comment |
1
$begingroup$
Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
$endgroup$
– KCd
Jan 13 '15 at 2:17
1
1
$begingroup$
Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
$endgroup$
– KCd
Jan 13 '15 at 2:17
$begingroup$
Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
$endgroup$
– KCd
Jan 13 '15 at 2:17
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.
The term $binom{-1/2}{n}$ is equal to
$$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
$$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
The rest of the calculation is straightforward.
$endgroup$
add a comment |
$begingroup$
$$begin{align*}
(-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
&=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
&=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
&=frac{2^nn!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!}{(n!)^2}\\
&=binom{2n}n
end{align*}$$
$endgroup$
add a comment |
$begingroup$
$$begin{align}
binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
/n! \
&= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
&= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
&= (-1)^n (2n)!/ (2^n n!)^2 \
&= (-1/4)^n binom{2n}{n}.
end{align}$$
Hence:
$$ (1-4X)^{-1/2} $$
generates (link)
$$ binom{2n}{n}.$$
$endgroup$
add a comment |
$begingroup$
I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.
begin{align}
(-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
end{align}
$endgroup$
add a comment |
Your Answer
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4 Answers
4
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oldest
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4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.
The term $binom{-1/2}{n}$ is equal to
$$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
$$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
The rest of the calculation is straightforward.
$endgroup$
add a comment |
$begingroup$
You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.
The term $binom{-1/2}{n}$ is equal to
$$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
$$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
The rest of the calculation is straightforward.
$endgroup$
add a comment |
$begingroup$
You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.
The term $binom{-1/2}{n}$ is equal to
$$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
$$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
The rest of the calculation is straightforward.
$endgroup$
You have not specified what is meant by the definition of the generalized binomial coefficient. We take one interpretation, which may not be the intended one.
The term $binom{-1/2}{n}$ is equal to
$$frac{1}{n!}left(-frac{1}{2}right)left(-frac{3}{2}right)cdots left(-frac{2n-1}{2}right).$$
Multiply the top by $(2)(4)cdots (2n)$ and the bottom by the same thing, in the form $2^n n!$. We get
$$(-1)^nfrac{1}{n!}cdot frac{1}{2^ncdot 2^n} frac{2n!}{n!}.$$
The rest of the calculation is straightforward.
edited Jan 13 '15 at 2:11
answered Jan 13 '15 at 2:03
André NicolasAndré Nicolas
453k36427813
453k36427813
add a comment |
add a comment |
$begingroup$
$$begin{align*}
(-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
&=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
&=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
&=frac{2^nn!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!}{(n!)^2}\\
&=binom{2n}n
end{align*}$$
$endgroup$
add a comment |
$begingroup$
$$begin{align*}
(-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
&=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
&=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
&=frac{2^nn!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!}{(n!)^2}\\
&=binom{2n}n
end{align*}$$
$endgroup$
add a comment |
$begingroup$
$$begin{align*}
(-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
&=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
&=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
&=frac{2^nn!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!}{(n!)^2}\\
&=binom{2n}n
end{align*}$$
$endgroup$
$$begin{align*}
(-4)^nbinom{-1/2}n&=(-4)^nfrac{(-1/2)^{underline n}}{n!}\\
&=frac{(-4)^n}{n!}left(-frac12right)left(-frac32right)ldotsleft(-frac{2n-1}2right)\\
&=frac{(-4)^n}{n!}cdotfrac{(2n-1)!!}{(-2)^n}\\
&=frac{2^nn!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!!(2n-1)!!}{(n!)^2}\\
&=frac{(2n)!}{(n!)^2}\\
&=binom{2n}n
end{align*}$$
answered Jan 13 '15 at 2:09
Brian M. ScottBrian M. Scott
458k38511913
458k38511913
add a comment |
add a comment |
$begingroup$
$$begin{align}
binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
/n! \
&= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
&= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
&= (-1)^n (2n)!/ (2^n n!)^2 \
&= (-1/4)^n binom{2n}{n}.
end{align}$$
Hence:
$$ (1-4X)^{-1/2} $$
generates (link)
$$ binom{2n}{n}.$$
$endgroup$
add a comment |
$begingroup$
$$begin{align}
binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
/n! \
&= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
&= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
&= (-1)^n (2n)!/ (2^n n!)^2 \
&= (-1/4)^n binom{2n}{n}.
end{align}$$
Hence:
$$ (1-4X)^{-1/2} $$
generates (link)
$$ binom{2n}{n}.$$
$endgroup$
add a comment |
$begingroup$
$$begin{align}
binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
/n! \
&= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
&= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
&= (-1)^n (2n)!/ (2^n n!)^2 \
&= (-1/4)^n binom{2n}{n}.
end{align}$$
Hence:
$$ (1-4X)^{-1/2} $$
generates (link)
$$ binom{2n}{n}.$$
$endgroup$
$$begin{align}
binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)ldots(-(2n-1)/2)
/n! \
&= (-1)^n 1cdot 3cdot 5cdotldotscdot(2n-1)/ (2^n n!) \
&= (-1)^n (2n)!/ (2^n cdot n! cdot 2 cdot 4 cdot 6cdot ldots cdot (2n)) \
&= (-1)^n (2n)!/ (2^n n!)^2 \
&= (-1/4)^n binom{2n}{n}.
end{align}$$
Hence:
$$ (1-4X)^{-1/2} $$
generates (link)
$$ binom{2n}{n}.$$
edited Dec 22 '18 at 3:03
Shaun
9,241113684
9,241113684
answered Jan 13 '15 at 2:08
ir7ir7
4,14311015
4,14311015
add a comment |
add a comment |
$begingroup$
I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.
begin{align}
(-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
end{align}
$endgroup$
add a comment |
$begingroup$
I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.
begin{align}
(-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
end{align}
$endgroup$
add a comment |
$begingroup$
I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.
begin{align}
(-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
end{align}
$endgroup$
I'll assume the generalized binomial coefficient is defined via the gamma function, with $binom{x}{y}=frac{Gamma(x+1)}{Gamma(y+1)Gamma(x-y+1)}$ This will match the normal definition for integers, and will follow most of the familiar properties. When dealing with the gamma function, the most important properties to remember are that $Gamma(x+1)=xGamma(x)$ and that the gamma function is defined for all real numbers, except for the negative integers and zero. In my view, this is by far the most natural extension for the binomial coefficients.
begin{align}
(-4)^nbinom{-frac{1}{2}}{n}&=(-4)^nfrac{Gamma(frac{1}{2})}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)Gamma(frac{1}{2}-n)}{Gamma(n+1)Gamma(frac{1}{2}-n)}\&=(-4)^nfrac{(prodlimits_{k=1}^nfrac{1}{2}-k)}{Gamma(n+1)}\&=2^nfrac{prodlimits_{k=1}^n2k-1}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^n(2k-1)(2k)}{prodlimits_{k-1}^n2k}}{n!}\&=2^nfrac{frac{prodlimits_{k=1}^{2n}k}{2^nprodlimits_{k=1}^nk}}{n!}=2^nfrac{frac{(2n)!}{2^nn!}}{n!}\&=frac{(2n)!}{(n!)^2}\&=frac{(2n)!}{n!(2n-n)!}\&=binom{2n}{n}
end{align}
answered Jan 13 '15 at 2:56
Benjamin RoycraftBenjamin Roycraft
40628
40628
add a comment |
add a comment |
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$begingroup$
Try to show $1cdot 3 cdot 5 cdots (2n-1)$, the product of the first $n$ odd numbers, is $(2n)!/(2^nn!)$ by inserting factors $2, 4, 6, dots, 2n$ to make the product $(2n)!$ and then dividing by $2 cdot 4 cdot 6 cdots (2n)$ and then pulling a factor of $2$ from each of those even numbers in the denominator.
$endgroup$
– KCd
Jan 13 '15 at 2:17