Mixing
Definition of $binom{frac{1}{2}}{1}$?
$begingroup$
How interpret the combination
$binom{frac{1}{2}}{n}$ when $n$ is a positive integer ?
combinatorics binomial-coefficients
edited Jan 13 at 22:53
Simply Beautiful Art
50.5k578182
asked Nov 16 '18 at 21:54
MedoMedo
632214
$endgroup$
add a comment |
$begingroup$
How interpret the combination
$binom{frac{1}{2}}{n}$ when $n$ is a positive integer ?
combinatorics binomial-coefficients
edited Jan 13 at 22:53
Simply Beautiful Art
50.5k578182
asked Nov 16 '18 at 21:54
MedoMedo
632214
$endgroup$
$begingroup$
In what context does this come up?
$endgroup$
– Michael Burr
Nov 16 '18 at 21:57
$begingroup$
@MichaelBurr: This comes up in the binomial theorem with fractional exponents allowed.
$endgroup$
– Cheerful Parsnip
Nov 16 '18 at 22:01
add a comment |
$begingroup$
How interpret the combination
$binom{frac{1}{2}}{n}$ when $n$ is a positive integer ?
combinatorics binomial-coefficients
edited Jan 13 at 22:53
Simply Beautiful Art
50.5k578182
asked Nov 16 '18 at 21:54
MedoMedo
632214
$endgroup$
How interpret the combination
$binom{frac{1}{2}}{n}$ when $n$ is a positive integer ?
combinatorics binomial-coefficients
combinatorics binomial-coefficients
edited Jan 13 at 22:53
Simply Beautiful Art
50.5k578182
asked Nov 16 '18 at 21:54
MedoMedo
632214
edited Jan 13 at 22:53
Simply Beautiful Art
50.5k578182
asked Nov 16 '18 at 21:54
MedoMedo
632214
edited Jan 13 at 22:53
Simply Beautiful Art
50.5k578182
edited Jan 13 at 22:53
Simply Beautiful Art
50.5k578182
edited Jan 13 at 22:53
Simply Beautiful Art
50.5k578182
50.5k578182
asked Nov 16 '18 at 21:54
MedoMedo
632214
asked Nov 16 '18 at 21:54
MedoMedo
632214
asked Nov 16 '18 at 21:54
MedoMedo
632214
632214
$begingroup$
In what context does this come up?
$endgroup$
– Michael Burr
Nov 16 '18 at 21:57
$begingroup$
@MichaelBurr: This comes up in the binomial theorem with fractional exponents allowed.
$endgroup$
– Cheerful Parsnip
Nov 16 '18 at 22:01
add a comment |
$begingroup$
In what context does this come up?
$endgroup$
– Michael Burr
Nov 16 '18 at 21:57
$begingroup$
@MichaelBurr: This comes up in the binomial theorem with fractional exponents allowed.
$endgroup$
– Cheerful Parsnip
Nov 16 '18 at 22:01
$begingroup$
In what context does this come up?
$endgroup$
– Michael Burr
Nov 16 '18 at 21:57
$begingroup$
In what context does this come up?
$endgroup$
– Michael Burr
Nov 16 '18 at 21:57
$begingroup$
@MichaelBurr: This comes up in the binomial theorem with fractional exponents allowed.
$endgroup$
– Cheerful Parsnip
Nov 16 '18 at 22:01
$begingroup$
@MichaelBurr: This comes up in the binomial theorem with fractional exponents allowed.
$endgroup$
– Cheerful Parsnip
Nov 16 '18 at 22:01
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The definition is
$$
binom{x}{n}=frac{x(x-1)dotsm(x-n+1)}{n!}
$$
where $x$ is any real number. In the numerator there are $n$ factors starting from $x$ and decreasing by $1$ at each step. This is zero if and only if $x$ is integer and $n>x$.
In the special case $x=1/2$, a different formula can be found. Indeed,
begin{align}
frac{1}{2}left(frac{1}{2}-1right)left(frac{1}{2}-2right)dotsmleft(frac{1}{2}-n+1right)
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr) \
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr)
frac{2cdot4cdotdotscdot(2n-2)}{2^{n-1}(1cdot 2cdotdotscdot(n-1))} \
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n-2)!}{(n-1)!}\
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n)!}{n!}frac{n}{2n(2n-1)}\
&=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{n!}
end{align}
and therefore
$$
binom{1/2}{n}=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{(n!)^2}
$$
answered Nov 16 '18 at 22:16
egregegreg
182k1485203
$endgroup$
add a comment |
$begingroup$
The general definition of $binomalpha k$ for $kinBbb N$ and $alphainBbb C$ is $$binomalpha k=frac1{k!}prod_{h=0}^{k-1} (alpha-h)$$
answered Nov 16 '18 at 21:57
Saucy O'PathSaucy O'Path
5,9691627
$endgroup$
add a comment |
$begingroup$
Same way as usual binomial. $binom{frac{1}{2}}{n}=frac{frac{1}{2}timesfrac{-1}{2}...times(frac{1}{2}-n+1)}{n!}$
answered Nov 16 '18 at 22:04
herb steinbergherb steinberg
2,7582310
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The definition is
$$
binom{x}{n}=frac{x(x-1)dotsm(x-n+1)}{n!}
$$
where $x$ is any real number. In the numerator there are $n$ factors starting from $x$ and decreasing by $1$ at each step. This is zero if and only if $x$ is integer and $n>x$.
In the special case $x=1/2$, a different formula can be found. Indeed,
begin{align}
frac{1}{2}left(frac{1}{2}-1right)left(frac{1}{2}-2right)dotsmleft(frac{1}{2}-n+1right)
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr) \
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr)
frac{2cdot4cdotdotscdot(2n-2)}{2^{n-1}(1cdot 2cdotdotscdot(n-1))} \
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n-2)!}{(n-1)!}\
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n)!}{n!}frac{n}{2n(2n-1)}\
&=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{n!}
end{align}
and therefore
$$
binom{1/2}{n}=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{(n!)^2}
$$
answered Nov 16 '18 at 22:16
egregegreg
182k1485203
$endgroup$
add a comment |
$begingroup$
The definition is
$$
binom{x}{n}=frac{x(x-1)dotsm(x-n+1)}{n!}
$$
where $x$ is any real number. In the numerator there are $n$ factors starting from $x$ and decreasing by $1$ at each step. This is zero if and only if $x$ is integer and $n>x$.
In the special case $x=1/2$, a different formula can be found. Indeed,
begin{align}
frac{1}{2}left(frac{1}{2}-1right)left(frac{1}{2}-2right)dotsmleft(frac{1}{2}-n+1right)
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr) \
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr)
frac{2cdot4cdotdotscdot(2n-2)}{2^{n-1}(1cdot 2cdotdotscdot(n-1))} \
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n-2)!}{(n-1)!}\
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n)!}{n!}frac{n}{2n(2n-1)}\
&=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{n!}
end{align}
and therefore
$$
binom{1/2}{n}=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{(n!)^2}
$$
answered Nov 16 '18 at 22:16
egregegreg
182k1485203
$endgroup$
add a comment |
$begingroup$
The definition is
$$
binom{x}{n}=frac{x(x-1)dotsm(x-n+1)}{n!}
$$
where $x$ is any real number. In the numerator there are $n$ factors starting from $x$ and decreasing by $1$ at each step. This is zero if and only if $x$ is integer and $n>x$.
In the special case $x=1/2$, a different formula can be found. Indeed,
begin{align}
frac{1}{2}left(frac{1}{2}-1right)left(frac{1}{2}-2right)dotsmleft(frac{1}{2}-n+1right)
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr) \
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr)
frac{2cdot4cdotdotscdot(2n-2)}{2^{n-1}(1cdot 2cdotdotscdot(n-1))} \
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n-2)!}{(n-1)!}\
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n)!}{n!}frac{n}{2n(2n-1)}\
&=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{n!}
end{align}
and therefore
$$
binom{1/2}{n}=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{(n!)^2}
$$
answered Nov 16 '18 at 22:16
egregegreg
182k1485203
$endgroup$
The definition is
$$
binom{x}{n}=frac{x(x-1)dotsm(x-n+1)}{n!}
$$
where $x$ is any real number. In the numerator there are $n$ factors starting from $x$ and decreasing by $1$ at each step. This is zero if and only if $x$ is integer and $n>x$.
In the special case $x=1/2$, a different formula can be found. Indeed,
begin{align}
frac{1}{2}left(frac{1}{2}-1right)left(frac{1}{2}-2right)dotsmleft(frac{1}{2}-n+1right)
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr) \
&=frac{(-1)^{n-1}}{2^n}bigl(1cdot3cdotdotscdot(2n-3)bigr)
frac{2cdot4cdotdotscdot(2n-2)}{2^{n-1}(1cdot 2cdotdotscdot(n-1))} \
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n-2)!}{(n-1)!}\
&=frac{(-1)^{n-1}}{2^{2n-1}}frac{(2n)!}{n!}frac{n}{2n(2n-1)}\
&=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{n!}
end{align}
and therefore
$$
binom{1/2}{n}=frac{(-1)^{n-1}}{2^{2n}(2n-1)}frac{(2n)!}{(n!)^2}
$$
answered Nov 16 '18 at 22:16
egregegreg
182k1485203
answered Nov 16 '18 at 22:16
egregegreg
182k1485203
answered Nov 16 '18 at 22:16
egregegreg
182k1485203
answered Nov 16 '18 at 22:16
egregegreg
182k1485203
182k1485203
add a comment |
add a comment |
$begingroup$
The general definition of $binomalpha k$ for $kinBbb N$ and $alphainBbb C$ is $$binomalpha k=frac1{k!}prod_{h=0}^{k-1} (alpha-h)$$
answered Nov 16 '18 at 21:57
Saucy O'PathSaucy O'Path
5,9691627
$endgroup$
add a comment |
$begingroup$
The general definition of $binomalpha k$ for $kinBbb N$ and $alphainBbb C$ is $$binomalpha k=frac1{k!}prod_{h=0}^{k-1} (alpha-h)$$
answered Nov 16 '18 at 21:57
Saucy O'PathSaucy O'Path
5,9691627
$endgroup$
add a comment |
$begingroup$
The general definition of $binomalpha k$ for $kinBbb N$ and $alphainBbb C$ is $$binomalpha k=frac1{k!}prod_{h=0}^{k-1} (alpha-h)$$
answered Nov 16 '18 at 21:57
Saucy O'PathSaucy O'Path
5,9691627
$endgroup$
The general definition of $binomalpha k$ for $kinBbb N$ and $alphainBbb C$ is $$binomalpha k=frac1{k!}prod_{h=0}^{k-1} (alpha-h)$$
answered Nov 16 '18 at 21:57
Saucy O'PathSaucy O'Path
5,9691627
answered Nov 16 '18 at 21:57
Saucy O'PathSaucy O'Path
5,9691627
answered Nov 16 '18 at 21:57
Saucy O'PathSaucy O'Path
5,9691627
answered Nov 16 '18 at 21:57
Saucy O'PathSaucy O'Path
5,9691627
5,9691627
add a comment |
add a comment |
$begingroup$
Same way as usual binomial. $binom{frac{1}{2}}{n}=frac{frac{1}{2}timesfrac{-1}{2}...times(frac{1}{2}-n+1)}{n!}$
answered Nov 16 '18 at 22:04
herb steinbergherb steinberg
2,7582310
$endgroup$
add a comment |
$begingroup$
Same way as usual binomial. $binom{frac{1}{2}}{n}=frac{frac{1}{2}timesfrac{-1}{2}...times(frac{1}{2}-n+1)}{n!}$
answered Nov 16 '18 at 22:04
herb steinbergherb steinberg
2,7582310
$endgroup$
add a comment |
$begingroup$
Same way as usual binomial. $binom{frac{1}{2}}{n}=frac{frac{1}{2}timesfrac{-1}{2}...times(frac{1}{2}-n+1)}{n!}$
answered Nov 16 '18 at 22:04
herb steinbergherb steinberg
2,7582310
$endgroup$
Same way as usual binomial. $binom{frac{1}{2}}{n}=frac{frac{1}{2}timesfrac{-1}{2}...times(frac{1}{2}-n+1)}{n!}$
answered Nov 16 '18 at 22:04
herb steinbergherb steinberg
2,7582310
answered Nov 16 '18 at 22:04
herb steinbergherb steinberg
2,7582310
answered Nov 16 '18 at 22:04
herb steinbergherb steinberg
2,7582310
answered Nov 16 '18 at 22:04
herb steinbergherb steinberg
2,7582310
2,7582310
add a comment |
add a comment |
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$begingroup$
In what context does this come up?
$endgroup$
– Michael Burr
Nov 16 '18 at 21:57
$begingroup$
@MichaelBurr: This comes up in the binomial theorem with fractional exponents allowed.
$endgroup$
– Cheerful Parsnip
Nov 16 '18 at 22:01