Generalizing rising and falling factorial to complex arguments preserving zeroes
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Falling factorials count injective functions. There are no injective functions $ A to B $ if $|A| gt |B|$ . Is there a way to extend the definition of the falling factorial, ideally to $mathbb{C} times mathbb{C} to mathbb{C}$, so that the zeroes are preserved? The most obvious way of informally generalizing the falling factorial involves taking ratios of the gamma function, but I'm not convinced that that works out.
When the right argument of the rising factorial is a natural number, it is defined in the following way:
$$ (x)^{overline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x + i) $$
$$ (x)^{overline{k}} = (x)(x+1)cdots(x+k-2)(x+k-1) $$
And the falling factorial is defined in the following way
$$ (x)^{underline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x - i) $$
$$ (x)^{underline{k}} = (x)(x-1)cdots (x-k+2)(x-k+1) $$
The falling factorial can be defined in terms of the rising factorial in two ways.
$$ (x)^{underline{k}} = (-1)^k (-x)^{overline{k}} $$
or
$$ (x)^{underline{k}} = (x-k+1) ^ {overline{k}} $$
The rising factorial satisfies the following identity:
$$ (x)^{overline{k}} = frac{(x+k-1)!}{(x-1)!} $$
Which suggests the generalization
$$ (x)^{overline{k}} = frac{Gamma(x+k)}{Gamma(x)} $$
and
$$ (x)^{underline{k}} = frac{Gamma(x + 1)}{Gamma(x-k + 1)} $$
This makes me nervous since $0 notin text{Im}(Gamma)$ and $Gamma$ has a pole when $ x le -1 $ .
Does a nice way of extending the rising/falling factorial to complex arguments exist?
factorial analytic-continuation
asked Nov 17 at 20:38
Gregory Nisbet
371211
add a comment |
up vote
0
down vote
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Falling factorials count injective functions. There are no injective functions $ A to B $ if $|A| gt |B|$ . Is there a way to extend the definition of the falling factorial, ideally to $mathbb{C} times mathbb{C} to mathbb{C}$, so that the zeroes are preserved? The most obvious way of informally generalizing the falling factorial involves taking ratios of the gamma function, but I'm not convinced that that works out.
When the right argument of the rising factorial is a natural number, it is defined in the following way:
$$ (x)^{overline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x + i) $$
$$ (x)^{overline{k}} = (x)(x+1)cdots(x+k-2)(x+k-1) $$
And the falling factorial is defined in the following way
$$ (x)^{underline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x - i) $$
$$ (x)^{underline{k}} = (x)(x-1)cdots (x-k+2)(x-k+1) $$
The falling factorial can be defined in terms of the rising factorial in two ways.
$$ (x)^{underline{k}} = (-1)^k (-x)^{overline{k}} $$
or
$$ (x)^{underline{k}} = (x-k+1) ^ {overline{k}} $$
The rising factorial satisfies the following identity:
$$ (x)^{overline{k}} = frac{(x+k-1)!}{(x-1)!} $$
Which suggests the generalization
$$ (x)^{overline{k}} = frac{Gamma(x+k)}{Gamma(x)} $$
and
$$ (x)^{underline{k}} = frac{Gamma(x + 1)}{Gamma(x-k + 1)} $$
This makes me nervous since $0 notin text{Im}(Gamma)$ and $Gamma$ has a pole when $ x le -1 $ .
Does a nice way of extending the rising/falling factorial to complex arguments exist?
factorial analytic-continuation
asked Nov 17 at 20:38
Gregory Nisbet
371211
I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
– G Cab
Nov 17 at 23:18
1
by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
– G Cab
Nov 17 at 23:24
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Falling factorials count injective functions. There are no injective functions $ A to B $ if $|A| gt |B|$ . Is there a way to extend the definition of the falling factorial, ideally to $mathbb{C} times mathbb{C} to mathbb{C}$, so that the zeroes are preserved? The most obvious way of informally generalizing the falling factorial involves taking ratios of the gamma function, but I'm not convinced that that works out.
When the right argument of the rising factorial is a natural number, it is defined in the following way:
$$ (x)^{overline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x + i) $$
$$ (x)^{overline{k}} = (x)(x+1)cdots(x+k-2)(x+k-1) $$
And the falling factorial is defined in the following way
$$ (x)^{underline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x - i) $$
$$ (x)^{underline{k}} = (x)(x-1)cdots (x-k+2)(x-k+1) $$
The falling factorial can be defined in terms of the rising factorial in two ways.
$$ (x)^{underline{k}} = (-1)^k (-x)^{overline{k}} $$
or
$$ (x)^{underline{k}} = (x-k+1) ^ {overline{k}} $$
The rising factorial satisfies the following identity:
$$ (x)^{overline{k}} = frac{(x+k-1)!}{(x-1)!} $$
Which suggests the generalization
$$ (x)^{overline{k}} = frac{Gamma(x+k)}{Gamma(x)} $$
and
$$ (x)^{underline{k}} = frac{Gamma(x + 1)}{Gamma(x-k + 1)} $$
This makes me nervous since $0 notin text{Im}(Gamma)$ and $Gamma$ has a pole when $ x le -1 $ .
Does a nice way of extending the rising/falling factorial to complex arguments exist?
factorial analytic-continuation
asked Nov 17 at 20:38
Gregory Nisbet
371211
Falling factorials count injective functions. There are no injective functions $ A to B $ if $|A| gt |B|$ . Is there a way to extend the definition of the falling factorial, ideally to $mathbb{C} times mathbb{C} to mathbb{C}$, so that the zeroes are preserved? The most obvious way of informally generalizing the falling factorial involves taking ratios of the gamma function, but I'm not convinced that that works out.
When the right argument of the rising factorial is a natural number, it is defined in the following way:
$$ (x)^{overline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x + i) $$
$$ (x)^{overline{k}} = (x)(x+1)cdots(x+k-2)(x+k-1) $$
And the falling factorial is defined in the following way
$$ (x)^{underline{k}} stackrel{df}{=} prod_{i = 0}^{k-1} (x - i) $$
$$ (x)^{underline{k}} = (x)(x-1)cdots (x-k+2)(x-k+1) $$
The falling factorial can be defined in terms of the rising factorial in two ways.
$$ (x)^{underline{k}} = (-1)^k (-x)^{overline{k}} $$
or
$$ (x)^{underline{k}} = (x-k+1) ^ {overline{k}} $$
The rising factorial satisfies the following identity:
$$ (x)^{overline{k}} = frac{(x+k-1)!}{(x-1)!} $$
Which suggests the generalization
$$ (x)^{overline{k}} = frac{Gamma(x+k)}{Gamma(x)} $$
and
$$ (x)^{underline{k}} = frac{Gamma(x + 1)}{Gamma(x-k + 1)} $$
This makes me nervous since $0 notin text{Im}(Gamma)$ and $Gamma$ has a pole when $ x le -1 $ .
Does a nice way of extending the rising/falling factorial to complex arguments exist?
factorial analytic-continuation
factorial analytic-continuation
asked Nov 17 at 20:38
Gregory Nisbet
371211
asked Nov 17 at 20:38
Gregory Nisbet
371211
edited Nov 17 at 23:50
asked Nov 17 at 20:38
Gregory Nisbet
371211
asked Nov 17 at 20:38
Gregory Nisbet
371211
asked Nov 17 at 20:38
Gregory Nisbet
371211
371211
I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
– G Cab
Nov 17 at 23:18
1
by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
– G Cab
Nov 17 at 23:24
add a comment |
I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
– G Cab
Nov 17 at 23:18
1
by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
– G Cab
Nov 17 at 23:24
I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
– G Cab
Nov 17 at 23:18
I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
– G Cab
Nov 17 at 23:18
1
1
by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
– G Cab
Nov 17 at 23:24
by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
– G Cab
Nov 17 at 23:24
add a comment |
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I do not know a better way than through the gamma ratio you show. Of course, the poles in the numerator will cancel with the denominator if both are negative integers
– G Cab
Nov 17 at 23:18
1
by the way, it is actually $$ x^{,underline {,k,} } = left( {x - k + 1} right)^{,overline {,k,} } = {{Gamma (x + 1)} over {Gamma (x - k + 1)}} $$
– G Cab
Nov 17 at 23:24