Mixing
Why is Euler's Gamma function the “best” extension of the factorial function to the reals?
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There are lots (an infinitude) of smooth functions that coincide with f(n)=n! on the integers. Is there a simple reason why Euler's Gamma function $Gamma (z) = int_0^infty t^{z-1} e^t dt$ is "best"? In particular, I'm looking for reasons that I can explain to first-year calculus students.
real-analysis education gamma-function special-functions
edited Aug 28 '10 at 9:52
J. M. is not a mathematician
61.4k5152290
asked Aug 4 '10 at 15:01
pbrookspbrooks
843276
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add a comment |
$begingroup$
There are lots (an infinitude) of smooth functions that coincide with f(n)=n! on the integers. Is there a simple reason why Euler's Gamma function $Gamma (z) = int_0^infty t^{z-1} e^t dt$ is "best"? In particular, I'm looking for reasons that I can explain to first-year calculus students.
real-analysis education gamma-function special-functions
edited Aug 28 '10 at 9:52
J. M. is not a mathematician
61.4k5152290
asked Aug 4 '10 at 15:01
pbrookspbrooks
843276
$endgroup$
$begingroup$
Good question. I'm not very familiar with the process of analytic continuation, but I suspect the answer lies there.
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– Noldorin
Aug 4 '10 at 15:05
5
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@Noldorin: I think analytic continuations are only unique if you have a (locally?) dense set of points, which positive integers are not. After all, you can have an entire function that is zero at every positive integer. Euler's Gamma is used in some important probability distributions, and also in the Riemann Zeta function, but not many first-year calculus students will care about this.
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– Larry Wang
Aug 4 '10 at 16:28
4
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It's also used in the fractional calculus (fractional differentiation and integration -- see en.wikipedia.org/wiki/Fractional_calculus), which is why I wanted some more intuitive understanding of its virtues.
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– pbrooks
Aug 4 '10 at 17:06
add a comment |
$begingroup$
There are lots (an infinitude) of smooth functions that coincide with f(n)=n! on the integers. Is there a simple reason why Euler's Gamma function $Gamma (z) = int_0^infty t^{z-1} e^t dt$ is "best"? In particular, I'm looking for reasons that I can explain to first-year calculus students.
real-analysis education gamma-function special-functions
edited Aug 28 '10 at 9:52
J. M. is not a mathematician
61.4k5152290
asked Aug 4 '10 at 15:01
pbrookspbrooks
843276
$endgroup$
There are lots (an infinitude) of smooth functions that coincide with f(n)=n! on the integers. Is there a simple reason why Euler's Gamma function $Gamma (z) = int_0^infty t^{z-1} e^t dt$ is "best"? In particular, I'm looking for reasons that I can explain to first-year calculus students.
real-analysis education gamma-function special-functions
real-analysis education gamma-function special-functions
edited Aug 28 '10 at 9:52
J. M. is not a mathematician
61.4k5152290
asked Aug 4 '10 at 15:01
pbrookspbrooks
843276
edited Aug 28 '10 at 9:52
J. M. is not a mathematician
61.4k5152290
asked Aug 4 '10 at 15:01
pbrookspbrooks
843276
edited Aug 28 '10 at 9:52
J. M. is not a mathematician
61.4k5152290
edited Aug 28 '10 at 9:52
J. M. is not a mathematician
61.4k5152290
edited Aug 28 '10 at 9:52
J. M. is not a mathematician
61.4k5152290
61.4k5152290
asked Aug 4 '10 at 15:01
pbrookspbrooks
843276
asked Aug 4 '10 at 15:01
pbrookspbrooks
843276
asked Aug 4 '10 at 15:01
pbrookspbrooks
843276
843276
$begingroup$
Good question. I'm not very familiar with the process of analytic continuation, but I suspect the answer lies there.
$endgroup$
– Noldorin
Aug 4 '10 at 15:05
5
$begingroup$
@Noldorin: I think analytic continuations are only unique if you have a (locally?) dense set of points, which positive integers are not. After all, you can have an entire function that is zero at every positive integer. Euler's Gamma is used in some important probability distributions, and also in the Riemann Zeta function, but not many first-year calculus students will care about this.
$endgroup$
– Larry Wang
Aug 4 '10 at 16:28
4
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It's also used in the fractional calculus (fractional differentiation and integration -- see en.wikipedia.org/wiki/Fractional_calculus), which is why I wanted some more intuitive understanding of its virtues.
$endgroup$
– pbrooks
Aug 4 '10 at 17:06
add a comment |
$begingroup$
Good question. I'm not very familiar with the process of analytic continuation, but I suspect the answer lies there.
$endgroup$
– Noldorin
Aug 4 '10 at 15:05
5
$begingroup$
@Noldorin: I think analytic continuations are only unique if you have a (locally?) dense set of points, which positive integers are not. After all, you can have an entire function that is zero at every positive integer. Euler's Gamma is used in some important probability distributions, and also in the Riemann Zeta function, but not many first-year calculus students will care about this.
$endgroup$
– Larry Wang
Aug 4 '10 at 16:28
4
$begingroup$
It's also used in the fractional calculus (fractional differentiation and integration -- see en.wikipedia.org/wiki/Fractional_calculus), which is why I wanted some more intuitive understanding of its virtues.
$endgroup$
– pbrooks
Aug 4 '10 at 17:06
$begingroup$
Good question. I'm not very familiar with the process of analytic continuation, but I suspect the answer lies there.
$endgroup$
– Noldorin
Aug 4 '10 at 15:05
$begingroup$
Good question. I'm not very familiar with the process of analytic continuation, but I suspect the answer lies there.
$endgroup$
– Noldorin
Aug 4 '10 at 15:05
5
5
$begingroup$
@Noldorin: I think analytic continuations are only unique if you have a (locally?) dense set of points, which positive integers are not. After all, you can have an entire function that is zero at every positive integer. Euler's Gamma is used in some important probability distributions, and also in the Riemann Zeta function, but not many first-year calculus students will care about this.
$endgroup$
– Larry Wang
Aug 4 '10 at 16:28
$begingroup$
@Noldorin: I think analytic continuations are only unique if you have a (locally?) dense set of points, which positive integers are not. After all, you can have an entire function that is zero at every positive integer. Euler's Gamma is used in some important probability distributions, and also in the Riemann Zeta function, but not many first-year calculus students will care about this.
$endgroup$
– Larry Wang
Aug 4 '10 at 16:28
4
4
$begingroup$
It's also used in the fractional calculus (fractional differentiation and integration -- see en.wikipedia.org/wiki/Fractional_calculus), which is why I wanted some more intuitive understanding of its virtues.
$endgroup$
– pbrooks
Aug 4 '10 at 17:06
$begingroup$
It's also used in the fractional calculus (fractional differentiation and integration -- see en.wikipedia.org/wiki/Fractional_calculus), which is why I wanted some more intuitive understanding of its virtues.
$endgroup$
– pbrooks
Aug 4 '10 at 17:06
add a comment |
7 Answers
7
active
oldest
votes
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The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties
$f(1)=1$;
$f(x+1)=xf(x)$ for every $xgeq 0$;
$log f$ is a convex function.
The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function.
By the way, the gamma function is not the only meromorphic function satisfying
$$f(z+1)=z f(z),qquad f(1)=1,$$
with no zeroes and no poles other than the points $z=-n$, $n=0,1,2dots$. There is a whole family of such functions, which, in general, have the form
$$f(z)=exp{(-g(z))}frac{1}{zprodlimits_{m=1}^{infty} left(1+frac{z}{m}right)e^{-z/m}},$$
where $g(z)$ is an entire function such that
$$g(z+1)-g(z)=gamma+2kpi i,quad kinmathbb Z, $$
($gamma$ is Euler's constant). The gamma function corresponds to the simplest choice
$g(z)=gamma z$.
Edit: corrected index in the product.
edited Feb 11 at 5:34
Community♦
1
answered Aug 4 '10 at 15:41
Andrey RekaloAndrey Rekalo
6,10323641
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Also note that the first two properties already imply log convexity for positive integers, so it is reasonable to ask that an extension be log convex for positive reals. I found this more convincing than some of the infinite product inequalities (assuming these are the ones Andrey is referring to).
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– Larry Wang
Aug 4 '10 at 16:59
10
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The logarithm $log Gamma(c)$ can be thought of as a "continuous sum" $log Gamma(x) = sum_{k=1}^x k$. Is there a compelling reason why we should postulate that it's convex?
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– Greg Graviton
Aug 4 '10 at 17:31
1
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Is there a function that satisfies the first 2 properties but is not convex?
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– Dan
Apr 18 '12 at 1:26
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@Dan: Infinitely many. Just choose arbitrary (nonconvex) values for f(x) on 0 < x < 1.
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– Charles
Sep 25 '12 at 20:05
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Regarding your second point: it is true, however there is Wielandt's theorem which says that $Gamma(z)$ is unique upon adding the stipulation that $Gamma(z)$ be bounded in every vertical strip of the complex plane of unit width. This can be weakened to having at most exponential growth of base $e^{2pi}$ on a single such strip.
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– The_Sympathizer
Dec 6 '15 at 10:08
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Actually there are other (less-frequently) used extensions to the factorial, with different properties from the gamma function which may be desirable in some contexts.
Euler's Gamma Function
Hadamard's Gamma function
Luschny's factorial function
See here for more information.
answered Aug 4 '10 at 16:11
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,86342943
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For whatever reason, Nature (by which I mean integrals) seems to prefer the Gamma function as the "correct" substitute for the factorial in various integrals, which seems to come more or less from its integral definition. For example, for non-negative integers $a, b$, it's not hard to show (and there's a really nice probabilistic argument) that
$displaystyle int_{0}^1 t^a (1 - t)^b , dt = frac{a! b!}{(a+b+1)!}.$
For (non-negative?) real values of $a$ and $b$ the correct generalization is
$displaystyle int_0^1 t^a (1 - t)^b , dt = frac{Gamma(a+1) Gamma(b+1)}{Gamma(a+b+2)}.$
And, of course, integrals are important, so the Gamma function must also be important. For example, the Gamma function appears in the general formula for the volume of an n-sphere. But the result that, for me, really forces us to take the Gamma function seriously is its appearance in the functional equation for the Riemann zeta function.
answered Aug 4 '10 at 16:55
Qiaochu YuanQiaochu Yuan
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Why is that the "correct generalization"?
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– Neil G
Aug 28 '10 at 9:30
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Also, please see my first question -- I guess you know the answer!
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– Neil G
Aug 28 '10 at 9:32
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@Neil G: I wasn't sure what you were asking at first, but now I see. It could be that there exists some other function G such that G(a-1) G(b-1)/G(a+b) is also the answer. I don't know how to rule out this possibility, but again, the result that really makes it clear to me that the Gamma function is important is the last one.
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– Qiaochu Yuan
Aug 28 '10 at 12:22
1
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@QiaochuYuan: In the end you say a good reason for you is the connection to the zeta function. But I feel the connection via the functional equation is on the same level as the connection via the Mellin transform representation, which you might write as $zeta(s) = dfrac{int_0^inftyfrac{x^{s}}{e^x-1}frac{{mathrm d}x}{x}}{int_0^infty frac{x^{s}}{e^x-0}frac{{mathrm d}x}{x}}$. So in the end, I'd rather argue that it's important because of it's connection to the hom between addition and multiplication - the exponential function. After all, gamma = int monomial exp. What do you say?
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– Nikolaj-K
Feb 6 '15 at 19:24
1
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@MathematicsStudent: pick $a + b + 1$ random numbers in the interval $[0, 1]$, and compute the probability that the last such random number is greater than $a$ of the others and less than $b$ of the others, in two ways.
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– Qiaochu Yuan
Aug 1 '17 at 21:55
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show 4 more comments
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This is a comment posted as an answer for lack of reputation.
Following Qiaochu Yuan, the gamma function shows up in the functional equation of the zeta function as the factor in the Euler product corresponding to the "prime at infinity", and it occurs there as the Mellin transform of some gaussian function. (Gaussian functions occur in turn as eigenvectors of the Fourier transform.)
This is at least as old as Tate's thesis, and a possible reference is Weil's Basic Number Theory.
EDIT. Artin was one of the first people to popularize the log-convexity property of the gamma function (see his book on the function in question), and also perhaps the first mathematician to fully understand this Euler-factor-at-infinity aspect of the same function (he was Tate's thesis advisor). I thought his name had to be mentioned in a discussion about the gamma function.
answered Aug 5 '10 at 10:44
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
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It is the Mellin transform of $e^{-t}$ that appears in the functional equation much more than "the best interpolation of $n!$"
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– reuns
Jan 24 at 0:55
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Wielandt's theorem says that the gamma-function is the only function $f$ that satisfies the properties:
- $f(1)=1$
- $f(z+1)=zf(z)$ for all $z>0$
- $f(z)$ is analytic for $operatorname{Re}z>0$
- $f(z)$ is bounded for $1leq operatorname{Re}zleq 2$
(See also the related MathOverflow thread Importance of Log Convexity of the Gamma Function, where I learned about the above theorem.)
edited Apr 13 '17 at 12:58
Community♦
1
answered Aug 5 '10 at 9:45
SamuelSamuel
4,79412033
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Looking for a difference that makes a difference.
Flipping the gamma function and looking at Newton interpolation provides another angle on the question:
Accepting the general binomial coefficient as a natural extension of the integral coefficient through the Taylor series, or binomial theorem, for $(1+x)^{s-1}$ and with the finite differences $bigtriangledown^{s-1}_{n}c_n=sum_{n=0}^{infty}(-1)^n binom{s-1}{n}c_n$, Newton interpolation gives
$$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j} frac{x^j}{j!}=frac{x^{s-1}}{(s-1)!}$$
for $Real(s)>0,$ so the values of the factorial at the nonnegative integers determine uniquely the generalized factorial, or gamma function, in the right-half of the complex plane through Newton interpolation.
Then with the sequence $a_j=1=int_0^infty e^{-x} frac{x^{j}}{j!}dx$,
$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}a_j=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}1=1=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}int_0^infty e^{-x} frac{x^{j}}{j!}dx=int_0^infty e^{-x} frac{x^{s-1}}{(s-1)!}dx$.
This last integral allows the interpolation of the gamma function to be analytically continued to the left half-plane as in MSE-Q132727, so the factorial can be uniquely extended from it's values at the non-negative integers to the entire complex domain as a meromorphic function.
This generalizes in a natural way; the integral, a modified Mellin transform, can be regarded as a means to interpolate the coefficients of an exponential generating function (in this particular case, exp(x)) and extrapolate the interpolation to the whole complex plane. (See my examples in MO-Q79868 and MSE-Q32692 and try the same with the exponential generating function of the Bernoulli numbers to obtain an interpolation to the Riemann zeta function.)
From these perspectives--its dual roles as a natural interpolation of a sequence itself and in interpolating others and also as an iconic meromorphic function--the gamma function provides the "best" generalization of the factorial from the nonneg integers to the complex plane.
To more sharply connect pbrooks interest in fractional calculus and the gamma function with Quiaochu's in the beta integral, it's better to look at a natural sinc interpolation of the binomial coefficient:
Consider the fractional integro-derivative
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=FPfrac{1}{2pi i}displaystyleoint_{|z-x|=|x|}frac{z^{alpha}}{alpha!}frac{beta!}{(z-x)^{beta+1}}dz=FPdisplaystyleint_{0}^{x}frac{z^{alpha}}{alpha!}frac{(x-z)^{-beta-1}}{(-beta-1)!} dz$
$=displaystylefrac{x^{alpha-beta}}{(alpha-beta)!}$
where FP denotes a Hadamard-type finite part, $x>0$, and $alpha$ and $beta$ are real.
For $alpha>0$ and $beta<0$, the finite part is not required for the beta integral, and it can be written as
$displaystyleint_{0}^{1}frac{(1-z)^{alpha}}{alpha!}frac{z^{-beta-1}}{(-beta-1)!} dz=sum_{n=0}^{infty } (-1)^n
binom{alpha }{n}frac{1}{n-beta}frac{1}{alpha!}frac{1}{(-beta-1)!}$
$=displaystylesum_{n=0}^{infty }frac{beta!}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!}, $ or
$$displaystylesum_{n=0}^{infty }frac{1}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!beta!},$$
where use has been made of $frac{sin (pi beta)}{pi beta}=frac{1}{beta!(-beta)!}$, and $alpha$, of course, can be a positive integer. The final sinc fct. interpolation holds for $Real(alpha)>-1$ and all complex $beta$.
Euler's motivation (update July 2014):
R. Hilfer on pg. 18 of "Threefold Introduction to Fractional Derivatives" states, "Derivatives of non-integer (fractional) order motivated Euler to introduce the Gamma function ...." Euler introduced in the same reference given by Hilfer essentially
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=frac{x^{alpha-beta}}{(alpha-beta)!}$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 17 '12 at 23:08
Tom CopelandTom Copeland
1,3201024
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The MO question Samuel alludes to in his answer shows that the Euler integral definition of the gamma function implies log convexity, and, of course, it also implies the gamma functional relation $z(z-1)!=z!$ through integration by parts.
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– Tom Copeland
May 2 '12 at 21:21
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Why is imposing log convexity important? It ensures uniqueness given the other two quite natural conditions in Andrey's answer, but unique doesn't equal optimal and optimal is defined only under given criteria. In the end, pragmatism rules and "the best" turns out to mean "the most useful" given a scenario. I've given an interpolation scenario that straightforwardly gives a unique extension, directly implies the three conditions in Andrey's answer, and relates the utility of the extended factorial within this scenario to the beta and zeta fcts. and fractional calculus.
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– Tom Copeland
May 10 '12 at 1:27
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See also "Construction and physical application of the fractional calculus" by N. Wheeler.
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– Tom Copeland
Jun 18 '15 at 3:28
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See also "Fractional derivatives and special functions" by Lavoie, Osler, and Trembley.
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– Tom Copeland
Jan 1 '16 at 17:12
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See tcjpn.wordpress.com/2015/11/21/… for differential generators for $x^alpha / alpha!$ incorporating $zeta(n)$ for $n>1$.
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– Tom Copeland
Aug 15 '16 at 18:26
add a comment |
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For me another argument is the convincing one.
Consider the log of the factorial resp the gamma-function; this is for the integer arguments a sum of logarithms of the integers. Now for the interpolation of sums to fractional indexes (which is required for the gamma to noninteger arguments) there exists the concept of "indefinite summation", and the operator for that indefinite summation can be expressed by a power series. we find, that the power series for the log of the Eulerian gamma-function matches exactly that of that operator for the indefinite summation of the sum of logarithms.
I've seen this argument elsewhere; I thought it has been here at mse before (by the used "anixx") but may be it is at MO; I'm not aware of a specific literature at the moment, but I've put that heuristic in a small amateurish article on my website; in the essence the representation of that indefinite summation is fairly elementary and should be existent in older mathematical articles. See "uncompleting the gamma" pg 13 if that seems interesting.
Conclusion: the Eulerian gamma-function is "the correct one", because it is coherent with the indefinite summation-formula for the sums of consecutive logarithms.
answered Sep 25 '12 at 21:06
Gottfried HelmsGottfried Helms
23.5k24599
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7 Answers
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7 Answers
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$begingroup$
The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties
$f(1)=1$;
$f(x+1)=xf(x)$ for every $xgeq 0$;
$log f$ is a convex function.
The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function.
By the way, the gamma function is not the only meromorphic function satisfying
$$f(z+1)=z f(z),qquad f(1)=1,$$
with no zeroes and no poles other than the points $z=-n$, $n=0,1,2dots$. There is a whole family of such functions, which, in general, have the form
$$f(z)=exp{(-g(z))}frac{1}{zprodlimits_{m=1}^{infty} left(1+frac{z}{m}right)e^{-z/m}},$$
where $g(z)$ is an entire function such that
$$g(z+1)-g(z)=gamma+2kpi i,quad kinmathbb Z, $$
($gamma$ is Euler's constant). The gamma function corresponds to the simplest choice
$g(z)=gamma z$.
Edit: corrected index in the product.
edited Feb 11 at 5:34
Community♦
1
answered Aug 4 '10 at 15:41
Andrey RekaloAndrey Rekalo
6,10323641
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Also note that the first two properties already imply log convexity for positive integers, so it is reasonable to ask that an extension be log convex for positive reals. I found this more convincing than some of the infinite product inequalities (assuming these are the ones Andrey is referring to).
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– Larry Wang
Aug 4 '10 at 16:59
10
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The logarithm $log Gamma(c)$ can be thought of as a "continuous sum" $log Gamma(x) = sum_{k=1}^x k$. Is there a compelling reason why we should postulate that it's convex?
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– Greg Graviton
Aug 4 '10 at 17:31
1
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Is there a function that satisfies the first 2 properties but is not convex?
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– Dan
Apr 18 '12 at 1:26
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@Dan: Infinitely many. Just choose arbitrary (nonconvex) values for f(x) on 0 < x < 1.
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– Charles
Sep 25 '12 at 20:05
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Regarding your second point: it is true, however there is Wielandt's theorem which says that $Gamma(z)$ is unique upon adding the stipulation that $Gamma(z)$ be bounded in every vertical strip of the complex plane of unit width. This can be weakened to having at most exponential growth of base $e^{2pi}$ on a single such strip.
$endgroup$
– The_Sympathizer
Dec 6 '15 at 10:08
add a comment |
$begingroup$
The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties
$f(1)=1$;
$f(x+1)=xf(x)$ for every $xgeq 0$;
$log f$ is a convex function.
The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function.
By the way, the gamma function is not the only meromorphic function satisfying
$$f(z+1)=z f(z),qquad f(1)=1,$$
with no zeroes and no poles other than the points $z=-n$, $n=0,1,2dots$. There is a whole family of such functions, which, in general, have the form
$$f(z)=exp{(-g(z))}frac{1}{zprodlimits_{m=1}^{infty} left(1+frac{z}{m}right)e^{-z/m}},$$
where $g(z)$ is an entire function such that
$$g(z+1)-g(z)=gamma+2kpi i,quad kinmathbb Z, $$
($gamma$ is Euler's constant). The gamma function corresponds to the simplest choice
$g(z)=gamma z$.
Edit: corrected index in the product.
edited Feb 11 at 5:34
Community♦
1
answered Aug 4 '10 at 15:41
Andrey RekaloAndrey Rekalo
6,10323641
$endgroup$
$begingroup$
Also note that the first two properties already imply log convexity for positive integers, so it is reasonable to ask that an extension be log convex for positive reals. I found this more convincing than some of the infinite product inequalities (assuming these are the ones Andrey is referring to).
$endgroup$
– Larry Wang
Aug 4 '10 at 16:59
10
$begingroup$
The logarithm $log Gamma(c)$ can be thought of as a "continuous sum" $log Gamma(x) = sum_{k=1}^x k$. Is there a compelling reason why we should postulate that it's convex?
$endgroup$
– Greg Graviton
Aug 4 '10 at 17:31
1
$begingroup$
Is there a function that satisfies the first 2 properties but is not convex?
$endgroup$
– Dan
Apr 18 '12 at 1:26
$begingroup$
@Dan: Infinitely many. Just choose arbitrary (nonconvex) values for f(x) on 0 < x < 1.
$endgroup$
– Charles
Sep 25 '12 at 20:05
$begingroup$
Regarding your second point: it is true, however there is Wielandt's theorem which says that $Gamma(z)$ is unique upon adding the stipulation that $Gamma(z)$ be bounded in every vertical strip of the complex plane of unit width. This can be weakened to having at most exponential growth of base $e^{2pi}$ on a single such strip.
$endgroup$
– The_Sympathizer
Dec 6 '15 at 10:08
add a comment |
$begingroup$
The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties
$f(1)=1$;
$f(x+1)=xf(x)$ for every $xgeq 0$;
$log f$ is a convex function.
The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function.
By the way, the gamma function is not the only meromorphic function satisfying
$$f(z+1)=z f(z),qquad f(1)=1,$$
with no zeroes and no poles other than the points $z=-n$, $n=0,1,2dots$. There is a whole family of such functions, which, in general, have the form
$$f(z)=exp{(-g(z))}frac{1}{zprodlimits_{m=1}^{infty} left(1+frac{z}{m}right)e^{-z/m}},$$
where $g(z)$ is an entire function such that
$$g(z+1)-g(z)=gamma+2kpi i,quad kinmathbb Z, $$
($gamma$ is Euler's constant). The gamma function corresponds to the simplest choice
$g(z)=gamma z$.
Edit: corrected index in the product.
edited Feb 11 at 5:34
Community♦
1
answered Aug 4 '10 at 15:41
Andrey RekaloAndrey Rekalo
6,10323641
$endgroup$
The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties
$f(1)=1$;
$f(x+1)=xf(x)$ for every $xgeq 0$;
$log f$ is a convex function.
The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function.
By the way, the gamma function is not the only meromorphic function satisfying
$$f(z+1)=z f(z),qquad f(1)=1,$$
with no zeroes and no poles other than the points $z=-n$, $n=0,1,2dots$. There is a whole family of such functions, which, in general, have the form
$$f(z)=exp{(-g(z))}frac{1}{zprodlimits_{m=1}^{infty} left(1+frac{z}{m}right)e^{-z/m}},$$
where $g(z)$ is an entire function such that
$$g(z+1)-g(z)=gamma+2kpi i,quad kinmathbb Z, $$
($gamma$ is Euler's constant). The gamma function corresponds to the simplest choice
$g(z)=gamma z$.
Edit: corrected index in the product.
edited Feb 11 at 5:34
Community♦
1
answered Aug 4 '10 at 15:41
Andrey RekaloAndrey Rekalo
6,10323641
edited Feb 11 at 5:34
Community♦
1
edited Feb 11 at 5:34
Community♦
1
edited Feb 11 at 5:34
Community♦
1
1
answered Aug 4 '10 at 15:41
Andrey RekaloAndrey Rekalo
6,10323641
answered Aug 4 '10 at 15:41
Andrey RekaloAndrey Rekalo
6,10323641
answered Aug 4 '10 at 15:41
Andrey RekaloAndrey Rekalo
6,10323641
6,10323641
$begingroup$
Also note that the first two properties already imply log convexity for positive integers, so it is reasonable to ask that an extension be log convex for positive reals. I found this more convincing than some of the infinite product inequalities (assuming these are the ones Andrey is referring to).
$endgroup$
– Larry Wang
Aug 4 '10 at 16:59
10
$begingroup$
The logarithm $log Gamma(c)$ can be thought of as a "continuous sum" $log Gamma(x) = sum_{k=1}^x k$. Is there a compelling reason why we should postulate that it's convex?
$endgroup$
– Greg Graviton
Aug 4 '10 at 17:31
1
$begingroup$
Is there a function that satisfies the first 2 properties but is not convex?
$endgroup$
– Dan
Apr 18 '12 at 1:26
$begingroup$
@Dan: Infinitely many. Just choose arbitrary (nonconvex) values for f(x) on 0 < x < 1.
$endgroup$
– Charles
Sep 25 '12 at 20:05
$begingroup$
Regarding your second point: it is true, however there is Wielandt's theorem which says that $Gamma(z)$ is unique upon adding the stipulation that $Gamma(z)$ be bounded in every vertical strip of the complex plane of unit width. This can be weakened to having at most exponential growth of base $e^{2pi}$ on a single such strip.
$endgroup$
– The_Sympathizer
Dec 6 '15 at 10:08
add a comment |
$begingroup$
Also note that the first two properties already imply log convexity for positive integers, so it is reasonable to ask that an extension be log convex for positive reals. I found this more convincing than some of the infinite product inequalities (assuming these are the ones Andrey is referring to).
$endgroup$
– Larry Wang
Aug 4 '10 at 16:59
10
$begingroup$
The logarithm $log Gamma(c)$ can be thought of as a "continuous sum" $log Gamma(x) = sum_{k=1}^x k$. Is there a compelling reason why we should postulate that it's convex?
$endgroup$
– Greg Graviton
Aug 4 '10 at 17:31
1
$begingroup$
Is there a function that satisfies the first 2 properties but is not convex?
$endgroup$
– Dan
Apr 18 '12 at 1:26
$begingroup$
@Dan: Infinitely many. Just choose arbitrary (nonconvex) values for f(x) on 0 < x < 1.
$endgroup$
– Charles
Sep 25 '12 at 20:05
$begingroup$
Regarding your second point: it is true, however there is Wielandt's theorem which says that $Gamma(z)$ is unique upon adding the stipulation that $Gamma(z)$ be bounded in every vertical strip of the complex plane of unit width. This can be weakened to having at most exponential growth of base $e^{2pi}$ on a single such strip.
$endgroup$
– The_Sympathizer
Dec 6 '15 at 10:08
$begingroup$
Also note that the first two properties already imply log convexity for positive integers, so it is reasonable to ask that an extension be log convex for positive reals. I found this more convincing than some of the infinite product inequalities (assuming these are the ones Andrey is referring to).
$endgroup$
– Larry Wang
Aug 4 '10 at 16:59
$begingroup$
Also note that the first two properties already imply log convexity for positive integers, so it is reasonable to ask that an extension be log convex for positive reals. I found this more convincing than some of the infinite product inequalities (assuming these are the ones Andrey is referring to).
$endgroup$
– Larry Wang
Aug 4 '10 at 16:59
10
10
$begingroup$
The logarithm $log Gamma(c)$ can be thought of as a "continuous sum" $log Gamma(x) = sum_{k=1}^x k$. Is there a compelling reason why we should postulate that it's convex?
$endgroup$
– Greg Graviton
Aug 4 '10 at 17:31
$begingroup$
The logarithm $log Gamma(c)$ can be thought of as a "continuous sum" $log Gamma(x) = sum_{k=1}^x k$. Is there a compelling reason why we should postulate that it's convex?
$endgroup$
– Greg Graviton
Aug 4 '10 at 17:31
1
1
$begingroup$
Is there a function that satisfies the first 2 properties but is not convex?
$endgroup$
– Dan
Apr 18 '12 at 1:26
$begingroup$
Is there a function that satisfies the first 2 properties but is not convex?
$endgroup$
– Dan
Apr 18 '12 at 1:26
$begingroup$
@Dan: Infinitely many. Just choose arbitrary (nonconvex) values for f(x) on 0 < x < 1.
$endgroup$
– Charles
Sep 25 '12 at 20:05
$begingroup$
@Dan: Infinitely many. Just choose arbitrary (nonconvex) values for f(x) on 0 < x < 1.
$endgroup$
– Charles
Sep 25 '12 at 20:05
$begingroup$
Regarding your second point: it is true, however there is Wielandt's theorem which says that $Gamma(z)$ is unique upon adding the stipulation that $Gamma(z)$ be bounded in every vertical strip of the complex plane of unit width. This can be weakened to having at most exponential growth of base $e^{2pi}$ on a single such strip.
$endgroup$
– The_Sympathizer
Dec 6 '15 at 10:08
$begingroup$
Regarding your second point: it is true, however there is Wielandt's theorem which says that $Gamma(z)$ is unique upon adding the stipulation that $Gamma(z)$ be bounded in every vertical strip of the complex plane of unit width. This can be weakened to having at most exponential growth of base $e^{2pi}$ on a single such strip.
$endgroup$
– The_Sympathizer
Dec 6 '15 at 10:08
add a comment |
$begingroup$
Actually there are other (less-frequently) used extensions to the factorial, with different properties from the gamma function which may be desirable in some contexts.
Euler's Gamma Function
Hadamard's Gamma function
Luschny's factorial function
See here for more information.
answered Aug 4 '10 at 16:11
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,86342943
$endgroup$
add a comment |
$begingroup$
Actually there are other (less-frequently) used extensions to the factorial, with different properties from the gamma function which may be desirable in some contexts.
Euler's Gamma Function
Hadamard's Gamma function
Luschny's factorial function
See here for more information.
answered Aug 4 '10 at 16:11
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,86342943
$endgroup$
add a comment |
$begingroup$
Actually there are other (less-frequently) used extensions to the factorial, with different properties from the gamma function which may be desirable in some contexts.
Euler's Gamma Function
Hadamard's Gamma function
Luschny's factorial function
See here for more information.
answered Aug 4 '10 at 16:11
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,86342943
$endgroup$
Actually there are other (less-frequently) used extensions to the factorial, with different properties from the gamma function which may be desirable in some contexts.
Euler's Gamma Function
Hadamard's Gamma function
Luschny's factorial function
See here for more information.
answered Aug 4 '10 at 16:11
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,86342943
edited Jun 26 '17 at 12:52
answered Aug 4 '10 at 16:11
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,86342943
answered Aug 4 '10 at 16:11
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,86342943
answered Aug 4 '10 at 16:11
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,86342943
5,86342943
add a comment |
add a comment |
$begingroup$
For whatever reason, Nature (by which I mean integrals) seems to prefer the Gamma function as the "correct" substitute for the factorial in various integrals, which seems to come more or less from its integral definition. For example, for non-negative integers $a, b$, it's not hard to show (and there's a really nice probabilistic argument) that
$displaystyle int_{0}^1 t^a (1 - t)^b , dt = frac{a! b!}{(a+b+1)!}.$
For (non-negative?) real values of $a$ and $b$ the correct generalization is
$displaystyle int_0^1 t^a (1 - t)^b , dt = frac{Gamma(a+1) Gamma(b+1)}{Gamma(a+b+2)}.$
And, of course, integrals are important, so the Gamma function must also be important. For example, the Gamma function appears in the general formula for the volume of an n-sphere. But the result that, for me, really forces us to take the Gamma function seriously is its appearance in the functional equation for the Riemann zeta function.
answered Aug 4 '10 at 16:55
Qiaochu YuanQiaochu Yuan
281k32592938
$endgroup$
1
$begingroup$
Why is that the "correct generalization"?
$endgroup$
– Neil G
Aug 28 '10 at 9:30
$begingroup$
Also, please see my first question -- I guess you know the answer!
$endgroup$
– Neil G
Aug 28 '10 at 9:32
$begingroup$
@Neil G: I wasn't sure what you were asking at first, but now I see. It could be that there exists some other function G such that G(a-1) G(b-1)/G(a+b) is also the answer. I don't know how to rule out this possibility, but again, the result that really makes it clear to me that the Gamma function is important is the last one.
$endgroup$
– Qiaochu Yuan
Aug 28 '10 at 12:22
1
$begingroup$
@QiaochuYuan: In the end you say a good reason for you is the connection to the zeta function. But I feel the connection via the functional equation is on the same level as the connection via the Mellin transform representation, which you might write as $zeta(s) = dfrac{int_0^inftyfrac{x^{s}}{e^x-1}frac{{mathrm d}x}{x}}{int_0^infty frac{x^{s}}{e^x-0}frac{{mathrm d}x}{x}}$. So in the end, I'd rather argue that it's important because of it's connection to the hom between addition and multiplication - the exponential function. After all, gamma = int monomial exp. What do you say?
$endgroup$
– Nikolaj-K
Feb 6 '15 at 19:24
1
$begingroup$
@MathematicsStudent: pick $a + b + 1$ random numbers in the interval $[0, 1]$, and compute the probability that the last such random number is greater than $a$ of the others and less than $b$ of the others, in two ways.
$endgroup$
– Qiaochu Yuan
Aug 1 '17 at 21:55
|
show 4 more comments
$begingroup$
For whatever reason, Nature (by which I mean integrals) seems to prefer the Gamma function as the "correct" substitute for the factorial in various integrals, which seems to come more or less from its integral definition. For example, for non-negative integers $a, b$, it's not hard to show (and there's a really nice probabilistic argument) that
$displaystyle int_{0}^1 t^a (1 - t)^b , dt = frac{a! b!}{(a+b+1)!}.$
For (non-negative?) real values of $a$ and $b$ the correct generalization is
$displaystyle int_0^1 t^a (1 - t)^b , dt = frac{Gamma(a+1) Gamma(b+1)}{Gamma(a+b+2)}.$
And, of course, integrals are important, so the Gamma function must also be important. For example, the Gamma function appears in the general formula for the volume of an n-sphere. But the result that, for me, really forces us to take the Gamma function seriously is its appearance in the functional equation for the Riemann zeta function.
answered Aug 4 '10 at 16:55
Qiaochu YuanQiaochu Yuan
281k32592938
$endgroup$
1
$begingroup$
Why is that the "correct generalization"?
$endgroup$
– Neil G
Aug 28 '10 at 9:30
$begingroup$
Also, please see my first question -- I guess you know the answer!
$endgroup$
– Neil G
Aug 28 '10 at 9:32
$begingroup$
@Neil G: I wasn't sure what you were asking at first, but now I see. It could be that there exists some other function G such that G(a-1) G(b-1)/G(a+b) is also the answer. I don't know how to rule out this possibility, but again, the result that really makes it clear to me that the Gamma function is important is the last one.
$endgroup$
– Qiaochu Yuan
Aug 28 '10 at 12:22
1
$begingroup$
@QiaochuYuan: In the end you say a good reason for you is the connection to the zeta function. But I feel the connection via the functional equation is on the same level as the connection via the Mellin transform representation, which you might write as $zeta(s) = dfrac{int_0^inftyfrac{x^{s}}{e^x-1}frac{{mathrm d}x}{x}}{int_0^infty frac{x^{s}}{e^x-0}frac{{mathrm d}x}{x}}$. So in the end, I'd rather argue that it's important because of it's connection to the hom between addition and multiplication - the exponential function. After all, gamma = int monomial exp. What do you say?
$endgroup$
– Nikolaj-K
Feb 6 '15 at 19:24
1
$begingroup$
@MathematicsStudent: pick $a + b + 1$ random numbers in the interval $[0, 1]$, and compute the probability that the last such random number is greater than $a$ of the others and less than $b$ of the others, in two ways.
$endgroup$
– Qiaochu Yuan
Aug 1 '17 at 21:55
|
show 4 more comments
$begingroup$
For whatever reason, Nature (by which I mean integrals) seems to prefer the Gamma function as the "correct" substitute for the factorial in various integrals, which seems to come more or less from its integral definition. For example, for non-negative integers $a, b$, it's not hard to show (and there's a really nice probabilistic argument) that
$displaystyle int_{0}^1 t^a (1 - t)^b , dt = frac{a! b!}{(a+b+1)!}.$
For (non-negative?) real values of $a$ and $b$ the correct generalization is
$displaystyle int_0^1 t^a (1 - t)^b , dt = frac{Gamma(a+1) Gamma(b+1)}{Gamma(a+b+2)}.$
And, of course, integrals are important, so the Gamma function must also be important. For example, the Gamma function appears in the general formula for the volume of an n-sphere. But the result that, for me, really forces us to take the Gamma function seriously is its appearance in the functional equation for the Riemann zeta function.
answered Aug 4 '10 at 16:55
Qiaochu YuanQiaochu Yuan
281k32592938
$endgroup$
For whatever reason, Nature (by which I mean integrals) seems to prefer the Gamma function as the "correct" substitute for the factorial in various integrals, which seems to come more or less from its integral definition. For example, for non-negative integers $a, b$, it's not hard to show (and there's a really nice probabilistic argument) that
$displaystyle int_{0}^1 t^a (1 - t)^b , dt = frac{a! b!}{(a+b+1)!}.$
For (non-negative?) real values of $a$ and $b$ the correct generalization is
$displaystyle int_0^1 t^a (1 - t)^b , dt = frac{Gamma(a+1) Gamma(b+1)}{Gamma(a+b+2)}.$
And, of course, integrals are important, so the Gamma function must also be important. For example, the Gamma function appears in the general formula for the volume of an n-sphere. But the result that, for me, really forces us to take the Gamma function seriously is its appearance in the functional equation for the Riemann zeta function.
answered Aug 4 '10 at 16:55
Qiaochu YuanQiaochu Yuan
281k32592938
edited Mar 23 '12 at 16:25
answered Aug 4 '10 at 16:55
Qiaochu YuanQiaochu Yuan
281k32592938
answered Aug 4 '10 at 16:55
Qiaochu YuanQiaochu Yuan
281k32592938
answered Aug 4 '10 at 16:55
Qiaochu YuanQiaochu Yuan
281k32592938
281k32592938
1
$begingroup$
Why is that the "correct generalization"?
$endgroup$
– Neil G
Aug 28 '10 at 9:30
$begingroup$
Also, please see my first question -- I guess you know the answer!
$endgroup$
– Neil G
Aug 28 '10 at 9:32
$begingroup$
@Neil G: I wasn't sure what you were asking at first, but now I see. It could be that there exists some other function G such that G(a-1) G(b-1)/G(a+b) is also the answer. I don't know how to rule out this possibility, but again, the result that really makes it clear to me that the Gamma function is important is the last one.
$endgroup$
– Qiaochu Yuan
Aug 28 '10 at 12:22
1
$begingroup$
@QiaochuYuan: In the end you say a good reason for you is the connection to the zeta function. But I feel the connection via the functional equation is on the same level as the connection via the Mellin transform representation, which you might write as $zeta(s) = dfrac{int_0^inftyfrac{x^{s}}{e^x-1}frac{{mathrm d}x}{x}}{int_0^infty frac{x^{s}}{e^x-0}frac{{mathrm d}x}{x}}$. So in the end, I'd rather argue that it's important because of it's connection to the hom between addition and multiplication - the exponential function. After all, gamma = int monomial exp. What do you say?
$endgroup$
– Nikolaj-K
Feb 6 '15 at 19:24
1
$begingroup$
@MathematicsStudent: pick $a + b + 1$ random numbers in the interval $[0, 1]$, and compute the probability that the last such random number is greater than $a$ of the others and less than $b$ of the others, in two ways.
$endgroup$
– Qiaochu Yuan
Aug 1 '17 at 21:55
|
show 4 more comments
1
$begingroup$
Why is that the "correct generalization"?
$endgroup$
– Neil G
Aug 28 '10 at 9:30
$begingroup$
Also, please see my first question -- I guess you know the answer!
$endgroup$
– Neil G
Aug 28 '10 at 9:32
$begingroup$
@Neil G: I wasn't sure what you were asking at first, but now I see. It could be that there exists some other function G such that G(a-1) G(b-1)/G(a+b) is also the answer. I don't know how to rule out this possibility, but again, the result that really makes it clear to me that the Gamma function is important is the last one.
$endgroup$
– Qiaochu Yuan
Aug 28 '10 at 12:22
1
$begingroup$
@QiaochuYuan: In the end you say a good reason for you is the connection to the zeta function. But I feel the connection via the functional equation is on the same level as the connection via the Mellin transform representation, which you might write as $zeta(s) = dfrac{int_0^inftyfrac{x^{s}}{e^x-1}frac{{mathrm d}x}{x}}{int_0^infty frac{x^{s}}{e^x-0}frac{{mathrm d}x}{x}}$. So in the end, I'd rather argue that it's important because of it's connection to the hom between addition and multiplication - the exponential function. After all, gamma = int monomial exp. What do you say?
$endgroup$
– Nikolaj-K
Feb 6 '15 at 19:24
1
$begingroup$
@MathematicsStudent: pick $a + b + 1$ random numbers in the interval $[0, 1]$, and compute the probability that the last such random number is greater than $a$ of the others and less than $b$ of the others, in two ways.
$endgroup$
– Qiaochu Yuan
Aug 1 '17 at 21:55
1
1
$begingroup$
Why is that the "correct generalization"?
$endgroup$
– Neil G
Aug 28 '10 at 9:30
$begingroup$
Why is that the "correct generalization"?
$endgroup$
– Neil G
Aug 28 '10 at 9:30
$begingroup$
Also, please see my first question -- I guess you know the answer!
$endgroup$
– Neil G
Aug 28 '10 at 9:32
$begingroup$
Also, please see my first question -- I guess you know the answer!
$endgroup$
– Neil G
Aug 28 '10 at 9:32
$begingroup$
@Neil G: I wasn't sure what you were asking at first, but now I see. It could be that there exists some other function G such that G(a-1) G(b-1)/G(a+b) is also the answer. I don't know how to rule out this possibility, but again, the result that really makes it clear to me that the Gamma function is important is the last one.
$endgroup$
– Qiaochu Yuan
Aug 28 '10 at 12:22
$begingroup$
@Neil G: I wasn't sure what you were asking at first, but now I see. It could be that there exists some other function G such that G(a-1) G(b-1)/G(a+b) is also the answer. I don't know how to rule out this possibility, but again, the result that really makes it clear to me that the Gamma function is important is the last one.
$endgroup$
– Qiaochu Yuan
Aug 28 '10 at 12:22
1
1
$begingroup$
@QiaochuYuan: In the end you say a good reason for you is the connection to the zeta function. But I feel the connection via the functional equation is on the same level as the connection via the Mellin transform representation, which you might write as $zeta(s) = dfrac{int_0^inftyfrac{x^{s}}{e^x-1}frac{{mathrm d}x}{x}}{int_0^infty frac{x^{s}}{e^x-0}frac{{mathrm d}x}{x}}$. So in the end, I'd rather argue that it's important because of it's connection to the hom between addition and multiplication - the exponential function. After all, gamma = int monomial exp. What do you say?
$endgroup$
– Nikolaj-K
Feb 6 '15 at 19:24
$begingroup$
@QiaochuYuan: In the end you say a good reason for you is the connection to the zeta function. But I feel the connection via the functional equation is on the same level as the connection via the Mellin transform representation, which you might write as $zeta(s) = dfrac{int_0^inftyfrac{x^{s}}{e^x-1}frac{{mathrm d}x}{x}}{int_0^infty frac{x^{s}}{e^x-0}frac{{mathrm d}x}{x}}$. So in the end, I'd rather argue that it's important because of it's connection to the hom between addition and multiplication - the exponential function. After all, gamma = int monomial exp. What do you say?
$endgroup$
– Nikolaj-K
Feb 6 '15 at 19:24
1
1
$begingroup$
@MathematicsStudent: pick $a + b + 1$ random numbers in the interval $[0, 1]$, and compute the probability that the last such random number is greater than $a$ of the others and less than $b$ of the others, in two ways.
$endgroup$
– Qiaochu Yuan
Aug 1 '17 at 21:55
$begingroup$
@MathematicsStudent: pick $a + b + 1$ random numbers in the interval $[0, 1]$, and compute the probability that the last such random number is greater than $a$ of the others and less than $b$ of the others, in two ways.
$endgroup$
– Qiaochu Yuan
Aug 1 '17 at 21:55
|
show 4 more comments
$begingroup$
This is a comment posted as an answer for lack of reputation.
Following Qiaochu Yuan, the gamma function shows up in the functional equation of the zeta function as the factor in the Euler product corresponding to the "prime at infinity", and it occurs there as the Mellin transform of some gaussian function. (Gaussian functions occur in turn as eigenvectors of the Fourier transform.)
This is at least as old as Tate's thesis, and a possible reference is Weil's Basic Number Theory.
EDIT. Artin was one of the first people to popularize the log-convexity property of the gamma function (see his book on the function in question), and also perhaps the first mathematician to fully understand this Euler-factor-at-infinity aspect of the same function (he was Tate's thesis advisor). I thought his name had to be mentioned in a discussion about the gamma function.
answered Aug 5 '10 at 10:44
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
$endgroup$
$begingroup$
It is the Mellin transform of $e^{-t}$ that appears in the functional equation much more than "the best interpolation of $n!$"
$endgroup$
– reuns
Jan 24 at 0:55
add a comment |
$begingroup$
This is a comment posted as an answer for lack of reputation.
Following Qiaochu Yuan, the gamma function shows up in the functional equation of the zeta function as the factor in the Euler product corresponding to the "prime at infinity", and it occurs there as the Mellin transform of some gaussian function. (Gaussian functions occur in turn as eigenvectors of the Fourier transform.)
This is at least as old as Tate's thesis, and a possible reference is Weil's Basic Number Theory.
EDIT. Artin was one of the first people to popularize the log-convexity property of the gamma function (see his book on the function in question), and also perhaps the first mathematician to fully understand this Euler-factor-at-infinity aspect of the same function (he was Tate's thesis advisor). I thought his name had to be mentioned in a discussion about the gamma function.
answered Aug 5 '10 at 10:44
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
$endgroup$
$begingroup$
It is the Mellin transform of $e^{-t}$ that appears in the functional equation much more than "the best interpolation of $n!$"
$endgroup$
– reuns
Jan 24 at 0:55
add a comment |
$begingroup$
This is a comment posted as an answer for lack of reputation.
Following Qiaochu Yuan, the gamma function shows up in the functional equation of the zeta function as the factor in the Euler product corresponding to the "prime at infinity", and it occurs there as the Mellin transform of some gaussian function. (Gaussian functions occur in turn as eigenvectors of the Fourier transform.)
This is at least as old as Tate's thesis, and a possible reference is Weil's Basic Number Theory.
EDIT. Artin was one of the first people to popularize the log-convexity property of the gamma function (see his book on the function in question), and also perhaps the first mathematician to fully understand this Euler-factor-at-infinity aspect of the same function (he was Tate's thesis advisor). I thought his name had to be mentioned in a discussion about the gamma function.
answered Aug 5 '10 at 10:44
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
$endgroup$
This is a comment posted as an answer for lack of reputation.
Following Qiaochu Yuan, the gamma function shows up in the functional equation of the zeta function as the factor in the Euler product corresponding to the "prime at infinity", and it occurs there as the Mellin transform of some gaussian function. (Gaussian functions occur in turn as eigenvectors of the Fourier transform.)
This is at least as old as Tate's thesis, and a possible reference is Weil's Basic Number Theory.
EDIT. Artin was one of the first people to popularize the log-convexity property of the gamma function (see his book on the function in question), and also perhaps the first mathematician to fully understand this Euler-factor-at-infinity aspect of the same function (he was Tate's thesis advisor). I thought his name had to be mentioned in a discussion about the gamma function.
answered Aug 5 '10 at 10:44
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
edited Aug 5 '10 at 11:13
answered Aug 5 '10 at 10:44
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
answered Aug 5 '10 at 10:44
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
answered Aug 5 '10 at 10:44
Pierre-Yves GaillardPierre-Yves Gaillard
13.4k23184
13.4k23184
$begingroup$
It is the Mellin transform of $e^{-t}$ that appears in the functional equation much more than "the best interpolation of $n!$"
$endgroup$
– reuns
Jan 24 at 0:55
add a comment |
$begingroup$
It is the Mellin transform of $e^{-t}$ that appears in the functional equation much more than "the best interpolation of $n!$"
$endgroup$
– reuns
Jan 24 at 0:55
$begingroup$
It is the Mellin transform of $e^{-t}$ that appears in the functional equation much more than "the best interpolation of $n!$"
$endgroup$
– reuns
Jan 24 at 0:55
$begingroup$
It is the Mellin transform of $e^{-t}$ that appears in the functional equation much more than "the best interpolation of $n!$"
$endgroup$
– reuns
Jan 24 at 0:55
add a comment |
$begingroup$
Wielandt's theorem says that the gamma-function is the only function $f$ that satisfies the properties:
- $f(1)=1$
- $f(z+1)=zf(z)$ for all $z>0$
- $f(z)$ is analytic for $operatorname{Re}z>0$
- $f(z)$ is bounded for $1leq operatorname{Re}zleq 2$
(See also the related MathOverflow thread Importance of Log Convexity of the Gamma Function, where I learned about the above theorem.)
edited Apr 13 '17 at 12:58
Community♦
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answered Aug 5 '10 at 9:45
SamuelSamuel
4,79412033
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add a comment |
$begingroup$
Wielandt's theorem says that the gamma-function is the only function $f$ that satisfies the properties:
- $f(1)=1$
- $f(z+1)=zf(z)$ for all $z>0$
- $f(z)$ is analytic for $operatorname{Re}z>0$
- $f(z)$ is bounded for $1leq operatorname{Re}zleq 2$
(See also the related MathOverflow thread Importance of Log Convexity of the Gamma Function, where I learned about the above theorem.)
edited Apr 13 '17 at 12:58
Community♦
1
answered Aug 5 '10 at 9:45
SamuelSamuel
4,79412033
$endgroup$
add a comment |
$begingroup$
Wielandt's theorem says that the gamma-function is the only function $f$ that satisfies the properties:
- $f(1)=1$
- $f(z+1)=zf(z)$ for all $z>0$
- $f(z)$ is analytic for $operatorname{Re}z>0$
- $f(z)$ is bounded for $1leq operatorname{Re}zleq 2$
(See also the related MathOverflow thread Importance of Log Convexity of the Gamma Function, where I learned about the above theorem.)
edited Apr 13 '17 at 12:58
Community♦
1
answered Aug 5 '10 at 9:45
SamuelSamuel
4,79412033
$endgroup$
Wielandt's theorem says that the gamma-function is the only function $f$ that satisfies the properties:
- $f(1)=1$
- $f(z+1)=zf(z)$ for all $z>0$
- $f(z)$ is analytic for $operatorname{Re}z>0$
- $f(z)$ is bounded for $1leq operatorname{Re}zleq 2$
(See also the related MathOverflow thread Importance of Log Convexity of the Gamma Function, where I learned about the above theorem.)
edited Apr 13 '17 at 12:58
Community♦
1
answered Aug 5 '10 at 9:45
SamuelSamuel
4,79412033
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
1
answered Aug 5 '10 at 9:45
SamuelSamuel
4,79412033
answered Aug 5 '10 at 9:45
SamuelSamuel
4,79412033
answered Aug 5 '10 at 9:45
SamuelSamuel
4,79412033
4,79412033
add a comment |
add a comment |
$begingroup$
Looking for a difference that makes a difference.
Flipping the gamma function and looking at Newton interpolation provides another angle on the question:
Accepting the general binomial coefficient as a natural extension of the integral coefficient through the Taylor series, or binomial theorem, for $(1+x)^{s-1}$ and with the finite differences $bigtriangledown^{s-1}_{n}c_n=sum_{n=0}^{infty}(-1)^n binom{s-1}{n}c_n$, Newton interpolation gives
$$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j} frac{x^j}{j!}=frac{x^{s-1}}{(s-1)!}$$
for $Real(s)>0,$ so the values of the factorial at the nonnegative integers determine uniquely the generalized factorial, or gamma function, in the right-half of the complex plane through Newton interpolation.
Then with the sequence $a_j=1=int_0^infty e^{-x} frac{x^{j}}{j!}dx$,
$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}a_j=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}1=1=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}int_0^infty e^{-x} frac{x^{j}}{j!}dx=int_0^infty e^{-x} frac{x^{s-1}}{(s-1)!}dx$.
This last integral allows the interpolation of the gamma function to be analytically continued to the left half-plane as in MSE-Q132727, so the factorial can be uniquely extended from it's values at the non-negative integers to the entire complex domain as a meromorphic function.
This generalizes in a natural way; the integral, a modified Mellin transform, can be regarded as a means to interpolate the coefficients of an exponential generating function (in this particular case, exp(x)) and extrapolate the interpolation to the whole complex plane. (See my examples in MO-Q79868 and MSE-Q32692 and try the same with the exponential generating function of the Bernoulli numbers to obtain an interpolation to the Riemann zeta function.)
From these perspectives--its dual roles as a natural interpolation of a sequence itself and in interpolating others and also as an iconic meromorphic function--the gamma function provides the "best" generalization of the factorial from the nonneg integers to the complex plane.
To more sharply connect pbrooks interest in fractional calculus and the gamma function with Quiaochu's in the beta integral, it's better to look at a natural sinc interpolation of the binomial coefficient:
Consider the fractional integro-derivative
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=FPfrac{1}{2pi i}displaystyleoint_{|z-x|=|x|}frac{z^{alpha}}{alpha!}frac{beta!}{(z-x)^{beta+1}}dz=FPdisplaystyleint_{0}^{x}frac{z^{alpha}}{alpha!}frac{(x-z)^{-beta-1}}{(-beta-1)!} dz$
$=displaystylefrac{x^{alpha-beta}}{(alpha-beta)!}$
where FP denotes a Hadamard-type finite part, $x>0$, and $alpha$ and $beta$ are real.
For $alpha>0$ and $beta<0$, the finite part is not required for the beta integral, and it can be written as
$displaystyleint_{0}^{1}frac{(1-z)^{alpha}}{alpha!}frac{z^{-beta-1}}{(-beta-1)!} dz=sum_{n=0}^{infty } (-1)^n
binom{alpha }{n}frac{1}{n-beta}frac{1}{alpha!}frac{1}{(-beta-1)!}$
$=displaystylesum_{n=0}^{infty }frac{beta!}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!}, $ or
$$displaystylesum_{n=0}^{infty }frac{1}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!beta!},$$
where use has been made of $frac{sin (pi beta)}{pi beta}=frac{1}{beta!(-beta)!}$, and $alpha$, of course, can be a positive integer. The final sinc fct. interpolation holds for $Real(alpha)>-1$ and all complex $beta$.
Euler's motivation (update July 2014):
R. Hilfer on pg. 18 of "Threefold Introduction to Fractional Derivatives" states, "Derivatives of non-integer (fractional) order motivated Euler to introduce the Gamma function ...." Euler introduced in the same reference given by Hilfer essentially
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=frac{x^{alpha-beta}}{(alpha-beta)!}$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 17 '12 at 23:08
Tom CopelandTom Copeland
1,3201024
$endgroup$
$begingroup$
The MO question Samuel alludes to in his answer shows that the Euler integral definition of the gamma function implies log convexity, and, of course, it also implies the gamma functional relation $z(z-1)!=z!$ through integration by parts.
$endgroup$
– Tom Copeland
May 2 '12 at 21:21
$begingroup$
Why is imposing log convexity important? It ensures uniqueness given the other two quite natural conditions in Andrey's answer, but unique doesn't equal optimal and optimal is defined only under given criteria. In the end, pragmatism rules and "the best" turns out to mean "the most useful" given a scenario. I've given an interpolation scenario that straightforwardly gives a unique extension, directly implies the three conditions in Andrey's answer, and relates the utility of the extended factorial within this scenario to the beta and zeta fcts. and fractional calculus.
$endgroup$
– Tom Copeland
May 10 '12 at 1:27
$begingroup$
See also "Construction and physical application of the fractional calculus" by N. Wheeler.
$endgroup$
– Tom Copeland
Jun 18 '15 at 3:28
$begingroup$
See also "Fractional derivatives and special functions" by Lavoie, Osler, and Trembley.
$endgroup$
– Tom Copeland
Jan 1 '16 at 17:12
$begingroup$
See tcjpn.wordpress.com/2015/11/21/… for differential generators for $x^alpha / alpha!$ incorporating $zeta(n)$ for $n>1$.
$endgroup$
– Tom Copeland
Aug 15 '16 at 18:26
add a comment |
$begingroup$
Looking for a difference that makes a difference.
Flipping the gamma function and looking at Newton interpolation provides another angle on the question:
Accepting the general binomial coefficient as a natural extension of the integral coefficient through the Taylor series, or binomial theorem, for $(1+x)^{s-1}$ and with the finite differences $bigtriangledown^{s-1}_{n}c_n=sum_{n=0}^{infty}(-1)^n binom{s-1}{n}c_n$, Newton interpolation gives
$$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j} frac{x^j}{j!}=frac{x^{s-1}}{(s-1)!}$$
for $Real(s)>0,$ so the values of the factorial at the nonnegative integers determine uniquely the generalized factorial, or gamma function, in the right-half of the complex plane through Newton interpolation.
Then with the sequence $a_j=1=int_0^infty e^{-x} frac{x^{j}}{j!}dx$,
$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}a_j=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}1=1=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}int_0^infty e^{-x} frac{x^{j}}{j!}dx=int_0^infty e^{-x} frac{x^{s-1}}{(s-1)!}dx$.
This last integral allows the interpolation of the gamma function to be analytically continued to the left half-plane as in MSE-Q132727, so the factorial can be uniquely extended from it's values at the non-negative integers to the entire complex domain as a meromorphic function.
This generalizes in a natural way; the integral, a modified Mellin transform, can be regarded as a means to interpolate the coefficients of an exponential generating function (in this particular case, exp(x)) and extrapolate the interpolation to the whole complex plane. (See my examples in MO-Q79868 and MSE-Q32692 and try the same with the exponential generating function of the Bernoulli numbers to obtain an interpolation to the Riemann zeta function.)
From these perspectives--its dual roles as a natural interpolation of a sequence itself and in interpolating others and also as an iconic meromorphic function--the gamma function provides the "best" generalization of the factorial from the nonneg integers to the complex plane.
To more sharply connect pbrooks interest in fractional calculus and the gamma function with Quiaochu's in the beta integral, it's better to look at a natural sinc interpolation of the binomial coefficient:
Consider the fractional integro-derivative
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=FPfrac{1}{2pi i}displaystyleoint_{|z-x|=|x|}frac{z^{alpha}}{alpha!}frac{beta!}{(z-x)^{beta+1}}dz=FPdisplaystyleint_{0}^{x}frac{z^{alpha}}{alpha!}frac{(x-z)^{-beta-1}}{(-beta-1)!} dz$
$=displaystylefrac{x^{alpha-beta}}{(alpha-beta)!}$
where FP denotes a Hadamard-type finite part, $x>0$, and $alpha$ and $beta$ are real.
For $alpha>0$ and $beta<0$, the finite part is not required for the beta integral, and it can be written as
$displaystyleint_{0}^{1}frac{(1-z)^{alpha}}{alpha!}frac{z^{-beta-1}}{(-beta-1)!} dz=sum_{n=0}^{infty } (-1)^n
binom{alpha }{n}frac{1}{n-beta}frac{1}{alpha!}frac{1}{(-beta-1)!}$
$=displaystylesum_{n=0}^{infty }frac{beta!}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!}, $ or
$$displaystylesum_{n=0}^{infty }frac{1}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!beta!},$$
where use has been made of $frac{sin (pi beta)}{pi beta}=frac{1}{beta!(-beta)!}$, and $alpha$, of course, can be a positive integer. The final sinc fct. interpolation holds for $Real(alpha)>-1$ and all complex $beta$.
Euler's motivation (update July 2014):
R. Hilfer on pg. 18 of "Threefold Introduction to Fractional Derivatives" states, "Derivatives of non-integer (fractional) order motivated Euler to introduce the Gamma function ...." Euler introduced in the same reference given by Hilfer essentially
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=frac{x^{alpha-beta}}{(alpha-beta)!}$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 17 '12 at 23:08
Tom CopelandTom Copeland
1,3201024
$endgroup$
$begingroup$
The MO question Samuel alludes to in his answer shows that the Euler integral definition of the gamma function implies log convexity, and, of course, it also implies the gamma functional relation $z(z-1)!=z!$ through integration by parts.
$endgroup$
– Tom Copeland
May 2 '12 at 21:21
$begingroup$
Why is imposing log convexity important? It ensures uniqueness given the other two quite natural conditions in Andrey's answer, but unique doesn't equal optimal and optimal is defined only under given criteria. In the end, pragmatism rules and "the best" turns out to mean "the most useful" given a scenario. I've given an interpolation scenario that straightforwardly gives a unique extension, directly implies the three conditions in Andrey's answer, and relates the utility of the extended factorial within this scenario to the beta and zeta fcts. and fractional calculus.
$endgroup$
– Tom Copeland
May 10 '12 at 1:27
$begingroup$
See also "Construction and physical application of the fractional calculus" by N. Wheeler.
$endgroup$
– Tom Copeland
Jun 18 '15 at 3:28
$begingroup$
See also "Fractional derivatives and special functions" by Lavoie, Osler, and Trembley.
$endgroup$
– Tom Copeland
Jan 1 '16 at 17:12
$begingroup$
See tcjpn.wordpress.com/2015/11/21/… for differential generators for $x^alpha / alpha!$ incorporating $zeta(n)$ for $n>1$.
$endgroup$
– Tom Copeland
Aug 15 '16 at 18:26
add a comment |
$begingroup$
Looking for a difference that makes a difference.
Flipping the gamma function and looking at Newton interpolation provides another angle on the question:
Accepting the general binomial coefficient as a natural extension of the integral coefficient through the Taylor series, or binomial theorem, for $(1+x)^{s-1}$ and with the finite differences $bigtriangledown^{s-1}_{n}c_n=sum_{n=0}^{infty}(-1)^n binom{s-1}{n}c_n$, Newton interpolation gives
$$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j} frac{x^j}{j!}=frac{x^{s-1}}{(s-1)!}$$
for $Real(s)>0,$ so the values of the factorial at the nonnegative integers determine uniquely the generalized factorial, or gamma function, in the right-half of the complex plane through Newton interpolation.
Then with the sequence $a_j=1=int_0^infty e^{-x} frac{x^{j}}{j!}dx$,
$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}a_j=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}1=1=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}int_0^infty e^{-x} frac{x^{j}}{j!}dx=int_0^infty e^{-x} frac{x^{s-1}}{(s-1)!}dx$.
This last integral allows the interpolation of the gamma function to be analytically continued to the left half-plane as in MSE-Q132727, so the factorial can be uniquely extended from it's values at the non-negative integers to the entire complex domain as a meromorphic function.
This generalizes in a natural way; the integral, a modified Mellin transform, can be regarded as a means to interpolate the coefficients of an exponential generating function (in this particular case, exp(x)) and extrapolate the interpolation to the whole complex plane. (See my examples in MO-Q79868 and MSE-Q32692 and try the same with the exponential generating function of the Bernoulli numbers to obtain an interpolation to the Riemann zeta function.)
From these perspectives--its dual roles as a natural interpolation of a sequence itself and in interpolating others and also as an iconic meromorphic function--the gamma function provides the "best" generalization of the factorial from the nonneg integers to the complex plane.
To more sharply connect pbrooks interest in fractional calculus and the gamma function with Quiaochu's in the beta integral, it's better to look at a natural sinc interpolation of the binomial coefficient:
Consider the fractional integro-derivative
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=FPfrac{1}{2pi i}displaystyleoint_{|z-x|=|x|}frac{z^{alpha}}{alpha!}frac{beta!}{(z-x)^{beta+1}}dz=FPdisplaystyleint_{0}^{x}frac{z^{alpha}}{alpha!}frac{(x-z)^{-beta-1}}{(-beta-1)!} dz$
$=displaystylefrac{x^{alpha-beta}}{(alpha-beta)!}$
where FP denotes a Hadamard-type finite part, $x>0$, and $alpha$ and $beta$ are real.
For $alpha>0$ and $beta<0$, the finite part is not required for the beta integral, and it can be written as
$displaystyleint_{0}^{1}frac{(1-z)^{alpha}}{alpha!}frac{z^{-beta-1}}{(-beta-1)!} dz=sum_{n=0}^{infty } (-1)^n
binom{alpha }{n}frac{1}{n-beta}frac{1}{alpha!}frac{1}{(-beta-1)!}$
$=displaystylesum_{n=0}^{infty }frac{beta!}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!}, $ or
$$displaystylesum_{n=0}^{infty }frac{1}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!beta!},$$
where use has been made of $frac{sin (pi beta)}{pi beta}=frac{1}{beta!(-beta)!}$, and $alpha$, of course, can be a positive integer. The final sinc fct. interpolation holds for $Real(alpha)>-1$ and all complex $beta$.
Euler's motivation (update July 2014):
R. Hilfer on pg. 18 of "Threefold Introduction to Fractional Derivatives" states, "Derivatives of non-integer (fractional) order motivated Euler to introduce the Gamma function ...." Euler introduced in the same reference given by Hilfer essentially
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=frac{x^{alpha-beta}}{(alpha-beta)!}$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 17 '12 at 23:08
Tom CopelandTom Copeland
1,3201024
$endgroup$
Looking for a difference that makes a difference.
Flipping the gamma function and looking at Newton interpolation provides another angle on the question:
Accepting the general binomial coefficient as a natural extension of the integral coefficient through the Taylor series, or binomial theorem, for $(1+x)^{s-1}$ and with the finite differences $bigtriangledown^{s-1}_{n}c_n=sum_{n=0}^{infty}(-1)^n binom{s-1}{n}c_n$, Newton interpolation gives
$$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j} frac{x^j}{j!}=frac{x^{s-1}}{(s-1)!}$$
for $Real(s)>0,$ so the values of the factorial at the nonnegative integers determine uniquely the generalized factorial, or gamma function, in the right-half of the complex plane through Newton interpolation.
Then with the sequence $a_j=1=int_0^infty e^{-x} frac{x^{j}}{j!}dx$,
$bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}a_j=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}1=1=bigtriangledown^{s-1}_{n} bigtriangledown^{n}_{j}int_0^infty e^{-x} frac{x^{j}}{j!}dx=int_0^infty e^{-x} frac{x^{s-1}}{(s-1)!}dx$.
This last integral allows the interpolation of the gamma function to be analytically continued to the left half-plane as in MSE-Q132727, so the factorial can be uniquely extended from it's values at the non-negative integers to the entire complex domain as a meromorphic function.
This generalizes in a natural way; the integral, a modified Mellin transform, can be regarded as a means to interpolate the coefficients of an exponential generating function (in this particular case, exp(x)) and extrapolate the interpolation to the whole complex plane. (See my examples in MO-Q79868 and MSE-Q32692 and try the same with the exponential generating function of the Bernoulli numbers to obtain an interpolation to the Riemann zeta function.)
From these perspectives--its dual roles as a natural interpolation of a sequence itself and in interpolating others and also as an iconic meromorphic function--the gamma function provides the "best" generalization of the factorial from the nonneg integers to the complex plane.
To more sharply connect pbrooks interest in fractional calculus and the gamma function with Quiaochu's in the beta integral, it's better to look at a natural sinc interpolation of the binomial coefficient:
Consider the fractional integro-derivative
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=FPfrac{1}{2pi i}displaystyleoint_{|z-x|=|x|}frac{z^{alpha}}{alpha!}frac{beta!}{(z-x)^{beta+1}}dz=FPdisplaystyleint_{0}^{x}frac{z^{alpha}}{alpha!}frac{(x-z)^{-beta-1}}{(-beta-1)!} dz$
$=displaystylefrac{x^{alpha-beta}}{(alpha-beta)!}$
where FP denotes a Hadamard-type finite part, $x>0$, and $alpha$ and $beta$ are real.
For $alpha>0$ and $beta<0$, the finite part is not required for the beta integral, and it can be written as
$displaystyleint_{0}^{1}frac{(1-z)^{alpha}}{alpha!}frac{z^{-beta-1}}{(-beta-1)!} dz=sum_{n=0}^{infty } (-1)^n
binom{alpha }{n}frac{1}{n-beta}frac{1}{alpha!}frac{1}{(-beta-1)!}$
$=displaystylesum_{n=0}^{infty }frac{beta!}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!}, $ or
$$displaystylesum_{n=0}^{infty }frac{1}{(alpha-n)! n!}frac{sin (pi (beta -n))}{pi (beta -n)}=frac{1}{(alpha-beta)!beta!},$$
where use has been made of $frac{sin (pi beta)}{pi beta}=frac{1}{beta!(-beta)!}$, and $alpha$, of course, can be a positive integer. The final sinc fct. interpolation holds for $Real(alpha)>-1$ and all complex $beta$.
Euler's motivation (update July 2014):
R. Hilfer on pg. 18 of "Threefold Introduction to Fractional Derivatives" states, "Derivatives of non-integer (fractional) order motivated Euler to introduce the Gamma function ...." Euler introduced in the same reference given by Hilfer essentially
$displaystylefrac{d^{beta}}{dx^beta}frac{x^{alpha}}{alpha!}=frac{x^{alpha-beta}}{(alpha-beta)!}$.
edited Apr 13 '17 at 12:58
Community♦
1
answered Apr 17 '12 at 23:08
Tom CopelandTom Copeland
1,3201024
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
1
answered Apr 17 '12 at 23:08
Tom CopelandTom Copeland
1,3201024
answered Apr 17 '12 at 23:08
Tom CopelandTom Copeland
1,3201024
answered Apr 17 '12 at 23:08
Tom CopelandTom Copeland
1,3201024
1,3201024
$begingroup$
The MO question Samuel alludes to in his answer shows that the Euler integral definition of the gamma function implies log convexity, and, of course, it also implies the gamma functional relation $z(z-1)!=z!$ through integration by parts.
$endgroup$
– Tom Copeland
May 2 '12 at 21:21
$begingroup$
Why is imposing log convexity important? It ensures uniqueness given the other two quite natural conditions in Andrey's answer, but unique doesn't equal optimal and optimal is defined only under given criteria. In the end, pragmatism rules and "the best" turns out to mean "the most useful" given a scenario. I've given an interpolation scenario that straightforwardly gives a unique extension, directly implies the three conditions in Andrey's answer, and relates the utility of the extended factorial within this scenario to the beta and zeta fcts. and fractional calculus.
$endgroup$
– Tom Copeland
May 10 '12 at 1:27
$begingroup$
See also "Construction and physical application of the fractional calculus" by N. Wheeler.
$endgroup$
– Tom Copeland
Jun 18 '15 at 3:28
$begingroup$
See also "Fractional derivatives and special functions" by Lavoie, Osler, and Trembley.
$endgroup$
– Tom Copeland
Jan 1 '16 at 17:12
$begingroup$
See tcjpn.wordpress.com/2015/11/21/… for differential generators for $x^alpha / alpha!$ incorporating $zeta(n)$ for $n>1$.
$endgroup$
– Tom Copeland
Aug 15 '16 at 18:26
add a comment |
$begingroup$
The MO question Samuel alludes to in his answer shows that the Euler integral definition of the gamma function implies log convexity, and, of course, it also implies the gamma functional relation $z(z-1)!=z!$ through integration by parts.
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– Tom Copeland
May 2 '12 at 21:21
$begingroup$
Why is imposing log convexity important? It ensures uniqueness given the other two quite natural conditions in Andrey's answer, but unique doesn't equal optimal and optimal is defined only under given criteria. In the end, pragmatism rules and "the best" turns out to mean "the most useful" given a scenario. I've given an interpolation scenario that straightforwardly gives a unique extension, directly implies the three conditions in Andrey's answer, and relates the utility of the extended factorial within this scenario to the beta and zeta fcts. and fractional calculus.
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– Tom Copeland
May 10 '12 at 1:27
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See also "Construction and physical application of the fractional calculus" by N. Wheeler.
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– Tom Copeland
Jun 18 '15 at 3:28
$begingroup$
See also "Fractional derivatives and special functions" by Lavoie, Osler, and Trembley.
$endgroup$
– Tom Copeland
Jan 1 '16 at 17:12
$begingroup$
See tcjpn.wordpress.com/2015/11/21/… for differential generators for $x^alpha / alpha!$ incorporating $zeta(n)$ for $n>1$.
$endgroup$
– Tom Copeland
Aug 15 '16 at 18:26
$begingroup$
The MO question Samuel alludes to in his answer shows that the Euler integral definition of the gamma function implies log convexity, and, of course, it also implies the gamma functional relation $z(z-1)!=z!$ through integration by parts.
$endgroup$
– Tom Copeland
May 2 '12 at 21:21
$begingroup$
The MO question Samuel alludes to in his answer shows that the Euler integral definition of the gamma function implies log convexity, and, of course, it also implies the gamma functional relation $z(z-1)!=z!$ through integration by parts.
$endgroup$
– Tom Copeland
May 2 '12 at 21:21
$begingroup$
Why is imposing log convexity important? It ensures uniqueness given the other two quite natural conditions in Andrey's answer, but unique doesn't equal optimal and optimal is defined only under given criteria. In the end, pragmatism rules and "the best" turns out to mean "the most useful" given a scenario. I've given an interpolation scenario that straightforwardly gives a unique extension, directly implies the three conditions in Andrey's answer, and relates the utility of the extended factorial within this scenario to the beta and zeta fcts. and fractional calculus.
$endgroup$
– Tom Copeland
May 10 '12 at 1:27
$begingroup$
Why is imposing log convexity important? It ensures uniqueness given the other two quite natural conditions in Andrey's answer, but unique doesn't equal optimal and optimal is defined only under given criteria. In the end, pragmatism rules and "the best" turns out to mean "the most useful" given a scenario. I've given an interpolation scenario that straightforwardly gives a unique extension, directly implies the three conditions in Andrey's answer, and relates the utility of the extended factorial within this scenario to the beta and zeta fcts. and fractional calculus.
$endgroup$
– Tom Copeland
May 10 '12 at 1:27
$begingroup$
See also "Construction and physical application of the fractional calculus" by N. Wheeler.
$endgroup$
– Tom Copeland
Jun 18 '15 at 3:28
$begingroup$
See also "Construction and physical application of the fractional calculus" by N. Wheeler.
$endgroup$
– Tom Copeland
Jun 18 '15 at 3:28
$begingroup$
See also "Fractional derivatives and special functions" by Lavoie, Osler, and Trembley.
$endgroup$
– Tom Copeland
Jan 1 '16 at 17:12
$begingroup$
See also "Fractional derivatives and special functions" by Lavoie, Osler, and Trembley.
$endgroup$
– Tom Copeland
Jan 1 '16 at 17:12
$begingroup$
See tcjpn.wordpress.com/2015/11/21/… for differential generators for $x^alpha / alpha!$ incorporating $zeta(n)$ for $n>1$.
$endgroup$
– Tom Copeland
Aug 15 '16 at 18:26
$begingroup$
See tcjpn.wordpress.com/2015/11/21/… for differential generators for $x^alpha / alpha!$ incorporating $zeta(n)$ for $n>1$.
$endgroup$
– Tom Copeland
Aug 15 '16 at 18:26
add a comment |
$begingroup$
For me another argument is the convincing one.
Consider the log of the factorial resp the gamma-function; this is for the integer arguments a sum of logarithms of the integers. Now for the interpolation of sums to fractional indexes (which is required for the gamma to noninteger arguments) there exists the concept of "indefinite summation", and the operator for that indefinite summation can be expressed by a power series. we find, that the power series for the log of the Eulerian gamma-function matches exactly that of that operator for the indefinite summation of the sum of logarithms.
I've seen this argument elsewhere; I thought it has been here at mse before (by the used "anixx") but may be it is at MO; I'm not aware of a specific literature at the moment, but I've put that heuristic in a small amateurish article on my website; in the essence the representation of that indefinite summation is fairly elementary and should be existent in older mathematical articles. See "uncompleting the gamma" pg 13 if that seems interesting.
Conclusion: the Eulerian gamma-function is "the correct one", because it is coherent with the indefinite summation-formula for the sums of consecutive logarithms.
answered Sep 25 '12 at 21:06
Gottfried HelmsGottfried Helms
23.5k24599
$endgroup$
add a comment |
$begingroup$
For me another argument is the convincing one.
Consider the log of the factorial resp the gamma-function; this is for the integer arguments a sum of logarithms of the integers. Now for the interpolation of sums to fractional indexes (which is required for the gamma to noninteger arguments) there exists the concept of "indefinite summation", and the operator for that indefinite summation can be expressed by a power series. we find, that the power series for the log of the Eulerian gamma-function matches exactly that of that operator for the indefinite summation of the sum of logarithms.
I've seen this argument elsewhere; I thought it has been here at mse before (by the used "anixx") but may be it is at MO; I'm not aware of a specific literature at the moment, but I've put that heuristic in a small amateurish article on my website; in the essence the representation of that indefinite summation is fairly elementary and should be existent in older mathematical articles. See "uncompleting the gamma" pg 13 if that seems interesting.
Conclusion: the Eulerian gamma-function is "the correct one", because it is coherent with the indefinite summation-formula for the sums of consecutive logarithms.
answered Sep 25 '12 at 21:06
Gottfried HelmsGottfried Helms
23.5k24599
$endgroup$
add a comment |
$begingroup$
For me another argument is the convincing one.
Consider the log of the factorial resp the gamma-function; this is for the integer arguments a sum of logarithms of the integers. Now for the interpolation of sums to fractional indexes (which is required for the gamma to noninteger arguments) there exists the concept of "indefinite summation", and the operator for that indefinite summation can be expressed by a power series. we find, that the power series for the log of the Eulerian gamma-function matches exactly that of that operator for the indefinite summation of the sum of logarithms.
I've seen this argument elsewhere; I thought it has been here at mse before (by the used "anixx") but may be it is at MO; I'm not aware of a specific literature at the moment, but I've put that heuristic in a small amateurish article on my website; in the essence the representation of that indefinite summation is fairly elementary and should be existent in older mathematical articles. See "uncompleting the gamma" pg 13 if that seems interesting.
Conclusion: the Eulerian gamma-function is "the correct one", because it is coherent with the indefinite summation-formula for the sums of consecutive logarithms.
answered Sep 25 '12 at 21:06
Gottfried HelmsGottfried Helms
23.5k24599
$endgroup$
For me another argument is the convincing one.
Consider the log of the factorial resp the gamma-function; this is for the integer arguments a sum of logarithms of the integers. Now for the interpolation of sums to fractional indexes (which is required for the gamma to noninteger arguments) there exists the concept of "indefinite summation", and the operator for that indefinite summation can be expressed by a power series. we find, that the power series for the log of the Eulerian gamma-function matches exactly that of that operator for the indefinite summation of the sum of logarithms.
I've seen this argument elsewhere; I thought it has been here at mse before (by the used "anixx") but may be it is at MO; I'm not aware of a specific literature at the moment, but I've put that heuristic in a small amateurish article on my website; in the essence the representation of that indefinite summation is fairly elementary and should be existent in older mathematical articles. See "uncompleting the gamma" pg 13 if that seems interesting.
Conclusion: the Eulerian gamma-function is "the correct one", because it is coherent with the indefinite summation-formula for the sums of consecutive logarithms.
answered Sep 25 '12 at 21:06
Gottfried HelmsGottfried Helms
23.5k24599
answered Sep 25 '12 at 21:06
Gottfried HelmsGottfried Helms
23.5k24599
answered Sep 25 '12 at 21:06
Gottfried HelmsGottfried Helms
23.5k24599
answered Sep 25 '12 at 21:06
Gottfried HelmsGottfried Helms
23.5k24599
23.5k24599
add a comment |
add a comment |
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$begingroup$
Good question. I'm not very familiar with the process of analytic continuation, but I suspect the answer lies there.
$endgroup$
– Noldorin
Aug 4 '10 at 15:05
5
$begingroup$
@Noldorin: I think analytic continuations are only unique if you have a (locally?) dense set of points, which positive integers are not. After all, you can have an entire function that is zero at every positive integer. Euler's Gamma is used in some important probability distributions, and also in the Riemann Zeta function, but not many first-year calculus students will care about this.
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– Larry Wang
Aug 4 '10 at 16:28
4
$begingroup$
It's also used in the fractional calculus (fractional differentiation and integration -- see en.wikipedia.org/wiki/Fractional_calculus), which is why I wanted some more intuitive understanding of its virtues.
$endgroup$
– pbrooks
Aug 4 '10 at 17:06