do you know about the "reciprocal gamma function"??
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- Опубликовано: 26 авг 2023
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The reciprocal funcion of the Γ function is literally his invert symbol L
Lmao thats good
lol - it looks like a reflection along the real axis or any line parallel to real axis
So, Waluigi and Luigi?
(x)⅃
(x)⌝
13:38 smooth
that is indeed a good place to
And that's a good place to s
It reminds me of trainee logician who could not say no and his master who could not say
A slightly different technique: Every L(n) except L(1) can be converted to a contour integral in the same way as here (actually this should work for L(1) too, but there one has to be a bit more careful to show that the integral along the half-circle vanishes in the limit). Then one notes a striking similarity between the integrand and that of the Cauchy integral formula, so that after pulling out a factor of i^n from the denominator:
L(n) = i/n! * (-i)^n * ((n-1)'th derivative of exp(1 + iz) at z=i) = 1/n!
It should be i/(n-1)!*i^(-n)
And thus L(n)=1/(n-1)! L(n+1)=1/n!
I don't think you needed the induction part at all, just evaluate the residue by performing the series expansion of the exponential around t=i instead of evaluating the residue with a limit. The exponential is even already set up in that way since it's evaluated for (1+it). That way you already get 1/n! straight from the coefficients.
I guess this is a simple form for the Bromwich contour integral, at least when S is large enough.
What’s really interesting about the reciprocal gamma function is that it’s pretty much the sequence of partial products of the unit fractions, instead of natural numbers
Great vid Michael!
michael you’re a legend, thank you for all the interesting and informative videos
Use residue from the beginning we can easily solve the integration, just care for the convergence
For the Gamma integral, I think it should be pointed out that there is convergence only if the real part of s > 0.
You can then use recursion to extend, but the original integral only converges if re{s} >0.
1/Gamma is an entire function.
Is that inverse Laplace transform of 1/s^n
Right! L(s) is simply the inverse Laplace transform when integration is over line 'Re(z) = 1' and substitution 'z = 1 + it' is performed
@@GiornoYoshikage In the definition, the inverse Laplace transform is the Bromwich's integral. Can z=it?
I have always felt that the reciprocal is the more "natural" representation of the gamma function. If you look at the graph of the function it has all these infinities, especially at the negative integers. These become neat zeros in the reciprocal, and the function itself is "entire" over the complex numbers.
17:07
Bro there is no stop 😂
And that's a good place to...
That is a good place
Hi. Please consider making a video on the inverse Gamma function, where e.g. x! = 12; solve for x. Thanks in advance
Pretty nice function, integration by part for recursive formula's always works! I'm a bit confused by L(1) cause it doesn't feel like it necessary converges how it's written, but 0! is not a big deal, also I think the analytic continuation of that function must be unique like it's unique for the factorial making it 1/Gamma probably?
The function is clearly analytic, but to show that it coincides with 1/Γ(s), I believe you need agreement in an open set, (rather than on a countably infinite set of isolated points, which is what was shown here)
@@leostein128if I may ask: how do you get capital gamma to appear as text in your comment?
You can use this to define transcendental derivatives because it completes the Taylor series for all n values. Is there an application for this? Could you use the Ith derivative coupled with quantum calculus and the Schrödinger equation to reveal hidden properties of the quantum wave equation?
12:21 i don't know why the module of the denominator is bounded by R²
Expand (1 + i t)^2 with t = R cos(theta) + i R sin(theta), take the modulus, and you'll see that it's always bounded by R^2.
|1+it|² = |1-t_i + it_r|² = (1-t_i)² + (t_r)²
I feel a super complicated representation of e^x emerging.
Nice, the Hankel integral respresentation of Γ. Almost... The Hankel contour is a bit different than the one here.
Complex residues are fascinating tool
I have a nice challenging number theory problem for you: Find all n natural such as 5^n-121 is a perfect square. I believe it uses complex numbers, but I have no idea how to prove that n=3 is the only solution.
Is there any motivation for the construction of the integral, or did someone just happened to guess at the correct form (which feels unlikely)?
In the integration by parts, why does the u*v part end up being zero? What does e^it mean when t goes towards positive or negative infinity?
He sort of tackles that. On it's own, e^it doesn't converge: it's modulus is always 1 but the phase oscillates for ever. However, when taking the denominator (which scales as 1/t^2) into account, then we can see the modulus of the whole ratio goes to 0. There's only one number whose modulus is 0 - regardless of its phase - and that's 0. So the ratio as a whole converges, even if the numerator is stuck in a cycle.
Hi,
17:08 : missing "stop".
"We can pull an eye out the denominator" Poor denominator.
Is L(z) = 1/Γ(z) for all z, or only for integers?
I would conjecture so, but you'd have to deal with appropriately evaluating (1+it)^s when s isn't a positive integer.
According to the wiki page "Reciprocal gamma function", it is for all z.
@12:20 why is the denominator bounded by R^2?
He didn't do that carefully and I think what he said is literally wrong, but it's anyway on the "order of" R^2 which is good enough and leads to the value 0 for this integral in the limit. Points t along that arc are of the form t = Re^{ia} where a is some angle. In absolute value, the expression (1 + it)^2 is bounded below by (|it| - 1)^2 = (R-1)^2 and bounded above by (|it| + 1)^2 = (R + 1)^2, by applications of the triangle inequality. So the integral overall is bounded above by integral_{Gamma_R} e/(R-1)^2 which tends to 0 anyway as R --> infinity.
@@toddtrimble2555 thanks can you pleas explain the inequalities here using the triangle inequality. Also why is |it|=R here?
@@Happy_Abe We have |iRe^{ia}| = |i| times R times |e^{ia}| (using |xy| = |x| |y|), where |i| = 1 = |e^{ia}| because both i and e^{ia} lie on the unit circle (use Euler's formula for the latter). The two triangle inequalities are |x| - |y|
@@toddtrimble2555 honestly amazing explanation, thank you so much. Perfectly clear!
Bello!
beyond me - as usual
KUM-BA-YAH!
I like it, sir
The Gamma Function is the most intesting function in math... after the zeta function ofc. I found a formula for 1.k!, where k HAS TO BE AN INTEGER: it equals \frac{2}{\pi}\int_{0}^{\pi}e^{\cos\left(t
ight)}\cos\left(\sin t
ight)\cos\left(xt
ight)dt
and that's a good place to s--