Ramanujan's Master Theorem: One to Rule Them All

One Theorem to RULE them all,
One Theorem to FIND them,
One Theorem to BRING them all,
And in the Darkness, BIND them.

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[GandalfTheWise]

I'd never heard of this before. Very interesting and an easily followed derivation. I'm not sure the video zooms are worth the extra editing effort. The choice of colors, consistent & legible handwriting, generally good parallel lines of writing, and clean arrows and boxes really carries these videos. They are easier to read and follow than many computer text videos and much easier to follow than most handwritten math channels. These videos are hard to improve on with regard to quality. Clean, elegant, easy to follow, and interesting variety of topics; I personally cannot ask for more. The only thing I've not seen on the chalkboard is some of the tricks us old-timers used to do for emphasis and variety like edge of the chalk shading and getting dotted lines by "pushing" on the chalk at the correct angle with the chalk pointed slightly forward so it "skips" along.

[TheLowstef]

The editor is having fun with these lately and I wholeheartedly approve.

[eatingnsleeping]

11:30 How do you integrate over the operator u? Spectral integration?

[skibelo]

The editing of this channel has got a lot better lately. Well done! Nice little poem in the description today.😊

[MooImABunny]

I'm a physicist and even I had a little pearl clutching moment when you set u = sigma*x
Like sure, I know this integral should obviously be from 0 to infty, what else would it be, but this is incredibly sketchy lol

[ManuelRacle]

Perhaps I really didn't understand the argument well, but around 8:00 it didn't make any sense for me to pass from "x^n . sigma^n" to "(x . sigma)^n". To my eyes it was totally cheating with notations: x^n denotes the nth power of x while sigma^n denotes applying sigma n times... I can't understand it works when the sigma operator is not simply "multiply by a constant" in other terms a_i is a geometric progression.

[rv]

I'm just noting that that integral is the Fourier transform of f(x) on the multiplicative Lie group of positive real numbers: the Haar measure is dx/x and the character is x^t, so integral of x^tf(x) against the measure dx/x.

[qingninghuo]

It seems that something is quite wrong in this video. At 6:21, the integral depends on f(x), which depends on a_n for n >= 0. But the end result at 12:00 depends on a single value of a_{-t}. So if this is correct, there must be some relationship between the a_n, which has not been clarified.

[jonathanbeeson]

Thank you Dr. Penn. For me this was a good companion video to the one by Maths 505 where he uses Ramanujan's Master Theorem to evaluate the first two Fresnel Integrals.

[normanstevens]

The "function" sigma is not a function and is not even well defined. if a_n is equal to a_m but a_n+1 does not equal a_m+1 then sigma maps the value a_n to two different values. Similarly with the inverse of sigma.

[digxx]

I see that sigma is a linear shift operator, but when making the change of variables u=x*sigma, u should be operator valued, so how can u be treated as a real number with limits 0 and infinity?

[davis.yoshida]

Can you explain why it doesn't cause problems to replace x^n sigma^n with (x sigma)^n? It seems like sigma and x should certainly not commute

[Nikolas_Davis]

Ooh, this reminds me of the "insanity with a method in it" us particle physicists routinely pull with quantum / thermal field theory 😛

As the saying goes, the only math you need for field theory is the gaussian integral (but with all stops pulled).

[rtheben]

question: I didn't get the substitution u=x*sigma, It seems to me u becomes an operator itself, how can you integrate it from 0 to +inf? thanks

[bennyloodts]

Very cool stuff! I have the feeling there are even more tricks in that area of Ramanujan.

[jplikesmaths]

Michael Penn is slowly becoming my favourite math channel

[TheMemesofDestruction]

Thank you so much for this Professor and Team Penn!!

[wyboo]

the sigma function reminds me of another function i used when i was playing around with a "generating vector" (a vector in an infinite-dimensional vector space where the e_n component is the a_n term in the sequence) called the left shift. the left shift is a linear transformation that brings e_n to e_(n+1)

[davidmitchell]

There is one minor quibble here. The reversal of the integration and summation requires that the a x is convergent. From the alternating.series theorem we know this is the case if ax tends to zero. In the worked example the ax term tends to zero so the theorem applies. In other cases it may not hold.
The worked example is nice as the result can be computed by hand and the result checked.

[tolberthobson]

VERY COOL. I've never heard of this concept and its pretty cool how it can be applied to solve otherwise complicated integrals.