Mod-04 Lec-09 Analytic continuation and the gamma function (Part I)

[@SpeedcoreDancecore]

I'm a french student in Superior and Special Mathematics, and apart from some notation differences those lectures are very good and instructive for me !

[@yimoawanardo]

Some naming differences in vectors are quite funny. They call le produit scalaire as the dot product. How they came up with that name, I'm quite not sure lol

[@MTahir-tn7tc]

Thats India bro!!

[@aniketsaha7455]

@@yimoawanardo lol you call dot product produit scalaria..we also call it scalar product..but dot products goes handy and easy..so we call it dot product

[@qwertyyoe3334]

Special maths

[@ndmaphy]

​@@yimoawanardo Word Scalar product is more informal than dot product bcz it is a map from V×V to Field and element of fields are called scalars so is this term.

[@elamvaluthis7268]

Beyond doubt he is a great professor indeed.He is not giving scope to find fault some find fault which is trivial.

[@wontpower]

I've been watching this video ever since I started learning calculus, and now everything is starting to make sense

[@AngioloHuaman]

the clarity of the professor is awesome

[@peterhall6656]

This is an exceptionally good presentation of analytic continuation in the context of the gamma function. Excellent teacher.

[@akarshnarasimhan]

Minor board mistake: While writing down the Stirling formula, Prof. Balakrishnan wrote O(n²), when he really meant O(1/n²), although it should be clear from his explanation that it is indeed O(1/n²)

[@carlossaa3193]

+akarshnarasimhan Indeed, a small mistake was made by the brilliant dr. Balakrishnan. The compact correction factor is e^( q/(12(n+1))) where 0

[@NikhilBhagyaRaj]

Man, that's actually considered as a Function, so no matter the variables are, the arbitrary function can be anything unless you have the same expansion

[@bachirblackers7299]

This is Maths i do love more than any thing ... Thanks much prof .

[@NomenNominandum]

Brilliant lectures !

[@dubey_ji]

amazing lectures . #nptelhrd you guys are the best .

[@Aman-tf8bt]

what an outsanding lecture of such an interesting topic.

[@cro-magnon8771]

At 18:46, I think there is a point mistake in integration by parts. Its great lecture. Thank u for making it free.

[@wronski11]

Excellent lecture and lecturer !

[@peteriring1721]

Perfect. Amazing. I love it. Fantastic.

[@GrandMoffTarkinsTeaDispenser]

Such a nice lecture, superb. Thank you very much for posting it.

[@rahmatkhan3982]

you are doing great job.keep up the good work..

[@imnimbusy2885]

The ending otro for this lecture slaps!

[@sergiosebastiani6045]

WOW. Thats Clear !

[@B.A.Gondal]

My Windows 7 Calculator can calculate up to 3248.77 factorial :)

[@pramod120895]

It is still very low for the order of avagadro number

[@sergiosebastiani6045]

Around minute 15. Is it right to say that immaginary part of z doesn't affect at all convergence? In fact the oscillations help the convergence so maybe its better to say that immaginary part helps convergence. The domain of convergence that we define step by step (Re z > 0, Re z > -1 .... ) is where we are sure that the integral converges but is it correct to say that those are the domain of the gamma representations? The oscillation determined by Im z 0 they drastically modify the behavoir of the integral so maybe we are making such assumptions on the convergence domains just because we don't have a well known theory for the intergration of oscillating functions ?

[@mrmo3340]

And around 5:4 Isn't the graph of t^n have to started from one.??

[@neelav4731]

excellent sir

[@wabisabi4944]

sir u r excellent your lecture is excellent bt movement of camera is creating disturbance to understand properly..so please try to fix the camera which can cover the blackboard constantly...it's moving according to the movement of teacher.

[@Pseudophysics1.5]

Exactly

[@bensnodgrass6548]

Interesting lecture but SURELY (dangerous word in maths i know) the part around 20min just isn't rigourous. When he makes the new gamma function with the 1/z at the start although this new function appears to be defined for z>-1 rather than z>0, but that's only because he has assumed that z>0 to make it work, so to say that it's defined for z>-1 too doesn't seem rigourous. Perhaps I've missed something but it appears he's just ignoring a subtlety to get it to work

[@hbatmit]

He specifically says "except for z=0" a couple of times, and says there's a pole at z=0 a few times. What is the issue with that in terms of rigor?

[@MushyMellowMenu]

A bit late, but hopefully I'll be able to clear this up a bit. If not for you, then maybe for someone else. The assumption of Re(z)>0 is used to prove, that the new formula is equal to the old one for all z on which the old formula is also defined (which is Re(z)>0). So for this domain, both formulae will define the same function. The new formula being also valid for -10 (as the old function was not defined anywhere else). This is the analytical continuation of our old function. The non-rigorous part is, that he also calls and denotes the analytical continuation Gamma, even though they are not strictly the same. Again, I hope this was helpful. I did my best, though writing on a tiny mobile youtube comment-box probably isn't helping.

[@parmalchouriya2365]

Best lecture

[@rabbit-ku1bn]

Muryokusho

[@KumarNishantRanjanSinha]

Brilliant !

[@lifejourney9308]

Can sme explaine to me why he refered to Zwhile calculate the first term of eq at 20:52

[@qwertyyoe3334]

Doer spek engrilsh