Cauchy Integral Formula

Proof of the famous Cauchy’s integral formula, which is the quintessential theorem that makes complex analysis work! For example, from this you can deduce Liouville’s Theorem which says that a bounded holomorphic function must be constant. The proof itself is very neat and analysis-y Enjoy!

[@mathunt1130]

Interestingly enough, the Cauchy integral formula is the basis for many numerical methods for the computation of many fully nonlinear water wave profiles. It's a very very nice technique.

[@drpeyam]

Interesting!!!

[@eduardodanielfarfanduran6346]

You are excellent Dr. Peyam

[@wompastompa3692]

Put me in, Cauchy. I'm ready to play, today.

[@ffggddss]

Look at me, I can be Center of the contour of integration! [ I fear I've just ruined a perfectly fine John Fogarty tune :-( ] Fred

[@ricardoguzman5014]

There's one drawback to Liouville's Theorem. A person with dyslexia might think they are reading about a city in Kentucky.

[@l.thiggins3406]

Great video Peyam, thanks for the proof!

[@foreachepsilon]

Proof by wishful thinking is so true.

[@danielaorozco9995]

Fave theorem by fave mathematician by fave person 😍😍😍😍 cant get any better than this

[@mustafakhedr8977]

Very cool! Excited to take complex analysis next semester!

[@TheNachoesuncapo]

I have to wait an entire year :(

[@shandyverdyo7688]

Same here :). I can't wait for next semester.

[@mathunt1130]

Using geometric algebra, it's possible to generalise Cauchy's integral theorem to higher dimensions.

[@alexmiernfnt]

Great job!

[@Myrslokstok]

Can't you prove the epsilon delta stuff of Weierstrass, thanks! 😃

[@channelnamechannel]

thank you!

[@MrRyanroberson1]

Have you done any videos on statistics and distributions? Things like making use of that Gaussian integral you did a dozen times. I heard somewhere that for any random* distribution f(x) that having multiple instances of f(x) will immediately produce the Gaussian. I really feel suspicious about that, but I can't really prove any of it

[@Matixcubix]

Nice video one of my favorite formulas. I note that on 9:19 you need |dz|.

[@shandyverdyo7688]

Dr... please to make a video about green function...

[@pedrocusinato1743]

Hey Peyam, using some induction on Cauchy's integral formula we can express the n'th derivative of a function f as an integral that depends on f and n. Can't we use this formula to express fractional derivatives?

[@drpeyam]

There’s a video on that

[@pedrocusinato1743]

@@drpeyam Nice, do u have the link?

[@faith3174]

eeee this is probably my favourite theorem/formula

[@cedricp.4941]

How do you prove that the integral I independant of the path ? This is key, since it requires that f is holomorphic (you only used continuous here)

[@arthurhenriquecaixeta9291]

You can just take an arbitrary simple (with no intersections) closed path C around z_0, and then construct an abitrarily small circle around z_0 entirely contained in the interior of C. After that, you construc two line segments in the interior of C, linking the small circle to C, obtaining two simple and closed paths. By Cauchy-Goursat's theorem, the integral on the both paths you just constructed should be zero. Draw a picture and you'll be able to find out yourself :P

[@cedricp.4941]

@@arthurhenriquecaixeta9291 I know that, but it doesn't make sense to use Cauchy's theorem to prove the more elementary Cauchy Integral formula !

[@arthurhenriquecaixeta9291]

At 12:18 you say that the integral of one is 2 pi R. But since the function 1 is analytic, and the path is simple and closed, that integral should be zero, shouldn't it?

[@giraculum9981]

That's true when you're weighting the integral with a dot product, like with a work integral. But in this case, the integral doesn't care about the relative direction of the path and the field. It's just adding up a bunch of 1's, which sums to form the length of the path.

[@Matixcubix]

Because he forgot to consider the absolute value of dz, you are integrating |dz| thus R*dtheta that over 2pi gives 2pi R

[@arthurhenriquecaixeta9291]

@@Matixcubix yeah, that makes more sense now. Thanks :P

[@tarunpurohit6522]

Wow u r my favorite

[@jacks.4390]

Can somebody please tell me what am I missing? Because I don't see how this proof works at all. What we want to show is that the formula holds for a fixed contour; when we are shrinking R in the integral that we want to show is small, we are also shrinking R in the integral where we actually want to have to original contour.

[@Mazsi1201]

it was assumed in the proof that the integral is independent of C as long as C is a simple connected path containing z0. So shrinking R does not change the integral. A lot of people pointed out that this assumption is not obvious and so I am expecting a followup video explaining it. Until that you should just accept it I guess

[@jeremy.N]

Interesting Video I have yet to learn more about Cauchy's work, so this was a helpful Video. btw. Do you have an e-mail?

[@alexanderbasler6259]

Dear Dr. Peyam, I have a remark concerning your proof. In your epsilon-delta argument, you assume epsilon stays constant and can hence be taken out of the integral. I believe this reasoning, however, for a general continuous function is erroneous. For a fixed R, which you assume in order to integrate along the boundary of the circle, the epsilon, the degree of precision with which f(z) approximates f(z_0), may vary because of the definition of a continuous function. This is similar to the fact that for a fixed epsilon, the radius R we need to consider may vary, though in this case the problem could easily be resolved by taking the infimum over all R. Since R is fixed, however, and epsilon varies with respect to z, we cannot simply integrate epsilon like a constant, nor is there an easy fix to the problem. Your reasoning works perfectly fine for uniformly continuous functions, which have the property that the epsilon does indeed stay constant for a fixed R and varying z. Am I missing something? Did I misunderstand a central concept? I would be very glad if you could enlighten me. Liebe Grüsse aus der Schweiz!

[@drpeyam]

By definition of epsilon it is fixed (that’s why you always start by saying let epsilon be given), it’s delta that depends on all the other parameters

[@alexanderbasler6259]

@@drpeyam But then if epsilon remains fixed, R probably varies for different z so that we cannot integrate along the border of a circle with radius R since R is not fixed, or can we?

[@drpeyam]

No, I think R is fixed, it’s the radius of the circle

[@Mazsi1201]

Only continuity at z0 was used, so there was only a single delta for any fixed epsilon. No need for uniform continuity, as only one continous point was needed to be examined

[@gonzalezm244]

This would’ve been nice last week for my Complex Analysis Midterm 😂

[@gonzalezm244]

Could you please do a video on residue theory? Thank you :)

[@gonzalezm244]

Awesome video btw!

[@rigorless6330]

14:34 i still struggle with these kinds of proofs... why is it ok to say "as epsilon goes to 0, these things are equal, therefore: they are equal"? wouldnt you have to put a lim ε->0 out front to make it true?

[@MisterSarcastic]

Once we show that a quantity is within epsilon of one another, we implicitly assumed that epsilon goes to zero, we just don’t bother writing it because we all understand that’s what’s going to happen (mathematicians being lazy and all that). Also keep in mind that epsilon can take on any value as well (For every ε > 0), the case where it goes to zero implies that the two numbers are equal to one another.

[@rigorless6330]

Mister Sarcastic ok but, what’s the problem with just saying “let the difference between f(z0) and f(z) be epsilon. Since epsilon goes to 0, the difference is 0. Therefore they are the same” I feel like there’s something I’m missing here

[@jacks.4390]

@@rigorless6330 Well you would never write “let the difference between f(z0) and f(z) be epsilon", because the only thing you have direct control over and are modifying is R (of course, you have control over the result of the integral as well; it is your puppet, but you are controlling it through the strings of R). Thus if you want, you can take the limit as R approaches 0 on both sides.

[@exo_01]

The point is, that the integral is strictly less than ε, for all ε, meaning that the integral is 0, because we could choose an arbitrary small number and the integral still would be less. Thus providing the equality, without any "real" limit argumentation.

[@rigorless6330]

​@Jack S. ahh, so basically, what we showed is that: lim ε->0 (this stuff) = lim R->0 (this stuff) and because of path independence, we can change R without changing the value, so its fine to apply lim R-> 0 on this stuff

[@dgrandlapinblanc]

Genial. I become with that more stronger that Ramanujan. I joke... Thank you very much.

[@LucaIlarioCarbonini]

Little after minute one you name Wills theorem, am I getting the spelling properly?

[@drpeyam]

Liouville’s Theorem

[@Czeckie]

0:30 no. you need the function to be holomorphic

[@datsmydab-minecraft-and-mo5666]

where is the lioville theorem video?

[@drpeyam]

There is none. It follows from this video just by calculating f’(x) and applying the triangle inequality

[@willnewman9783]

There are a few things I think are wrong with this proof: 1. You say that this works for all continuous functions, but it does not. It only works for holomorphic functions. Moreau's theorem says that if the integral of a function is independent of path for all paths, then it is holomorphic. You also know that CIF can"t be true for all continuous functions because in the end you show that any function satisfying CIF is differentiable. 2. The whole idea that Cauchy Gourset implies that the integral is independent of path is a little shakey. Cauchy Gourset says that the integral of a holomorphic function should be independent of path in a simplely connected region. I have heard others say that this implies what it is independent of path in this case, as long as you go around z_0, but I have never seen a proof. If you know one, you should do a video on it, along with Cauchy Gourset. 3. This is not exactly a problem, but you show that the absolute value of the integral is less than epsilon for all epsilon. Therefore, it must be zero, so there is no need to take a limit at the end. I would not have mentioned this, but it seems like this confused a few other people in the comments.

[@jacks.4390]

Umm can you tell me what am I missing? Because I think the proof just does not work at all. What we want to show is that the formula holds for a fixed contour; when we are shrinking R in the integral that we want to show is small, we are also shrinking R in the integral where we actually want to have to original contour.

[@willnewman9783]

@@jacks.4390 He is using a theorem that says that any integrating the function around ANY path will give the same value, as long as z_0 is in the inside of the path. This is definitely true, but I do not know how to justify it without proving CIF in a different way.

[@jacks.4390]

@@willnewman9783 Ah okay. But to apply the theorem, do you need to use that f(z) - f(z_0) / z - z_0 has a removable discontinuity at z_0, so that the function is holomorphic (almost) everywhere in the interior of the path? Or I think it doesn't matter, right?

[@user-vq8on7dh1y]

Before watching this lesson, I broke up with my girlfriend. After finishing the lesson, I am independent of the past 🙂

[@jeremy.N]

thank you for the Heart, but you have not answered the questions about an Email or something

[@amyfalconer1660]

don't be creepy dude lol

[@jeremy.N]

it is just a little annoying, I mean He mist have read it, so i would even be okay with a no, but nö answer at all is just so unsetteling