Mr. I cannot thank you enough, your video series for Measure Theory and Complex Analysis saved me from failing my Analysis III exam. Keep up the good work!!
@brightsideofmaths
Жыл бұрын
You're very welcome!
@riadsouissi
Жыл бұрын
I know that was not the point of the video, but leaving sin(pi/6) like that and not replacing it with 1/2 left me hanging with a feeling of emptiness...
@brightsideofmaths
Жыл бұрын
Sorry! :)
@alicebobson2868
10 ай бұрын
best video on this ever, way better than my lectures!
@brightsideofmaths
10 ай бұрын
Glad you think so! :)
@smoosq9501
Жыл бұрын
thank you, really detailed explained, helped me a lot.
@ohjoshrules
Жыл бұрын
Thank you for your videos. They have been so so so helpful in my undergraduate studies
@mortimertz6660
Жыл бұрын
such a good play list. please consider Lie groups, Lie algebra, Haar measure.
@StratosFair
Жыл бұрын
Thank you for this series, I learned a lot !! It would be great if you could consider extending it by covering some more advanced topics (residue at infinity, conformal maps, Rouché's theorem...) :)
@brightsideofmaths
Жыл бұрын
Great suggestion! I will definitely do that!
@numb2023
10 ай бұрын
Hi, I have a problem understanding Rouche's theorem, but I couldn't find any videos on it in your videos. I was wondering if you have covered that topic. Thank you!
@brightsideofmaths
10 ай бұрын
Great suggestion!
@wonjonghyeon
Жыл бұрын
Why did not you compute the value of sin(π/6) at the end?
@이재경-l3e
Жыл бұрын
여기서 한국인을 보네 ㅋㅋ
@wonjonghyeon
Жыл бұрын
@@이재경-l3e 안녕하세요. :)
@brightsideofmaths
Жыл бұрын
Oh, I didn't think that this was important for this demonstration. I wanted to show things that can be generally used to calculate similar integrals.
@wonjonghyeon
Жыл бұрын
@@brightsideofmaths I see. Thank you for replying.
@TheSandkastenverbot
Жыл бұрын
@@brightsideofmaths Good decision. Such calculations would just distract us from the central points.
@lauraponti5998
Жыл бұрын
very nice videos! Thanks for the explainations! I just do not understand, how you find so quick the residues for z1,z2 and z3.
@brightsideofmaths
Жыл бұрын
Thanks you very much! Did you watch the previous videos about residues?
@darcash1738
9 ай бұрын
Complex analysis seems awesome but what does real analysis do?
@brightsideofmaths
9 ай бұрын
Thanks! Real Analysis is calculus together with the theory :)
@darcash1738
9 ай бұрын
@@brightsideofmaths ah ok. So if we know calculus already, it’s not gonna really add any cool integral techniques like this, just proofs?
@brightsideofmaths
9 ай бұрын
That depends what you already know. You should definitely watch it :D@@darcash1738
@TKanal3
Жыл бұрын
Großartig. Endlich verstanden. Klausur hoffentlich gerettet. Jetzt hoffe ich das es drankommt, das Gegenteil von vor diesem Video.
@brightsideofmaths
Жыл бұрын
Good luck :)
@Independent_Man3
Жыл бұрын
5:44 you are a mathematician so I wanted to ask you if it's correct practice to write the inequality as
@brightsideofmaths
Жыл бұрын
I would never use inequalities with the O notation. Maybe you give an additional meaning for this??
@kehoerg
Жыл бұрын
Perhaps not the right place to ask this question you might think, but as the people who are watching this are so up to date with your work, they're likely to be someone in the same shoes as myself.... Having finished a degree 10 years ago, and not being proud of my result, I am attempting to learn things again. Without exams to sit, there is seemingly no pressure to get it done, and if I'm honest I am failing. I can't but help think things like, why am I even doing this? What am I trying to prove. Has anyone any advice, maybe even the content creator himself. Thanks in advance.... am I alone in this I wonder?
@squarerootofpi
Жыл бұрын
I guess, the very fact that you have no exams to sit, actually frees you up to learn what YOU want, not what the SYLLABUS wants. So, you may not have learned this particular topic as well as it ought to be, but if it is as good as you'd like yourself to be, why not?
@GeoffryGifari
Жыл бұрын
on the last part when we discard the cos( ), do we implicitly take Re( ) of the entire contour integral? so that the real part of a contour integral must equal the integral we want?
@whalep
Жыл бұрын
The contour integral *is* what we want, no real-part-taking required. He's saying the result of evaluating all the sines and cosines will be real at the end regardless (otherwise, something's gone wrong). In other words, notice how there's an i out front and, by Euler's formula, exp(it) = cos(t) + i sin(t). This means i exp(it) = i cos(t) - sin(t), so if we're expecting a real result out of all the sines and cosines being added there, then all the cosines better go away - so just make them go away by ignoring them because they all will cancel out anyways.
@GeoffryGifari
Жыл бұрын
@@whalep thanks!
@brightsideofmaths
Жыл бұрын
@@GeoffryGifari You can definitely take the real part on both sides because the left-hand side will not change. It is just a matter of taste how you structure the calculation.
@karinablanchard9511
6 ай бұрын
Where can I find more info on standard estimates? I am lost on that part
@brightsideofmaths
6 ай бұрын
It's discusses in this series :)
@marko_duvnjak
Жыл бұрын
Thank you.
@marytheraspberry3145
Жыл бұрын
Hello. Your videos are very useful. I thank you for all the efforts you put in them. I have a request: can you make a video series on Fourrier and Laplace transforms? Thank you.
@brightsideofmaths
Жыл бұрын
Great suggestion!
@funnyvideo8983
Жыл бұрын
Why we take only upper half of the circle?
@brightsideofmaths
Жыл бұрын
You could also choose the lower half.
@johnlazarus146
Жыл бұрын
Thanks for your videos! I have an issue with the formula for simple poles (h(z) /g(z) ) used. When can we use it? I tried solving with the formula lim z-z0 [(z-z0) f(z)] to calculate the limit and the answers are different. Also tried it for the improper integral for f(x) = x^2/(1+x^4) and had a different answer also.
@johnlazarus146
Жыл бұрын
The formula for simple poles I referred to from the video is res(h/g, zi) = h(z) /g'(z)
@brightsideofmaths
Жыл бұрын
Yes, you need a simple pole for this :)
@StratosFair
Жыл бұрын
This directly follows from the general formula to compute the residue at a pole. Note that if z_0 is a simple pole of h/g, then g(z_0)=0 hence lim_{z\to z_0} (z-z_0) h(z)/g(z) = lim_{z\to z_0} d/dz h(z)/[g(z)-g(z_0)/(z-z_0)] = h(z_0)/g'(z_0)
@edztyMKWII
Жыл бұрын
Is this the final part of this series or will there be more? :)
@brightsideofmaths
Жыл бұрын
At the moment, the series has an end here but I will continue it next year!
@kamalsaleh6497
Жыл бұрын
But how do we calculate the roots so quickly? Any videos you recommend? Also, I really love your videos!
@brightsideofmaths
Жыл бұрын
I have videos about calculating complex roots: tbsom.de/s/ov
@MrChicken1joe
Жыл бұрын
05:03 Why is there no Pi in the power? Since its a halfcircle the power should be i*pi*t right? I think it does not affect the estimation however.
@brightsideofmaths
Жыл бұрын
It depends what domain your t has, doesn't it?
@MrChicken1joe
Жыл бұрын
@@brightsideofmaths thanks for your answer! do you mean because of i*pi*t=i*t' for some t' from domain? If that: But how do you know t' is inside [-R,R]? For example t=R, then pi*t=t' isnt inside the domain.
@brightsideofmaths
Жыл бұрын
@@MrChicken1joe t does not have to be inside [-R,R] for the delta curve :)
@MrChicken1joe
Жыл бұрын
@@brightsideofmaths ah I can see, thanks!!
@franciscopereira2993
Жыл бұрын
Amazing!
@brightsideofmaths
Жыл бұрын
Thanks :)
@theone4782
Ай бұрын
What is g prime?
@brightsideofmaths
Ай бұрын
Derivative of g :)
@theone4782
Ай бұрын
@@brightsideofmaths thanks allot. 👍
@richardpatove4587
9 ай бұрын
bro is not solving for a 8 munit
@brightsideofmaths
9 ай бұрын
What?
@richardpatove4587
9 ай бұрын
@@brightsideofmaths now that i got your attraction, we love you :D
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