Contour Integrals: A Complex Proof of Reflection Formula

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Contour Integrals: A Complex Proof of Reflection Formula

Рет қаралды 4,333

Күн бұрын

Hope everyone enjoyed! Longer style with this one but very rewarding complex proof - please comment with any questions or suggestions for new topics, and as always, subscribe to stay updated.
~ Thanks for watching!
Lets keep going to 2k!
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Пікірлер: 33

@ADDiOUMAARIR 10 күн бұрын

Very nice ❤, keep going my friend ❤❤❤

@OscgrMaths 10 күн бұрын

@@ADDiOUMAARIR Thank you!

@pokemil5705 Ай бұрын

I'm so glad you turned the volume down slightly when the marker was squeaky. It's so nice, thanks :) And overall, it is a really nice video

@user-cd9dd1mx4n Ай бұрын

I have encountered many proofs of this before, but struggled to grasp them fully. Your explanation has finally clarified it for me. It remains rigorous yet you have made it remarkably engaging. I truly appreciate your clear and enjoyable explanation. Thank you very much. -------- We know that |a+b| ≤ |a| + |b| Replacing b with -b, we get |a-b| ≤ |a| + |-b|, but |'b|=|b|, so |a-b| ≤ |a| + |b|. Rearranging: |a-b| - |b| ≤ |a|. [*] With the same argument, by replacing a with -a, we get |a-b| - |a| ≤ |b|. [**] Combining * and **, we get ||a|-|b|| ≤ |a-b| ----- Well, I am not good at complex analysis, but I think the answer for (why) at 22:12, is because otherwise the function will no longer be analytic, and then we cannot integrate. ---- Thanks again for this great video.

@OscgrMaths Ай бұрын

@@user-cd9dd1mx4n Thanks for the comment!

@franolich3 Ай бұрын

Very nice indeed! Well worth the wait!!!

@OscgrMaths Ай бұрын

Thanks so much! Glad you enjoyed.

@DavidMFChapman Ай бұрын

Good job!

@OscgrMaths Ай бұрын

@@DavidMFChapman Thank you!!

@dakcom-mk6mp Ай бұрын

Nice

@OscgrMaths Ай бұрын

@@dakcom-mk6mp Thanks!

@Georgeclassified Ай бұрын

Awesome and rigorous Video as always! By the way I would like to make a suggestion about a future video you can consider making on the relationship between the zeta function and the dirichlet eta function OR on the analetic continuation of the zeta function and its contribution to the studying of primes and number theory(euler product formula for the zeta function)❤!

@OscgrMaths Ай бұрын

Thanks so much! I'll definitely take a look at these thanks for the suggestion.

@Georgeclassified Ай бұрын

@@OscgrMaths My pleasure😄

@leofoxpro2841 Ай бұрын

Hello, great videos!!

@OscgrMaths Ай бұрын

@@leofoxpro2841 Thank you!

@user-jm6rm2xn3z Ай бұрын

hi sir i really enjoyed the video but i have a question concerning the branch cut: why the pole -1 (is a real number ) isn't branch cut i mean for example in the integral proposed by Math 505 int from 0 to inf cos(x))/(π^2 - 4x^2) the poles when are real he made a branch cut i hope that you understand what i mean

@OscgrMaths Ай бұрын

@@user-jm6rm2xn3z I know what you mean - my branch cut was only positive real axis so since the pole was negative it's okay. I could have chosen imaginary which would have changed the contour but I find real easier normslly

@niom-nx7kb Ай бұрын

13:44 is always going to be...negative :)

@gregoriuswillson4153 Ай бұрын

such a nice applications of contour integrations , btw may i know what book you use to learn this method

@OscgrMaths Ай бұрын

To learn complex analysis and contour integration, Introduction To Complex Analysis by Priestley is a classic! Thanks for the comment.

@gregoriuswillson4153 Ай бұрын

@@OscgrMaths Thanks for the reply bro , i would love to try

@girianshiido Ай бұрын

You act as if the two horizontal line segments of the keyhole contour are on the real axis: they’re not.

@OscgrMaths Ай бұрын

@@girianshiido Great question! Thats the why I parametrised with 0 and 2pi. Also, as epsilon reaches 0, they do approach the real axis given that the circle they are coming from essentially approaches a point at the origin. Hope that clarifies!