Complex Analysis L13: Bromwich Integrals and the Inverse Laplace Transform

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Steve Brunton

Steve Brunton

Күн бұрын

Пікірлер: 49
@gean7917
@gean7917 Жыл бұрын
It's unbelievable an amazing course like this available completely free on KZbin. The guy is really good!
@tolkienfan1972
@tolkienfan1972 Жыл бұрын
I really appreciate you making these lectures public. It's dense material, but that's exactly what I'm looking for. Thank you
@ElMalikHydaspes
@ElMalikHydaspes 10 ай бұрын
Bravo! What an excellent set of lectures on complex numbers! Really well taught by Dr Brunton.
@guiliangzheng5704
@guiliangzheng5704 Жыл бұрын
What a ride in the complex analysis world! Thank you so much for putting it together! What a ride it is
@saeida.alghamdi1671
@saeida.alghamdi1671 25 күн бұрын
Very valuable lecture on the basics of finding the Inv. LT using the theory of complex integral … Thank you 🌿🌴
@andresfeliperamirezgaviria8976
@andresfeliperamirezgaviria8976 10 ай бұрын
Excelent course, greetings and congratulations.
@timepass4783
@timepass4783 7 ай бұрын
thanks a lot, completed the whole complex analysis 10 hours before my finals, You're a brilliant teacher!!!
@byronwatkins2565
@byronwatkins2565 Жыл бұрын
At 34:50, the way theta and the contour are defined requires integration from pi/2 to -pi/2; but the limit is still zero. Jordan's lemma proves this in general...
@chrislubs1341
@chrislubs1341 Жыл бұрын
Saw this, but noted he redefines theta to (PI - theta) to get the corect integral, which is important to keep inequalities from reversing due to a sign error. This video might serve to suggest useful FOURIER TAUBERIAN THEOREMS.
@tariqandrea398
@tariqandrea398 9 ай бұрын
This isn't mere mathematics. It is a work of community service, a work of kindness, and a work of charity.
@ralvarezb78
@ralvarezb78 5 ай бұрын
at 44:00 I think this inequality can be deduced using Taylor series expansion up to 2 terms for cosine
@andrej5861
@andrej5861 2 жыл бұрын
I believe those "tricks" for showing that parts of integral in complex plane are 0 etc...are called Jordans lemmas (theorems)....unfortunately I do not have my textbook with me and it has been over 20 years...
@quantum4everyone
@quantum4everyone Жыл бұрын
Yes, that is correct. Jordan’s lemma holds for a broader set of integrands, and his formulation is slightly odd because everything is rotated by 90 degrees because his exponential in the integrand has no i in it. But the essence of the argument would be very similar.
@ralvarezb78
@ralvarezb78 5 ай бұрын
minute 41:00 I think the trick is |R-a| = sqrt((R-a)^2) wich leads to sqrt(R^2 - 2aR + a^2) If theta = pi, the inequality becomes equal, so true, but if theta is different from pi, the term 2aR cos(theta) < 2aR which is also true. Then you're right
@eng4529
@eng4529 Жыл бұрын
Question: why would ML bound work? At 31:48, he assumes that exp(gamm*t)*gamma does not go to infinity, but it may, if gamma >1 and t -> infinity?
@papawhiskeybravo
@papawhiskeybravo Жыл бұрын
Thanks for a great series. I was very well taught.
@aram9167
@aram9167 7 ай бұрын
42:30 Is all the following gymnastics necessary? Since -Rcos(theta)t is always negative between -pi and pi, as we tend R to infinity, the integrand goes to 0, so the integral goes to 0
@basics5427
@basics5427 2 жыл бұрын
How many hours do you have in one day? 70? More than 70? I just scrolled xN speed (with N huge) this series about complex analysis. Very well done. Lots of students in Engineering dealing with dynamical systems and control (so, almost every student in Engineering) curious about some detail about the math behind them and coming across these lectures should be so thankful to you. Obviously they're not enough without personal effort and study, but they're a good point to start for sure. Anyone who wants a concise and quite precise introduction to complex analysis and many other mathematical topics useful in engineering, could have on Schaum's Outlines, Advanced Mathematics for Engineers and Scientists: 10-15 pages of theory for every topic, and proofs left as an exercise to the reader.
@Eigensteve
@Eigensteve 2 жыл бұрын
Awesome, thanks for the kind words -- glad you like them!
@eig_himanshu
@eig_himanshu Жыл бұрын
superb sir!!!!!! you way of explanation is fantabulous
@leonardoalcayaga6640
@leonardoalcayaga6640 Жыл бұрын
Your lectures are really great, thanks a lot! Is it possible to follow a similar apporach to obtain Fourier transforms?
@Alan-zf2tt
@Alan-zf2tt 10 ай бұрын
About 38:20 or thereabouts am I right in thinking: Given R² + 2Racosθ + a² then holding R and a fixed while theta varies satisfies (R- a)² ⩽ R² + 2Racosθ + a² ⩽ (R+a)² since -1 ⩽ cosθ ⩽ 1 hence |R-a| ⩽ √(R² + 2Racosθ + a²) ⩽ |R+a|
@danielhoven570
@danielhoven570 2 жыл бұрын
Thank You! I just finished a dynamics homework with no reference to a Laplace transform table. Un-necessary for sure, but I feel like a boss lol. Anyway the only bit I had to dig for myself was finding residues of higher order poles, but without your introduction I'd have struggled to make sense of the literature.
@hoseinzahedifar1562
@hoseinzahedifar1562 Жыл бұрын
Great great great lecture...Thank you so much.
@pkiwan93
@pkiwan93 2 ай бұрын
Good lecture. Thank you for your lecture. However, I still have a question that makes me unhappy. Originally, f(s) is defined in s > gamma (Laplace transform). In the region of s < gamma, f(s) diverges so that f(s) itself is not defined there. However, this contour uses this diverging area. How can this be solved? Some textbooks explain that analytic continuation justifies it. However, this does not seem to solve the paradox. Is it possible because exp(st) converges to zero faster than f(s) diverges?
@p1ll4r
@p1ll4r 8 ай бұрын
*stops to e^at @ t=021.140*
@JarogniewBorkowski
@JarogniewBorkowski 3 ай бұрын
Nice video but how to derive general Laplace inverse transform formula?
@PakkoT
@PakkoT 17 күн бұрын
My class is more focused on the fourier transform. Thanks for these explainations
@xenofurmi
@xenofurmi 6 ай бұрын
ATHF reference in last video of a Complex Analysis lecture... the future is now!
@hoseinzahedifar1562
@hoseinzahedifar1562 Жыл бұрын
In time 25:22: I think ds = -dx. isn't it correct?
@byronwatkins2565
@byronwatkins2565 Жыл бұрын
Yes. But, the limit is still zero.
@TnF-x2n
@TnF-x2n 29 күн бұрын
@@byronwatkins2565 by definition s=x + iR, therefore ds = dx
@marekw4353
@marekw4353 2 жыл бұрын
Thank you for the course, I appreciate the conceive form. You've been mentioning, that in good ol' days there would have been a whole semester course on complex analysis. Could you maybe recommend any sources to dive deeper into the topic?
@eduardocarmona8157
@eduardocarmona8157 9 ай бұрын
Great class!
@GreenMeansGOF
@GreenMeansGOF Жыл бұрын
Would a semicircular contour be harder? Like if it was just a vertical line and a semicircle without C+ and C-?
@sakethvirupakshi5961
@sakethvirupakshi5961 Жыл бұрын
Hello Steve, it will be nice if you make some videos related to statistics and probability theory
@Jean-BaptisteMalagnoux
@Jean-BaptisteMalagnoux Жыл бұрын
Does someone now what marker pen this guys use ?
@TheTimeRay
@TheTimeRay Жыл бұрын
so, this 50 mins to prove C+, C-, Cr are 0 - is this proof just for the simplest f(s)=1/(s-a) ? ... do we need to re-do this math proof for each other possible f(s) ?
@chrislubs1341
@chrislubs1341 Жыл бұрын
Be aware different inverse transforms f(t) corrispond to F[s] as distinquished by domain of F[s], so pick a desired f(t) by considering the Bromwich integral.
@Justin-dk9rl
@Justin-dk9rl Жыл бұрын
The (reverse) triangle inequality directly states |s-a| >= ||s|-|a|| = |R-a|.
@kov1ub
@kov1ub Жыл бұрын
Hey Steve, I really like your videos, and I'm curious - are you writing in reverse or did you flip the image? Either way, it's a cool effect!🤔
@bird9172
@bird9172 Ай бұрын
You, sir, deserve the best cheeseburgers in the world.
@belwizdadimed3967
@belwizdadimed3967 10 ай бұрын
Lack of consistency. Before talking about inverse Laplace transform, it is wise to first define the Laplace transform and how it is a generalization of the Fourier transform? Time to frequency domain? People with no background of signal theory will be confused.
@tobiassugandi
@tobiassugandi Жыл бұрын
what a ride..
@tomoki-v6o
@tomoki-v6o Жыл бұрын
complex step finite difference
@insainsin
@insainsin Жыл бұрын
Why is the default inverse definition the infinite integral and not the Cauchy integral formula?
@dantetheblizzard
@dantetheblizzard Жыл бұрын
You call this an integral, you don't have bacon on the curve?
@younique9710
@younique9710 7 ай бұрын
At 27:04, I wonder how the norm of a complex variable is just the real part. Should we regard Cauchy inequality, || e^x+e^iR ||
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