Complex Analysis L13: Bromwich Integrals and the Inverse Laplace Transform

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  • Опубліковано 20 кві 2023
  • This video is a culmination of this series on complex analysis, where we show how to compute the Bromwich integral used in the inverse Laplace transform.
    @eigensteve on Twitter
    eigensteve.com
    databookuw.com
    This video was produced at the University of Washington
  • Наука та технологія

КОМЕНТАРІ • 43

  • @gean7917
    @gean7917 11 місяців тому +11

    It's unbelievable an amazing course like this available completely free on UA-cam. The guy is really good!

  • @tolkienfan1972
    @tolkienfan1972 Рік тому +11

    I really appreciate you making these lectures public. It's dense material, but that's exactly what I'm looking for. Thank you

  • @guiliangzheng5704
    @guiliangzheng5704 Рік тому +6

    What a ride in the complex analysis world! Thank you so much for putting it together! What a ride it is

  • @ElMalikHydaspes
    @ElMalikHydaspes 6 місяців тому +3

    Bravo! What an excellent set of lectures on complex numbers! Really well taught by Dr Brunton.

  • @tariqandrea398
    @tariqandrea398 4 місяці тому +4

    This isn't mere mathematics. It is a work of community service, a work of kindness, and a work of charity.

  • @andresfeliperamirezgaviria8976
    @andresfeliperamirezgaviria8976 5 місяців тому +2

    Excelent course, greetings and congratulations.

  • @timepass4783
    @timepass4783 2 місяці тому

    thanks a lot, completed the whole complex analysis 10 hours before my finals, You're a brilliant teacher!!!

  • @andrej5861
    @andrej5861 Рік тому +9

    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 Рік тому

      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.

  • @papawhiskeybravo
    @papawhiskeybravo 8 місяців тому +1

    Thanks for a great series. I was very well taught.

  • @hoseinzahedifar1562
    @hoseinzahedifar1562 Рік тому +1

    Great great great lecture...Thank you so much.

  • @eig_himanshu
    @eig_himanshu Рік тому +1

    superb sir!!!!!! you way of explanation is fantabulous

  • @xenofurmi
    @xenofurmi 2 місяці тому

    ATHF reference in last video of a Complex Analysis lecture... the future is now!

  • @eduardocarmona8157
    @eduardocarmona8157 4 місяці тому

    Great class!

  • @basics5427
    @basics5427 Рік тому +14

    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  Рік тому +5

      Awesome, thanks for the kind words -- glad you like them!

  • @danielhoven570
    @danielhoven570 Рік тому +2

    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.

  • @leonardoalcayaga6640
    @leonardoalcayaga6640 Рік тому +4

    Your lectures are really great, thanks a lot! Is it possible to follow a similar apporach to obtain Fourier transforms?

  • @byronwatkins2565
    @byronwatkins2565 Рік тому +2

    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 Рік тому

      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.

  • @marekw4353
    @marekw4353 Рік тому +1

    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?

  • @ralvarezb78
    @ralvarezb78 Місяць тому

    at 44:00 I think this inequality can be deduced using Taylor series expansion up to 2 terms for cosine

  • @tobiassugandi
    @tobiassugandi Рік тому +1

    what a ride..

  • @ralvarezb78
    @ralvarezb78 Місяць тому

    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

  • @sakethvirupakshi5961
    @sakethvirupakshi5961 Рік тому +1

    Hello Steve, it will be nice if you make some videos related to statistics and probability theory

  • @kov1ub
    @kov1ub 7 місяців тому

    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!🤔

  • @eng4529
    @eng4529 10 місяців тому +1

    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?

  • @aram9167
    @aram9167 2 місяці тому

    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

  • @Alan-zf2tt
    @Alan-zf2tt 6 місяців тому

    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|

  • @GreenMeansGOF
    @GreenMeansGOF 8 місяців тому

    Would a semicircular contour be harder? Like if it was just a vertical line and a semicircle without C+ and C-?

  • @Justin-dk9rl
    @Justin-dk9rl Рік тому

    The (reverse) triangle inequality directly states |s-a| >= ||s|-|a|| = |R-a|.

  • @user-eo7ll4yz6i
    @user-eo7ll4yz6i 10 місяців тому

    Does someone now what marker pen this guys use ?

  • @PillarArt
    @PillarArt 4 місяці тому +1

    *stops to e^at @ t=021.140*

  • @TheTimeRay
    @TheTimeRay Рік тому +1

    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 Рік тому +1

      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.

  • @hoseinzahedifar1562
    @hoseinzahedifar1562 Рік тому

    In time 25:22: I think ds = -dx. isn't it correct?

  • @meguellatiyounes8659
    @meguellatiyounes8659 Рік тому

    complex step finite difference

  • @belwizdadimed3967
    @belwizdadimed3967 5 місяців тому

    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.

  • @insainsin
    @insainsin Рік тому

    Why is the default inverse definition the infinite integral and not the Cauchy integral formula?

  • @dantetheblizzard
    @dantetheblizzard Рік тому +1

    You call this an integral, you don't have bacon on the curve?

  • @younique9710
    @younique9710 3 місяці тому

    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 ||