Heat equation + Fourier series

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  • Опубліковано 19 гру 2024

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  • @laserjohn
    @laserjohn 10 років тому

    Final TOMORROW.... due to English speaking issues, I simply could not comprehend my professor. You just gave me more in 20:53 minutes than I have received in weeks. Thanks so much!

  • @theminu09
    @theminu09 12 років тому +1

    Dr Chris Tisdell you are a legend!!! thank you for uploading all these helpful videos on you tube.. really helps students pass mathematics.. good job

  • @victoriafong1234
    @victoriafong1234 12 років тому

    My differential equations final in 2 days and I haven't been able to understand anything about Fourier series until I found these videos yesterday. Thank you so much!!!

    • @bihugadgets9267
      @bihugadgets9267 2 роки тому

      And now After 9 years
      I am here before 2 weaks of my differential equation finals 😇❤️

  • @DrChrisTisdell
    @DrChrisTisdell  12 років тому

    "OR, do we have to look for ONLY sin(xyz) = 0, since B cannot =0 and solve for xyz": - that's exactly right! Well done!
    Very good to hear you are finding the ebook useful.

  • @周宏駿-k5y
    @周宏駿-k5y 11 років тому

    i watch the video when i`m prepare my final exam ,it`s great to study with your video ,thx a lot !!!!

  • @Borewiczes
    @Borewiczes 11 років тому

    Man, You're great teacher. This helped me a lot! Cheers!

  • @oppositeofh20
    @oppositeofh20 13 років тому +1

    Thank you for this! You've helped me a lot.

  • @changbroot
    @changbroot 12 років тому +1

    By the way, I forgot to thank you for such an amazing videos and ebook. You have done a great job and it's very easy to follow and you're very clear on all the matters. Thank you very much for your hard work.
    So back to my question. I know that we want B not = 0. But isn't B*sin(xyz) = 0 yields zero, if we divide both sides by sin(xyz)? OR, do we have to look for ONLY sin(xyz) = 0, since B cannot =0 and solve for xyz? Maybe I'm tired of studying this week and can't think straight.

  • @DrChrisTisdell
    @DrChrisTisdell  12 років тому

    Hi - if we follow your suggestion then we obtain the zero (trivial) solution. We are interested in the non-trivial solutions and so that's we we disregard B = 0.

  • @DrChrisTisdell
    @DrChrisTisdell  12 років тому

    Thanks! Good luck with your studies.

  • @blurginouliz
    @blurginouliz 8 років тому

    You showed an example where the temperature at the end points of the bar are zero, which makes it possible to determine some of the coefficients, but it take it that in order for this to work inphysiscs, you have to use some absolute scale of temperature such as Kelvin, so it's a bit weird to say the end points will be at absolute zero.. On a more realistic problem, where the end points are at constant temperatures, but non zero, I wouldn't be able to eliminate the coefficient of the Cosine function.. How would I proceed then? Could you please show an example, or point me to a vide where you've already done it maybe? Thank you very much! You're a great teacher, and your videos are awesome!

  • @seggyRK
    @seggyRK 9 років тому +1

    what if my initial temperature was f(x)=T_0=50degreesC? How would that afffect my final solution?

  • @DinoDudeDillon
    @DinoDudeDillon 4 роки тому

    Where does (-1)^(1+n) come from in b sub k? I can get b sub k is -2/n by doing the inner product with sin(nx). You mention another video but I'm not sure where to look

  • @YourAverageHater
    @YourAverageHater 9 років тому

    might be a dumb question, but how did determine the constant to be "bn", which is applicable to the bn-sine-fourier summation, opposed to, say, the an-cos-fourier summation? Was that deliberate on your part, so that it would fit with the sine in your solution of the heat equation?

  • @Novak2611
    @Novak2611 5 років тому

    The fourier series does not converge to pi when x=pi but it converges to 0. Shouldn't you choose an initial condition which is compatible with boundary condition?

  • @DrChrisTisdell
    @DrChrisTisdell  13 років тому

    @Reyder93 Integrate by parts.

  • @MatrixOfDynamism
    @MatrixOfDynamism 12 років тому

    Dear Dr Tisdell, There is a question that I did really like to ask you if I could meet you in person but I shall post it here anyway. Trignometric functions Sin, Cos come out from working out ratios of right angled of triangles, historically. It so happens that they end up in Fourier series and thus in Fourier Transform and FFT. Without Fourier Series and Transform, communication engineering would not exist (no concept of bandwidth), No IT age. Where would mankind be without Sin & Cos than?

  • @kinio00
    @kinio00 12 років тому

    Good Job! It Was very helpful

  • @DrChrisTisdell
    @DrChrisTisdell  12 років тому

    You are welcome.

  • @DrChrisTisdell
    @DrChrisTisdell  12 років тому

    Sin and Cos are one of the most important aspects of mathematics and its applications. You may be surprised to learn that some mathematicians are questioning the validity of these concepts. Check out my colleague's channel njwildberger for more!

  • @seggyRK
    @seggyRK 9 років тому

    The two ends of the bar are at T1 and T2 and constant for all time

  • @MatrixOfDynamism
    @MatrixOfDynamism 12 років тому

    there are so many videos. I don't know where to look at. What I know is that a fundamental concept in communications is that of bandwidth. There cannot be a concept of bandwidth without fourier series. So I wonder what about all the mobile phones and internet, would they exist now?

  • @DrChrisTisdell
    @DrChrisTisdell  12 років тому

    hope you aced the test!

  • @ayakamal8634
    @ayakamal8634 9 років тому +1

    Thanks, now i get it :)

  • @DrChrisTisdell
    @DrChrisTisdell  12 років тому +1

    The pleasure is mine.

  • @changbroot
    @changbroot 12 років тому

    At 9 min, you say A*pi = 0, the then A = 0, so why don't you use the same way to say B*sin(....) = 0 (at 11 min), then B = 0? No matter what happens to sine, but B*sin(...) = 0 should yield B = 0/sin(...) = 0. Am I doing something wrong here?

  • @DrChrisTisdell
    @DrChrisTisdell  12 років тому

    Thanks!

  • @ashowmega
    @ashowmega 13 років тому

    Great! Its all clear.