Complex Fourier Series (fourier series engineering mathematics)

Comments · 178

• BL

  @blackpenredpen · 7 years ago · pinned

  At <a href="https://www.youtube.com/watch?v=aC0j8CW58AM&amp;t=495">8:15</a>, n should go from negative inf to postive inf. Sorry I missed the negative sign.

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• AN

  @AndrewDotsonvideos · 7 years ago

  The fast and the fouriers.

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• AM

  @amritas2400 · 5 years ago

  His positivity and enthusiasm is contagious! I find myself smiling while he explains everything clearly and simply. Love him.❤

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• RO

  @robsoncarlosdemourajunior6818 · 1 month ago

  I struggled to understand the issue of coefficients with negative indices. Your video was fundamental in helping me understand this part. Thank you very much from Campinas, Brazil.

• BU

  @bumpyturtle127 · 2 years ago

  For anyone wondering, if you were to evaluate the integral at <a href="https://www.youtube.com/watch?v=aC0j8CW58AM&amp;t=639">10:39</a>, it would evaluate to 2sin(pi(n-m))/(n-m). This is defined for all values of n and m except where n = m. For every other value, n-m becomes a whole number, and sin(any whole number*pi) = 0.

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• NI

  @nimmira · 7 years ago

  you deserve a 1M subscriber by the end of the year, not just 400K.

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• TI

  @tickettomathsisland7138 · 4 years ago

  Wow, here comes heavenly sent lecturer. Thank you very much

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• OZ

  @ozzyfromspace · 6 years ago

  It&apos;s January 1st, 2020 and, yes, we met his goal of 400k subs (he has 409k today) ☺️ I&apos;m very happy for you bro!

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• HA

  @harryjohnson9643 · 2 years ago

  I was trying to find this derivation for ages after my university lecturers seemed to ignore it. Thanks, you absolutely nailed it!

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• IA

  @iabervon · 7 years ago

  Complex is better; you don&apos;t need to remember which coefficients have a 1/π and which have 1/2π. It also just works if f is a complex-valued function, and you just don&apos;t get all conjugate pairs.

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• 10

  @100MWSmile · 7 years ago (edited)

  With this formula, the Fourier series for e^x in the interval (-π, π) is more obvious, since the coefficient sequence is easier to calculate. One has that the integrand is e^[(1 - im)x] which is anti-differentiated to [(1 + im)/(1 + m^2)]e^[(1 - im)x]. Evaluate this at the boundaries and subtract to get [(1 + im)/(1 + m^2)][(e^π - e^(-π))][(-1)^m]. The real and imaginary parts gives you the cosine and sine coefficients of the real Fourier series, respectively. Yet this is neater and easier to obtain.<br><br>You should make a video on the Fourier transform and its inverse. It would complete this series, and it also relates to the Laplace transform as well, which you have covered in this channel already.

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• AN

  @andreapaps · 5 years ago

  Watching all these maths videos to relax is making me want to return to university :D ... So much better than Netflix &lt;3 &lt;3 &lt;3

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