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Physical derivation of the Sin series

  1. Aug 15, 2012 #1
    I was reading feynman's lectures on physics and i came across the 23-2 in volume two where he is talking about a capcitor at high frequencies. He then uses the equations of E and M to come up with an approximation of the electric field between the two plates as the field oscillates at a high frequency. In order to make the approximation better he accounts for the fact that a changing electric field generates a magenetic field. then he corrects the magnetic field according to the corrected electric field, so on and so forth.

    In the end he has an infinite series J[itex]_{0}[/itex] that is the bessel funciton.

    I was wondering if there would be any way to come up with the sine expansion in a similar way. I was first looking at a circle with arc length [itex]\vartheta[/itex] and trying to show that the length of the chord was rSin([itex]\vartheta[/itex]) but im not sure how to go about it. Maybe someone has some suggestion of how to do it for a more physical situation similar to how feynman did it.
  2. jcsd
  3. Aug 15, 2012 #2
    Ive been thinking that one possible way would be looking at a harmonic oscillator. if we have a k constant of 1 and initial conditions, x(0)=1 and x'(0)=0. then we have a solution of x(t)=cos(t). so at time t=0 we have x=1, so we have the beginning of the cos expansion. i dont know how to go forward from there just yet.
  4. Aug 15, 2012 #3
    I figure we will have to look at the higher terms of the potential because that way we can better approximate the motion. similar to how adding terms of higher power to a taylor expansion sort of bends the line out to curve more closely to the function in question. The potential is U=1/2kx^2 or ,in our case with k=1, U=1/2*x^2 or x^2/2!. This is exactly the next term we need for the next piece of the cos expansion. Is this mere coincidence or is it signifigant?
  5. Aug 15, 2012 #4
    My last post I realize now was completely wrong. I just got excited about seeing the 1/2!x^2 term. however it also shows up if you calculate the distance traveled from the acceleration at time t=0. Perhaps this is in the right track? I'm not sure how to go on from here yet though .
    Last edited: Aug 15, 2012
  6. Aug 15, 2012 #5
    So here's how I worked it out. After everything above, I remembered that since at t=0 the acceleration is a max so it's derivative (the jerk) is 0. Moving on to the next derivative of motion (the snap?) we find the distance traveled is the integral of the jerk or 1/4!x^4. Which is just what we are looking for! I then just continued the argument, assuming all the odd powered terms are zero due to being the derivative of some maximum quantity. Then I found the contribution of the new term to the distance. So I come up with the series 1-1/2!x^2+1/4!x^4-1/6!x^6... Which is the cos ! Does this work? Or did I go wrong somewhere?
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