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Taylor expansion of ODE?

  1. Apr 7, 2014 #1
    I'm confused by problem 2.31 in mathematical tools for physics.

    Problem:
    2.31 The Doppler effect for sound with a moving source and for a moving observer have different formulas. The Doppler
    effect for light, including relativistic effects is different still. Show that for low speeds they are all about the same.

    [itex]f' = f \frac{v - v_0}{v}[/itex], [itex]f' = f \frac{v}{v+v_s}[/itex], [itex]f' = f \sqrt{\frac{1-v/c}{1+v/c}}[/itex]

    The symbols have various meanings: v is the speed of sound in the first two, with the other terms being the velocity
    of the observer and the velocity of the source. In the third equation c is the speed of light and v is the velocity of the
    observer. And no, 1 = 1 isn't good enough; you should get these at least to first order in the speed.


    Solution:
    From the selected solutions:
    [itex]f' = f(1-v_0/v)[/itex], [itex]f' = f(1-v_s/v)[/itex], [itex]f'=f(1-v/c)[/itex]

    Question:
    Clearly I'm supposed to do a tailor expansion of something, but I'm unsure of which part of the original differential equation I'm supposed to expand. Also, whichever part I do expand I end up with a different result than the given solution, which makes me think I'm interpreting the equation wrong. My interpretation is:
    [itex] f'(x) = \frac{v - v_0}{v} f(x)[/itex]

    Thanks for any help clearing this up.
     
  2. jcsd
  3. Apr 7, 2014 #2
    Yes, a Taylor expansion is the way to go. These are not differential equations though. The primes merely denote frequencies in different reference frames.
     
  4. Apr 7, 2014 #3
    Thanks Jilang, that cleared things up for me. I suppose this is a difference between a "Math for Physics" text and a pure mathematics one.

    I'm new to the forum, is there a standard for marking questions and posts as [solved] ?
     
  5. Apr 7, 2014 #4

    SammyS

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    Hello tssuser. Welcome to PF !
     
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