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Linearizing a function

  1. Jan 18, 2006 #1
    Well this is a physics related question but I think putting it in math is best. I am trying to linearize the doppler shift function of light. I know that means to expand it and take the linear terms but don't really know what to do. I was guessing a taylor series expansion but f[x]=f[a]+f'[a](x-a) has me confused as I am not sure what to use as a, as 0 doesn't work too well.
    The equation is [tex] v0 \frac{1+\frac{v}{c}} {1-\frac{v}{c}} [/tex] where v0 is on the order of 10^10, c is 3*10^8, and v is around 45.
     
    Last edited: Jan 18, 2006
  2. jcsd
  3. Jan 18, 2006 #2

    hotvette

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    Maybe this will help:

    http://www.chass.utoronto.ca/~krybakov/teaching_files/math_econ/problem_sets/ps4.pdf [Broken]
     
    Last edited by a moderator: May 2, 2017
  4. Jan 18, 2006 #3

    benorin

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    Shouldn't it be [tex] f_0\sqrt{ \frac{1+\frac{v}{c}} {1-\frac{v}{c}}} [/tex] ?

    Check-out the Doppler Effect page at www.scienceworld.com[/url], near the bottom they give such an expansion (cf. [url=http://scienceworld.wolfram.com/physics/RelativisticRedshift.html] Their Red Shift page[/URL] for more detail.)
     
    Last edited by a moderator: Apr 21, 2017
  5. Jan 19, 2006 #4

    benorin

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    In case I'm wrong, (I'm no physicist) try this:

    [tex] v_0 \frac{1+\frac{v}{c}} {1-\frac{v}{c}} = v_0 \left( 1+\frac{v}{c} \right) \frac{1} {1-\frac{v}{c}} = v_0 \left( 1+\frac{v}{c} \right) \left[ 1+ \frac{v}{c} + \left( \frac{v}{c}\right) ^2 + \cdots \right] = v_0 \left[ 1+ 2\frac{v}{c} + 2\left( \frac{v}{c}\right) ^2 + \cdots \right] \approx v_0 \left( 1+ 2\frac{v}{c} \right)[/tex]

    for [tex]v\ll c[/tex]
     
    Last edited: Jan 19, 2006
  6. Jan 19, 2006 #5
    Thanks, just incase anyone is wondering it is a special case(should have said that) so it doesn't have the square root.
     
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