1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook