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Taylor Series Expansion for the Relativistic Factor of Momentum

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Using the technique of Taylor expansion, find an approximate expression for the relativistic factor γ for small v (i.e., expanded around v = 0) that is correct to order v2.


    2. Relevant equations
    γ=1/SQRT(1+ V2/C2). But in class, my professor just substituted X=V/C, so the equation became γ=1/SQRT(1-X2).


    3. The attempt at a solution
    I tried to expand the function γ(v) with v at 0.....
    γ(v)=1+γ'(0)(X-0)+(1/2)γ''(0)(X-0)2
    However, this expression wasn't correct. I'm not really sure how else to expand it! Any help would be much, much appreciated, thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 26, 2010 #2

    Mark44

    Staff: Mentor

    With that substitution, you get y = 1/sqrt(1 + x2).
    This is really y(x) with x at 0. If v = 0, then x = 0. It's very likely that the sign error above is affecting what you got for your Maclaurin series.
     
  4. Jan 26, 2010 #3
    Sorry for that typo! The actual function is y=1/SQRT(1-V2/C2).

    In finding the parts of this maclaurin series, I have...
    y(0)=1/SQRT(1-X2)(X-0)0=1
    y'(0)=(1/2)(1-X2)(-3/2)*X2
    y''(0)=(1/2)(1-X2)(-3/2)*X2+(1-X2)(-3/2)*(1-X2)(-5/2)*3X4

    Sorry for my lack of showing my work, but I believe those expressions are right
    when using the yn(v)=[f(n)(X0(X-X0)n]/n!
     
  5. Jan 26, 2010 #4
    how to Put the Boolean function q'r + p'q' + pq'r' into DNF and find a simpler representation for the function using a Karnaugh map. Show the output of this simpler function is the same as in question 1 for p = 1, q = 1, r = 0.
     
  6. Jan 26, 2010 #5

    vela

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    Staff Emeritus
    Science Advisor
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    Your derivatives are wrong because you're mixing in stuff from the series expansion. You should have:

    [tex]y(X)=1/\sqrt{1-X^2} \rightarrow y(0)=1[/tex]
    [tex]y'(X)=X/(1-X^2)^{-3/2} \rightarrow y'(0)=0[/tex]

    and so on. You then plug these values for y(0), y'(0), y''(0) into the series expansion.
     
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