Taylor Series Expansion for the Relativistic Factor of Momentum

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Homework Statement


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.


Homework 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).


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!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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Homework Statement


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.


Homework 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).
With that substitution, you get y = 1/sqrt(1 + x2).

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
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.
However, this expression wasn't correct. I'm not really sure how else to expand it! Any help would be much, much appreciated, thanks!

Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #3
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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!
 
  • #4
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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.
 
  • #5
vela
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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!
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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