# Taylor Series Expansion for the Relativistic Factor of Momentum

1. Jan 26, 2010

### leejqs

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. Jan 26, 2010

### 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.

3. Jan 26, 2010

### leejqs

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. Jan 26, 2010

### anagonga

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. Jan 26, 2010

### vela

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

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

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