Taylor Series Expansion for the Relativistic Factor of Momentum

[leejqs]

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!

 

[Mark44]

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!

 

[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!

 

[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.

 

[vela]

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:

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

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