Linearizing Doppler Shift Function: How to Expand and Take Linear Terms?

mewmew
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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 v0 \frac{1+\frac{v}{c}} {1-\frac{v}{c}} where v0 is on the order of 10^10, c is 3*10^8, and v is around 45.
 
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Maybe this will help:

http://www.chass.utoronto.ca/~krybakov/teaching_files/math_econ/problem_sets/ps4.pdf
 
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Shouldn't it be f_0\sqrt{ \frac{1+\frac{v}{c}} {1-\frac{v}{c}}} ?

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.)
 
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In case I'm wrong, (I'm no physicist) try this:

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)

for v\ll c
 
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Thanks, just incase anyone is wondering it is a special case(should have said that) so it doesn't have the square root.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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