# Linearizing a function

1. Jan 18, 2006

### mewmew

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.

Last edited: Jan 18, 2006
2. Jan 18, 2006

### hotvette

Maybe this will help:

http://www.chass.utoronto.ca/~krybakov/teaching_files/math_econ/problem_sets/ps4.pdf [Broken]

Last edited by a moderator: May 2, 2017
3. Jan 18, 2006

### benorin

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.)

Last edited by a moderator: Apr 21, 2017
4. Jan 19, 2006

### benorin

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 $$v\ll c$$

Last edited: Jan 19, 2006
5. Jan 19, 2006

### mewmew

Thanks, just incase anyone is wondering it is a special case(should have said that) so it doesn't have the square root.