Expanding Fraction with Infinite Value: Simple Expansion Homework Solution

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SUMMARY

The discussion focuses on expanding the expression \(\frac{1}{(1+\frac{a^2}{z^2})^{1/2}}\) under the condition that \(z >> a\). Participants clarify that the expansion can be approached by substituting \(x = \frac{a}{z}\) and expanding \((1 + x^2)^{-1/2}\) for small \(x\). The first term simplifies to 1, while the second term involves the second derivative for further accuracy. This method provides a systematic approach to handle the expansion for small values of \(x\).

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Homework Statement


I am trying to expand [itex]\frac{1}{(1+\frac{a^2}{z^2})^{1/2}}[/itex] for z>>a.

Homework Equations

The Attempt at a Solution


First, I rearranged the equation to [itex](1+\frac{a^2}{z^2})^{-1/2}[/itex]. After this, since z>>a, can I assume z takes a value of infinity and say the first term is 1+0=1? And I am not sure what to do for the second term. I take the first derivative which is [itex]-\frac{1}{2}(1+\frac{a^2}{z^2})^{-3/2}(\frac{-2a^2}{z^3})[/itex] and not sure what to do with it.
Any help will be appreciated.
 
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spacetimedude said:

Homework Statement


I am trying to expand [itex]\frac{1}{(1+\frac{a^2}{z^2})^{1/2}}[/itex] for z>>a.

Homework Equations

The Attempt at a Solution


First, I rearranged the equation to [itex](1+\frac{a^2}{z^2})^{-1/2}[/itex]. After this, since z>>a, can I assume z takes a value of infinity and say the first term is 1+0=1? And I am not sure what to do for the second term. I take the first derivative which is [itex]-\frac{1}{2}(1+\frac{a^2}{z^2})^{-3/2}(\frac{-2a^2}{z^3})[/itex] and not sure what to do with it.
Any help will be appreciated.

So, are you not just trying to expand ##(1 + x^2)^{-1/2}## for small ##x = a/z##?
 
Ray Vickson said:
So, are you not just trying to expand ##(1 + x^2)^{-1/2}## for small ##x = a/z##?
I'm having difficulty understanding how to expand for small x. I've only come across questions that ask something like "expand this function around x= some number". Do I take x=0?

EDIT: Ah, so do I take x=0 and is the second non-zero term the term using second derivative?
 

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