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# Series expansion watch

1. To find the series expansion of , why does expanding the binomial first (as ) give a different answer to expanding the exponential first, as and then expanding binomially?
2. What result do you get in each case?
3. Note that when x = 0, your expression is undefined. So it's going to be impossible to find a power series expansion that works.
4. (Original post by nuodai)
What result do you get in each case?
I'm looking at the coefficient of so the first expression will give this as 0 whereas the second will give this as -1 (disregarding higher order terms).
5. (Original post by DFranklin)
Note that when x = 0, your expression is undefined. So it's going to be impossible to find a power series expansion that works.
Can I not find a Laurent expansion in a punctured disc around 0?
6. (Original post by thatguyoverthere)
Can I not find a Laurent expansion in a punctured disc around 0?
If you meant a Laurent expansion you should have said so.

Anyhow, the same point holds: What is the range of |z| such that the binomial expansion of 1/(1+z) is valid? So, if z = e^x, is the expansion going to be valid when x = 0?
7. (Original post by DFranklin)
If you meant a Laurent expansion you should have said so.

Anyhow, the same point holds: What is the range of |z| such that the binomial expansion of 1/(1+z) is valid? So, if z = e^x, is the expansion going to be valid when x = 0?
Sorry!

Ok, so z=e^x expansion doesn't work because the binomial expansion is not valid in a punctured open disc around 0. But the alternative expansion does work because can be within the range required for the expansion to work.

Is that right?
8. Yes. (Note that in the latter case the value of 'z' is x/2!+x^2/3! + ..., which is small when x is small.
9. (Original post by DFranklin)
Yes. (Note that in the latter case the value of 'z' is x/2!+x^2/3! + ..., which is small when x is small.
Great, thank you so much!

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