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    To find the series expansion of \[

\dfrac{1}{{1 - e^x }}

\]

, why does expanding the binomial first (as \[

1 + e^x  + e^{2x}  + ...

\]

) give a different answer to expanding the exponential first, as \[

\dfrac{{ - 1}}{x}\left( {\frac{1}{{1 + {\textstyle{x \over {2!}}} + {\textstyle{{x^2 } \over {3!}}} + ...}}} \right)

\]



and then expanding binomially?
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    What result do you get in each case?
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    Note that when x = 0, your expression is undefined. So it's going to be impossible to find a power series expansion that works.
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    (Original post by nuodai)
    What result do you get in each case?
    I'm looking at the coefficient of \[

\dfrac{1}{x}

\]

so the first expression will give this as 0 whereas the second will give this as -1 (disregarding higher order terms).
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    (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?
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    (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?
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    (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 \[

{1 + {\textstyle{x \over {2!}}} + {\textstyle{{x^2 } \over {3!}}} + ...}

\]

can be within the range required for the expansion to work.

    Is that right?
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    Yes. (Note that in the latter case the value of 'z' is x/2!+x^2/3! + ..., which is small when x is small.
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    (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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