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Is there any way to expand this expression

a+b(1-Exp[-T/(b c)]/(z-Exp[-T/ (b c)])

(where a, b, c, and T are constants) as a series in negative powers of $z$? The result should be in the form

a0 + a1 z^(-1) + a2 z^(-2) + a3 z^(-3) + ... + an z^(-n)

I tried solutions like Series[a + b (1 - k/(z - k)), {z, 0, -5}] but this did not work.

Thank you.

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An expansion around zero Series[a + b (1 - k/(z - k)), {z, 0, 5}] gives $(a+2 b)+\frac{b z}{k}+\frac{b z^2}{k^2}+\frac{b z^3}{k^3}+\frac{b z^4}{k^4}+\frac{b z^5}{k^5}+O\left(z^6\right)$ –  image_doctor Jan 9 '13 at 12:45
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@image_doctor thank you for the answer. but i want a serie with negative power. Series[a + b (1 - k/(z - k)), {z, 0, -5}] don't work –  Mag Num Jan 9 '13 at 14:03
    
I was going to suggest expanding at infinity, then noticed @whuber said it better. –  Daniel Lichtblau Jan 9 '13 at 15:18
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1 Answer

You want first to fix any typographical errors (such as the unbalanced parentheses) and it's also wise to avoid symbol names beginning with capital letters. Then, to obtain a series expansion in powers of $1/z$, expand the expression around infinity, not zero:

Series[a + b (1 - Exp[-t/(b c)]/(z - Exp[-t/(b c)])) , {z, Infinity, 5}] 

$(a+b)-\frac{b e^{-\frac{t}{b c}}}{z}-\frac{b e^{-\frac{2 t}{b c}}}{z^2}-\frac{b e^{-\frac{3 t}{b c}}}{z^3}-\frac{b e^{-\frac{4 t}{b c}}}{z^4}-\frac{b e^{-\frac{5 t}{b c}}}{z^5}+O\left[\frac{1}{z}\right]^6$

To confirm this, we could also replace $z$ by $1/z$, expand the series in non-negative powers of $z$, and then substitute $1/z$ back for $z$ once more:

Series[a + b (1 - Exp[-t/(b c)]/(z - Exp[-t/(b c)])) /. z -> (1/z), {z, 0, 5}] /. z -> 1/z

$(a+b)-\frac{b e^{-\frac{t}{b c}}}{z}-b e^{-\frac{2 t}{b c}} \left(\frac{1}{z}\right)^2-b e^{-\frac{3 t}{b c}} \left(\frac{1}{z}\right)^3-b e^{-\frac{4 t}{b c}} \left(\frac{1}{z}\right)^4-b e^{-\frac{5 t}{b c}} \left(\frac{1}{z}\right)^5+O\left[\frac{1}{z}\right]^6$

The two results are clearly equivalent expressions of the same series. If the terminal O[] term is undesirable, remove it by applying Normal to the output.

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Thank you whuber, thats what am looking for :-) –  Mag Num Jan 10 '13 at 8:31
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