# How to expand a function into a power series with negative powers?

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
Welcome to Mathematica.SE! In the current form, your question does not fit the standards of this site and will most probably be downvoted or closes. Would you please consider to read the faq! and improve your answer by giving some background information? Furthermore when you see good answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer by clicking the checkmark sign! –  halirutan Jan 9 '13 at 13:02
@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

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