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Im trying to find the Laurent expansion of the function

$$f(z):=\frac{a-b}{(z-a)(z-b)},\quad\text{for }0<|a|<|b|$$

around $z=0$ in the annulus defined by $A:=\{z\in\Bbb C:|a|<|z|<|b|\}$. What I did by now is trying to write a function for the coefficients defined by

$$c_n=\frac1{2\pi i}\int_{r\partial\Bbb D}f(z)z^{-n-1}\, dz,\quad n\in\Bbb Z,\, r\in(|a|,|b|)$$

thus I write

f[z_] := (a - b)/((z - a) (z - b))
c[k_, f_] := Integrate[With[{z = r E^(I t)}, f[z]/(2 Pi z^k)], {t, 0, 2 Pi}]

However Im not sure how to set $r\in(|a|,|b|)$ in the above code. Of course setting values for $a$ and $b$ I can get an expansion, but Im interested in the symbolic expansion in terms of $a$ and $b$ with the restrictions said above, that is, that $0<|a|<|b|$.

I had read almost all topics related to Laurent series but I dont find something to obtain symbolic expressions of these kind.

Some help will be appreciated, thank you.

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  • $\begingroup$ Check this prior thread. $\endgroup$ – Daniel Lichtblau Apr 2 '18 at 16:33
  • $\begingroup$ @Daniel Lichtblau: I tried that with no success (see my answer). $\endgroup$ – user64494 Apr 2 '18 at 16:39
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A standard approach is to use a partial fraction decomposition, rearranged appropriately for the region of interest, that is, so that the corresponding infinite sum will converge.

In this case the pfd is as below.

In[209]:= Apart[(a - b)/((z - a) (z - b)), z]

(* Out[209]= 1/(-a + z) - 1/(-b + z) *)

Since |z|>|a| we write the first term as

(1/z)/(1 - a/z)

This expands to `1/z (1+a/z+(a/z)^2+(a/z)^3+...)

Since |z|<|b| the second terms should be written as

-(1/(-b) b)/(1 - z/b)

This expands to `(1/b) (1+z/b+(z/b)^2+(z/b)^3+...)

The Laurent expansion is simply the sum of these.

Here is an explicit way to get a finite truncation of specific order in the Wolfram Language. Expand the part involving a at infinity, and the part involving b at the origin, then sum.

In[211]:= 
Normal[Series[1/(-a + z), {z, Infinity, 3}]] + 
 Normal[Series[-(1/(-b + z)), {z, 0, 3}]]

(* Out[211]= 1/b + a^2/z^3 + a/z^2 + 1/z + z/b^2 + z^2/b^3 + z^3/b^4 *)
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This is a variation on the theme of that topic.

With[{r = (Abs[a] + Abs[b])/2, n = 8}, 
 Table[FourierCoefficient[With[{z=r Exp[I t]},(a-b)/((z-a)*(z - b))], t, k ,Assumptions -> Abs[a] > 0 && Abs[a] < Abs[b]]/r^k, {k,-n,n}]]

$$\left\{0,0,0,0,0,0,0,0,\frac{1}{b}-\frac{1}{a},\frac{a^2-b^2}{a^2 b^2},\frac{a^3-b^3}{a^3 b^3},\frac{a^4-b^4}{a^4 b^4},\frac{a^5-b^5}{a^5 b^5},\frac{a^6-b^6}{a^6 b^6},\frac{a^7-b^7}{a^7 b^7},\frac{a^8-b^8}{a^8 b^8},\frac{a^9-b^9}{a^9 b^9}\right\} $$

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  • $\begingroup$ there is something wrong there. In the defined annulus we have that $c_{-n}=a^{n-1}\neq 0$, however for $c_n$ your table seems correct. $\endgroup$ – Masacroso Apr 2 '18 at 15:23
  • $\begingroup$ @Masacroso: Sorry, the result coinsides with the result of the Subho95's answer. Can you base your doubt? $\endgroup$ – user64494 Apr 2 '18 at 15:37
  • $\begingroup$ because $$c_{-n}=\frac1{2\pi i}\int_{r\partial\Bbb D}\frac{a-b}{(z-b)}z^{n-1}\cdot\frac1{z-a}\, dz=a^{n-1},\quad\text{for } n\ge 1$$ for $r\in(|a|,|b|)$. This is a consequence of $\frac{z^{n-1}}{z-b}$ being holomorphic in the described annulus. $\endgroup$ – Masacroso Apr 2 '18 at 15:42
  • $\begingroup$ @Masacroso: You are right. I have to think about that. The assumptions on $a$ and $b$ in the FourierCoefficient command change nothing in the output. $\endgroup$ – user64494 Apr 2 '18 at 16:22
  • $\begingroup$ @Masacroso: Likely a bug in FourierCoefficient or a gap in the help to this command. $\endgroup$ – user64494 Apr 2 '18 at 17:55
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A more mechanical way of finding Laurant series can be done as follows

$Assumptions = {0 < a < r, 0 < r < b};
f[a_, b_, n_][z_] = 1/(2 \[Pi] I) (a - b)/((z - a) (z - b))  z^(-n - 1)
g[a_, b_, r_, n_][\[Theta]_] = 
f[a, b, n][ r Exp[I \[Theta]]] I r  Exp[ I \[Theta]] // Simplify
Table[ Integrate[ 
   g[a, b, r, n][\[Theta]], {\[Theta], 0, 2 \[Pi]}], {n, -10, 10}]
(* {a^9, a^8, a^7, a^6, a^5, a^4, a^3, a^2, a, 1, 1/b, 1/b^2, 1/b^3, \
 1/b^4, 1/b^5, 1/b^6, 1/b^7, 1/b^8, 1/b^9, 1/b^10, 1/b^11} *)

Unfortunately, Mathematica cannot evaluate the integral for arbitrary integers even with the assumption Assuming[ {n \[Element] Integers}, ....

This is basically a similar repetition of Laurent series expansion, which has been mentioned in this post several times.

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  • $\begingroup$ This is not it. You additionally assume that $a$ and $b$ are positive real numbers. $\endgroup$ – user64494 Apr 2 '18 at 17:21
  • $\begingroup$ @user64494 Depending on what you want, it can be both wrong and right. As I mentioned already, Mathematica cannot handle the full generality of the problem, so restricting the problem to solvable subproblems is not necessarily a bad thing. But I totally understand your point and this point was not clear enough in the post. $\endgroup$ – Sungmin Apr 2 '18 at 17:32
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Series with Assumptions option is an easy way:

In[11]:= f[z_] := (a - b)/((z - a) (z - b))


In[12]:= Series[f[z], {z, 0, 10}, 
  Assumptions -> (Abs[a] < Abs[z] < Abs[b])] // Normal

Out[12]= (a - b)/(a b) + ((a - b) (a + b) z)/(
 a^2 b^2) + ((a - b) (a^2 + a b + b^2) z^2)/(
 a^3 b^3) + ((a - b) (a^3 + a^2 b + a b^2 + b^3) z^3)/(
 a^4 b^4) + (1/(a b^5) + 1/(a^2 b^4) + 1/(a^3 b^3) + 1/(a^4 b^2) + 1/(
    a^5 b)) (a - b) z^4 + (1/(a b^6) + 1/(a^2 b^5) + 1/(a^3 b^4) + 1/(
    a^4 b^3) + 1/(a^5 b^2) + 1/(a^6 b)) (a - b) z^5 + (1/(a b^7) + 1/(
    a^2 b^6) + 1/(a^3 b^5) + 1/(a^4 b^4) + 1/(a^5 b^3) + 1/(
    a^6 b^2) + 1/(a^7 b)) (a - b) z^6 + (1/(a b^8) + 1/(a^2 b^7) + 1/(
    a^3 b^6) + 1/(a^4 b^5) + 1/(a^5 b^4) + 1/(a^6 b^3) + 1/(
    a^7 b^2) + 1/(a^8 b)) (a - b) z^7 + (1/(a b^9) + 1/(a^2 b^8) + 1/(
    a^3 b^7) + 1/(a^4 b^6) + 1/(a^5 b^5) + 1/(a^6 b^4) + 1/(
    a^7 b^3) + 1/(a^8 b^2) + 1/(a^9 b)) (a - b) z^8 + (1/(a b^10) + 
    1/(a^2 b^9) + 1/(a^3 b^8) + 1/(a^4 b^7) + 1/(a^5 b^6) + 1/(
    a^6 b^5) + 1/(a^7 b^4) + 1/(a^8 b^3) + 1/(a^9 b^2) + 1/(
    a^10 b)) (a - b) z^9 + (1/(a b^11) + 1/(a^2 b^10) + 1/(a^3 b^9) + 
    1/(a^4 b^8) + 1/(a^5 b^7) + 1/(a^6 b^6) + 1/(a^7 b^5) + 1/(
    a^8 b^4) + 1/(a^9 b^3) + 1/(a^10 b^2) + 1/(a^11 b)) (a - b) z^10
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  • $\begingroup$ but this is not a Laurent expansion of the function in the annulus, this is just the Taylor expansion of the function $\endgroup$ – Masacroso Apr 2 '18 at 15:36
  • $\begingroup$ Series does give the Laurent expansion. For your example, there is simply no negative power term in the Laurent expansion. $\endgroup$ – Subho Apr 2 '18 at 15:43
  • $\begingroup$ this is not correct, the negative coefficients are $c_{-n}=a^{n-1}$. See the comment that I give in the other answer. $\endgroup$ – Masacroso Apr 2 '18 at 15:44
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    $\begingroup$ The assumptions are not compatible with expanding at the origin. Which can mess things up in an internal use of Refine inside Series. $\endgroup$ – Daniel Lichtblau Apr 2 '18 at 16:27
  • $\begingroup$ @Masacroso Yeah, you are right $\endgroup$ – Subho Apr 2 '18 at 16:50

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