13
$\begingroup$

I've tried the following but it didn't work:

Residue[Exp[z - 1/z], {z, 0}]

not even this:

Residue[Exp[1/z], {z, 0}]

Manually, I've computed the residue in the following way:

$$ f(z)=e^{z-\frac{1}{z}}=e^ze^{-\frac{1}{z}}=\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\left(\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}z^{-n}\right)= $$ $$ = \sum_{n=-\infty}^{\infty}\left(\sum_{k=\max(0,n)}^{\infty}\frac{(-1)^{k-n}}{k!(k-n)!}\right)z^n $$

Since the residue is the coefficient of $z^{-1}$, it's equal to:

$$ \sum_{k=0}^{\infty}\frac{(-1)^{k + 1}}{k! (k + 1)!} $$

Going back to Mathematica, I get:

Sum[(-1)^(k + 1)/(k! (k + 1)!), {k, 0, Infinity}]
(* -BesselJ[1, 2] *)

N[-BesselJ[1, 2], 30]
(* -0.576724807756873387202448242269 *)

How can I get that result without manual computations?

Of course, the symbolic result is much preferred.

$\endgroup$
  • 5
    $\begingroup$ You can get the numeric result by using Cauchy's theorem : NIntegrate[ Exp[z - 1/z], {z, 1 + I, -1 + I, -1 - I, 1 - I, 1 + I}]/(2 Pi I). $\endgroup$ – b.gates.you.know.what Feb 5 '14 at 21:32
11
$\begingroup$

You can use b.gatessucks idea to make it analytically:

f[t_] = Exp[z - 1/z] /. z -> E^(I t) // FullSimplify

then integrate:

1/(2 Pi) Integrate[E^(I t) f[t], {t, 0, 2 Pi}]
(* -BesselJ[1, 2] *)
$\endgroup$
1
$\begingroup$

You can use:

$\exp \left[\frac{1}{2} t\left(z-\frac{1}{z}\right)\right]=\sum_{k=-\infty }^{\infty } z^k J_k(t)$

with $t=2$. So rhs is basically the Laurent series of f[t] in $z=0$, so the coefficient $a_{-1}$ of the series is the Residueso $J_{-1}(2)=-J_1(2)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.