- #1

- 661

- 16

[tex]G(x,t)=e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=-\infty}J_n(x)t^n[/tex]

Why here we have Laurent series, even in case when functions are of real variables?

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- A
- Thread starter LagrangeEuler
- Start date

- #1

- 661

- 16

[tex]G(x,t)=e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=-\infty}J_n(x)t^n[/tex]

Why here we have Laurent series, even in case when functions are of real variables?

- #2

- 17,244

- 7,063

$$

\nabla^2 e^{r(e^{i\varphi}- e^{-i\varphi})/2} = \nabla^2 e^{r\sin(\varphi)} = e^{r\sin(\varphi)}.

$$

Correspondingly, you will find that

$$

(\nabla^2 + 1) f(r,\varphi) = \sum_{n=-\infty}^\infty e^{in\varphi} \left( \frac{1}{r} \partial_r r \partial_r J_n(r) + \left(1 - \frac{n^2}{r^2}\right) J_n(r)\right) = 0.

$$

Since each ##e^{in\varphi}## is linearly independent, this can only be satisfied if

$$

\frac{1}{r} \partial_r r \partial_r J_n(r) + \left(1 - \frac{n^2}{r^2}\right) J_n(r) = 0,

$$

which is Bessel's differential equation.

- #3

- 17,244

- 7,063

$$

\nabla^2 e^{r(e^{i\varphi}- e^{-i\varphi})/2} = \nabla^2 e^{r\sin(\varphi)} = e^{r\sin(\varphi)}.

$$

Correspondingly, you will find that

$$

(\nabla^2 + 1) f(r,\varphi) = \sum_{n=-\infty}^\infty e^{in\varphi} \left( \frac{1}{r} \partial_r r \partial_r J_n(r) + \left(1 - \frac{n^2}{r^2}\right) J_n(r)\right) = 0.

$$

Since each ##e^{in\varphi}## is linearly independent, this can only be satisfied if

$$

\frac{1}{r} \partial_r r \partial_r J_n(r) + \left(1 - \frac{n^2}{r^2}\right) J_n(r) = 0,

$$

which is Bessel's differential equation.

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