# Recurrence relations for Bessel functions

Related topics: Bessel functions

## Consecutive orders

$J_{\nu}\!\left(z\right) = \frac{z}{2 \nu} \left(J_{\nu - 1}\!\left(z\right) + J_{\nu + 1}\!\left(z\right)\right)$
$Y_{\nu}\!\left(z\right) = \frac{z}{2 \nu} \left(Y_{\nu - 1}\!\left(z\right) + Y_{\nu + 1}\!\left(z\right)\right)$
$I_{\nu}\!\left(z\right) = \frac{z}{2 \nu} \left(I_{\nu - 1}\!\left(z\right) - I_{\nu + 1}\!\left(z\right)\right)$
$K_{\nu}\!\left(z\right) = -\frac{z}{2 \nu} \left(K_{\nu - 1}\!\left(z\right) - K_{\nu + 1}\!\left(z\right)\right)$

## Derivatives in terms of the direct functions

$J'_{0}\!\left(z\right) = -J_{1}\!\left(z\right)$
$Y'_{0}\!\left(z\right) = -Y_{1}\!\left(z\right)$
$I'_{0}\!\left(z\right) = I_{1}\!\left(z\right)$
$K'_{0}\!\left(z\right) = -K_{1}\!\left(z\right)$
$J'_{\nu}\!\left(z\right) = \frac{J_{\nu - 1}\!\left(z\right) - J_{\nu + 1}\!\left(z\right)}{2}$
$Y'_{\nu}\!\left(z\right) = \frac{Y_{\nu - 1}\!\left(z\right) - Y_{\nu + 1}\!\left(z\right)}{2}$
$I'_{\nu}\!\left(z\right) = \frac{I_{\nu - 1}\!\left(z\right) + I_{\nu + 1}\!\left(z\right)}{2}$
$K'_{\nu}\!\left(z\right) = -\frac{K_{\nu - 1}\!\left(z\right) + K_{\nu + 1}\!\left(z\right)}{2}$
$J^{(r)}_{\nu}\!\left(z\right) = \frac{1}{{2}^{r}} \sum_{k=0}^{r} {\left(-1\right)}^{k} {r \choose k} J_{\nu + 2 k - r}\!\left(z\right)$
$Y^{(r)}_{\nu}\!\left(z\right) = \frac{1}{{2}^{r}} \sum_{k=0}^{r} {\left(-1\right)}^{k} {r \choose k} Y_{\nu + 2 k - r}\!\left(z\right)$
$I^{(r)}_{\nu}\!\left(z\right) = \frac{1}{{2}^{r}} \sum_{k=0}^{r} {r \choose k} I_{\nu + 2 k - r}\!\left(z\right)$
$K^{(r)}_{\nu}\!\left(z\right) = \frac{{\left(-1\right)}^{r}}{{2}^{r}} \sum_{k=0}^{r} {r \choose k} K_{\nu + 2 k - r}\!\left(z\right)$

## Consecutive derivatives

$\left({r}^{2} + 4 r - {n}^{2} + 4\right) J^{(r + 2)}_{n}\!\left(0\right) + \left(r + 1\right) \left(r + 2\right) J^{(r)}_{n}\!\left(0\right) = 0$
$\left({r}^{2} + 4 r - {n}^{2} + 4\right) I^{(r + 2)}_{n}\!\left(0\right) - \left(r + 1\right) \left(r + 2\right) I^{(r)}_{n}\!\left(0\right) = 0$
${z}^{2} \left({r}^{2} + 7 r + 12\right) \frac{J^{(r + 4)}_{\nu}\!\left(z\right)}{\left(r + 4\right)!} + z \left(2 {r}^{2} + 11 r + 15\right) \frac{J^{(r + 3)}_{\nu}\!\left(z\right)}{\left(r + 3\right)!} + \left(r \left(r + 4\right) + {z}^{2} - {\nu}^{2} + 4\right) \frac{J^{(r + 2)}_{\nu}\!\left(z\right)}{\left(r + 2\right)!} + 2 z \frac{J^{(r + 1)}_{\nu}\!\left(z\right)}{\left(r + 1\right)!} + \frac{J^{(r)}_{\nu}\!\left(z\right)}{r !} = 0$
${z}^{2} \left({r}^{2} + 7 r + 12\right) \frac{Y^{(r + 4)}_{\nu}\!\left(z\right)}{\left(r + 4\right)!} + z \left(2 {r}^{2} + 11 r + 15\right) \frac{Y^{(r + 3)}_{\nu}\!\left(z\right)}{\left(r + 3\right)!} + \left(r \left(r + 4\right) + {z}^{2} - {\nu}^{2} + 4\right) \frac{Y^{(r + 2)}_{\nu}\!\left(z\right)}{\left(r + 2\right)!} + 2 z \frac{Y^{(r + 1)}_{\nu}\!\left(z\right)}{\left(r + 1\right)!} + \frac{Y^{(r)}_{\nu}\!\left(z\right)}{r !} = 0$
${z}^{2} \left({r}^{2} + 7 r + 12\right) \frac{I^{(r + 4)}_{\nu}\!\left(z\right)}{\left(r + 4\right)!} + z \left(2 {r}^{2} + 11 r + 15\right) \frac{I^{(r + 3)}_{\nu}\!\left(z\right)}{\left(r + 3\right)!} + \left(r \left(r + 4\right) - {z}^{2} - {\nu}^{2} + 4\right) \frac{I^{(r + 2)}_{\nu}\!\left(z\right)}{\left(r + 2\right)!} - 2 z \frac{I^{(r + 1)}_{\nu}\!\left(z\right)}{\left(r + 1\right)!} - \frac{I^{(r)}_{\nu}\!\left(z\right)}{r !} = 0$
${z}^{2} \left({r}^{2} + 7 r + 12\right) \frac{K^{(r + 4)}_{\nu}\!\left(z\right)}{\left(r + 4\right)!} + z \left(2 {r}^{2} + 11 r + 15\right) \frac{K^{(r + 3)}_{\nu}\!\left(z\right)}{\left(r + 3\right)!} + \left(r \left(r + 4\right) - {z}^{2} - {\nu}^{2} + 4\right) \frac{K^{(r + 2)}_{\nu}\!\left(z\right)}{\left(r + 2\right)!} - 2 z \frac{K^{(r + 1)}_{\nu}\!\left(z\right)}{\left(r + 1\right)!} - \frac{K^{(r)}_{\nu}\!\left(z\right)}{r !} = 0$

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2021-03-15 19:12:00.328586 UTC