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Recurrence relations for Bessel functions

Table of contents: Consecutive orders - Derivatives in terms of the direct functions - Consecutive derivatives

Related topics: Bessel functions

Consecutive orders

d56914
Jν ⁣(z)=z2ν(Jν1 ⁣(z)+Jν+1 ⁣(z))J_{\nu}\!\left(z\right) = \frac{z}{2 \nu} \left(J_{\nu - 1}\!\left(z\right) + J_{\nu + 1}\!\left(z\right)\right)
b6d600
Yν ⁣(z)=z2ν(Yν1 ⁣(z)+Yν+1 ⁣(z))Y_{\nu}\!\left(z\right) = \frac{z}{2 \nu} \left(Y_{\nu - 1}\!\left(z\right) + Y_{\nu + 1}\!\left(z\right)\right)
4fb391
Iν ⁣(z)=z2ν(Iν1 ⁣(z)Iν+1 ⁣(z))I_{\nu}\!\left(z\right) = \frac{z}{2 \nu} \left(I_{\nu - 1}\!\left(z\right) - I_{\nu + 1}\!\left(z\right)\right)
9d98f8
Kν ⁣(z)=z2ν(Kν1 ⁣(z)Kν+1 ⁣(z))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

f1afc0
J0 ⁣(z)=J1 ⁣(z)J'_{0}\!\left(z\right) = -J_{1}\!\left(z\right)
8b6264
Y0 ⁣(z)=Y1 ⁣(z)Y'_{0}\!\left(z\right) = -Y_{1}\!\left(z\right)
c0247f
I0 ⁣(z)=I1 ⁣(z)I'_{0}\!\left(z\right) = I_{1}\!\left(z\right)
81ffcd
K0 ⁣(z)=K1 ⁣(z)K'_{0}\!\left(z\right) = -K_{1}\!\left(z\right)
5aceb9
Jν ⁣(z)=Jν1 ⁣(z)Jν+1 ⁣(z)2J'_{\nu}\!\left(z\right) = \frac{J_{\nu - 1}\!\left(z\right) - J_{\nu + 1}\!\left(z\right)}{2}
40aeb6
Yν ⁣(z)=Yν1 ⁣(z)Yν+1 ⁣(z)2Y'_{\nu}\!\left(z\right) = \frac{Y_{\nu - 1}\!\left(z\right) - Y_{\nu + 1}\!\left(z\right)}{2}
58d91f
Iν ⁣(z)=Iν1 ⁣(z)+Iν+1 ⁣(z)2I'_{\nu}\!\left(z\right) = \frac{I_{\nu - 1}\!\left(z\right) + I_{\nu + 1}\!\left(z\right)}{2}
a0ff0b
Kν ⁣(z)=Kν1 ⁣(z)+Kν+1 ⁣(z)2K'_{\nu}\!\left(z\right) = -\frac{K_{\nu - 1}\!\left(z\right) + K_{\nu + 1}\!\left(z\right)}{2}
2488bb
Jν(r) ⁣(z)=12rk=0r(1)k(rk)Jν+2kr ⁣(z)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)
68cc2f
Yν(r) ⁣(z)=12rk=0r(1)k(rk)Yν+2kr ⁣(z)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)
e284d7
Iν(r) ⁣(z)=12rk=0r(rk)Iν+2kr ⁣(z)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)
807f3f
Kν(r) ⁣(z)=(1)r2rk=0r(rk)Kν+2kr ⁣(z)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

15ac84
(r2+4rn2+4)Jn(r+2) ⁣(0)+(r+1)(r+2)Jn(r) ⁣(0)=0\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
f303c9
(r2+4rn2+4)In(r+2) ⁣(0)(r+1)(r+2)In(r) ⁣(0)=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
9b2f38
z2(r2+7r+12)Jν(r+4) ⁣(z)(r+4)!+z(2r2+11r+15)Jν(r+3) ⁣(z)(r+3)!+(r(r+4)+z2ν2+4)Jν(r+2) ⁣(z)(r+2)!+2zJν(r+1) ⁣(z)(r+1)!+Jν(r) ⁣(z)r!=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
e85dee
z2(r2+7r+12)Yν(r+4) ⁣(z)(r+4)!+z(2r2+11r+15)Yν(r+3) ⁣(z)(r+3)!+(r(r+4)+z2ν2+4)Yν(r+2) ⁣(z)(r+2)!+2zYν(r+1) ⁣(z)(r+1)!+Yν(r) ⁣(z)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
e233b0
z2(r2+7r+12)Iν(r+4) ⁣(z)(r+4)!+z(2r2+11r+15)Iν(r+3) ⁣(z)(r+3)!+(r(r+4)z2ν2+4)Iν(r+2) ⁣(z)(r+2)!2zIν(r+1) ⁣(z)(r+1)!Iν(r) ⁣(z)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
7377c8
z2(r2+7r+12)Kν(r+4) ⁣(z)(r+4)!+z(2r2+11r+15)Kν(r+3) ⁣(z)(r+3)!+(r(r+4)z2ν2+4)Kν(r+2) ⁣(z)(r+2)!2zKν(r+1) ⁣(z)(r+1)!Kν(r) ⁣(z)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

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2019-10-05 13:11:19.856591 UTC