Bessel Functions - 10.6 Recurrence Relations and Derivatives

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DLMF Formula Constraints Maple Mathematica Symbolic
Maple 		Symbolic
Mathematica 		Numeric
Maple 		Numeric
Mathematica
10.6#E3X J 0 ′ ⁡ ( z ) = - J 1 ⁡ ( z ) diffop Bessel-J 0 1 𝑧 Bessel-J 1 𝑧 {\displaystyle{\displaystyle\displaystyle J_{0}'\left(z\right)=-J_{1}\left(z% \right)}}
\displaystyle\BesselJ{0}'@{z} = -\BesselJ{1}@{z}		 ℜ ⁡ ( 0 + k + 1 ) > 0 , ℜ ⁡ ( 1 + k + 1 ) > 0 formulae-sequence 0 𝑘 1 0 1 𝑘 1 0 {\displaystyle{\displaystyle\Re(0+k+1)>0,\Re(1+k+1)>0}} 
diff( BesselJ(0, z), z$(1) ) = - BesselJ(1, z)
 
D[BesselJ[0, z], {z, 1}] == - BesselJ[1, z]
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10.6#E3X Y 0 ′ ⁡ ( z ) = - Y 1 ⁡ ( z ) diffop Bessel-Y-Weber 0 1 𝑧 Bessel-Y-Weber 1 𝑧 {\displaystyle{\displaystyle\displaystyle Y_{0}'\left(z\right)=-Y_{1}\left(z% \right)}}
\displaystyle\BesselY{0}'@{z} = -\BesselY{1}@{z}		 ℜ ⁡ ( 0 + k + 1 ) > 0 , ℜ ⁡ ( 1 + k + 1 ) > 0 , ℜ ⁡ ( ( - 0 ) + k + 1 ) > 0 , ℜ ⁡ ( ( - 1 ) + k + 1 ) > 0 formulae-sequence 0 𝑘 1 0 formulae-sequence 1 𝑘 1 0 formulae-sequence 0 𝑘 1 0 1 𝑘 1 0 {\displaystyle{\displaystyle\Re(0+k+1)>0,\Re(1+k+1)>0,\Re((-0)+k+1)>0,\Re((-1)% +k+1)>0}} 
diff( BesselY(0, z), z$(1) ) = - BesselY(1, z)
 
D[BesselY[0, z], {z, 1}] == - BesselY[1, z]
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10.6#E3Xa H 0 ( 1 ) ′ ⁡ ( z ) = - H 1 ( 1 ) ⁡ ( z ) diffop Hankel-H-1-Bessel-third-kind 0 1 𝑧 Hankel-H-1-Bessel-third-kind 1 𝑧 {\displaystyle{\displaystyle\displaystyle{H^{(1)}_{0}}'\left(z\right)=-{H^{(1)% }_{1}}\left(z\right)}}
\displaystyle\HankelH{1}{0}'@{z} = -\HankelH{1}{1}@{z}		 

diff( HankelH1(0, z), z$(1) ) = - HankelH1(1, z)
 
D[HankelH1[0, z], {z, 1}] == - HankelH1[1, z]
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10.6#E3Xa H 0 ( 2 ) ′ ⁡ ( z ) = - H 1 ( 2 ) ⁡ ( z ) diffop Hankel-H-2-Bessel-third-kind 0 1 𝑧 Hankel-H-2-Bessel-third-kind 1 𝑧 {\displaystyle{\displaystyle\displaystyle{H^{(2)}_{0}}'\left(z\right)=-{H^{(2)% }_{1}}\left(z\right)}}
\displaystyle\HankelH{2}{0}'@{z} = -\HankelH{2}{1}@{z}		 

diff( HankelH2(0, z), z$(1) ) = - HankelH2(1, z)
 
D[HankelH2[0, z], {z, 1}] == - HankelH2[1, z]
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10.6#Ex5 f ν - 1 ⁢ ( z ) + f ν + 1 ⁢ ( z ) = ( 2 ⁢ ν / λ ) ⁢ z - q ⁢ f ν ⁢ ( z ) subscript 𝑓 𝜈 1 𝑧 subscript 𝑓 𝜈 1 𝑧 2 𝜈 𝜆 superscript 𝑧 𝑞 subscript 𝑓 𝜈 𝑧 {\displaystyle{\displaystyle f_{\nu-1}(z)+f_{\nu+1}(z)=(2\nu/\lambda)z^{-q}f_{% \nu}(z)}}
f_{\nu-1}(z)+f_{\nu+1}(z) = (2\nu/\lambda)z^{-q}f_{\nu}(z)		 

f[nu - 1](z)+ f[nu + 1](z) = (2*nu/lambda)*(z)^(- q)* f[nu](z)
 
Subscript[f, \[Nu]- 1][z]+ Subscript[f, \[Nu]+ 1][z] == (2*\[Nu]/\[Lambda])*(z)^(- q)* Subscript[f, \[Nu]][z]
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10.6#Ex15 p ν + 1 - p ν - 1 = - 2 ⁢ ν a ⁢ q ν - 2 ⁢ ν b ⁢ r ν subscript 𝑝 𝜈 1 subscript 𝑝 𝜈 1 2 𝜈 𝑎 subscript 𝑞 𝜈 2 𝜈 𝑏 subscript 𝑟 𝜈 {\displaystyle{\displaystyle p_{\nu+1}-p_{\nu-1}=-\frac{2\nu}{a}q_{\nu}-\frac{% 2\nu}{b}r_{\nu}}}
p_{\nu+1}-p_{\nu-1} = -\frac{2\nu}{a}q_{\nu}-\frac{2\nu}{b}r_{\nu}		 

p[nu + 1]- p[nu - 1] = -(2*nu)/(a)*q[nu]-(2*nu)/(b)*r[nu]
 
Subscript[p, \[Nu]+ 1]- Subscript[p, \[Nu]- 1] == -Divide[2*\[Nu],a]*Subscript[q, \[Nu]]-Divide[2*\[Nu],b]*Subscript[r, \[Nu]]
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10.6#Ex16 q ν + 1 + r ν = ν a ⁢ p ν - ν + 1 b ⁢ p ν + 1 subscript 𝑞 𝜈 1 subscript 𝑟 𝜈 𝜈 𝑎 subscript 𝑝 𝜈 𝜈 1 𝑏 subscript 𝑝 𝜈 1 {\displaystyle{\displaystyle q_{\nu+1}+r_{\nu}=\frac{\nu}{a}p_{\nu}-\frac{\nu+% 1}{b}p_{\nu+1}}}
q_{\nu+1}+r_{\nu} = \frac{\nu}{a}p_{\nu}-\frac{\nu+1}{b}p_{\nu+1}		 

q[nu + 1]+ r[nu] = (nu)/(a)*p[nu]-(nu + 1)/(b)*p[nu + 1]
 
Subscript[q, \[Nu]+ 1]+ Subscript[r, \[Nu]] == Divide[\[Nu],a]*Subscript[p, \[Nu]]-Divide[\[Nu]+ 1,b]*Subscript[p, \[Nu]+ 1]
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10.6#Ex17 r ν + 1 + q ν = ν b ⁢ p ν - ν + 1 a ⁢ p ν + 1 subscript 𝑟 𝜈 1 subscript 𝑞 𝜈 𝜈 𝑏 subscript 𝑝 𝜈 𝜈 1 𝑎 subscript 𝑝 𝜈 1 {\displaystyle{\displaystyle r_{\nu+1}+q_{\nu}=\frac{\nu}{b}p_{\nu}-\frac{\nu+% 1}{a}p_{\nu+1}}}
r_{\nu+1}+q_{\nu} = \frac{\nu}{b}p_{\nu}-\frac{\nu+1}{a}p_{\nu+1}		 

r[nu + 1]+ q[nu] = (nu)/(b)*p[nu]-(nu + 1)/(a)*p[nu + 1]
 
Subscript[r, \[Nu]+ 1]+ Subscript[q, \[Nu]] == Divide[\[Nu],b]*Subscript[p, \[Nu]]-Divide[\[Nu]+ 1,a]*Subscript[p, \[Nu]+ 1]
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10.6#Ex18 s ν = 1 2 ⁢ p ν + 1 + 1 2 ⁢ p ν - 1 - ν 2 a ⁢ b ⁢ p ν subscript 𝑠 𝜈 1 2 subscript 𝑝 𝜈 1 1 2 subscript 𝑝 𝜈 1 superscript 𝜈 2 𝑎 𝑏 subscript 𝑝 𝜈 {\displaystyle{\displaystyle s_{\nu}=\tfrac{1}{2}p_{\nu+1}+\tfrac{1}{2}p_{\nu-% 1}-\frac{\nu^{2}}{ab}p_{\nu}}}
s_{\nu} = \tfrac{1}{2}p_{\nu+1}+\tfrac{1}{2}p_{\nu-1}-\frac{\nu^{2}}{ab}p_{\nu}		 ℜ ⁡ ( ν + k + 1 ) > 0 , ℜ ⁡ ( ( - ν ) + k + 1 ) > 0 formulae-sequence 𝜈 𝑘 1 0 𝜈 𝑘 1 0 {\displaystyle{\displaystyle\Re(\nu+k+1)>0,\Re((-\nu)+k+1)>0}} 
(diff( BesselJ(nu, a), a$(1) )*diff( BesselY(nu, b), b$(1) )- diff( BesselJ(nu, b), b$(1) )*diff( BesselY(nu, a), a$(1) )) = (1)/(2)*p[nu + 1]+(1)/(2)*p[nu - 1]-((nu)^(2))/(a*b)*p[nu]
 
(D[BesselJ[\[Nu], a], {a, 1}]*D[BesselY[\[Nu], b], {b, 1}]- D[BesselJ[\[Nu], b], {b, 1}]*D[BesselY[\[Nu], a], {a, 1}]) == Divide[1,2]*Subscript[p, \[Nu]+ 1]+Divide[1,2]*Subscript[p, \[Nu]- 1]-Divide[\[Nu]^(2),a*b]*Subscript[p, \[Nu]]
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10.6.E10 p ν ⁢ s ν - q ν ⁢ r ν = 4 / ( π 2 ⁢ a ⁢ b ) subscript 𝑝 𝜈 subscript 𝑠 𝜈 subscript 𝑞 𝜈 subscript 𝑟 𝜈 4 superscript 𝜋 2 𝑎 𝑏 {\displaystyle{\displaystyle p_{\nu}s_{\nu}-q_{\nu}r_{\nu}=4/(\pi^{2}ab)}}
p_{\nu}s_{\nu}-q_{\nu}r_{\nu} = 4/(\pi^{2}ab)		 ℜ ⁡ ( ν + k + 1 ) > 0 , ℜ ⁡ ( ( - ν ) + k + 1 ) > 0 formulae-sequence 𝜈 𝑘 1 0 𝜈 𝑘 1 0 {\displaystyle{\displaystyle\Re(\nu+k+1)>0,\Re((-\nu)+k+1)>0}} 
p[nu]*(diff( BesselJ(nu, a), a$(1) )*diff( BesselY(nu, b), b$(1) )- diff( BesselJ(nu, b), b$(1) )*diff( BesselY(nu, a), a$(1) ))- q[nu]*r[nu] = 4/((Pi)^(2)* a*b)
 
Subscript[p, \[Nu]]*(D[BesselJ[\[Nu], a], {a, 1}]*D[BesselY[\[Nu], b], {b, 1}]- D[BesselJ[\[Nu], b], {b, 1}]*D[BesselY[\[Nu], a], {a, 1}])- Subscript[q, \[Nu]]*Subscript[r, \[Nu]] == 4/((Pi)^(2)* a*b)
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