Mathieu Functions and Hill’s Equation - 28.4 Fourier Series

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DLMF Formula Constraints Maple Mathematica Symbolic
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Mathematica 		Numeric
Maple 		Numeric
Mathematica
28.4.E1 ce 2 ⁢ n ⁡ ( z , q ) = ∑ m = 0 ∞ A 2 ⁢ m 2 ⁢ n ⁢ ( q ) ⁢ cos ⁡ 2 ⁢ m ⁢ z Mathieu-ce 2 𝑛 𝑧 𝑞 superscript subscript 𝑚 0 subscript superscript 𝐴 2 𝑛 2 𝑚 𝑞 2 𝑚 𝑧 {\displaystyle{\displaystyle\mathrm{ce}_{2n}\left(z,q\right)=\sum_{m=0}^{% \infty}A^{2n}_{2m}(q)\cos 2mz}}
\Mathieuce{2n}@{z}{q} = \sum_{m=0}^{\infty}A^{2n}_{2m}(q)\cos@@{2mz}		 

MathieuCE(2*n, q, z) = sum((A[2*m])^(2*n)(q)* cos(2*m*z), m = 0..infinity)

MathieuC[2*n, q, z] == Sum[(Subscript[A, 2*m])^(2*n)[q]* Cos[2*m*z], {m, 0, Infinity}, GenerateConditions->None]
Failure Failure Skipped - Because timed out Skipped - Because timed out
28.4.E2 ce 2 ⁢ n + 1 ⁡ ( z , q ) = ∑ m = 0 ∞ A 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) ⁢ cos ⁡ ( 2 ⁢ m + 1 ) ⁢ z Mathieu-ce 2 𝑛 1 𝑧 𝑞 superscript subscript 𝑚 0 subscript superscript 𝐴 2 𝑛 1 2 𝑚 1 𝑞 2 𝑚 1 𝑧 {\displaystyle{\displaystyle\mathrm{ce}_{2n+1}\left(z,q\right)=\sum_{m=0}^{% \infty}A^{2n+1}_{2m+1}(q)\cos(2m+1)z}}
\Mathieuce{2n+1}@{z}{q} = \sum_{m=0}^{\infty}A^{2n+1}_{2m+1}(q)\cos@@{(2m+1)z}		 

MathieuCE(2*n + 1, q, z) = sum((A[2*m + 1])^(2*n + 1)(q)* cos((2*m + 1)*z), m = 0..infinity)

MathieuC[2*n + 1, q, z] == Sum[(Subscript[A, 2*m + 1])^(2*n + 1)[q]* Cos[(2*m + 1)*z], {m, 0, Infinity}, GenerateConditions->None]
Failure Failure Skipped - Because timed out Skipped - Because timed out
28.4.E3 se 2 ⁢ n + 1 ⁡ ( z , q ) = ∑ m = 0 ∞ B 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) ⁢ sin ⁡ ( 2 ⁢ m + 1 ) ⁢ z Mathieu-se 2 𝑛 1 𝑧 𝑞 superscript subscript 𝑚 0 subscript superscript 𝐵 2 𝑛 1 2 𝑚 1 𝑞 2 𝑚 1 𝑧 {\displaystyle{\displaystyle\mathrm{se}_{2n+1}\left(z,q\right)=\sum_{m=0}^{% \infty}B^{2n+1}_{2m+1}(q)\sin(2m+1)z}}
\Mathieuse{2n+1}@{z}{q} = \sum_{m=0}^{\infty}B^{2n+1}_{2m+1}(q)\sin@@{(2m+1)z}		 

MathieuSE(2*n + 1, q, z) = sum((B[2*m + 1])^(2*n + 1)(q)* sin((2*m + 1)*z), m = 0..infinity)

MathieuS[2*n + 1, q, z] == Sum[(Subscript[B, 2*m + 1])^(2*n + 1)[q]* Sin[(2*m + 1)*z], {m, 0, Infinity}, GenerateConditions->None]
Failure Failure Skipped - Because timed out Skipped - Because timed out
28.4.E4 se 2 ⁢ n + 2 ⁡ ( z , q ) = ∑ m = 0 ∞ B 2 ⁢ m + 2 2 ⁢ n + 2 ⁢ ( q ) ⁢ sin ⁡ ( 2 ⁢ m + 2 ) ⁢ z Mathieu-se 2 𝑛 2 𝑧 𝑞 superscript subscript 𝑚 0 subscript superscript 𝐵 2 𝑛 2 2 𝑚 2 𝑞 2 𝑚 2 𝑧 {\displaystyle{\displaystyle\mathrm{se}_{2n+2}\left(z,q\right)=\sum_{m=0}^{% \infty}B^{2n+2}_{2m+2}(q)\sin(2m+2)z}}
\Mathieuse{2n+2}@{z}{q} = \sum_{m=0}^{\infty}B^{2n+2}_{2m+2}(q)\sin@@{(2m+2)z}		 

MathieuSE(2*n + 2, q, z) = sum((B[2*m + 2])^(2*n + 2)(q)* sin((2*m + 2)*z), m = 0..infinity)

MathieuS[2*n + 2, q, z] == Sum[(Subscript[B, 2*m + 2])^(2*n + 2)[q]* Sin[(2*m + 2)*z], {m, 0, Infinity}, GenerateConditions->None]
Failure Failure Skipped - Because timed out Skipped - Because timed out
28.4#Ex1 a ⁢ A 0 - q ⁢ A 2 = 0 𝑎 subscript 𝐴 0 𝑞 subscript 𝐴 2 0 {\displaystyle{\displaystyle aA_{0}-qA_{2}=0}}
aA_{0}-qA_{2} = 0		 

a*A[0]- q*A[2] = 0
 
a*Subscript[A, 0]- q*Subscript[A, 2] == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex2 ( a - 4 ) ⁢ A 2 - q ⁢ ( 2 ⁢ A 0 + A 4 ) = 0 𝑎 4 subscript 𝐴 2 𝑞 2 subscript 𝐴 0 subscript 𝐴 4 0 {\displaystyle{\displaystyle(a-4)A_{2}-q(2A_{0}+A_{4})=0}}
(a-4)A_{2}-q(2A_{0}+A_{4}) = 0		 

(a - 4)*A[2]- q*(2*A[0]+ A[4]) = 0
 
(a - 4)*Subscript[A, 2]- q*(2*Subscript[A, 0]+ Subscript[A, 4]) == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex3 ( a - 4 ⁢ m 2 ) ⁢ A 2 ⁢ m - q ⁢ ( A 2 ⁢ m - 2 + A 2 ⁢ m + 2 ) = 0 𝑎 4 superscript 𝑚 2 subscript 𝐴 2 𝑚 𝑞 subscript 𝐴 2 𝑚 2 subscript 𝐴 2 𝑚 2 0 {\displaystyle{\displaystyle(a-4m^{2})A_{2m}-q(A_{2m-2}+A_{2m+2})=0}}
(a-4m^{2})A_{2m}-q(A_{2m-2}+A_{2m+2}) = 0		 a = a 2 ⁢ n ⁡ ( q ) , A 2 ⁢ m = A 2 ⁢ m 2 ⁢ n ⁢ ( q ) formulae-sequence 𝑎 Mathieu-eigenvalue-a 2 𝑛 𝑞 subscript 𝐴 2 𝑚 superscript subscript 𝐴 2 𝑚 2 𝑛 𝑞 {\displaystyle{\displaystyle a=a_{2n}\left(q\right),A_{2m}=A_{2m}^{2n}(q)}} 
(a - 4*(m)^(2))*A[2*m]- q*(A[2*m - 2]+ A[2*m + 2]) = 0
 
(a - 4*(m)^(2))*Subscript[A, 2*m]- q*(Subscript[A, 2*m - 2]+ Subscript[A, 2*m + 2]) == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex4 ( a - 1 - q ) ⁢ A 1 - q ⁢ A 3 = 0 𝑎 1 𝑞 subscript 𝐴 1 𝑞 subscript 𝐴 3 0 {\displaystyle{\displaystyle(a-1-q)A_{1}-qA_{3}=0}}
(a-1-q)A_{1}-qA_{3} = 0		 

(a - 1 - q)*A[1]- q*A[3] = 0
 
(a - 1 - q)*Subscript[A, 1]- q*Subscript[A, 3] == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex5 ( a - ( 2 ⁢ m + 1 ) 2 ) ⁢ A 2 ⁢ m + 1 - q ⁢ ( A 2 ⁢ m - 1 + A 2 ⁢ m + 3 ) = 0 𝑎 superscript 2 𝑚 1 2 subscript 𝐴 2 𝑚 1 𝑞 subscript 𝐴 2 𝑚 1 subscript 𝐴 2 𝑚 3 0 {\displaystyle{\displaystyle\left(a-(2m+1)^{2}\right)A_{2m+1}-q(A_{2m-1}+A_{2m% +3})=0}}
\left(a-(2m+1)^{2}\right)A_{2m+1}-q(A_{2m-1}+A_{2m+3}) = 0		 a = a 2 ⁢ n + 1 ⁡ ( q ) , A 2 ⁢ m + 1 = A 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) formulae-sequence 𝑎 Mathieu-eigenvalue-a 2 𝑛 1 𝑞 subscript 𝐴 2 𝑚 1 superscript subscript 𝐴 2 𝑚 1 2 𝑛 1 𝑞 {\displaystyle{\displaystyle a=a_{2n+1}\left(q\right),A_{2m+1}=A_{2m+1}^{2n+1}% (q)}} 
(a -(2*m + 1)^(2))*A[2*m + 1]- q*(A[2*m - 1]+ A[2*m + 3]) = 0
 
(a -(2*m + 1)^(2))*Subscript[A, 2*m + 1]- q*(Subscript[A, 2*m - 1]+ Subscript[A, 2*m + 3]) == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex6 ( a - 1 + q ) ⁢ B 1 - q ⁢ B 3 = 0 𝑎 1 𝑞 subscript 𝐵 1 𝑞 subscript 𝐵 3 0 {\displaystyle{\displaystyle(a-1+q)B_{1}-qB_{3}=0}}
(a-1+q)B_{1}-qB_{3} = 0		 

(a - 1 + q)*B[1]- q*B[3] = 0
 
(a - 1 + q)*Subscript[B, 1]- q*Subscript[B, 3] == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex7 ( a - ( 2 ⁢ m + 1 ) 2 ) ⁢ B 2 ⁢ m + 1 - q ⁢ ( B 2 ⁢ m - 1 + B 2 ⁢ m + 3 ) = 0 𝑎 superscript 2 𝑚 1 2 subscript 𝐵 2 𝑚 1 𝑞 subscript 𝐵 2 𝑚 1 subscript 𝐵 2 𝑚 3 0 {\displaystyle{\displaystyle\left(a-(2m+1)^{2}\right)B_{2m+1}-q(B_{2m-1}+B_{2m% +3})=0}}
\left(a-(2m+1)^{2}\right)B_{2m+1}-q(B_{2m-1}+B_{2m+3}) = 0		 a = b 2 ⁢ n + 1 ⁡ ( q ) , B 2 ⁢ m + 1 = B 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) formulae-sequence 𝑎 Mathieu-eigenvalue-b 2 𝑛 1 𝑞 subscript 𝐵 2 𝑚 1 superscript subscript 𝐵 2 𝑚 1 2 𝑛 1 𝑞 {\displaystyle{\displaystyle a=b_{2n+1}\left(q\right),B_{2m+1}=B_{2m+1}^{2n+1}% (q)}} 
(a -(2*m + 1)^(2))*B[2*m + 1]- q*(B[2*m - 1]+ B[2*m + 3]) = 0
 
(a -(2*m + 1)^(2))*Subscript[B, 2*m + 1]- q*(Subscript[B, 2*m - 1]+ Subscript[B, 2*m + 3]) == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex8 ( a - 4 ) ⁢ B 2 - q ⁢ B 4 = 0 𝑎 4 subscript 𝐵 2 𝑞 subscript 𝐵 4 0 {\displaystyle{\displaystyle(a-4)B_{2}-qB_{4}=0}}
(a-4)B_{2}-qB_{4} = 0		 

(a - 4)*B[2]- q*B[4] = 0
 
(a - 4)*Subscript[B, 2]- q*Subscript[B, 4] == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex9 ( a - 4 ⁢ m 2 ) ⁢ B 2 ⁢ m - q ⁢ ( B 2 ⁢ m - 2 + B 2 ⁢ m + 2 ) = 0 𝑎 4 superscript 𝑚 2 subscript 𝐵 2 𝑚 𝑞 subscript 𝐵 2 𝑚 2 subscript 𝐵 2 𝑚 2 0 {\displaystyle{\displaystyle(a-4m^{2})B_{2m}-q(B_{2m-2}+B_{2m+2})=0}}
(a-4m^{2})B_{2m}-q(B_{2m-2}+B_{2m+2}) = 0		 a = b 2 ⁢ n + 2 ⁡ ( q ) , B 2 ⁢ m + 2 = B 2 ⁢ m + 2 2 ⁢ n + 2 ⁢ ( q ) . formulae-sequence 𝑎 Mathieu-eigenvalue-b 2 𝑛 2 𝑞 subscript 𝐵 2 𝑚 2 superscript subscript 𝐵 2 𝑚 2 2 𝑛 2 𝑞 {\displaystyle{\displaystyle a=b_{2n+2}\left(q\right),B_{2m+2}=B_{2m+2}^{2n+2}% (q).}} 
(a - 4*(m)^(2))*B[2*m]- q*(B[2*m - 2]+ B[2*m + 2]) = 0
 
(a - 4*(m)^(2))*Subscript[B, 2*m]- q*(Subscript[B, 2*m - 2]+ Subscript[B, 2*m + 2]) == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4.E9 2 ⁢ ( A 0 2 ⁢ n ⁢ ( q ) ) 2 + ∑ m = 1 ∞ ( A 2 ⁢ m 2 ⁢ n ⁢ ( q ) ) 2 = 1 2 superscript subscript superscript 𝐴 2 𝑛 0 𝑞 2 superscript subscript 𝑚 1 superscript subscript superscript 𝐴 2 𝑛 2 𝑚 𝑞 2 1 {\displaystyle{\displaystyle 2\left(A^{2n}_{0}(q)\right)^{2}+\sum_{m=1}^{% \infty}\left(A^{2n}_{2m}(q)\right)^{2}=1}}
2\left(A^{2n}_{0}(q)\right)^{2}+\sum_{m=1}^{\infty}\left(A^{2n}_{2m}(q)\right)^{2} = 1		 

2*((A[0])^(2*n)(q))^(2)+ sum(((A[2*m])^(2*n)(q))^(2), m = 1..infinity) = 1
 
2*((Subscript[A, 0])^(2*n)[q])^(2)+ Sum[((Subscript[A, 2*m])^(2*n)[q])^(2), {m, 1, Infinity}, GenerateConditions->None] == 1
 Skipped - no semantic math Skipped - no semantic math - -
28.4.E10 ∑ m = 0 ∞ ( A 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) ) 2 = 1 superscript subscript 𝑚 0 superscript subscript superscript 𝐴 2 𝑛 1 2 𝑚 1 𝑞 2 1 {\displaystyle{\displaystyle\sum_{m=0}^{\infty}\left(A^{2n+1}_{2m+1}(q)\right)% ^{2}=1}}
\sum_{m=0}^{\infty}\left(A^{2n+1}_{2m+1}(q)\right)^{2} = 1		 

sum(((A[2*m + 1])^(2*n + 1)(q))^(2), m = 0..infinity) = 1
 
Sum[((Subscript[A, 2*m + 1])^(2*n + 1)[q])^(2), {m, 0, Infinity}, GenerateConditions->None] == 1
 Skipped - no semantic math Skipped - no semantic math - -
28.4.E11 ∑ m = 0 ∞ ( B 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) ) 2 = 1 superscript subscript 𝑚 0 superscript subscript superscript 𝐵 2 𝑛 1 2 𝑚 1 𝑞 2 1 {\displaystyle{\displaystyle\sum_{m=0}^{\infty}\left(B^{2n+1}_{2m+1}(q)\right)% ^{2}=1}}
\sum_{m=0}^{\infty}\left(B^{2n+1}_{2m+1}(q)\right)^{2} = 1		 

sum(((B[2*m + 1])^(2*n + 1)(q))^(2), m = 0..infinity) = 1
 
Sum[((Subscript[B, 2*m + 1])^(2*n + 1)[q])^(2), {m, 0, Infinity}, GenerateConditions->None] == 1
 Skipped - no semantic math Skipped - no semantic math - -
28.4.E12 ∑ m = 0 ∞ ( B 2 ⁢ m + 2 2 ⁢ n + 2 ⁢ ( q ) ) 2 = 1 superscript subscript 𝑚 0 superscript subscript superscript 𝐵 2 𝑛 2 2 𝑚 2 𝑞 2 1 {\displaystyle{\displaystyle\sum_{m=0}^{\infty}\left(B^{2n+2}_{2m+2}(q)\right)% ^{2}=1}}
\sum_{m=0}^{\infty}\left(B^{2n+2}_{2m+2}(q)\right)^{2} = 1		 

sum(((B[2*m + 2])^(2*n + 2)(q))^(2), m = 0..infinity) = 1
 
Sum[((Subscript[B, 2*m + 2])^(2*n + 2)[q])^(2), {m, 0, Infinity}, GenerateConditions->None] == 1
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex10 A 0 0 ⁢ ( 0 ) = 1 / 2 , A 2 ⁢ n 2 ⁢ n ⁢ ( 0 ) subscript superscript 𝐴 0 0 0 1 2 subscript superscript 𝐴 2 𝑛 2 𝑛 0 {\displaystyle{\displaystyle A^{0}_{0}(0)=1/\sqrt{2},\quad A^{2n}_{2n}(0)}}
A^{0}_{0}(0) = 1/\sqrt{2},\quad A^{2n}_{2n}(0)		 n > 0 𝑛 0 {\displaystyle{\displaystyle n>0}} 
(A[0])^(0)(0) = 1/(sqrt(2))
 
(Subscript[A, 0])^(0)[0] == 1/(Sqrt[2])
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex11 A 2 ⁢ m 2 ⁢ n ⁢ ( 0 ) = 0 subscript superscript 𝐴 2 𝑛 2 𝑚 0 0 {\displaystyle{\displaystyle A^{2n}_{2m}(0)=0}}
A^{2n}_{2m}(0) = 0		 n ≠ m 𝑛 𝑚 {\displaystyle{\displaystyle n\neq m}} 
(A[2*m])^(2*n)(0) = 0
 
(Subscript[A, 2*m])^(2*n)[0] == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex12 A 2 ⁢ n + 1 2 ⁢ n + 1 ⁢ ( 0 ) = 1 subscript superscript 𝐴 2 𝑛 1 2 𝑛 1 0 1 {\displaystyle{\displaystyle A^{2n+1}_{2n+1}(0)=1}}
A^{2n+1}_{2n+1}(0) = 1		 

(A[2*n + 1])^(2*n + 1)(0) = 1
 
(Subscript[A, 2*n + 1])^(2*n + 1)[0] == 1
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex13 A 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( 0 ) = 0 subscript superscript 𝐴 2 𝑛 1 2 𝑚 1 0 0 {\displaystyle{\displaystyle A^{2n+1}_{2m+1}(0)=0}}
A^{2n+1}_{2m+1}(0) = 0		 n ≠ m 𝑛 𝑚 {\displaystyle{\displaystyle n\neq m}} 
(A[2*m + 1])^(2*n + 1)(0) = 0
 
(Subscript[A, 2*m + 1])^(2*n + 1)[0] == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex14 B 2 ⁢ n + 1 2 ⁢ n + 1 ⁢ ( 0 ) = 1 subscript superscript 𝐵 2 𝑛 1 2 𝑛 1 0 1 {\displaystyle{\displaystyle B^{2n+1}_{2n+1}(0)=1}}
B^{2n+1}_{2n+1}(0) = 1		 

(B[2*n + 1])^(2*n + 1)(0) = 1
 
(Subscript[B, 2*n + 1])^(2*n + 1)[0] == 1
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex15 B 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( 0 ) = 0 subscript superscript 𝐵 2 𝑛 1 2 𝑚 1 0 0 {\displaystyle{\displaystyle B^{2n+1}_{2m+1}(0)=0}}
B^{2n+1}_{2m+1}(0) = 0		 n ≠ m 𝑛 𝑚 {\displaystyle{\displaystyle n\neq m}} 
(B[2*m + 1])^(2*n + 1)(0) = 0
 
(Subscript[B, 2*m + 1])^(2*n + 1)[0] == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex16 B 2 ⁢ n + 2 2 ⁢ n + 2 ⁢ ( 0 ) = 1 subscript superscript 𝐵 2 𝑛 2 2 𝑛 2 0 1 {\displaystyle{\displaystyle B^{2n+2}_{2n+2}(0)=1}}
B^{2n+2}_{2n+2}(0) = 1		 

(B[2*n + 2])^(2*n + 2)(0) = 1
 
(Subscript[B, 2*n + 2])^(2*n + 2)[0] == 1
 Skipped - no semantic math Skipped - no semantic math - -
28.4#Ex17 B 2 ⁢ m + 2 2 ⁢ n + 2 ⁢ ( 0 ) = 0 subscript superscript 𝐵 2 𝑛 2 2 𝑚 2 0 0 {\displaystyle{\displaystyle B^{2n+2}_{2m+2}(0)=0}}
B^{2n+2}_{2m+2}(0) = 0		 n ≠ m 𝑛 𝑚 {\displaystyle{\displaystyle n\neq m}} 
(B[2*m + 2])^(2*n + 2)(0) = 0
 
(Subscript[B, 2*m + 2])^(2*n + 2)[0] == 0
 Skipped - no semantic math Skipped - no semantic math - -
28.4.E17 A 2 ⁢ m 2 ⁢ n ⁢ ( - q ) = ( - 1 ) n - m ⁢ A 2 ⁢ m 2 ⁢ n ⁢ ( q ) subscript superscript 𝐴 2 𝑛 2 𝑚 𝑞 superscript 1 𝑛 𝑚 subscript superscript 𝐴 2 𝑛 2 𝑚 𝑞 {\displaystyle{\displaystyle A^{2n}_{2m}(-q)=(-1)^{n-m}A^{2n}_{2m}(q)}}
A^{2n}_{2m}(-q) = (-1)^{n-m}A^{2n}_{2m}(q)		 

(A[2*m])^(2*n)(- q) = (- 1)^(n - m)* (A[2*m])^(2*n)(q)
 
(Subscript[A, 2*m])^(2*n)[- q] == (- 1)^(n - m)* (Subscript[A, 2*m])^(2*n)[q]
 Skipped - no semantic math Skipped - no semantic math - -
28.4.E18 B 2 ⁢ m + 2 2 ⁢ n + 2 ⁢ ( - q ) = ( - 1 ) n - m ⁢ B 2 ⁢ m + 2 2 ⁢ n + 2 ⁢ ( q ) subscript superscript 𝐵 2 𝑛 2 2 𝑚 2 𝑞 superscript 1 𝑛 𝑚 subscript superscript 𝐵 2 𝑛 2 2 𝑚 2 𝑞 {\displaystyle{\displaystyle B^{2n+2}_{2m+2}(-q)=(-1)^{n-m}B^{2n+2}_{2m+2}(q)}}
B^{2n+2}_{2m+2}(-q) = (-1)^{n-m}B^{2n+2}_{2m+2}(q)		 

(B[2*m + 2])^(2*n + 2)(- q) = (- 1)^(n - m)* (B[2*m + 2])^(2*n + 2)(q)
 
(Subscript[B, 2*m + 2])^(2*n + 2)[- q] == (- 1)^(n - m)* (Subscript[B, 2*m + 2])^(2*n + 2)[q]
 Skipped - no semantic math Skipped - no semantic math - -
28.4.E19 A 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( - q ) = ( - 1 ) n - m ⁢ B 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) subscript superscript 𝐴 2 𝑛 1 2 𝑚 1 𝑞 superscript 1 𝑛 𝑚 subscript superscript 𝐵 2 𝑛 1 2 𝑚 1 𝑞 {\displaystyle{\displaystyle A^{2n+1}_{2m+1}(-q)=(-1)^{n-m}B^{2n+1}_{2m+1}(q)}}
A^{2n+1}_{2m+1}(-q) = (-1)^{n-m}B^{2n+1}_{2m+1}(q)		 

(A[2*m + 1])^(2*n + 1)(- q) = (- 1)^(n - m)* (B[2*m + 1])^(2*n + 1)(q)
 
(Subscript[A, 2*m + 1])^(2*n + 1)[- q] == (- 1)^(n - m)* (Subscript[B, 2*m + 1])^(2*n + 1)[q]
 Skipped - no semantic math Skipped - no semantic math - -
28.4.E20 B 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( - q ) = ( - 1 ) n - m ⁢ A 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) subscript superscript 𝐵 2 𝑛 1 2 𝑚 1 𝑞 superscript 1 𝑛 𝑚 subscript superscript 𝐴 2 𝑛 1 2 𝑚 1 𝑞 {\displaystyle{\displaystyle B^{2n+1}_{2m+1}(-q)=(-1)^{n-m}A^{2n+1}_{2m+1}(q)}}
B^{2n+1}_{2m+1}(-q) = (-1)^{n-m}A^{2n+1}_{2m+1}(q)		 

(B[2*m + 1])^(2*n + 1)(- q) = (- 1)^(n - m)* (A[2*m + 1])^(2*n + 1)(q)
 
(Subscript[B, 2*m + 1])^(2*n + 1)[- q] == (- 1)^(n - m)* (Subscript[A, 2*m + 1])^(2*n + 1)[q]
 Skipped - no semantic math Skipped - no semantic math - -
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