Mathieu Functions and Hill’s Equation - 28.11 Expansions in Series of Mathieu Functions

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DLMF Formula Constraints Maple Mathematica Symbolic
Maple 		Symbolic
Mathematica 		Numeric
Maple 		Numeric
Mathematica
28.11.E3 1 = 2 ⁢ ∑ n = 0 ∞ A 0 2 ⁢ n ⁢ ( q ) ⁢ ce 2 ⁢ n ⁡ ( z , q ) 1 2 superscript subscript 𝑛 0 superscript subscript 𝐴 0 2 𝑛 𝑞 Mathieu-ce 2 𝑛 𝑧 𝑞 {\displaystyle{\displaystyle 1=2\sum_{n=0}^{\infty}A_{0}^{2n}(q)\mathrm{ce}_{2% n}\left(z,q\right)}}
1 = 2\sum_{n=0}^{\infty}A_{0}^{2n}(q)\Mathieuce{2n}@{z}{q}		 

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

1 == 2*Sum[(Subscript[A, 0])^(2*n)[q]* MathieuC[2*n, q, z], {n, 0, Infinity}, GenerateConditions->None]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
28.11.E4 cos ⁡ 2 ⁢ m ⁢ z = ∑ n = 0 ∞ A 2 ⁢ m 2 ⁢ n ⁢ ( q ) ⁢ ce 2 ⁢ n ⁡ ( z , q ) 2 𝑚 𝑧 superscript subscript 𝑛 0 superscript subscript 𝐴 2 𝑚 2 𝑛 𝑞 Mathieu-ce 2 𝑛 𝑧 𝑞 {\displaystyle{\displaystyle\cos 2mz=\sum_{n=0}^{\infty}A_{2m}^{2n}(q)\mathrm{% ce}_{2n}\left(z,q\right)}}
\cos@@{2mz} = \sum_{n=0}^{\infty}A_{2m}^{2n}(q)\Mathieuce{2n}@{z}{q}		 m ≠ 0 𝑚 0 {\displaystyle{\displaystyle m\neq 0}} 
cos(2*m*z) = sum((A[2*m])^(2*n)(q)* MathieuCE(2*n, q, z), n = 0..infinity)

Cos[2*m*z] == Sum[(Subscript[A, 2*m])^(2*n)[q]* MathieuC[2*n, q, z], {n, 0, Infinity}, GenerateConditions->None]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
28.11.E5 cos ⁡ ( 2 ⁢ m + 1 ) ⁢ z = ∑ n = 0 ∞ A 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) ⁢ ce 2 ⁢ n + 1 ⁡ ( z , q ) 2 𝑚 1 𝑧 superscript subscript 𝑛 0 superscript subscript 𝐴 2 𝑚 1 2 𝑛 1 𝑞 Mathieu-ce 2 𝑛 1 𝑧 𝑞 {\displaystyle{\displaystyle\cos(2m+1)z=\sum_{n=0}^{\infty}A_{2m+1}^{2n+1}(q)% \mathrm{ce}_{2n+1}\left(z,q\right)}}
\cos@@{(2m+1)z} = \sum_{n=0}^{\infty}A_{2m+1}^{2n+1}(q)\Mathieuce{2n+1}@{z}{q}		 

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

Cos[(2*m + 1)*z] == Sum[(Subscript[A, 2*m + 1])^(2*n + 1)[q]* MathieuC[2*n + 1, q, z], {n, 0, Infinity}, GenerateConditions->None]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
28.11.E6 sin ⁡ ( 2 ⁢ m + 1 ) ⁢ z = ∑ n = 0 ∞ B 2 ⁢ m + 1 2 ⁢ n + 1 ⁢ ( q ) ⁢ se 2 ⁢ n + 1 ⁡ ( z , q ) 2 𝑚 1 𝑧 superscript subscript 𝑛 0 superscript subscript 𝐵 2 𝑚 1 2 𝑛 1 𝑞 Mathieu-se 2 𝑛 1 𝑧 𝑞 {\displaystyle{\displaystyle\sin(2m+1)z=\sum_{n=0}^{\infty}B_{2m+1}^{2n+1}(q)% \mathrm{se}_{2n+1}\left(z,q\right)}}
\sin@@{(2m+1)z} = \sum_{n=0}^{\infty}B_{2m+1}^{2n+1}(q)\Mathieuse{2n+1}@{z}{q}		 

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

Sin[(2*m + 1)*z] == Sum[(Subscript[B, 2*m + 1])^(2*n + 1)[q]* MathieuS[2*n + 1, q, z], {n, 0, Infinity}, GenerateConditions->None]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
28.11.E7 sin ⁡ ( 2 ⁢ m + 2 ) ⁢ z = ∑ n = 0 ∞ B 2 ⁢ m + 2 2 ⁢ n + 2 ⁢ ( q ) ⁢ se 2 ⁢ n + 2 ⁡ ( z , q ) 2 𝑚 2 𝑧 superscript subscript 𝑛 0 superscript subscript 𝐵 2 𝑚 2 2 𝑛 2 𝑞 Mathieu-se 2 𝑛 2 𝑧 𝑞 {\displaystyle{\displaystyle\sin(2m+2)z=\sum_{n=0}^{\infty}B_{2m+2}^{2n+2}(q)% \mathrm{se}_{2n+2}\left(z,q\right)}}
\sin@@{(2m+2)z} = \sum_{n=0}^{\infty}B_{2m+2}^{2n+2}(q)\Mathieuse{2n+2}@{z}{q}		 

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

Sin[(2*m + 2)*z] == Sum[(Subscript[B, 2*m + 2])^(2*n + 2)[q]* MathieuS[2*n + 2, q, z], {n, 0, Infinity}, GenerateConditions->None]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
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