马丢函数

马丟函数（法語：Équation de Mathieu）是1868年法國數學家以米里迂·拉·馬丢（法语：Émile Mathieu）因研究数学物理所推得的特殊函數，下列马丟方程的解析解： d 2 y d x 2 + [ a − 2 q cos ⁡ ( 2 x ) ] y = 0. {\displaystyle {\frac {d^{2}y}{dx^{2}}}+[a-2q\cos(2x)]y=0.} 马丟方程有两个线性无关的解： 奇数解 MathieuCE(n, q, x)，或记为 w I ( n , q , x ) {\displaystyle w_{I}(n,q,x)} , 偶数解 MathieuSE(n, q, x).或记为 w I I ( n , q , x ) {\displaystyle w_{II}(n,q,x)} 称为基本解 马丟函数 MathieuC(a,q,z) 或 MathieuS(a,q,z) 只有一个是周期为 π {\displaystyle \pi } 或 2 π {\displaystyle 2\pi } 的周期解，另一个不是。 马丟函数 MathieuC(a,q,z) 和 MathieuS(a,q,z) 两者都有是周期为 2 n π {\displaystyle 2n\pi } （n≥2)的周期函数。 ∫ 0 2 ∗ π c e m ( x , q ) ∗ c e n ( x , q ) d x = 0 {\displaystyle \int _{0}^{2*\pi }\!ce_{m}(x,q)*ce_{n}(x,q)\,dx=0} ∫ 0 2 ∗ π c e m ( x , q ) ∗ s e n ( x , q ) d x = 0 {\displaystyle \int _{0}^{2*\pi }\!ce_{m}(x,q)*se_{n}(x,q)\,dx=0} ∫ 0 2 ∗ π s e m ( x , q ) ∗ s e n ( x , q ) d x = 0 {\displaystyle \int _{0}^{2*\pi }\!se_{m}(x,q)*se_{n}(x,q)\,dx=0} 马丟方程的特征方程是 c o s ( π ∗ v ) = w I ( a , q , π ) {\displaystyle cos(\pi *v)=w_{I}(a,q,\pi )} c o s ( π ∗ v ) = w I I ( b , q , π ) {\displaystyle cos(\pi *v)=w_{II}(b,q,\pi )} 对于给定的v，q, 上列特征方程给出无穷多个a、b解称为特征值。 马丟函数体特征值可展开成级数： a 0 ( q ) = − ( 1 / 2 ) ∗ z 2 + ( 7 / 128 ) ∗ z 4 − ( 29 / 2304 ) ∗ z 6 + ( 68687 / 18874368 ) ∗ z 8 + O ( z 10 ) {\displaystyle a_{0}(q)={-(1/2)*z^{2}+(7/128)*z^{4}-(29/2304)*z^{6}+(68687/18874368)*z^{8}+{\mathcal {O}}(z^{10})}} a 1 ( q ) = 1 + z − ( 1 / 8 ) ∗ z 2 − ( 1 / 64 ) ∗ z 3 − ( 1 / 1536 ) ∗ z 4 + ( 11 / 36864 ) ∗ z 5 + ( 49 / 589824 ) ∗ z 6 + ( 55 / 9437184 ) ∗ z 7 − ( 83 / 35389440 ) ∗ z 8 − ( 12121 / 15099494400 ) ∗ z 9 + O ( z 10 ) {\displaystyle a_{1}(q)={1+z-(1/8)*z^{2}-(1/64)*z^{3}-(1/1536)*z^{4}+(11/36864)*z^{5}+(49/589824)*z^{6}+(55/9437184)*z^{7}-(83/35389440)*z^{8}-(12121/15099494400)*z^{9}+{\mathcal {O}}(z^{10})}} a 2 ( q ) = 4 + ( 5 / 12 ) ∗ z 2 − ( 763 / 13824 ) ∗ z 4 + ( 1002401 / 79626240 ) ∗ z 6 − ( 1669068401 / 458647142400 ) ∗ z 8 + O ( z 10 ) {\displaystyle a_{2}(q)={4+(5/12)*z^{2}-(763/13824)*z^{4}+(1002401/79626240)*z^{6}-(1669068401/458647142400)*z^{8}+{\mathcal {O}}(z^{10})}} a 3 ( q ) = 9 + ( 1 / 16 ) ∗ z 2 + ( 1 / 64 ) ∗ z 3 + ( 13 / 20480 ) ∗ z 4 − ( 5 / 16384 ) ∗ z 5 − ( 1961 / 23592960 ) ∗ z 6 − ( 609 / 104857600 ) ∗ z 7 + ( 4957199 / 2113929216000 ) ∗ z 8 + ( 872713 / 1087163596800 ) ∗ z 9 + O ( z 10 ) {\displaystyle a_{3}(q)={9+(1/16)*z^{2}+(1/64)*z^{3}+(13/20480)*z^{4}-(5/16384)*z^{5}-(1961/23592960)*z^{6}-(609/104857600)*z^{7}+(4957199/2113929216000)*z^{8}+(872713/1087163596800)*z^{9}+{\mathcal {O}}(z^{10})}} b 1 ( q ) = 1 − z − ( 1 / 8 ) ∗ z 2 + ( 1 / 64 ) ∗ z 3 − ( 1 / 1536 ) ∗ z 4 − ( 11 / 36864 ) ∗ z 5 + ( 49 / 589824 ) ∗ z 6 − ( 55 / 9437184 ) ∗ z 7 − ( 83 / 35389440 ) ∗ z 8 + ( 12121 / 15099494400 ) ∗ z 9 + O ( z 10 ) {\displaystyle b_{1}(q)={1-z-(1/8)*z^{2}+(1/64)*z^{3}-(1/1536)*z^{4}-(11/36864)*z^{5}+(49/589824)*z^{6}-(55/9437184)*z^{7}-(83/35389440)*z^{8}+(12121/15099494400)*z^{9}+{\mathcal {O}}(z^{10})}} b 2 ( q ) = 4 − ( 1 / 12 ) ∗ z 2 + ( 5 / 13824 ) ∗ z 4 − ( 289 / 79626240 ) ∗ z 6 + ( 21391 / 458647142400 ) ∗ z 8 + O ( z 10 ) {\displaystyle b_{2}(q)={4-(1/12)*z^{2}+(5/13824)*z^{4}-(289/79626240)*z^{6}+(21391/458647142400)*z^{8}+{\mathcal {O}}(z^{10})}} b 3 ( q ) = 9 + ( 1 / 16 ) ∗ z 2 − ( 1 / 64 ) ∗ z 3 + ( 13 / 20480 ) ∗ z 4 + ( 5 / 16384 ) ∗ z 5 − ( 1961 / 23592960 ) ∗ z 6 + ( 609 / 104857600 ) ∗ z 7 + ( 4957199 / 2113929216000 ) ∗ z 8 − ( 872713 / 1087163596800 ) ∗ z 9 + O ( z 10 ) {\displaystyle b_{3}(q)={9+(1/16)*z^{2}-(1/64)*z^{3}+(13/20480)*z^{4}+(5/16384)*z^{5}-(1961/23592960)*z^{6}+(609/104857600)*z^{7}+(4957199/2113929216000)*z^{8}-(872713/1087163596800)*z^{9}+{\mathcal {O}}(z^{10})}} b 4 ( q ) = 16 + ( 1 / 30 ) ∗ z 2 − ( 317 / 864000 ) ∗ z 4 + ( 10049 / 2721600000 ) ∗ z 6 − ( 93824197 / 2006581248000000 ) ∗ z 8 + O ( z 10 ) {\displaystyle b_{4}(q)={16+(1/30)*z^{2}-(317/864000)*z^{4}+(10049/2721600000)*z^{6}-(93824197/2006581248000000)*z^{8}+{\mathcal {O}}(z^{10})}} b 5 ( q ) = 25 + ( 1 / 48 ) ∗ z 2 + ( 11 / 774144 ) ∗ z 4 − ( 1 / 147456 ) ∗ z 5 + ( 37 / 891813888 ) ∗ z 6 + ( 7 / 339738624 ) ∗ z 7 + ( 63439 / 201364441399296 ) ∗ z 8 + ( 1 / 2130840649728 ) ∗ z 9 + O ( z 10 ) {\displaystyle b_{5}(q)={25+(1/48)*z^{2}+(11/774144)*z^{4}-(1/147456)*z^{5}+(37/891813888)*z^{6}+(7/339738624)*z^{7}+(63439/201364441399296)*z^{8}+(1/2130840649728)*z^{9}+{\mathcal {O}}(z^{10})}} 马丟函数ce,se的级数展开 c e 0 ( z , q ) = 1 − ( 1 / 2 ) ∗ c o s ( 2 ∗ z ) ∗ q + ( − 1 / 16 + ( 1 / 32 ) ∗ c o s ( 4 ∗ z ) ) ∗ q 2 + ( ( 11 / 128 ) ∗ c o s ( 2 ∗ z ) − ( 1 / 1152 ) ∗ c o s ( 6 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle ce_{0}(z,q)={1-(1/2)*cos(2*z)*q+(-1/16+(1/32)*cos(4*z))*q^{2}+((11/128)*cos(2*z)-(1/1152)*cos(6*z))*q^{3}+O(q^{4})}} c e 1 ( z , q ) = c o s ( z ) − ( 1 / 8 ) ∗ c o s ( 3 ∗ z ) ∗ q + ( − ( 1 / 128 ) ∗ c o s ( z ) − ( 1 / 64 ) ∗ c o s ( 3 ∗ z ) + ( 1 / 192 ) ∗ c o s ( 5 ∗ z ) ) ∗ q 2 + ( − ( 1 / 512 ) ∗ c o s ( z ) + ( 1 / 3072 ) ∗ c o s ( 3 ∗ z ) + ( 1 / 1152 ) ∗ c o s ( 5 ∗ z ) − ( 1 / 9216 ) ∗ c o s ( 7 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle ce_{1}(z,q)={cos(z)-(1/8)*cos(3*z)*q+(-(1/128)*cos(z)-(1/64)*cos(3*z)+(1/192)*cos(5*z))*q^{2}+(-(1/512)*cos(z)+(1/3072)*cos(3*z)+(1/1152)*cos(5*z)-(1/9216)*cos(7*z))*q^{3}+O(q^{4})}} c e 2 ( z , q ) = c o s ( 2 ∗ z ) + ( 1 / 4 − ( 1 / 12 ) ∗ c o s ( 4 ∗ z ) ) ∗ q + ( − ( 19 / 288 ) ∗ c o s ( 2 ∗ z ) + ( 1 / 384 ) ∗ c o s ( 6 ∗ z ) ) ∗ q 2 + ( − 49 / 1152 + ( 11 / 4608 ) ∗ c o s ( 4 ∗ z ) − ( 1 / 23040 ) ∗ c o s ( 8 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle ce_{2}(z,q)={cos(2*z)+(1/4-(1/12)*cos(4*z))*q+(-(19/288)*cos(2*z)+(1/384)*cos(6*z))*q^{2}+(-49/1152+(11/4608)*cos(4*z)-(1/23040)*cos(8*z))*q^{3}+O(q^{4})}} c e 3 ( z , q ) = c o s ( 3 ∗ z ) + ( ( 1 / 8 ) ∗ c o s ( z ) − ( 1 / 16 ) ∗ c o s ( 5 ∗ z ) ) ∗ q + ( − ( 5 / 512 ) ∗ c o s ( 3 ∗ z ) + ( 1 / 64 ) ∗ c o s ( z ) + ( 1 / 640 ) ∗ c o s ( 7 ∗ z ) ) ∗ q 2 + ( − ( 1 / 512 ) ∗ c o s ( 3 ∗ z ) − ( 1 / 4096 ) ∗ c o s ( z ) + ( 11 / 40960 ) ∗ c o s ( 5 ∗ z ) − ( 1 / 46080 ) ∗ c o s ( 9 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle ce_{3}(z,q)={cos(3*z)+((1/8)*cos(z)-(1/16)*cos(5*z))*q+(-(5/512)*cos(3*z)+(1/64)*cos(z)+(1/640)*cos(7*z))*q^{2}+(-(1/512)*cos(3*z)-(1/4096)*cos(z)+(11/40960)*cos(5*z)-(1/46080)*cos(9*z))*q^{3}+O(q^{4})}} c e 4 ( z , q ) = c o s ( 4 ∗ z ) + ( ( 1 / 12 ) ∗ c o s ( 2 ∗ z ) − ( 1 / 20 ) ∗ c o s ( 6 ∗ z ) ) ∗ q + ( − ( 17 / 3600 ) ∗ c o s ( 4 ∗ z ) + 1 / 192 + ( 1 / 960 ) ∗ c o s ( 8 ∗ z ) ) ∗ q 2 + ( ( 7 / 28800 ) ∗ c o s ( 2 ∗ z ) + ( 29 / 288000 ) ∗ c o s ( 6 ∗ z ) − ( 1 / 80640 ) ∗ c o s ( 10 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle ce_{4}(z,q)={cos(4*z)+((1/12)*cos(2*z)-(1/20)*cos(6*z))*q+(-(17/3600)*cos(4*z)+1/192+(1/960)*cos(8*z))*q^{2}+((7/28800)*cos(2*z)+(29/288000)*cos(6*z)-(1/80640)*cos(10*z))*q^{3}+O(q^{4})}} s e 1 ( z , q ) = s i n ( z ) − ( 1 / 8 ) ∗ s i n ( 3 ∗ z ) ∗ q + ( − ( 1 / 128 ) ∗ s i n ( z ) + ( 1 / 64 ) ∗ s i n ( 3 ∗ z ) + ( 1 / 192 ) ∗ s i n ( 5 ∗ z ) ) ∗ q 2 + ( ( 1 / 512 ) ∗ s i n ( z ) + ( 1 / 3072 ) ∗ s i n ( 3 ∗ z ) − ( 1 / 1152 ) ∗ s i n ( 5 ∗ z ) − ( 1 / 9216 ) ∗ s i n ( 7 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle se_{1}(z,q)={sin(z)-(1/8)*sin(3*z)*q+(-(1/128)*sin(z)+(1/64)*sin(3*z)+(1/192)*sin(5*z))*q^{2}+((1/512)*sin(z)+(1/3072)*sin(3*z)-(1/1152)*sin(5*z)-(1/9216)*sin(7*z))*q^{3}+O(q^{4})}} s e 2 ( z , q ) = s i n ( 2 ∗ z ) − ( 1 / 12 ) ∗ s i n ( 4 ∗ z ) ∗ q + ( − ( 1 / 288 ) ∗ s i n ( 2 ∗ z ) + ( 1 / 384 ) ∗ s i n ( 6 ∗ z ) ) ∗ q 2 + ( ( 1 / 1536 ) ∗ s i n ( 4 ∗ z ) − ( 1 / 23040 ) ∗ s i n ( 8 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle se_{2}(z,q)={sin(2*z)-(1/12)*sin(4*z)*q+(-(1/288)*sin(2*z)+(1/384)*sin(6*z))*q^{2}+((1/1536)*sin(4*z)-(1/23040)*sin(8*z))*q^{3}+O(q^{4})}} s e 3 ( z , q ) = s i n ( 3 ∗ z ) + ( ( 1 / 8 ) ∗ s i n ( z ) − ( 1 / 16 ) ∗ s i n ( 5 ∗ z ) ) ∗ q + ( − ( 5 / 512 ) ∗ s i n ( 3 ∗ z ) − ( 1 / 64 ) ∗ s i n ( z ) + ( 1 / 640 ) ∗ s i n ( 7 ∗ z ) ) ∗ q 2 + ( ( 1 / 512 ) ∗ s i n ( 3 ∗ z ) − ( 1 / 4096 ) ∗ s i n ( z ) + ( 11 / 40960 ) ∗ s i n ( 5 ∗ z ) − ( 1 / 46080 ) ∗ s i n ( 9 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle se_{3}(z,q)={sin(3*z)+((1/8)*sin(z)-(1/16)*sin(5*z))*q+(-(5/512)*sin(3*z)-(1/64)*sin(z)+(1/640)*sin(7*z))*q^{2}+((1/512)*sin(3*z)-(1/4096)*sin(z)+(11/40960)*sin(5*z)-(1/46080)*sin(9*z))*q^{3}+O(q^{4})}} s e 4 ( z , q ) = s i n ( 4 ∗ z ) + ( ( 1 / 12 ) ∗ s i n ( 2 ∗ z ) − ( 1 / 20 ) ∗ s i n ( 6 ∗ z ) ) ∗ q + ( − ( 17 / 3600 ) ∗ s i n ( 4 ∗ z ) + ( 1 / 960 ) ∗ s i n ( 8 ∗ z ) ) ∗ q 2 + ( − ( 1 / 1600 ) ∗ s i n ( 2 ∗ z ) + ( 29 / 288000 ) ∗ s i n ( 6 ∗ z ) − ( 1 / 80640 ) ∗ s i n ( 10 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle se_{4}(z,q)={sin(4*z)+((1/12)*sin(2*z)-(1/20)*sin(6*z))*q+(-(17/3600)*sin(4*z)+(1/960)*sin(8*z))*q^{2}+(-(1/1600)*sin(2*z)+(29/288000)*sin(6*z)-(1/80640)*sin(10*z))*q^{3}+O(q^{4})}} s e 5 ( z , q ) = s i n ( 5 ∗ z ) + ( ( 1 / 16 ) ∗ s i n ( 3 ∗ z ) − ( 1 / 24 ) ∗ s i n ( 7 ∗ z ) ) ∗ q + ( − ( 13 / 4608 ) ∗ s i n ( 5 ∗ z ) + ( 1 / 384 ) ∗ s i n ( z ) + ( 1 / 1344 ) ∗ s i n ( 9 ∗ z ) ) ∗ q 2 + ( − ( 7 / 73728 ) ∗ s i n ( 3 ∗ z ) + ( 13 / 258048 ) ∗ s i n ( 7 ∗ z ) − ( 1 / 9216 ) ∗ s i n ( z ) − ( 1 / 129024 ) ∗ s i n ( 11 ∗ z ) ) ∗ q 3 + O ( q 4 ) {\displaystyle se_{5}(z,q)={sin(5*z)+((1/16)*sin(3*z)-(1/24)*sin(7*z))*q+(-(13/4608)*sin(5*z)+(1/384)*sin(z)+(1/1344)*sin(9*z))*q^{2}+(-(7/73728)*sin(3*z)+(13/258048)*sin(7*z)-(1/9216)*sin(z)-(1/129024)*sin(11*z))*q^{3}+O(q^{4})}} 马丟函数的傅立叶展开： M a t h i e u C E ( 2 n , q , x ) = ∑ m = 0 ∞ A 2 m 2 n ( q ) c o s ( 2 m x ) {\displaystyle MathieuCE(2n,q,x)=\sum _{m=0}^{\infty }A_{2m}^{2n}(q)cos(2mx)} M a t h i e u C E ( 2 n + 1 , q , x ) = ∑ m = 0 ∞ A 2 m + 1 2 n + 1 ( q ) c o s [ ( 2 m + 1 ) x ] {\displaystyle MathieuCE(2n+1,q,x)=\sum _{m=0}^{\infty }A_{2m+1}^{2n+1}(q)cos[(2m+1)x]} M a t h i e u S E ( 2 n + 1 , q , x ) = ∑ m = 0 ∞ B 2 m + 1 2 n + 1 ( q ) s i n [ ( 2 m + 1 ) x ] {\displaystyle MathieuSE(2n+1,q,x)=\sum _{m=0}^{\infty }B_{2m+1}^{2n+1}(q)sin[(2m+1)x]} M a t h i e u S E ( 2 n + 2 , q , x ) = ∑ m = 0 ∞ B 2 m + 2 2 n + 2 ( q ) s i n [ ( 2 m + 2 ) x ] {\displaystyle MathieuSE(2n+2,q,x)=\sum _{m=0}^{\infty }B_{2m+2}^{2n+2}(q)sin[(2m+2)x]} 其中系数A,B满足下列递归关系： a A 0 = q A 2 {\displaystyle aA_{0}=qA_{2}} ( a − 4 ) A 2 = q ( 2 A 0 + A 4 ) {\displaystyle (a-4)A_{2}=q(2A_{0}+A_{4})} ( a − 4 m 2 ) A 2 m = q ( A 2 m − 2 + A 2 m + 2 ) {\displaystyle (a-4m^{2})A_{2m}=q(A_{2m-2}+A_{2m+2})} ( a − 1 + q ) B 1 = q B 3 {\displaystyle (a-1+q)B_{1}=qB_{3}} ( a − ( 2 m + 1 ) 2 ) B 2 m + 1 = q ( B 2 m − 1 + B 2 m + 1 ) {\displaystyle (a-(2m+1)^{2})B_{2m+1}=q(B_{2m-1}+B_{2m+1})} 马丟方程的基本解 W I W I I {\displaystyle W_{I}W_{II}} 满足下列关系：: | w I ( n , q , 0 ) w I I ( n , q , 0 ) w i ′ ( n , q , 0 ) w I I ′ ( n , q , 0 ) | {\displaystyle {\begin{vmatrix}w_{I}(n,q,0)&w_{II}(n,q,0)\\w_{i}^{'}(n,q,0)&w_{II}^{'}(n,q,0)\end{vmatrix}}} = | 1 0 0 1 | {\displaystyle {\begin{vmatrix}1&0\\0&1\end{vmatrix}}} 郎斯基行列式： W [ w I , w I I ] = 1 {\displaystyle W[w_{I},w_{II}]=1} w I ( a , q , z + π ) = w I ( a , q , π ) ∗ w I ( a , q , z ) + w I ′ ( a , q , π ) ∗ w I I ( a , q , z ) {\displaystyle w_{I}(a,q,z+\pi )=w_{I}(a,q,\pi )*w_{I}(a,q,z)+w_{I}^{'}(a,q,\pi )*w_{II}(a,q,z)} w I ( a , q , z − π ) = w I ( a , q , π ) ∗ w I ( a , q , z ) − w I ′ ( a , q , π ) ∗ w I I ( a , q , z ) {\displaystyle w_{I}(a,q,z-\pi )=w_{I}(a,q,\pi )*w_{I}(a,q,z)-w_{I}^{'}(a,q,\pi )*w_{II}(a,q,z)} w I I ( a , q , z + π ) = w I I ( a , q , π ) ∗ w I I ( a , q , z ) + w I I ′ ( a , q , π ) ∗ w I I ( a , q , z ) {\displaystyle w_{II}(a,q,z+\pi )=w_{II}(a,q,\pi )*w_{II}(a,q,z)+w_{II}^{'}(a,q,\pi )*w_{II}(a,q,z)} w I I ( a , q , z − π ) = w I I ( a , q , π ) ∗ w I I ( a , q , z ) − w I ′ ( a , q , π ) ∗ w I I ( a , q , z ) {\displaystyle w_{II}(a,q,z-\pi )=w_{II}(a,q,\pi )*w_{II}(a,q,z)-w_{I}^{'}(a,q,\pi )*w_{II}(a,q,z)} w I ( − z ) = w I ( z ) {\displaystyle w_{I}(-z)=w_{I}(z)} w I I ( − z ) = − w I I ( z ) {\displaystyle w_{II}(-z)=-w_{II}(z)} {\displaystyle } {\displaystyle } {\displaystyle } {\displaystyle } C E ( a , 0 , z ) = c o s ( a z ) {\displaystyle CE(a,0,z)=cos(az)} S E ( a , 0 , z ) = s i n ( a z ) {\displaystyle SE(a,0,z)=sin(az)} M a t h i e u A ( 1 , 0 ) = 1 {\displaystyle MathieuA(1,0)=1} M a t h i e u A ( a , 0 ) = a 2 {\displaystyle MathieuA(a,0)=a^{2}} M a t h i e u B ( a , 0 ) = a 2 {\displaystyle MathieuB(a,0)=a^{2}} M a t h i e u F l o q u e t ( a , 0 , z ) = e x p ( I ∗ s q r t ( a ) ∗ z ) {\displaystyle MathieuFloquet(a,0,z)=exp(I*sqrt(a)*z)} {\displaystyle } 马丟函数中，如果 f ( x ) {\displaystyle f(x)} 是一个周期为 ω {\displaystyle \omega } 的解，并满足下列条件 f ( x + ω ) = σ ∗ f ( x ) {\displaystyle f(x+\omega )=\sigma *f(x)} ,其中 σ {\displaystyle \sigma } 与x 无关，则此解称为夫洛开解。 级数展开 M F ( 1 , 1 , z ) = .7992 − .5734 ∗ I + ( − .9134 + .6553 ∗ I ) ∗ z + ( .3996 − .2867 ∗ I ) ∗ z 2 + ( − .1523 + .1092 ∗ I ) ∗ z 3 + ( − .2331 + .1673 ∗ I ) ∗ z 4 + O ( z 5 ) {\displaystyle MF(1,1,z)={.7992-.5734*I+(-.9134+.6553*I)*z+(.3996-.2867*I)*z^{2}+(-.1523+.1092*I)*z^{3}+(-.2331+.1673*I)*z^{4}+O(z^{5})}} M F ( 1 , 2 , z ) = .7643 − .4526 ∗ I + ( − 1.167 + .6910 ∗ I ) ∗ z + ( 1.146 − .6789 ∗ I ) ∗ z 2 + ( − .5835 + .3455 ∗ I ) ∗ z 3 + ( − .2229 + .1320 ∗ I ) ∗ z 4 + O ( z 5 ) {\displaystyle MF(1,2,z)={.7643-.4526*I+(-1.167+.6910*I)*z+(1.146-.6789*I)*z^{2}+(-.5835+.3455*I)*z^{3}+(-.2229+.1320*I)*z^{4}+O(z^{5})}} M F ( 1 , 3 , z ) = .6841 − .3703 ∗ I + ( − 1.318 + .7135 ∗ I ) ∗ z + ( 1.710 − .9258 ∗ I ) ∗ z 2 + ( − 1.098 + .5946 ∗ I ) ∗ z 3 + ( 0.2851 e − 1 − 0.1543 e − 1 ∗ I ) ∗ z 4 + O ( z 5 ) {\displaystyle MF(1,3,z)={.6841-.3703*I+(-1.318+.7135*I)*z+(1.710-.9258*I)*z^{2}+(-1.098+.5946*I)*z^{3}+(0.2851e-1-0.1543e-1*I)*z^{4}+O(z^{5})}} 王竹溪 郭敦仁 603 引用错误：带有name属性“W”的<ref>标签用不同内容定义了多次 Frank p659 Frank p660 引用错误：带有name属性“F”的<ref>标签用不同内容定义了多次 王竹溪 郭敦仁 《特殊函数概论》 第十二章 马丟函数 北京大学出版社 2000 Frank J Oliver NIST Handbook of Mathematical Functions,Cambridge University PRESS, 2010