applying the theorem (7) to evaluate this expression, we have
(
−
1
)
n
r
n
+
1
n
!
∂
n
∂
z
n
1
r
=
(
2
n
)
!
2
n
n
!
n
!
1
r
n
{
1
−
r
2
∇
2
2.2
n
−
1
+
r
4
∇
4
2.4.2
n
−
1.2
n
−
3
−
}
z
n
=
(
2
n
)
!
2
n
n
!
n
!
{
μ
n
−
n
(
n
−
1
)
2.2
n
−
1
μ
n
−
2
+
…
}
,
{\displaystyle {\begin{aligned}{\frac {(-1)^{n}r^{n+1}}{n!}}{\frac {\partial ^{n}}{\partial z^{n}}}{\frac {1}{r}}&={\frac {(2n)!}{2^{n}n!n!}}{\frac {1}{r^{n}}}\left\{1-{\frac {r^{2}\nabla ^{2}}{2.2n-1}}+{\frac {r^{4}\nabla ^{4}}{2.4.2n-1.2n-3}}-\right\}z^{n}\\&={\frac {(2n)!}{2^{n}n!n!}}\left\{\mu ^{n}-{\frac {n(n-1)}{2.2n-1}}\mu ^{n-2}+\dots \right\},\end{aligned}}}
the expression on the right side is
P
n
(
μ
)
{\displaystyle P^{n}(\mu )}
P
n
(
μ
)
=
(
−
1
)
n
r
n
+
1
n
!
∂
n
∂
z
n
1
r
.
{\displaystyle P_{n}(\mu )={\frac {(-1)^{n}r^{n+1}}{n!}}{\frac {\partial ^{n}}{\partial z^{n}}}{\frac {1}{r}}.}
The zonal harmonic has therefore all its poles coincident with
the z axis. Next, suppose n - m axes coincide with the z axis,
and that the remaining m axes are distributed symmetrically in
the plane of x , y at intervals
π
/
m
{\displaystyle \pi /m}
cos
α
,
sin
α
,
0
{\displaystyle \cos \alpha ,\sin \alpha ,0}
∏
0
m
−
1
{
cos
(
α
+
r
π
m
)
∂
∂
x
+
sin
(
α
+
r
π
m
)
∂
∂
y
}
=
1
2
m
∏
{
e
i
(
α
+
r
π
m
)
(
∂
∂
x
−
i
∂
∂
y
)
+
e
−
i
(
α
+
r
π
m
)
(
∂
∂
x
+
i
∂
∂
y
)
}
.
{\displaystyle {\begin{aligned}\prod _{0}^{m-1}\left\{\cos \left(\alpha +{\frac {r\pi }{m}}\right){\frac {\partial }{\partial x}}+\sin \left(\alpha +{\frac {r\pi }{m}}\right){\frac {\partial }{\partial y}}\right\}&={\frac {1}{2^{m}}}\prod {\Bigl \{}e^{i\left(\alpha +{\frac {r\pi }{m}}\right)}\left({\frac {\partial }{\partial x}}-i{\frac {\partial }{\partial y}}\right)\\&+e^{-i\left(\alpha +{\frac {r\pi }{m}}\right)}\left({\frac {\partial }{\partial x}}+i{\frac {\partial }{\partial y}}\right){\Bigr \}}.\end{aligned}}}
Let
ξ
=
x
+
i
y
,
η
=
x
−
i
y
{\displaystyle \xi =x+iy,\eta =x-iy}
∏
0
m
−
1
{
e
i
(
α
+
r
π
m
)
∂
∂
ξ
+
e
−
i
(
α
+
r
π
m
)
∂
∂
η
}
{\displaystyle \prod _{0}^{m-1}\left\{e^{i\left(\alpha +{\frac {r\pi }{m}}\right)}{\frac {\partial }{\partial \xi }}+e^{-i\left(\alpha +{\frac {r\pi }{m}}\right)}{\frac {\partial }{\partial \eta }}\right\}}
which is equal to
e
(
m
−
1
)
i
π
2
∏
{
e
m
i
α
(
∂
∂
ξ
)
m
−
e
−
m
i
α
(
∂
∂
η
)
m
}
{\displaystyle e^{(m-1){\frac {i\pi }{2}}}\prod \left\{e^{mi\alpha }\left({\frac {\partial }{\partial \xi }}\right)^{m}-e^{-mi\alpha }\left({\frac {\partial }{\partial \eta }}\right)^{m}\right\}}
α
=
0
,
π
2
m
{\displaystyle \alpha =0,{\frac {\pi }{2m}}}
this becomes
e
(
m
−
1
)
i
π
2
{
(
∂
∂
ξ
)
m
−
(
−
1
)
m
(
∂
∂
η
)
m
}
{\displaystyle e^{(m-1){\frac {i\pi }{2}}}\left\{\left({\frac {\partial }{\partial \xi }}\right)^{m}-(-1)^{m}\left({\frac {\partial }{\partial \eta }}\right)^{m}\right\}}
e
(
m
−
1
)
i
π
2
[
(
∂
∂
ξ
)
m
+
(
−
1
)
m
(
∂
∂
η
)
m
]
{\displaystyle e^{(m-1){\frac {i\pi }{2}}}\left[\left({\frac {\partial }{\partial \xi }}\right)^{m}+(-1)^{m}\left({\frac {\partial }{\partial \eta }}\right)^{m}\right]}
From (7), we find
∂
n
−
m
∂
z
n
−
m
(
∂
∂
x
±
i
∂
∂
y
)
m
1
r
=
(
2
n
)
!
2
n
n
!
1
r
2
n
+
1
[
1
−
r
2
Δ
2
2.2
n
−
1
+
…
]
z
n
−
m
(
x
±
i
y
)
m
=
(
−
1
)
n
(
2
n
)
!
2
n
n
!
1
r
n
+
1
(
cos
m
ϕ
±
i
sin
m
ϕ
)
sin
m
θ
{
cos
n
−
m
θ
−
(
n
−
m
)
(
n
−
m
−
1
)
2.2
n
−
1
cos
n
−
m
−
2
+
…
θ
}
{\displaystyle {\begin{aligned}{\frac {\partial ^{n-m}}{\partial z^{n-m}}}\left({\frac {\partial }{\partial x}}\pm i{\frac {\partial }{\partial y}}\right)^{m}{\frac {1}{r}}&={\frac {(2n)!}{2^{n}n!}}{\frac {1}{r^{2n+1}}}\left[1-{\frac {r^{2}\Delta ^{2}}{2.2n-1}}+\dots \right]z^{n-m}(x\pm iy)^{m}\\&=(-1)^{n}{\frac {(2n)!}{2^{n}n!}}{\frac {1}{r^{n+1}}}(\cos m\phi \pm i\sin m\phi )\sin ^{m}\theta {\Bigl \{}\cos ^{n-m}\theta \\&-{\frac {(n-m)(n-m-1)}{2.2n-1}}\cos ^{n-m-2}+\dots \theta {\Bigr \}}\end{aligned}}}
hence
∂
n
−
m
∂
z
n
−
m
(
∂
∂
x
±
i
∂
∂
y
)
m
1
r
=
(
−
1
)
n
(
n
−
m
)
!
r
n
+
1
(
cos
m
ϕ
±
i
sin
m
ϕ
)
P
n
m
(
cos
θ
)
{\displaystyle {\frac {\partial ^{n-m}}{\partial z^{n-m}}}\left({\frac {\partial }{\partial x}}\pm i{\frac {\partial }{\partial y}}\right)^{m}{\frac {1}{r}}=(-1)^{n}{\frac {(n-m)!}{r^{n+1}}}(\cos m\phi \pm i\sin m\phi )P_{n}^{m}(\cos \theta )}
as we see on referring to (4); we thus obtain the formulae
∂
n
−
m
∂
z
n
−
m
{
(
∂
∂
ξ
)
m
+
(
∂
∂
η
)
m
}
1
r
=
(
−
1
)
n
(
n
−
m
)
!
2
m
−
1
r
n
+
1
cos
m
ϕ
P
n
m
(
cos
θ
)
{\displaystyle {\frac {\partial ^{n-m}}{\partial z^{n-m}}}\left\{\left({\frac {\partial }{\partial \xi }}\right)^{m}+\left({\frac {\partial }{\partial \eta }}\right)^{m}\right\}{\frac {1}{r}}=(-1)^{n}{\frac {(n-m)!}{2^{m-1}r^{n+1}}}\cos m\phi P_{n}^{m}(\cos \theta )}
∂
n
−
m
∂
z
n
−
m
{
(
∂
∂
ξ
)
m
−
(
∂
∂
η
)
m
}
1
r
=
(
−
1
)
n
(
n
−
m
)
!
2
m
−
1
r
n
+
1
sin
m
ϕ
P
n
m
(
cos
θ
)
{\displaystyle {\frac {\partial ^{n-m}}{\partial z^{n-m}}}\left\{\left({\frac {\partial }{\partial \xi }}\right)^{m}-\left({\frac {\partial }{\partial \eta }}\right)^{m}\right\}{\frac {1}{r}}=(-1)^{n}{\frac {(n-m)!}{2^{m-1}r^{n+1}}}\sin m\phi P_{n}^{m}(\cos \theta )}
It is thus seen that the tesseral harmonics of degree n and order
m are those which have
n
−
m
{\displaystyle n-m}
z axis, and
the other m axis distributed in the equatorial plane, at angular
intervals
π
/
m
{\displaystyle \pi /m}
8. Determination of the Poles of a given Harmonic. —It has been
shown that a spherical harmonic
Y
n
(
x
,
y
,
z
)
{\displaystyle Y_{n}(x,y,z)}
f
n
(
∂
∂
x
,
∂
∂
y
,
∂
∂
z
)
{\displaystyle f_{n}\left({\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}\right)}
1
r
{\displaystyle {\frac {1}{r}}}
the function
f
n
{\displaystyle f_{n}}
Y
n
(
x
,
y
,
z
)
=
(
−
1
)
n
(
2
n
)
!
2
n
n
!
{
1
−
r
2
∇
2
2.2
n
−
1
+
…
}
f
n
(
x
,
y
,
z
)
;
{\displaystyle Y_{n}(x,y,z)=(-1)^{n}{\frac {(2n)!}{2^{n}n!}}\left\{1-{\frac {r^{2}\nabla ^{2}}{2.2n-1}}+\dots \right\}f_{n}(x,y,z);}
this relation shows that if an expression of the form
(
x
2
+
y
2
+
2
)
f
n
−
2
(
x
,
y
,
z
)
{\displaystyle (x^{2}+y^{2}+^{2})f_{n-2}(x,y,z)}
is added to
f
n
(
x
,
y
,
z
)
{\displaystyle f_{n}(x,y,z)}
Y
n
(
x
,
y
,
z
)
{\displaystyle Y_{n}(x,y,z)}
Y
n
{\displaystyle Y_{n}}
f
n
(
x
,
y
,
z
)
=
0
{\displaystyle f_{n}(x,y,z)=0}
x
2
+
y
2
+
z
2
{\displaystyle x^{2}+y^{2}+z^{2}}
f
n
{\displaystyle f_{n}}
If x , y , z are such as to satisfy the two equations
Y
n
(
x
,
y
,
z
)
=
0
{\displaystyle Y_{n}(x,y,z)=0}
x
2
+
y
2
+
z
2
=
0
{\displaystyle x^{2}+y^{2}+z^{2}=0}
f
n
(
x
,
y
,
z
)
{\displaystyle f_{n}(x,y,z)}
Y
n
{\displaystyle Y_{n}}
x
2
+
y
2
+
z
2
=
0
{\displaystyle x^{2}+y^{2}+z^{2}=0}
Y
n
(
x
,
y
,
z
)
=
A
∏
s
=
1
m
(
l
s
x
+
m
s
y
+
n
s
z
)
+
(
x
2
+
y
2
+
z
2
)
V
n
−
2
(
x
,
y
,
z
)
,
{\displaystyle Y_{n}(x,y,z)=A\prod _{s=1}^{m}(l_{s}x+m_{s}y+n_{s}z)+(x^{2}+y^{2}+z^{2})V_{n-2}(x,y,z),}
we see that the plane
l
s
x
+
m
s
y
+
n
s
z
=
0
{\displaystyle l_{s}x+m_{s}y+n_{s}z=0}
2n generating lines of the imaginary cone
x
2
+
y
2
+
z
2
=
0
,
{\displaystyle x^{2}+y^{2}+z^{2}=0,}
Y
n
(
x
,
y
,
z
)
=
0
{\displaystyle Y_{n}(x,y,z)=0}
(
l
s
,
m
s
,
n
s
)
{\displaystyle (l_{s},m_{s},n_{s})}
x
2
+
y
2
+
z
2
=
0
{\displaystyle x^{2}+y^{2}+z^{2}=0}
n
(
2
n
−
1
)
{\displaystyle n(2n-1)}
2n generating lines in pairs. Of these systems of poles,
however, only one is real, viz. that in which the lines in each pair
correspond to conjugate complex roots of the equations
Y
n
=
0
{\displaystyle Y_{n}=0}
x
2
+
y
2
+
z
2
=
0
{\displaystyle x^{2}+y^{2}+z^{2}=0}
x
α
1
+
i
β
1
=
y
α
2
+
i
β
2
=
z
α
2
+
i
β
2
{\displaystyle {\frac {x}{\alpha _{1}+i\beta _{1}}}={\frac {y}{\alpha _{2}+i\beta _{2}}}={\frac {z}{\alpha _{2}+i\beta _{2}}}}
gives one generating line, then the conjugate one is given by
x
α
1
−
i
β
1
=
y
α
2
−
i
β
2
=
z
α
2
−
i
β
2
{\displaystyle {\frac {x}{\alpha _{1}-i\beta _{1}}}={\frac {y}{\alpha _{2}-i\beta _{2}}}={\frac {z}{\alpha _{2}-i\beta _{2}}}}
and the corresponding factor
l
x
+
m
y
+
n
z
{\displaystyle lx+my+nz}
|
x
y
z
α
1
+
i
β
1
α
2
+
i
β
2
α
3
+
i
β
3
α
1
−
i
β
1
α
2
−
i
β
2
α
3
−
i
β
3
|
{\displaystyle {\begin{vmatrix}x&y&z\\\alpha _{1}+i\beta _{1}&\alpha _{2}+i\beta _{2}&\alpha _{3}+i\beta _{3}\\\alpha _{1}-i\beta _{1}&\alpha _{2}-i\beta _{2}&\alpha _{3}-i\beta _{3}\\\end{vmatrix}}}
which is real. It is obvious that if any non-conjugate pair of
roots is taken, the corresponding factor, and therefore the pole, is
imaginary. There is therefore only one system of real poles of a
given harmonic, and its determination requires the solution of an
equation of degree 2n . This, theorem is due to Sylvester (Phil. Mag.
(1876), 5th series, vol. ii., "A Note on Spherical Harmonics ").
9. Expression for the Zonal Harmonic with any Axis .—The zonal
surface harmonic, whose axis is in the direction
x
′
r
′
,
y
′
r
′
,
z
′
r
′
{\displaystyle {\frac {x'}{r'}},{\frac {y'}{r'}},{\frac {z'}{r'}}}
P
n
(
x
x
′
+
y
y
′
+
z
z
′
r
r
′
)
{\displaystyle P_{n}\left({\frac {xx'+yy'+zz'}{rr'}}\right)}
or
P
n
(
cos
θ
cos
θ
′
+
sin
θ
sin
θ
′
cos
ϕ
−
ϕ
′
)
{\displaystyle P_{n}(\cos \theta \cos \theta '+\sin \theta \sin \theta '\cos \phi -\phi ')}
(
x
,
y
,
z
)
{\displaystyle (x,y,z)}
(
x
′
,
y
′
,
z
′
)
{\displaystyle (x',y',z')}
a
0
P
n
(
cos
θ
)
P
n
(
cos
θ
′
)
+
∑
1
n
a
m
P
n
m
(
cos
θ
)
P
n
m
(
cos
θ
′
)
cos
m
(
ϕ
−
ϕ
′
)
{\displaystyle a_{0}P_{n}(\cos \theta )P_{n}(\cos \theta ')+\sum _{1}^{n}a_{m}P_{n}^{m}(\cos \theta )P_{n}^{m}(\cos \theta ')\cos m(\phi -\phi ')}
and it only remains to determine the co-efficients
a
0
,
a
1
,
…
,
a
m
,
…
a
n
{\displaystyle a_{0},a_{1},\dots ,a_{m},\dots a_{n}}
(
x
′
x
+
y
′
y
+
z
′
z
)
n
{\displaystyle (x'x+y'y+z'z)^{n}}
x , y , z satisfy the condition
x
2
+
y
2
+
z
2
=
0
{\displaystyle x^{2}+y^{2}+z^{2}=0}
ξ
=
x
+
i
y
,
η
=
x
−
i
y
,
ξ
′
=
x
′
+
i
y
′
,
η
′
=
x
′
−
i
y
′
{\displaystyle \xi =x+iy,\eta =x-iy,\xi '=x'+iy',\eta '=x'-iy'}
(
x
x
′
+
y
y
′
+
z
z
′
)
m
=
(
1
2
η
′
ξ
+
1
2
ξ
′
η
+
z
z
′
)
n
{\displaystyle (xx'+yy'+zz')^{m}=({\frac {1}{2}}\eta '\xi +{\frac {1}{2}}\xi '\eta +zz')^{n}}
which equals
(
x
x
′
+
y
y
′
+
z
z
′
)
n
=
(
z
z
′
)
n
+
∑
∑
n
!
a
!
b
!
(
n
−
a
−
b
)
!
{
η
′
a
ξ
′
b
η
b
ξ
b
+
η
′
b
ξ
′
a
ξ
b
η
a
2
a
+
b
}
(
z
z
′
)
n
−
a
−
b
{\displaystyle (xx'+yy'+zz')^{n}=(zz')^{n}+\sum \sum {\frac {n!}{a!b!(n-a-b)!}}\left\{{\frac {\eta '^{a}\xi '^{b}\eta ^{b}\xi ^{b}+\eta '^{b}\xi '^{a}\xi ^{b}\eta ^{a}}{2^{a+b}}}\right\}(zz')^{n-a-b}}
the summation being taken for all values of a and b , such that
a
+
b
≤
n
{\displaystyle a+b\leq n}
a
>
b
{\displaystyle a>b}
a
=
0
,
b
=
0
{\displaystyle a=0,b=0}
(
z
z
′
)
n
{\displaystyle (zz')^{n}}
ξ
η
=
−
z
2
{\displaystyle \xi \eta =-z^{2}}
(
x
x
′
+
y
y
′
+
z
z
′
)
n
=
(
z
z
′
)
n
+
∑
∑
(
−
1
)
b
2
a
+
b
n
!
a
!
b
!
(
n
−
a
−
b
)
!
(
η
′
ξ
′
)
b
z
′
n
−
a
−
b
{
(
η
′
ξ
)
a
−
b
+
(
ξ
′
η
)
a
−
b
}
z
n
−
a
+
b
,
{\displaystyle {\begin{aligned}(xx'+yy'+zz')^{n}=(zz')^{n}+\sum \sum &{\frac {(-1)^{b}}{2^{a+b}}}{\frac {n!}{a!b!(n-a-b)!}}(\eta '\xi ')^{b}z'^{n-a-b}\\&\{(\eta '\xi )^{a-b}+(\xi '\eta )^{a-b}\}z^{n-a+b},\end{aligned}}}
putting
a
−
b
=
m
{\displaystyle a-b=m}
ξ
m
z
n
−
m
{\displaystyle \xi ^{m}z^{n-m}}
∑
(
−
1
)
b
2
m
+
2
b
n
!
b
!
(
m
+
b
)
!
(
n
−
m
−
2
b
)
!
(
ξ
′
η
′
)
b
η
′
m
z
′
n
−
m
−
2
b
,
{\displaystyle \sum {\frac {(-1)^{b}}{2^{m+2b}}}{\frac {n!}{b!(m+b)!(n-m-2b)!}}(\xi '\eta ')^{b}\eta '^{m}z'^{n-m-2b},}
from
b
=
0
{\displaystyle b=0}
b
=
1
2
(
n
−
m
)
{\displaystyle b={\frac {1}{2}}(n-m)}
1
2
(
n
−
m
−
1
)
{\displaystyle {\frac {1}{2}}(n-m-1)}
n
−
m
{\displaystyle n-m}
n
!
2
m
m
!
(
n
−
m
)
!
(
x
′
−
i
y
′
)
m
{
z
′
n
−
m
−
(
n
−
m
)
(
n
−
m
−
1
)
2.2
m
+
2
z
′
n
−
m
−
2
(
x
′
2
+
y
′
2
)
+
(
n
−
m
)
(
n
−
m
−
1
)
(
n
−
m
−
2
)
(
n
−
m
−
3
)
2.4.2
m
+
2.2
m
+
3
z
′
n
−
m
−
4
(
x
′
2
+
y
′
2
)
2
−
}
{\displaystyle {\begin{aligned}&{\frac {n!}{2^{m}m!(n-m)!}}(x'-iy')^{m}{\Bigl \{}z'^{n-m}-{\frac {(n-m)(n-m-1)}{2.2m+2}}z'^{n-m-2}(x'^{2}+y'^{2})\\&+{\frac {(n-m)(n-m-1)(n-m-2)(n-m-3)}{2.4.2m+2.2m+3}}z'^{n-m-4}(x'^{2}+y'^{2})^{2}-{\Bigr \}}\end{aligned}}}
in order to evaluate this coefficient, put
z
=
1
{\displaystyle z=1}
x
′
=
i
cos
α
{\displaystyle x'=i\cos \alpha }
y
′
=
i
sin
α
{\displaystyle y'=i\sin \alpha }
(
i
cos
α
+
sin
α
)
m
{\displaystyle (i\cos \alpha +\sin \alpha )^{m}}
i
m
e
−
m
i
α
{\displaystyle i^{m}e^{-mi\alpha }}
(
z
′
+
i
x
′
cos
α
+
i
y
′
sin
α
)
n
{\displaystyle (z'+ix'\cos \alpha +iy'\sin \alpha )^{n}}
e
−
i
α
{\displaystyle e^{-i\alpha }}
e
i
α
{\displaystyle e^{i\alpha }}
n
!
(
n
+
m
)
!
e
−
i
m
ϕ
′
P
n
m
(
cos
θ
′
)
r
′
n
.
{\displaystyle {\frac {n!}{(n+m)!}}e^{-im\phi '}P_{n}^{m}(\cos \theta ')r'^{n}.}
Similarly the coefficient of
η
m
z
n
−
m
{\displaystyle \eta ^{m}z^{n-m}}
n
!
(
n
+
m
!
)
e
+
i
m
ϕ
′
P
n
m
(
cos
θ
′
)
r
′
n
;
{\displaystyle {\frac {n!}{(n+m!)}}e^{+im\phi '}P_{n}^{m}(\cos \theta ')r'^{n};}
hence we have
1
r
′
n
(
x
x
′
+
y
y
′
+
z
z
′
)
n
=
z
n
P
n
m
(
cos
θ
′
)
+
n
!
∑
1
n
P
n
m
(
cos
θ
′
)
{
cos
m
ϕ
′
(
ξ
m
+
η
m
)
+
i
sin
m
ϕ
′
(
η
m
−
ξ
m
)
}
z
n
−
m
(
n
+
m
!
)
{\displaystyle {\begin{aligned}{\frac {1}{r'^{n}}}(xx'+yy'+zz')^{n}=z^{n}P_{n}^{m}(\cos \theta ')+n!\sum _{1}^{n}&P_{n}^{m}(\cos \theta ')\{\cos m\phi '(\xi ^{m}+\eta ^{m})\\&+i\sin m\phi '(\eta ^{m}-\xi ^{m})\}{\frac {z^{n-m}}{(n+m!)}}\end{aligned}}}
In this result, change x , y , z into
∂
∂
x
,
∂
∂
y
,
∂
∂
z
{\displaystyle {\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}}
![{\displaystyle e^{(m-1){\frac {i\pi }{2}}}\left[\left({\frac {\partial }{\partial \xi }}\right)^{m}+(-1)^{m}\left({\frac {\partial }{\partial \eta }}\right)^{m}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08fe20759a0364060bd01c7d545e55fa5d9fb36e)
![{\displaystyle {\begin{aligned}{\frac {\partial ^{n-m}}{\partial z^{n-m}}}\left({\frac {\partial }{\partial x}}\pm i{\frac {\partial }{\partial y}}\right)^{m}{\frac {1}{r}}&={\frac {(2n)!}{2^{n}n!}}{\frac {1}{r^{2n+1}}}\left[1-{\frac {r^{2}\Delta ^{2}}{2.2n-1}}+\dots \right]z^{n-m}(x\pm iy)^{m}\\&=(-1)^{n}{\frac {(2n)!}{2^{n}n!}}{\frac {1}{r^{n+1}}}(\cos m\phi \pm i\sin m\phi )\sin ^{m}\theta {\Bigl \{}\cos ^{n-m}\theta \\&-{\frac {(n-m)(n-m-1)}{2.2n-1}}\cos ^{n-m-2}+\dots \theta {\Bigr \}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19b6617ed425f253296f9b640f4dfe8e28835a1d)
