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whence, if m-i be not = 0,

1

21

m

IS Y, Yududø= 0.

f SS." vidudp will be investigated here

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The value of

after.

1

11. We may hence prove that if a function of u and can be developed in a series of surface harmonics, such development is possible in only one way.

For suppose, if possible, that there are two such developments, so that

F(d, $)= Y+ Y, + ... + Y, + ...

Y. and also

F (u,) = Y.' + Y,' +...+Y/' + ... Then subtracting, we have 0= Y. - Y/'+Y,- Y' + ... + Y,-Y! + ... identically,

Now, each of the expressions Yo-Yo, Y.-Y...Yi-Y being the difference of two surface harmonics of the degree 0, 1, ... i ... is itself a surface harmonic of the degree 0, 1, ... i .... Denote these expressions for shortness by 2, 2... Z.., so that

0 =2+ZA+ ... + 2.+... identically.
Z.

Z: Then, multiplying by Z, and integrating all over the surface of the sphere, we have

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That is, the sum of an infinite number of essentially positive quantities is = 0. This can only take place when each of the quantities is separately = 0. Hence Z, is identically = 0, or Y/' = Y, and therefore the two developments are identical.

We have not assumed here that such a development is always possible. That it is so, will be shewn hereafter,

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12. By referring to the expression for a surface harmonic given in Art. 4, 'we see that each of the Tesseral and Sectorial Harmonics involves (1 - M*)!, or some power of (1 – re?)t, as a factor, and therefore is equal to 0 when u=+1. From this it follows that when u=+1, the value of the Surface Harmonic is independent of $, or that if y(u, 6) represent a general surface harmonic, Y (+ 1, ) is independent of $, and may therefore be written as Y (+ i). Or Y (1) is the value of Ý (u, ) at the pole of the zonál harmonic Pilu), Y(-1) at the other extremity of the axis of Pi().

We may now prove that

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+

For, recurring to the fundamental equation,
d
dY1 1 d'Y

+ili+1)Y; = 0.
du

dus 1-re? do Now, if we integrate this equation with respect to , between the limits 0 and 27, we see that, since

d'Y dY,

do do and the value of Y, only involves $ under the form of cosines or sines of $ and its multiples, and therefore the values of dY,

are the same at both limits, it follows that

i

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dørdø=0.

Hence

21

som
dpt {(1 – 4) ()* 7,2$)}+i (i +1) UT Y db) = 0.

,
Hence ("Ydp is a function of ye which satisfies the

M

)

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fundamental equation for a zonal harmonic, and we therefore have

*

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$** Ydø= CP.6-), C being a constant, as yet unknown.

To determine C, put u=1, then by the remark just made, Y, becomes Y (1), and is independent of $. Hence, when v=1, $* Y,dø= 2Y,(1). Also P(x)=1. We have therefore

27 Y (1)= C,

.. Y.de - 2-Y. (1)P.6). It follows from this that

21

=

=,

21

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13. We may now enquire what will be the value of

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Y, Z, being two general surface harmonics of the degree i. Suppose each to be arranged in a series consisting of the zonal harmonic P, whose axis is the axis of z, and the system of tesseral and sectorial harmonics deduced from it. Let us represent them as follows: Y,=

АР. +C_T_(1) cos 0 + T12) cos 20 + ... + C T (6) cos 00 + ... C

+ CAT6 cos id + SAT>(1) sin $ + S,T,2) sin 20 + ... + SGT () sin op +...

+ S,T,(6) sin id; Z=

+CT() cos +c, T:12) cos 26+ ... + CT() cos o$ + ... c10

6

CoT;() cos ip +s, T.(1) sin ø+s,T! 2) sin 20 + ... +s 7,6) sin 00 + $

+od

+8,7, sin id. Hence the product Y, Z, will consist of a series of terms, in which $ will enter under the form cos o$ cos o't, or cos op sin o'p. This expression when integrated between

=

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a

the limits 0 and 2 vanishes in all cases, except when o=o and the expression consequently becomes equal to cos o$, or sin? op. In these cases we know that, o being any positive integer,

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Hence the question is reduced to the determination of the value of

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o

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do P, Now T() = (1 –

du" 1

(1 – 4)* dito (w?–1).

?? 2.1.2.3...

duito But, by the theorem of Rodrigues, proved in Chap. II. Art. 8, we know that dito (u* - 1)

di-o (Ma - 1) =(-1)

(1 ?)

duito

duino Hence T (o) may also be expressed under the form

1

ito "

Lito

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(-1) 24.1.2.3...i

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q

duino

1

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duito

whence it follows that

ito dito (?-1)' di-o (u?– 1)" (T~)*=(-1)0 (2.

2.1.2.3...) i-o
i

duino Now, putting (2 - 1)' = M for the moment, and integrating by parts, dito M di-M dito-1 M di-o M

du
duito duiro duito-1 duino

di+o-1 M di-o+1 M

άμ. duito-1 dpi-o+1

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S

di-M
The factor vanishes at both limits, hence

dui-o
dito M di-M

dito-1M di-o+1 M

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du

du

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di +0-2 M di-o+2 M
dpi+o-2 dui-o+2

du,

duito by a repetition of the same process. And by repeating this process o times, we see that

pi dito M di-o M

dp=(-1)

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Hence

and therefore

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0 lito 27

2

¿-0 21 +1• It will be observed that this result does not hold when o=0, in which case we have

47
2i +1

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47
21 +1

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* In this case

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