CHAPTER IV.

:

DIFFERENTIATION OF QUATERNIONS.

128.] IN Chapter I we have already considered as a special case the differentiation of a vector function of a scalar independent variable and it is easy to see at once that a similar process is applicable to a quaternion function of a scalar independent variable. The differential, or differential coefficient, thus found, is in general another function of the same scalar variable; and can therefore be differentiated anew by a second, third, &c. application of the same process. And precisely similar remarks apply to partial differentiation of a quaternion function of any number of scalar independent variables. In fact, this process is identical with ordinary differentiation.

129.] But when we come to differentiate a function of a vector, or of a quaternion, some caution is requisite; there is, in general, nothing which can be called a differential coefficient; and in fact we require (as already hinted in § 33) to employ a definition of a differential, somewhat different from the ordinary one but, coinciding with it when applied to functions of mere scalar variables.

130.] If r=F(q) be a function of a quaternion q,

where n is a scalar which is ultimately to be made infinite, is defined to be the differential of r or Fq.

Here dq may be any quaternion whatever, and the right-hand member may be written

f(q, dq),

where ƒ is a new function, depending on the form of F; homogeneous and of the first degree in dq; but not, in general, capable of being put in the form f (q) dq.

131.] To make more clear these last remarks, we may observe that the function

f(q, dq),

thus derived as the differential of F(q), is distributive with respect to dq. That is f (q, r + 8) = ƒ (q, r) +ƒ (q, 8),

And, as a particular case, it is obvious that if a be any scalar

x

r+

where q, r, s,... dq, dr, ds,

n

n

...... are any quaternions whatever; we

shall obviously arrive at a result which may be written

f(q, r, s,... dq, dr, ds,......),

where f is homogeneous and linear in the system of quaternions dq, dr, ds, ...... and distributive with respect to each of them. Thus, in differentiating any power, product, &c. of one or more quaternions, each factor is to be differentiated as if it alone were variable; and the terms corresponding to these are to be added for the complete differential. This differs from the ordinary process of scalar differentiation solely in the fact that, on account of the non-commutative property of quaternion multiplication, each factor must in general be differentiated in situ. Thus

d (qr) = dq.r+qdr, but not generally = rdq+qdr.

133.] As Examples we take chiefly those which lead to results which will be of constant use to us in succeeding Chapters. Some of the work will be given at full length as an exercise in quaternion transformations.

(1) (Tp)2 = p2.

The differential of the left-hand side is simply, since Tp is a scalar,

2 Tp dTp.

dq

n

This may be transformed into

Hence

since

=

p2 Tp2

Kq],

2TqdTq = d (qKq) = £= "[(q + 2) K (2 + 1/2 ) − q kg] ·

n

n

dTq

= S.UKqdq = S.Uq 1dq

Tq = TKq, and UKq=Uq-.

If q=p, a vector, Kq=Kp =—p, and the formula becomes

If q be a vector, as p, Sq and Sdq vanish, and we have

whence dq, i. e. dr3, is at once found in terms of dr. This process is given by Hamilton, Lectures, p. 628.

Since Kq=Sq-Vq, we find by a similar process

dKq = Kdq.

134.] Successive differentiation of course presents no new difficulty.

Thus, we have seen that

d(q2) = dq.q+qdq.

1

2dTp 2Spdp

d

and

=

Tp2 Tp3 Tp1

d.Vpdp=V.pd2p.

Up

Tp4

+

· ((Vpdp)2 +p2 Vpd2p—2V pdpSpdp) *.

135.] If the first differential of q be considered as a constant quaternion, we have, of course,

and the preceding formulæ become considerably simplified.

Hamilton has shewn that in this case Taylor's Theorem admits of an easy extension to quaternions. That is, we may write

ƒ(q+xdq) = ƒ (q)+xdƒ (q) + 22 d2 ƒ (q) + ......

1.2

if d2q = 0; subject, of course, to particular exceptions and limitations as in the ordinary applications to functions of scalar variables. Thus, let ƒ (q) = q3, and we have

df (q) = q2 dq+qdq.q+dq.q2,

d2 ƒ (q) = 2dq.qdq+2q (dq)2+2(dq)2q,

d3f (q) = 6 (dq)3,

and it is easy to verify by multiplication that we have rigorously (q+xdq)3=q3+x(q2dq+qdq.q+dq.q2)+x2 (dq.qdq+q(dq)2 + (dq)2q) +x3(dq)3; which is the value given by the application of the above form of Taylor's Theorem.

As we shall not have occasion to employ this theorem, and as the demonstrations which have been found are all too laborious for an elementary treatise, we refer the reader to Hamilton's works, where he will find several of them.

* This may be farther simplified; but it may be well to caution the student that we cannot, for such a purpose, write the above expression as