Since Kq=Sq-Vq, we find by a similar process and so on for higher orders. If q be a vector, as p, we have, § 133 (1), d(p2) = 2 Spdp. Hence d' (p2)=2(dp)2+2 Spd2p, and so on. [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 135. If the first differential of q be considered as a constant quaternion, we have, of course, d2q = 0, d3q = 0, &c., and the preceding formula become considerably simplified. Hamilton has shown that in this case Taylor's Theorem admits of an easy extension to quaternions. That is, we may write x2 ƒ(q+xdq) = ƒ (q)+xdƒ (q)+ · d2ƒ (q) + ...... 1.2 if d'q=0; subject, of course, to particular exceptions and limita- f(q) = q3, and we have df (q) = q2 dq+qdqq+dqq2, d2 ƒ (q) = 2 dq qdq+2q(dq)2 +2(dq)2q, d3 ƒ (q) = 6 (dq)3, and it is easy to verify by multiplication that we have rigorously (q + x dq)3 = q3 +x(q2dq+qdqq+dqq2)+x2 (dqqdq+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. 136. To differentiate a function of a function of a quaternion we proceed as with scalar variables, attending to the peculiarities already pointed out. 137. A case of considerable importance in geometrical applications of quaternions is the differentiation of a scalar function of p, the vector of any point in space. where F is a scalar function and C an arbitrary constant, be ƒ (p, dp) = 0, is, of course, a scalar function: and, being homogeneous and linear in dp, § 130, may be thus written, Svdp = 0, where v is a vector, in general a function of p. This vector, v, is easily seen to have the direction of the normal to the given surface at the extremity of p; being, in fact, perpendicular to every tangent line dp, §§ 36, 98. Its length, when F is a surface of the second degree, is the reciprocal of the distance of the tangent-plane from the origin. And we will show, later, that if (e.) d2.Tq = {S2.dqq ̄' — S. (dqq ̃')2 } Tq — — TqV ́2 ! dq CHAPTER V THE SOLUTION OF EQUATIONS OF THE FIRST DEGREE. 138. WE E have seen that the differentiation of any function whatever of a quaternion, q, leads to an where fis linear and homogeneous in dq. To complete the process of differentiation, we must have the means of solving this equation so as to be able to exhibit directly the value of dq. This general question is not of so much practical importance as the particular case in which q is a vector; and, besides, as we proceed to show, the solution of the general question may easily be made to depend upon that of the particular case; so that we shall commence with the latter. The most general expression for the function f is easily seen to be dr = ƒ (q, dq) = ΣV.adqb+S.cdq, where a, b, and c may be any quaternion functions of q whatever. Every possible term of a linear and homogeneous function is reducible to this form, as the reader may easily see by writing down all the forms he can devise. Taking the scalars of both sides, we have Sdr = S.cdq= Sdq Sc+S. Vdq Vc. But we have also, by taking the vector parts, Vdr = V.adqb = Sdq.Σ Vab + ΣV.a(Vdq)b, Eliminating Sdq between the equations for Sdr and Vdr it is obvious that a linear and vector expression in Vdq will remain. Such an expression, so far as it contains Vdq, may always be reduced to the form of a sum of terms of the type aS.ẞ Vdq, by the help of formulæ like those in §§ 90, 91. Solving this, we have Vdq, and Sdq is then found from the preceding equation. 139. The problem may now be stated thus. Find the value of p from the equation aSẞp+a, SB1p+....aSẞp = Y, = Σ.αδβρ γ, where a, ẞ, a,, B1, ...y are given vectors. [It will be shown later that the most general form requires but three terms, i. e. six vector constants a, ß, a,, ẞ1, a,, ß, in all.] and the object of our investigation is to find the value of the inverse function -1. 140. We have seen that any vector whatever may be expressed in terms of any three non-coplanar vectors. Hence, we should expect à priori that a vector such as pop, or p3p, for instance, should be capable of expression in terms of p, øp, and p2p. [This is, of course, on the supposition that p, øp, and 43p are not generally coplanar. But it may easily be seen to extend to this case also. For if these vectors be generally coplanar, so are op, op, and p3p, since they may be written σ, do, and po. And thus, of course, p3p can be expressed as above. If in a particular case, we should have, for some definite vector p, op = gp where g is a scalar, we shall obviously have 42p = g2p and p3p g3p, so that the equation will still subsist. And a φρ |