[math-fun] Fibo-quaternion sequences
Given 2 quaternions q0,q1, we can define the traditional vector cross product operation: q0 x q1 = (q0 q1 - q1 q0)/2 [Note that this result is *independent* of the scalar parts of q0,q1, so q0 x q1 = V(q0) x V(q1) = (V(q0)V(q1) - V(q0)V(q1))/2] Given two arbitrary quaternions a,b, we can form the infinite sequence: q0=a, q1=b, q2=q0xq1, q3=q1xq2, ..., q(n)=q(n-2)xq(n-1).
From q2 forward, the qi are all 'pure vector' quaternions having no scalar part.
For such pure vector quaternions qi, qi^2=-(qi.qi)=-|qi|^2, so we can normalize all non-zero pure vectors into pure 'unit' vectors: u(v) = v/|v| = v/sqrt(-v^2) Unless the vector part of our arbitrary quaternion b is a scalar multiple of the vector part of our arbitrary quaternion a, then our qi sequence describes a spiral from the origin about the axis vector u(q2)+u(q3)+u(q4) and q(n)=M*q(n-3), for some real scalar M. If we consider the normalized sequence u(q2),u(q3),u(q4),u(q5),..., then this sequence cycles through 3 pure unit vectors: u(q2)->u(q3)->u(q4)->u(q1). Note that u(q2)^2=u(q3)^2=u(q4)^2=u(q2)u(q3)u(q4)=-1, so I=u(q2),J=u(q3),K=u(q4) make perfectly good basis vectors for our quaternion space. Finally, we can define q' = -(q+IqI+JqJ+KqK)/2 which gives us a perfectly good 'conjugate' operation for our quaternion space. Using our conjugate operation, we can now extract the scalar and vector parts of an arbitrary quaternion: s(q) = (q+q')/2 V(q) = q-s(q) Thus, our quaternion sequence can construct a 'coordinate-free' representation of quaternions from the minimal raw materials of the scalar field, the non- commutative product operation, and two given arbitrary quaternions with linearly independent vector parts. ---- We could also consider other 'fibo-quaternion' sequences, e.g., qn=q(n-2)q(n-1), as well as all such sequences over other fields -- e.g., GF(q^k).
participants (1)
-
Henry Baker