Quaternion division and hyperbolic tangent tanh

Question

Quaternion multiplication is well-defined, and is known to me as "Hamilton product":

// hamilton product
vec4 qmul(in vec4 q1, in vec4 q2) {
 return vec4(
 q1.w * q2.xyz + q2.w * q1.xyz - cross(q1.xyz, q2.xyz),
 q1.w*q2.w - dot(q1.xyz, q2.xyz)
 );
}

However, for implementing qtanh() quaternionic function, we need division. So far I've found this, and it is working OK. Could you help me to undestand, where does this comes from?

// division
// https://www.boost.org/doc/libs/1_67_0/boost/math/quaternion.hpp
vec4 qdiv(in vec4 q1, in vec4 q2) {
float denominator = dot(q2,q2);
return vec4( 
 vec3(
 -q1.w*q2.x+q1.x*q2.w-q1.y*q2.z+q1.z*q2.y,
 -q1.w*q2.y+q1.x*q2.z+q1.y*q2.w-q1.z*q2.x,
 -q1.w*q2.z-q1.x*q2.y+q1.y*q2.x+q1.z*q2.w
 ),
 q1.w*q2.w + dot(q1.xyz, q2.xyz)
) / denominator;
}

Also, as far as I am trying to implement tanh().. are you aware of more computationally vat, than dividing sinh and cosh? For reals I've used to use following formula: tanh(x)=-1+2/(1+exp(-x)). And that involves only single exponential calculus, instead of two..

Answer 1

1. Division

Dividing a quaternion named p by a quaternion named q is nothing more than multiplying p by the reciprocal of q.

This is equivalent to multiplying p by the conjugation of q (which by definition equals a – bi – cj – dk) and dividing the product by a scalar equalling q norm squared:

From here it`s obvious where that denominator part comes from:

Now let`s rearrange the terms in vec3 sums for better readability:

vec3(
 -q1.w*q2.x + q1.x*q2.w - (q1.y*q2.z - q1.z*q2.y),
 -q1.w*q2.y + q1.y*q2.w - (q1.z*q2.x - q1.x*q2.z),
 -q1.w*q2.z + q1.z*q2.w - (q1.x*q2.y - q1.y*q2.x)
)

And now it suddenly gets clear what`s going on:

vec3(
 -q1.w * q2.x + q1.x * q2.w - (q1.y*q2.z - q1.z*q2.y),
 -q1.w * q2.y + q1.y * q2.w - (q1.z*q2.x - q1.x*q2.z),
 -q1.w * q2.z + q1.z * q2.w - (q1.x*q2.y - q1.y*q2.x)
)
...
 -q1.w * q2.xyz + q1.xyz * q2.w - (cross(q1.xyz, q2.xyz))

So yeah, quaternion division is just the regular run-of-the-mill multiplication, with the multiplicand being a reciprocal. That`s where the minuses come from, see the definition above.

2. Hyperbolic tangent

First, the definitions. For each q = a + bi + cj + dk = a + v̅:

So, to get both e^q and e^–q you only need to compute the following values: e^a, ||v̅||, sin(||v̅||), cos(||v̅||).

To compute e^–q you should take a reciprocal of e^a and multiply it by the rest of the equation with signs inverted. That won`t take much time, for all the values it consists of have already been computed. One exp() call, as requested =)

• xakepp35: Could a tanh be implemented with only one exp() call, eg, for reals I used tanh(x)==-1+2/(1+exp(-x))
• xakepp35: Could I cheat and somehow lower its computational complexity?
• hidefromkgb: @xakepp35 Seems like yes, but the comment section is not very convenient to post the solution. Please amend your question and I`ll add it to my answer.
• xakepp35: Added to the end
• hidefromkgb: @xakepp35 Done.