Quaternion Derivative: Product Rule Explained

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SUMMARY

The discussion centers on the application of the product rule in quaternion derivatives, specifically for the function f(q) = q* a q, where q is a quaternion and a is a constant quaternion. The participants clarify the definitions of right and left quaternion derivatives, concluding that f(q) does not possess either type of derivative due to its form. The correct derivative expression derived is d[f(q)] / dq = q* a + aq, with specific emphasis on the importance of choosing the correct multiplication direction in the limit definition.

PREREQUISITES
  • Understanding of quaternion algebra and operations
  • Familiarity with the concepts of left and right derivatives
  • Knowledge of limit definitions in calculus
  • Basic principles of quaternionic analysis
NEXT STEPS
  • Study the implications of the Gateaux derivative for quaternions
  • Explore quaternionic analysis as presented in A. Sudbery's paper
  • Learn about quaternion differentiation techniques in robotic applications
  • Investigate the mathematical properties of quaternion products and their derivatives
USEFUL FOR

Mathematicians, physicists, and engineers working with quaternionic analysis, particularly in fields such as robotics and mathematical physics, will benefit from this discussion.

abhiroop_k
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How does the quaternion derivative work in the presence of a quaternion product.
More specifically, does the standard product rule apply for quaternion derivatives?

Say, I have a function f(q) = q* x a x q [where q -> quaternion, a -> const vector x-> quat prod]

what is the result of the operation d[f(q)] / dq ?
 
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For quaternion, we have two kind of derivatives, the right derivative respect ## q## defined as

## \left(\frac{d}{dq}\right)_{r}f=\lim_{h\rightarrow 0}\frac{f(q+h)-f(q)}{h} ##

and in similar way the left derivative...
 
Thanks!
Could you please elaborate a little more w.r.t the example I mentioned?

I read a few articles and have come up with the following result for the equation i posed:

d [f(q)] / dq = [-(a x q)/2] + [q* x a]

Is this right?
 
yes it is possible, I don't understand the ##\frac{1}{2}##, using the definition

## \left(\frac{d}{dq}f\right)_{r}=\lim_{h\rightarrow 0}\frac{(q+h)^{*}a(q+h)-q^{*}aq}{h}=\lim_{h\rightarrow 0}\frac{q^{*}ah+haq+h^{2}a}{h}=q^{*}a+aq ##

I don't know if it is what you need because depends from the definition (other authors can use a different definition). Quaternionic analysis is useful in different fields as robotic, mathematical physics ... I wrote one chapter on my book of analysis about quaternionic analysis

http://www.lulu.com/it/it/shop/simone-camosso/analisi-matematica/paperback/product-20723617.html

( is in Italian ... )
 
Ssnow said:
For quaternion, we have two kind of derivatives, the right derivative respect ## q## defined as

## \left(\frac{d}{dq}\right)_{r}f=\lim_{h\rightarrow 0}\frac{f(q+h)-f(q)}{h} ##

and in similar way the left derivative...

I'm confused. What is "right" in this formula? What would the "left" version be? (I mean on the right-hand side of the equals sign.)

(Also, the division by h in the formula is not well-defined unless it is specified to be either left- or right-multiplication by 1/h.)
 
yes, right stands for right multiplication for ##\frac{1}{h}##, now that I see there is also an error in the previous example ... ## \lim_{h\rightarrow 0}\frac{q^{*}ah +h^{*}aq +h^{*}ha}{h}=q^{*}a+\lim_{h\rightarrow 0} h^{*}aqh^{-1}+ \|h\|^{2}ah^{-1}=q^{*}a -\frac{1}{2}aq## so the result agree with abhiroop_k ...
 
Given the function1

f(q) := q* a q​

(where a is a constant quaternion, and q* denotes the conjugate of q), what is the derivative d[f(q)] / dq ?
----------------------------------------------------------------------------------

As mentioned, the usual definition of the derivative needs to choose whether to multiply the difference quotient on the right or on the left by 1/h before taking the limit.

We can try to use either

d[f(q)] / dq := limh→0 (1/h)(f(q+h) - f(q))​

or

d[f(q)] / dq := limh→0 (1/h)(f(q+h) - f(q))(1/h).​

Suppose we decide to try to use the top definition, the quaternion derivative on the left.

Then Theorem 1 from the paper Quaternionic analysis by A. Sudbery, Math. Proc. Camb. Phil. Soc. (1979), vol. 85 states:

If the left quaternionic derivative exists in some open set of the quaternions, then

f(q) = c + q d

where c and d are constant quaternions.​

(Of course the hypothesis means that the increment h can approach 0 from any direction.)

Likewise, if we assume the existence of the right quaternionic derivative as in the second definition above, then

f(q) = c + d q​

for some constant quaternions c and d.

Finally, if the function f is both left- and right-quaternionically differentiable, then we can conclude that

f(q) = c + r q​

for some constant quaternion c and real number r.
Because the function f(q) = q* a q is not (and cannot be expressed) in the form

f(q) = c + q d​

or

f(q) = c + d q,​

that means it does not possesses either a left or right derivative.

______________________________________________________________________
1 The original question has been edited and reformatted for clarity.
 
Quaternion functions as ## f(q)=q^{2} ## have some problem with differentiation in the sense that I used above, the Theorem 1 cited by Zinq is a classical result in quaternionic analysis. Possible results can be obtained focusing in a particular directions using the Gateaux derivative for quaternions

## Df(q)(h)=\lim_{t\rightarrow 0}\frac{f(q+th)-f(q)}{t} ##

with ##t\in\mathbb{R}##.
 

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