# Quaternion Derivative

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1. Sep 28, 2015

### abhiroop_k

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 ?

2. Sep 29, 2015

### Ssnow

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...

3. Sep 29, 2015

### abhiroop_k

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?

4. Sep 29, 2015

### Ssnow

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 ... )

5. Oct 7, 2015

### zinq

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.)

6. Oct 8, 2015

### Ssnow

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 ...

7. Oct 11, 2015

### zinq

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 ?
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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 possess either a left or right derivative.

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1 The original question has been edited and reformatted for clarity.

8. Oct 12, 2015

### Ssnow

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 $t\in\mathbb{R}$.