Question

Quaternion differentiation

Answer 1

We have complex differentiation defined as \[
f'(z) = \lim_{|h|\to 0}\frac{f(z+h)-f(z)}{h}
\]Where \(z,h\) are complex numbers, I believe

So the idea here is whether this has any merit for Quaternions

Answer 2

@Kainui @dan815

Answer 3

All of the operations here are defined for quaternions, the only issue is whether things like anti commutativity causes issues.

Answer 4

\[
f(q) = q^2
\]

Answer 5

Actually, I think we should start with: \[
f(q) = q
\]

Answer 6

I'll denote \[
q=q_1+q_ii+q_jj+q_kk
\]

Answer 7

why cant we think of differentiation as this 4D vector difference then

Answer 8

But without really focusing on the exponents, I think we can say: \[
\frac{f(q+h)-f(q)}{h} = \frac{(q+h)-q}{h} = \frac{h}{h} =1
\]So we don't even needs limits here, this will work: \[
\frac{d}{dq} q =1
\]

Answer 9

Now we'll do \[
f(q) =q^2
\]

Answer 10

\[
\frac{(q+h)^2-q^2}{h} = \frac{q^2+qh+hq+h^2 - q^2}{h}
\]

Answer 11

The thing is, if we say \(qh = -hq\), then they'll cancel out instead of add up to \(2qh\).

Answer 12

Now that I think about it... \(qq = -qq \implies qq=0 = q^2\).

Answer 13

Actually, that isn't right, it applies to the basis, but not all \(q\)

Answer 14

I'm pretty sure it's really hard to take the derivative of quaternions.

Answer 15

Is it hard to divide quarternions?

Answer 16

It's easy to divide them I think there are no problems there. Here, scroll down to page 7 Definition 4.4 http://www.itpa.lt/~acus/Knygos/Clifford_articles/Clifford_Authors/05_DeLeo_Quaternions/quaternions5/Buchmann.pdf

Answer 17

It says \[
q^{-1}=\frac{q^*}{\|q\|^2}
\]And we know \(\|q\|= \sqrt{qq^*}\) so:\[
\frac{q^*}{\|q\|^2}=\frac{q^*}{qq^*}

\]

Answer 18

I think that is a direct result of the anti-commutativity of their multiplication.

Answer 19

wait u know how he only gives i^2=j^2=k^2=ijk=-1

what rules are allowed here to determine the other multiplications

Answer 20

like can i say jk=-1/i and -1/i =?

Answer 21

=1?

Answer 22

@dan815 Write it in it's matrix form, then multiply the matricies.

Answer 23

Make them as matrices, like this:\[
a+bi+cj+dk = \begin{bmatrix}
a&b&c&d\\
-b&a&-d&c\\
-c&d&a&-b\\
-d&-c&b&a
\end{bmatrix}
\]Multiply them, and then take off the top row of the result.

Answer 24

Since: \[
ijk=-1
\]You would say \[
i^{-1}ijk = i^{-1}(-1) \implies jk=-i^{-1}
\]I believe.

Answer 25

but to get that matrix id have to know what ij and ik was wudnt i

Answer 26

ya i was thinking about inverse too

Answer 27

Okay, start with :\[
ijk = -1
\]Multiply both sides by \(i\): \[
iijk=(-1)jk = i(-1)
\]

Answer 28

okay yep i see u can isolate for all just from his given rules

Answer 29

jk=i
-k=ji
j*-k=-i

Answer 30

i will be back -.- lemme try some actual 3d rotations in space with this stuff

Answer 31

Okay, here is how I would extend what I did before with \(j,k\) into \(R^4\).

Answer 32

\[
1=\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}
\]This will always be the case. We make our first basis by shifting every element to the right and making the bottom row negative: \[
i=\begin{bmatrix}
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
-1&0&0&0
\end{bmatrix}
\]Our remaining basis are just products of this shift. \[
j=i^2=\begin{bmatrix}
0&0&1&0\\
0&0&0&1\\
-1&0&0&0\\
0&-1&0&0
\end{bmatrix}\\
k=i^3=\begin{bmatrix}
0&0&0&1\\
-1&0&0&0\\
0&-1&0&0\\
0&0&-1&0
\end{bmatrix}\\
-1=i^4=\begin{bmatrix}
-1&0&0&0\\
0&-1&0&0\\
0&0&-1&0\\
0&0&0&-1
\end{bmatrix}
\]So this can be generalized into \(\mathbf R^{n}\)

Answer 33

It just so happens that in \(\mathbb R^2\), the complex numbers follow this set up.