Quaternionic analysis: Difference between revisions

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An important example of a function of a quaternion variable is

:<math>f_1(q) = u q u^{-1}</math>

which [[quaternions and spatial rotation|rotates the vector part of ''q'']] by twice the angle represented by the versor ''u''.

The quaternion [[multiplicative inverse]] <math>f_2(q) = q^{-1}</math> is another fundamental function, but as with other number systems, <math>f_2(0)</math> and related problems are generally excluded due to the nature of [[dividing by zero]].

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== The derivative for quaternions ==

Since the time of Hamilton, it has been realized that requiring the independence of the [[derivative]] from the path that a differential follows toward zero is too restrictive: it excludes even <math>\ f(q) = q^2\ </math> from differentiation. Therefore, a direction-dependent derivative is necessary for functions of a quaternion variable.<ref>{{harvharvp|Hamilton|1866|loc=Chapter &nbsp;II, On differentials and developments of functions of quaternions, pp. &nbsp;391–495 }}</ref><ref>{{harvharvp|Laisant|1881|loc=Chapitre &nbsp;5: Différentiation des Quaternions, pp. &nbsp;104–117 }}</ref>

Considering the increment of [[polynomial function]] of quaternionic argument shows that the increment is a linear map of increment of the argument. {{Dubious|date=March 2019}} From this, a definition can be made:

A continuous mapfunction

<math>\ f: \mathbb H \rightarrow \mathbb H\ </math>

is called ''differentiable on the set'' <math>\ U \subset \mathbb H\ ,</math>, if, at every point <math>\ x \in U\ ,</math>, thean increment of the mapfunction <math>\ f\ </math> corresponding to a quaternion increment <math>\ h\ </math> of its argument, can be represented as

: <math> f(x+h) - f(x) = \frac{\operatorname d f(x)}{\operatorname d x} \circ h + o(h)</math>

where

: <math>\frac{\operatorname d f(x)}{\operatorname d x}:\mathbb H\rightarrow\mathbb H</math>

is [[linear map]] of quaternion algebra <math>\ \mathbb H\ ,</math> and

<math>\ o:\mathbb H\rightarrow \mathbb H\ </math>

isrepresents asome continuous map such that

: <math>\lim_{a \rightarrow 0} \frac{\ \left|\ o(a)\ \right|\ }{ \left|\ a\ \right| } = 0\ ,</math>

and the notation <math>\ \circ h\ </math> denotes ...{{explain|date=September 2024}}

The linear map

<math>\frac{\operatorname d f(x)}{\operatorname d x}</math>

is called the derivative of the map <math>\ f ~.</math>.

On the quaternions, the derivative may be expressed as

: <math>\frac{\operatorname d f(x)}{\operatorname d x} = \sum_s \frac{d_\operatorname{d}_{s0} f(x)}{\operatorname d x} \otimes \frac{d_\operatorname{d}_{s1} f(x)}{\operatorname d x} </math>

Therefore, the differential of the map <math>\ f\ </math> may be expressed as follows, with brackets on either side.

:<math>\frac{\operatorname d f(x)}{\operatorname d x}\circ dx\operatorname d x = \left(\sum_s \frac{d_\operatorname{d}_{s0} f(x)}{\operatorname d x} \otimes \frac{d_\operatorname{d}_{s1} f(x)}{\operatorname d x}\right)\circ dx\operatorname d x = \sum_s \frac{d_\operatorname{d}_{s0} f(x)}{\operatorname d x} dx\left( \operatorname d x \right) \frac{d_\operatorname{d}_{s1} f(x)}{\operatorname d x}</math>

The number of terms in the sum will depend on the function ''<math>\ f'' ~.</math> The expressions

<math>~~ \frac{d_\operatorname{d}_{sp}\operatorname d f(x)}{\operatorname d x}, ~~ \mathsf{\ for\ } ~~ p = 0, 1 ~~</math> are called

components of derivative.

The derivative of a quaternionic function holdsis thedefined followingby the equalitiesexpression

: <math>\frac{df\operatorname d f(x)}{\operatorname d x}\circ h = \lim_{t\to 0}\left(t^\ \frac{-1}(\ f(x +th t\ h) - f(x))\ }{ t }\ \right)</math>

where the variable <math>\ t\ </math> is a real scalar.

The following equations then hold:

: <math>\frac{\operatorname d\left( f(x) + g(x) \right)}{\operatorname d x} = \frac{df\operatorname d f(x)}{\operatorname d x} + \frac{dg\operatorname d g(x)}{\operatorname d x}</math>

: <math>\frac{df\operatorname d\left( f(x)\ g(x)\right)}{\operatorname d x} = \frac{df\operatorname d f(x) }{\operatorname d x}\ g(x) + f(x)\ \frac{dg\operatorname d g(x)}{\operatorname d x}</math>

: <math>\frac{df\operatorname d \left( f(x)\ g(x)\right)}{\operatorname d x} \circ h = \left(\frac{df\operatorname d f(x)}{\operatorname d x}\circ h\right )\ g(x) + f(x) \left(\frac{dg\operatorname d g(x)}{\operatorname d x}\circ h\right)</math>

: <math>\frac{daf\operatorname d \left( a\ f(x)\ b\right)}{\operatorname d x} = a\ \frac{df\operatorname d f(x)}{\operatorname d x}\ b</math>

: <math>\frac{daf\operatorname d \left( a\ f(x)\ b\right)}{\operatorname d x}\circ h = a \left(\frac{df\operatorname d f(x)}{\operatorname d x}\circ h\right) b</math>

For the function ''<math>\ f''(''x'') = ''axb''a\ x\ b\ ,</math> where <math>\ a\ </math> and <math>\ b\ </math> are constant quaternions, the derivative is

{| class="wikitable"

|-

| <math> \frac{daxb\operatorname d \left( a\ x\ b \right)}{\operatorname d x} = a \otimes b </math>

|style="background:white;"| &emsp;

|

| <math>dy \operatorname d y=\frac{daxb\operatorname d \left(a\ x\ b\right)}{\operatorname d x} \circ dx\operatorname d x = a\,dx \,left(\operatorname d x\right)\ b</math>

|}

Line 111 ⟶ 115:

{| class="wikitable"

|-

| <math> \frac{d_\operatorname{d}_{10} axb\left(a\ x\ b\right)}{\operatorname d x} = a </math>

|style="background:white;"| &emsp;

|

| <math> \frac{d_\operatorname{d}_{11} axb\left(a\ x\ b\right)}{\operatorname d x} = b </math>

|}

Similarly, for the function ''<math>\ f''(''x'') = ''x<sup>^2\ ,</supmath>'', the derivative is

{| class="wikitable"

|-

| <math>\frac{dx\operatorname d x^2}{\operatorname d x}=x \otimes 1 + 1 \otimes x</math>

|style="background:white;"| &emsp;

|

| <math>dy\operatorname d y=\frac{dx\operatorname d x^2}{\operatorname d x}\circ dx\operatorname d x = x\,dx \operatorname d x +dx (\,operatorname d x)\ x </math>

|}

Line 127 ⟶ 131:

{| class="wikitable"

|-

| <math> \frac{d_\operatorname{d}_{10} x^2 }{\operatorname d x} = x </math>

|style="background:white;"| &emsp;

|

| <math> \frac{d_\operatorname{d}_{11} x^2}{\operatorname d x} = 1 </math>

|-

| <math>\frac{d_\operatorname{d}_{20} x^2 }{\operatorname d x} = 1 </math>

|style="background:white;"| &emsp;

|

| <math>\frac{d_\operatorname{d}_{21} x^2}{\operatorname d x} = x </math>

|}

Finally, for the function '' <math>\ f''(''x'') = ''x''<sup>−1^{-1}\ ,</supmath>, the derivative is

{| class="wikitable"

|-

| <math> \frac{dx\operatorname d x^{-1} }{\operatorname d x} = -x^{-1} \otimes x^{-1}</math>

|style="background:white;"| &emsp;

|

| <math>dy \operatorname d y = \frac{dx\operatorname d x^{-1} }{\operatorname d x} \circ dx\operatorname d x = -x^{-1}dx(\,operatorname d x)\ x^{-1}</math>

|}

Line 147 ⟶ 151:

{| class="wikitable"

|-

| <math>\frac{d_\operatorname{d}_{10} x^{-1} }{\operatorname d x} = -x^{-1} </math>

|style="background:white;"| &emsp;

|

| <math>\frac{d_\operatorname{d}_{11} x^{-1} }{\operatorname d x} = x^{-1} </math>

|}

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| first = A.

| title = Quaternionic analysis

| journal = [[Mathematical Proceedings of the Cambridge Philosophical Society]]

| volume = 85

| issue = 2