Quaternionic analysis: Difference between revisions

Revision as of 08:14, 18 September 2024

Revision as of 08:56, 18 September 2024

Line 61:

== 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>{{harvp|Hamilton|1866|loc=Chapter &nbsp;II, On differentials and developments of functions of quaternions, pp. &nbsp;391–495 }}</ref><ref>{{harvp|Laisant|1881|loc=Chapitre &nbsp;5: Différentiation des Quaternions, pp.&nbsp;104–117 104–117}}</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 that the direction-dependent quaternionic derivative is oriented in the direction of the quaternion ...{{mvarexplain|hdate=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{\operatorname{d}_{s0} f(x)}{\operatorname d x} \otimes \frac{\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 \operatorname d x = \left(\sum_s \frac{\operatorname{d}_{s0} f(x)}{\operatorname d x} \otimes \frac{\operatorname{d}_{s1} f(x)}{\operatorname d x}\right)\circ \operatorname d x = \sum_s \frac{\operatorname{d}_{s0} f(x)}{\operatorname d x} \left( \operatorname d x \right) \frac{\operatorname{d}_{s1} f(x)}{\operatorname d x}</math>

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

<math>~~ \frac{\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{\operatorname d f(x)}{\operatorname d x}\circ h = \lim_{t\to 0}\Biglleft(\ t^\frac{-1} \bigl(\ f(x + t\ h) - f(x)\ \bigr)}{ t }\ \Bigrright)</math>

:: where the variable {{mvar|<math>\ t}}\ </math> is reala /real scalar.

The following equations then hold:

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

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

Line 103 ⟶ 104:

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

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

{| class="wikitable"

|-

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

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

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

|}

Line 116 ⟶ 117:

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

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

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

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

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

|}

Line 139 ⟶ 140:

|}

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

{| class="wikitable"

|-

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

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

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

|}