Skip to main content

functions - Working with derivative of conjugate of a complex number


I have a complex function, lets say $g(x)$. I want to take its and its conjugate's derivative. I need the solution of derivative which must be symbolically and computationally efficient.
Lets take an example:


Derivative[1][g][x_] := d[g[x]]

Derivative[1][Conjugate][g[x_]] := Conjugate[d[g[x]]]/d[g[x]];

Derivative[1][Conjugate][d[x_]] := Conjugate[d[d[x]]]/d[d[x]]

Derivative[1][d][x_] := d[d[x]]/d[x];
Derivative[1][d][x_Symbol] := d[d[x]]

This will give me an effective symbolic representation of the derivative of $g(x)$ and $Conjugate(g(x))$ as d[g[x]] and Conjugate[d[g[x]]] but when I have to plug the analytical complex expression of $g(x)$ in d[g[x]], it will not compute the derivative of $g(x)$ instead gives only the symbolic representation of $d(g(x))$, which is computationally inefficient.
Is something can be done which is capable of symbolic as well as algebraic computation.


P.S. I do need the symbolic representation of the derivatives of conjugate in the above format only.



Answer



Maybe you can use the following two constructs to your advantage, which will keep the Conjugate, but evaluate and simplify the derivative inside. Using ReleaseHold, you can then evaluate even the Conjugate.



Note that I left out the divisor in the Conjugate-case for clarity, but you can easily add that into the second function's definition.


d[g_] := Derivative[1][g]
d[Conjugate[g_]] := With[{dg = d[g]@# // Simplify},
HoldForm[Conjugate[dg]] &]

(* example function *)
g[x_] := TrigToExp@Sin[x]

(* evaluation *)
d[g][x]

(* \[ExponentialE]^(-\[ImaginaryI] x)/2+
\[ExponentialE]^(\[ImaginaryI] x)/2 *)

d[Conjugate[g]][x]
(* Conjugate[1/2 \[ExponentialE]^(-\[ImaginaryI] x)
(1+\[ExponentialE]^(2 \[ImaginaryI] x))] *)



Update


If you want further derivatives, you can instead use this slight expansion of the idea above:



d[g_, n_:1] := Derivative[n][g]
d[Conjugate[g_], n_:1] := With[{dg = d[g, n]@# // Simplify}, HoldForm[Conjugate[dg]] &]

n gives the order of derivation you want. If left out, the first derivative is generated.


Interesting sidenote: n can even be negative, giving you the integral of your function. Observe e.g.:


h[x_]:=Sin[x]
d[h,0][x] (* Sin[x] *)

d[h,-1][x] (* -Cos[x] *)


d[h,1][x] (* Cos[x] *)
d[h][x] (* Cos[x] *)

Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...