Skip to main content

calculus and analysis - Write a function that returns the logarithmic derivative


How can we write a function that if we input an expression f, it returns the log derivative 1fdfdx? We have to use a conditional or pattern test so that the function accepts any symbols as input except for lists.


I will show all my attempts and I have some questions about why certain ways are wrong and why are they wrong. I am new to Mathematica and still not very familiar about how it works.


Here are all my attempts:

1).
In: f=x2;
In: func[f_] = 1ff′
Out: (x2)′x2
This attempt is wrong because first we are not using conditional or pattern test. Also, why cannot we differentiate the function by using the ' sign?


2).
In: f=x2;
In: func[f_] := 1ff′
Out: No output
Why is there no output when I use delayed assignment?



3).
In: f=x2;
In: func[f_,x_] = 1ff′
Out: (x2)′x2
Is adding another argument x_ redundant, or does it make any difference?


4).
In: f=x2;
In: func[f_] = Module[{f}, Switch[f, _Symbol, 1ff′]] Output: f$8344′f$8344
The syntax might not make any sense, but I am struggling to find ways to fix it. And I might have used _Symbol incorrectly, what is _Symbol after all?


Now I am trying to write the function in a less general way.

5).
Now In: f=x2;
In: func[f_] = 1fD[f,x]
Output: 2x
This method of course works but if I change the input to f = y2, the function will return 0 as output.
One way to solve this is by changing to func[f_] = 1fD[f,y], but I think this is considered as cheating, since we are not really answering the question and meeting the requirements set by the question.


What really makes me struggle is to compute the derivative, how can we compute the derivative of the input function with respect to its independent variable.


Also, I don't fully understand the Switch condition:
It is mentioned in Mathematica Help, that Switch evaluates expri, then compares it with each of the formi in turn, evaluating and returning the valuei. What is form and what is value?
Suppose I write the function:

In: f[x_] := Switch[x, x>0, 1, x<=0, 0] In: {f[5], f[-5], f[x]}
Out: {0,0,0}
Am I using the wrong form?


Any detailed help is greatly appreciated. Thanks!



Answer



Here's another way to proceed, using Derivative[], and sidestepping the use of a dummy variable:


LogDerivative[f_] := Derivative[1][Composition[Log, f]]

Test:


LogDerivative[Sin][x]

   Cot[x]

LogDerivative[Gamma][x]
   PolyGamma[0, x]

LogDerivative[#^3 &][x]
   3/x

Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more