0
$\begingroup$

I have coefficients in a list:

allAs = {1,2,3,4,5}

And another list that is created with the Table function, with an "x" in each element which I want to find the derivative with respect to:

expTable = Table[Exp[-I*2 Pi*x*k], {k, 1, 5}]; 

My function is formed from the two lists above in the following way

function = Abs[1 - Total[allAs*expTable]]^2;

I would like to find the derivative of this function with Derivative, but the "x" is not in the function explicitly. How should I go about finding the derivative of function with respect to x.

I have tried to define the function as below before taking the derivative but it doesn't work:

function[_x] := Abs[1 - Total[allAs*expTable]]^2;
$\endgroup$
1
$\begingroup$
Clear["Global`*"]

allAs = {1, 2, 3, 4, 5};

expTable = Table[Exp[-I*2 Pi*x*k], {k, 1, 5}];

Use Dot and assuming that x is real

function[x_] = Abs[1 - allAs.expTable]^2 // ComplexExpand // Simplify

(* 2 (28 + 39 Cos[2 π x] + 24 Cos[4 π x] + 11 Cos[6 π x] + 
   Cos[8 π x] - 5 Cos[10 π x]) *)

The derivative is

function'[x] // Simplify

(* -4 π (39 Sin[2 π x] + 48 Sin[4 π x] + 33 Sin[6 π x] + 
   4 Sin[8 π x] - 25 Sin[10 π x]) *)

Column[Plot[#, {x, -5, 5}, ImageSize -> Medium] & /@ {function[x], 
   function'[x]}]

enter image description here

$\endgroup$
0
$\begingroup$

The syntax of func[var_] := somestuff is used when you want to define a function that you evaluate over and over. The := means SetDelayed and the right-hand side is not evaluated until you call func[] with some argument. Also, the underscore should come after the x.

In this case, function already contains an expression that explicitly relies on x after your 3rd line of code. I would normally recommend D[function, x]. However, Abs is not differentiable so you will get a function that cannot be evaluated.

deriv = D[function, x]
deriv /. x -> 1
(* 2*((2*I*Pi)/E^(2*I*Pi*x) + (8*I*Pi)/E^(4*I*Pi*x) + 
   (18*I*Pi)/E^(6*I*Pi*x) + (32*I*Pi)/E^(8*I*Pi*x) + 
   (50*I*Pi)/E^(10*I*Pi*x))*Abs[1 - E^(-2*I*Pi*x) - 
   2/E^(4*I*Pi*x) - 3/E^(6*I*Pi*x) - 4/E^(8*I*Pi*x) - 
   5/E^(10*I*Pi*x)]*Derivative[1][Abs][-E^(-2*I*Pi*x) - 
   2/E^(4*I*Pi*x) - 3/E^(6*I*Pi*x) - 4/E^(8*I*Pi*x) - 
   5/E^(10*I*Pi*x) + 1]

   3080 I Pi Abs'[-14]
 *)

EDIT:

Also, my usual solution to define a function based on something previously evaluated, assuming deriv = D[function, x] yields something that can be evaluated, would be to write something like:

myfunc[x_] := deriv

and before evaluating, highlight deriv only, select the Evaluation menu > Evaluate in Place, or else

myfunc[x_] := Evaluate[derive]

There may be other, better ways.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.