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From the post, I know how to linear a simple polynomial function of several independent variables.

For example, for two variables a1=q1+eps*q1 and a2=q2+eps*q2, the function f[a1_, a2_] = a1^2 a2 can be linearized by expanding in power series to first order with Series.

Now, I would like to linearize some functions of functions. For example, given a function F of two functions g[x] and h[x]:

F[g_, h_] = (1 - g[x]*h[x]*Log[g[x]/b])/(c*Log[g[x]/a] - d*Log[g[x]/b]);

For example, how do I linearize its first derivative and square?

dFdx[g_, h_] = D[F[g, h], x]; Fsquare[g_, h_] = F[g, h]^2;

The simple extension of the method in that link did not work.

(Series[dFdx[g0 + ϵ*dg, h0 + ϵ*dh], {ϵ, 0, 1}] // Normal) /. ϵ -> 1

(Series[Fsquare[g0 + ϵ*dg, h0 + ϵ*dh], {ϵ, 0, 1}] // Normal) /. ϵ -> 1

In the linearization, I need to eliminate any variable preceded by ϵ with power higher than 1 and any product of two variables preceded by ϵ. Thank you!

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  • $\begingroup$ Do not set \[Epsilon] -> 1. Instead, use ``Coefficient[%, \[Epsilon]]`. $\endgroup$
    – bbgodfrey
    Commented Feb 16, 2022 at 5:30
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    $\begingroup$ I would define F as a function that also takes the independent variable x. That can make tasks like this easier. And it's also good practice in terms of coding. $\endgroup$ Commented Mar 18, 2022 at 15:02
  • $\begingroup$ @DanielLichtblau, did you mean to define F[x_]=... or F[g_, h_, x_]=..., I'm just confused by that "also" in your comment. Thank you! $\endgroup$
    – user95273
    Commented Mar 18, 2022 at 15:40
  • $\begingroup$ By "also" I had in mind that latter, F[g_,h_,x_]:=.... $\endgroup$ Commented Mar 18, 2022 at 21:43

3 Answers 3

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modified

Assuming g[x]-g0[x], h[x]-h0[x]to be small of equal order try

Fx = D[F[g, h],x] /. {g -> (g0[#] + eps dg[#] &), h -> (h0[#] + eps dh[#] &)}   
Normal[Series[Fx, {eps, 0, 1}]] /. eps -> 1 /. { dg[x] -> g[x] - g0[x],dh[x] -> h[x] - h0[x]} // Simplify
(* lengthy output...*)



F2 = F[g, h]^2 /. {g -> (g0[#] + eps dg[#] &), h -> (h0[#] + eps dh[#] &)}  
Normal[Series[F2, {eps, 0, 1}]] /. eps -> 1 /. { dg[x] -> g[x] - g0[x], dh[x] -> h[x] - h0[x]} // Simplify

(* (1/((c Log[g0[x]/a] - d Log[g0[x]/b])^3))(-((
2 (c - d) (g[x] - g0[x]) (-1 + g0[x] h0[x] Log[g0[x]/b])^2)/
g0[x]) + (c Log[g0[x]/a] - d Log[g0[x]/b]) (-1 +g0[x] h0[x]Log[g0[x]/b])^2 + 
2 (c Log[g0[x]/a] - d Log[g0[x]/b]) (1 - g0[x] h0[x] Log[g0[x]/b]) (-((g[x] - g0[x]) h0[x]) - (g0[x] (h[x] - 2 h0[x]) + g[x] h0[x]) Log[g0[x]/b]))*)

Hope it's what you are looking for!

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I think you're confusing the terms "function" and "expression," which I used to struggle with. In strict, pedantic math, a function $f$ can be applied to an argument like $f(a)$ or $f(3)$; and an expression like $x^2$ cannot: $x^2(a)$ and $x^2(3)$ do not mean $a^2$ or $3^2$. It does not help that in colloquial math, we say $x^2$ is a function and know what it means to evaluate it at $3$. Basic common sense and experience tell us that $x^2(3)$ is nonsense and not what the speaker meant and that they must mean to substitute $x=3$ into the expression $x^2$.

Likewise in Mathematica, there are two notions, function and expression, similar to the mathematical concepts. If an argument to F[g_, h_] is g = g0+ε*dg, then g[x] is (g0+ε*dg)[x], which is nonsense, albeit syntactically correct nonsense. That it is syntactically acceptable to Mathematica does not imply there is any built-in meaning to it. In the g[x] usage here, g is an expression being (mis)treated as a function.

One other thing before we get started. To linearize the derivative of a function requires the quadratic terms of the series expansion of the function. This is different from taking the derivative of the linearization. I assume the first is meant.


The first order of business then is to sort out whether you want F[] to operate on functions g, h or on expressions g, h. Since the linked post operates on expressions, then let's do that. As @DanielLichtblau said in the comments, it's best in this case to identify the independent variable that the expressions are functions of.

An "expressional" approach:

(* Better to use SetDelayed (:=) instead of Set (=)
   unless there is a specific reason not to.        *)
F // ClearAll;
F[g_, h_] := (1 - g*h*Log[g/b])/(c*Log[g/a] - d*Log[g/b]);

dFdx // ClearAll;
dFdx[g_, h_, x_] := D[F[g, h], x];
Fsquare // ClearAll;
Fsquare[g_, h_] := F[g, h]^2;
    
s1 = (Series[dFdx[
      g0 + ε*dg + ε^2/2*d2g, 
      h0 + ε*dh + ε^2/2*d2h, ε], {ε, 0, 1}] // 
    Normal) /. ε -> 1
s2 = (Series[Fsquare[g0 + ε*dg, h0 + ε*dh], {ε, 0, 1}] //
    Normal) /. ε -> 1

They have rather complicated output. If one wants to check the coefficients, then one could use CoefficientList[lin, {dg, dh}] or Coefficient[lin, dg] etc., where lin is one of the outputs.

A functional approach:

Here's a function-based approach in case that is desired or of interest. We can redefine things as follows:

F // ClearAll;
F[g_, h_] =  Function[ x,
   (1 - g[x]*h[x]*Log[g[x]/b])/(c*Log[g[x]/a] - d*Log[g[x]/b])];

dFdx // ClearAll;
dFdx[g_, h_] = F[g, h]';
Fsquare // ClearAll;
Fsquare[g_, h_] = Function[x, F[g, h][x]^2];

sub = {g[x0] -> g0, g'[x0] -> dg, g''[x0] -> d2g,
       h[x0] -> h0, h'[x0] -> dh, h''[x0] -> d2h};
ss1 = (Series[dFdx[g, h][x0 + ε, {ε, 0, 1}] // 
    Normal) /. ε -> 1 /. sub
ss2 = (Series[Fsquare[g, h][x0 + ε], {ε, 0, 1}] // 
    Normal) /. ε -> 1 /. sub

Check both approaches give the same output:

s1 == ss1 // Simplify
s2 == ss2 // Simplify
(*
True
True
*)
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To expand F around x==x0 you would write:

F[g_, h_] = (1 - g[x]*h[x]*Log[g[x]/b])/(c*Log[g[x]/a] - 
     d*Log[g[x]/b]);
Series[F[g, h], {x, x0, 1}] // Normal 

enter image description here

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  • $\begingroup$ thanks! it is the expansion of F, but how do I linearize its first derivative and square? $\endgroup$
    – user95273
    Commented Mar 12, 2022 at 3:32
  • $\begingroup$ Simple, instead of F in the expansion, you use F'. And for the square, you use F[..]^2 $\endgroup$ Commented Mar 15, 2022 at 15:52

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