Why does Mathematica simplify $x/x\to1$?

Question

If I enter x/x, I get 1. Such behavior leads to this:

Simplify[D[Sqrt[x^2], x, x]]

0

The same would be even if I use Together instead of Simplify.

One could then think that $\sqrt{x^2}$ is doubly differentiable at least $\forall x\in\mathbb R$, but if we remove Simplify call, we would reveal that it's not:

D[Sqrt[x^2], x, x]

-(x^2/(x^2)^(3/2)) + 1/Sqrt[x^2]

Even more ridiculous is this (which I guess is because x/x is simplified before feeding to Assuming):

Assuming[x == 0, x/x]

1

Why does Mathematica assume $x\ne0$? Is there a way to make it not cancel out such terms?

Answer 1

fun = Sqrt[x^2];

dd = D[fun, x, x];

With V10 we can explicitly define:

{dd, FunctionDomain[dd, x]}

Or

{Simplify @ dd, FunctionDomain[dd, x]}

{0, x < 0 || x > 0}

Also with V10 you might consider Inactivate:

di = D[Inactivate@Sqrt[x^2], x, x] // Together

which prevents Together from evaluating to 0

di // Activate

0

Answer 2

Let's define

f[x_] := Sqrt[x^2]

FullSimplify[D[f[x],x], x ∈ Reals]

(* ==> Sign[x] *)

You can use f'[x] or D[f[x],x] interchangeably here. Then the second derivative is simply

FullSimplify[D[Sign[x], x], x ∈ Reals]

(* ==> Derivative[1][Sign][x] *)

So doing the second derivative in two steps yields a mathematically correct answer. The derivative of the Sign function is not defined, but we can define it in the sense of distributions by replacing Sign[x] with 2HeavisideTheta[x]-1. Then one would obtain this nice result for the second derivative:

FullSimplify[D[2 HeavisideTheta[x] - 1, x], 
 x ∈ Reals]

(* ==> 2 DiracDelta[x] *)

Here is my initial way to get a mathematically expected result by making the domain explicit over which the function is defined:

f[x_] := Piecewise[{{Sqrt[x^2], x > 0}, {Undefined, True}}]

Simplify[D[f[x], x, x]]

(* ==> ConditionalExpression[0, x > 0] *)

However, this is only a band-aid.

Answer 3

Here are three approaches to the function:

f1[x_] := Piecewise[{{Sqrt[x^2], x != 0}}, 0]

f2[x_] := Piecewise[{{Sqrt[x^2], x < 0}, {Sqrt[x^2], x > 0}}, 0]

f3[x_] := Piecewise[{{-x, x < 0}, {x, x > 0}}, 0]

Their second derivatives.

Simplify@D[#[x], x, x] & /@ {f1, f2, f3}

It seems that f2 or f3 might be used, but Simplify[f2[x]] results in f1[x].

Simplify[f2[x]] // InputForm
(* Piecewise[{{Sqrt[x^2], x != 0}}, 0] *)

So if the functions are simplified, only f3 gives the correct result:

Simplify@D[Simplify@#[x], x, x] & /@ {f1, f2, f3}