Question about the derivative of Abs

[Sam Takoy]

Hi,

I kind of asked this question before, but in a more confusing context,
so now I would like to ask it by itself.

In the course of my computations, I get Abs' and Abs''. It is applied to
a positive number, so I think those values should be 1 and 0. However,
Mathematica fails to simplify it.

For example

In[34]:= Abs'[2.0]

Out[34]=
\!\(\*SuperscriptBox["Abs", "\[Prime]",
MultilineFunction->None]\)[2.]

These expressions really mess up my answers. How do I get Mathematica to
do these simplifications?

Thanks

Sam

[dh]

Hi Sam,
mathematica does not know the derivative of Abs, but it is easy to add
a corresponding rule:

Abs'[x_?Positive] = 1
Abs''[x_?Positive] = 0

Daniel

[Leonid Shifrin]

Sam,

The problem seems to be that Abs is defined generally on complex numbers.
Once you specify that your argument is real, it works:

In[1]:= FullSimplify[Abs'[x], Assumptions -> {Element[x, Reals]}]

Out[1]= Sign[x]

In fact, it works also for specific integer arguments (although not
automatically)

In[2]:= FullSimplify[Abs'[2]]

Out[2]= 1

For a reason I don't understand, this does not work for specific real
arguments however:

In[3]:= FullSimplify[Abs'[2.0]]

Out[3]= (Abs^\[Prime])[2.]

I don't see a good reason for this and think that it should be fixed to also
simplify just like in the case of integers. On the other hand, the above
simplification with FullSimplify seems to be rather a point-like solution -
a result of some specific rule that has been added to the kernel. For
example, the second derivative would not simplify:

In[4]:= FullSimplify[Abs''[2]]

Out[4]= (Abs^\[Prime]\[Prime])[2]

So, coming back to your specific question, here is what I would do:

1. Define your own function <abs>, for example like this:

Clear[abs];
abs[x_]:=Piecewise[{{x,x>0},{-x,x<=0}}]

you may check that it works for derivatives without any extra simplification
effort:

In[5]:= abs'[1.5]

Out[5]= 1

In[6]:= abs''[1.5]

Out[6]= 0

2. Use replacement rules to replace Abs with <abs>, in places where you need
this:

In[7]:= Abs''[2.0]/.Abs-> abs

Out[7]= 0

Hope this helps.

Regards,
Leonid

[Patrick Scheibe]

Hi,

PiecewiseExpand can do the trick. In your original post the Abs are
results from the Norm-function in ni[t_,a_,b_].
If you are sure the parameters there are Reals then the following is
Abs-free. To decide whether this is correct is your turn.

Clear["Global`*"]
Simp[a_, b_][expr_] :=
Simplify[expr,
Assumptions -> a > 0 && a < Pi/2 && b > 0 && b < Pi/2];
FSimp[a_, b_][expr_] :=
FullSimplify[expr,
Assumptions -> a > 0 && a < Pi/2 && b > 0 && b < Pi/2];

ComputeCs3D[zi_] := (ddt = Derivative[1, 0, 0];
dd1 = Derivative[0, 1, 0];
dd2 = Derivative[0, 0, 1];
vi = ddt[zi];
zialpha[t_, a_, b_] := {dd1[zi][t, a, b], dd2[zi][t, a, b]};
salphabeta =
Dot[zialpha[#1, #2, #3], Transpose[zialpha[#1, #2, #3]]] &;
sAlphaBeta = Inverse[salphabeta[#1, #2, #3]] &;
ni[t_, a_, b_] :=
Cross[zialpha[t, a, b][[1]], zialpha[t, a, b][[2]]]/
Norm[Cross[zialpha[t, a, b][[1]], zialpha[t, a, b][[2]]]] /.
Abs[val__] :> PiecewiseExpand[Abs[val], Reals];
c[t_, a_, b_] := Dot[ni[t, a, b], vi[t, a, b]] // FSimp[a, b];
c1[t_, a_, b_] :=
ddt[c][t, a, b] -
Dot[zialpha[t, a, b], vi[t, a, b],
sAlphaBeta[t, a, b], {dd1[c][t, a, b], dd2[c][t, a, b]}] //
Simp[a, b];
c2[t_, a_, b_] :=
ddt[c1][t, a, b] -
Dot[zialpha[t, a, b], vi[t, a, b],
sAlphaBeta[t, a, b], {dd1[c1][t, a, b], dd2[c1][t, a, b]}] //
Simp[a, b];
{c[#1, #2, #3], c1[#1, #2, #3], c2[#1, #2, #3]} &)
zi[t_, theta_, phi_] := {(1 + \[Epsilon] t)*Sin[theta] Cos[phi],
Sin[theta] Sin[phi], Cos[theta]};
ComputeCs3D[zi][0, \[Theta], \[Phi]] //
FSimp[\[Theta], \[Phi]] // MatrixForm

Cheers
Patrick

[Bob Hanlon]

sub = {
Abs'[x_?NumericQ /; x != 0] -> Sign[x],
Abs''[x_?NumericQ /; x != 0] -> 0};

Abs'[2.0] + Abs''[2.0] /. sub

1

Bob Hanlon

---- Sam Takoy <samt...@yahoo.com> wrote:

=============

[David W. Cantrell]

I don't know why you're getting Abs' and Abs'' in your answers, but perhaps
it's just because you have asked for derivatives of expressions involving
Abs itself. If that is the case, then you might consider avoiding Abs in
your input by instead using Sqrt[x^2] when x is real.

In[1]:= D[Sqrt[x^2], x]

Out[1]= x/Sqrt[x^2]

In[2]:= D[%, x]

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

In[3]:= Simplify[{%%, %}, x > 0]

Out[3]= {1, 0}

David

[David Bailey]

You could just clean up the answer by applying a suitable transformation:

expr /. {Abs'[y_]->1,Abs''[y_]->0}

David Bailey
http://www.dbaileyconsultancy.co.uk