Services & ResourcesWolfram ForumsMathGroup Archive

[Date Index] [Thread Index] [Author Index]

Re: Problems with differentiating Piecewise functions

• To: mathgroup at smc.vnet.net
• Subject: [mg86992] Re: Problems with differentiating Piecewise functions
• From: hlovatt <howard.lovatt at gmail.com>
• Date: Fri, 28 Mar 2008 03:16:33 -0500 (EST)
• References: <200803260955.EAA09634@smc.vnet.net> <fsg6r6$j9q$1@smc.vnet.net>

On Mar 28, 12:18 am, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote:
> On 26 Mar 2008, at 10:55, hlovatt wrote:
>
> > If I set up a piecewise function and differentiate it:
>
> > In[112]:= pw1 = Piecewise[{{x^2, x <= 0}, {x, x > 0}}]
>
> > Out[112]= \[Piecewise] {
> >  {x^2, x <= 0},
> >  {x, x > 0}
> > }
>
> > In[113]:= pw1 /. x -> 0
>
> > Out[113]= 0
>
> > In[114]:= pw1d = D[pw1, x]
>
> > Out[114]= \[Piecewise] {
> >  {2 x, x < 0},
> >  {1, x > 0},
> >  {Indeterminate, \!\(\*
> >     TagBox["True",
> >      "PiecewiseDefault",
> >      AutoDelete->False,
> >      DeletionWarning->True]\)}
> > }
>
> > In[115]:= pw1d /. x -> 0
>
> > Out[115]= Indeterminate
>
> > Then at the joins between the pieces I get Indeterminate values,
> > because the limit x <= 0 has become x < 0 after differentiation. Does
> > anyone know a solution to this problem?
>
> > Thanks,
>
> > Howard.
>
>   What do you mean by "a solution to this problem"? You have a
> function that is not differentiable at 0 and you would like it's
> derivative to have a value there? You can't expect a "solution to a
> problem" when you do not tell us what is the problem (except the fact
> that not all functions are differentiable - but that's life).
> Note that is your pieceise function is actually differentiable than
> the derivative is defined everywhere:
>
> pw2 = Piecewise[{{x^2, x <= 0}, {x^3, x > 0}}];
>
> pw2d = D[pw2, x]
> Piecewise[{{2*x, x < 0}, {0, x == 0}}, 3*x^2]
>
> This is also as it should be. What else would you expect?
>
> Andrzej Kozlowski

Thanks to everyone who replied. I have to apologise for the bad
example that I posted (I simplified my problem by cutting and pasting
an example from the help file to keep the post short). I am actually
fitting cubic splines and the functions are continuous up to the
second derivative (at least to the accuracy of machine precision). A
better example is:

In[54]:= pw[x_] :=
 Piecewise[{{0.+ 0.007508378277320685 x + 7.561460342471517*10^-7 x^3,
     x < 50}, {-4.8729206849430454*10^-6 (-125.76959597633721 +
       x) (1148.1044516606876- 47.50636365246156 x + x^2), 50 <= x}}]

In[55]:= pw[x]

Out[55]= \[Piecewise] {
  {0.+ 0.00750838 x + 7.56146*10^-7 x^3, x < 50},
  {-4.87292*10^-6 (-125.77 + x) (1148.1- 47.5064 x + x^2), 50 <= x}
 }

In[56]:= pw[50]

Out[56]= 0.469937

In[57]:= pw[50 + 10^-30]

Out[57]= 0.469937

In[58]:= pw[50 - 10^-30]

Out[58]= 0.469937

In[60]:= pw'[x]

Out[60]= \[Piecewise] {
  {0.00750838+ 2.26844*10^-6 x^2, x < 50},
  {-4.87292*10^-6 (-125.77 + x) (-47.5064 + 2 x) -
    4.87292*10^-6 (1148.1- 47.5064 x + x^2), x > 50},
  {Indeterminate, \!\(\*
     TagBox["True",
      "PiecewiseDefault",
      AutoDelete->False,
      DeletionWarning->True]\)}
 }

In[61]:= pw'[50]

Out[61]= Indeterminate

In[62]:= pw'[50 + 10^-30]

Out[62]= 0.0131795

In[63]:= pw'[50 - 10^-30]

Out[63]= 0.0131795

Also if you Plot pw or pw' you get an annoying gap in the plot at the
join (but strangely not pw''). My guess is that Mathematica is too
pedantic about machine precision and is treating each piece as an
algebraic equation. However this does not explain why Plot behaves
funnily and doesn't help if you are trying to do numerical analysis.

• Follow-Ups:
  • Re: Re: Problems with differentiating Piecewise functions
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
  • Re: Re: Problems with differentiating Piecewise functions
    • From: Curtis Osterhoudt <cfo@lanl.gov>
  • Re: Re: Problems with differentiating Piecewise functions
    • From: Carl Woll <carlw@wolfram.com>

• References:
  • Problems with differentiating Piecewise functions
    • From: hlovatt <howard.lovatt@gmail.com>