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MathGroup Archive 2005

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Re: Interpreting output

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56680] Re: Interpreting output
  • From: Maxim <ab_def at prontomail.com>
  • Date: Tue, 3 May 2005 05:26:31 -0400 (EDT)
  • References: <200505010446.AAA13098@smc.vnet.net> <d51vpv$frp$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On Sun, 1 May 2005 07:17:19 +0000 (UTC), Andrzej Kozlowski  
<akoz at mimuw.edu.pl> wrote:

>
> On 1 May 2005, at 13:46, MC wrote:
>
>> Hi everybody....
>>
>> I'm having troubles in interpreting the output of Mathematica.
>>
>> My problem is to define a function F(x,y,z) such that:
>> F = (whatever) if x>c
>> F=(whatever) if x<=c
>>
>> I achievied this using /;
>>
>> When I derive F with respect to one of the variables, the result is a
>> mess.
>> Just an example with 1 variable:
>>
>> f=x/;x>0;
>> f=-x^2/;x<0;
>>
>> D[f,x]=-2Condition(1,0)[x^2,x<0]+Condition(1,0)[x^2,x<0]Less(1,0)[x,0]
>>
>> Even in this very simple case, the result is a real mess:
>> I really do not know how should I read that...and I'm wondering if
>> I have defined the function in the right way.
>>
>> Anybody so kind to explain me what's goung on?
>>
>> Thank you very much for your patience
>>
>>
>
> What is going on is simply that you that Mathematica's notion of
> derivative does not work with piecewise functions defined by means of
> pattern matching. (Actually, there is nothing in the documentation that
> suggests that it might.)
> In order for differentiation to be possible you have to construct a
> piecewise function using a construction that "understands" the notion
> of a derivative. In Mathematica 5.1 this is easy:
>
> f[x_] := Piecewise[{{x^2, x < 2}, {x^3, x >= 2}}]
>
> f'[1]
> 2
>
>
> f'[3]
> 27
>
>
> f'[2]
> Indeterminate
>
>
> In earlier version you could try doing this by means of the UnitStep
> function:
>
> g[x_] := x^2*UnitStep[2 - x] + x^3*UnitStep[x - 2]
>
> this will work almost the same except for the singular value
>
>
> g'[1]
> 2
>
>
> g'[3]
>
> 27
>
>
>
> g'[1]
>
> 2
>
>
> g'[3]
>
> 27
>
>
> g'[2]
>
>
> 4*DiracDelta[0] + 16
>
>
> Note also that both approaches work well when you construct a
> differentiable everywhere piecewise function:
>
>
> f[x_] := Piecewise[{{(x-1)^2, x < 1}, {(x-1)^3, x >= 1}}]
>
>
> f'[1]
> 0
>
>
> g[x_]:=(x-1)^2*UnitStep[1-x]+(x-1)^3*UnitStep[x-1]
>
>
> g'[1]
>
> 0
>
> However, this is achieved in a quite different way, which we can see
> from
>
>
> f'[x]
>
> Piecewise[{{2*(x - 1), x < 1}, {0, x == 1}}, 3*(x - 1)^2]
>
>
> and
>
>
> g'[x]
>
> DiracDelta[x - 2]*(x - 2)^3 - DiracDelta[x - 2]*
>     (x - 2)^2 + 3*UnitStep[x - 2]*(x - 2)^2 +
>    2*UnitStep[2 - x]*(x - 2)
>
>
> The second answer is given as a "generalized function" and although it
> seems "correct" I would not rely on this type of use  of generalised
> functions (distributions) in anything but the most simple types of
> situations. The Piecewise approach looks more promising.
>
>
>
> Andrzej Kozlowski
> Chiba, Japan
> http://www.akikoz.net/andrzej/index.html
> http://www.mimuw.edu.pl/~akoz/
>

But Mathematica wouldn't be Mathematica if there weren't some accompanying  
errors with this new feature:

In[1]:=
D[Piecewise[{{Sin[x]^2/x, x != 0}}], x]

Out[1]=
Piecewise[{{(2*Cos[x]*Sin[x])/x - Sin[x]^2/x^2, x != 0}}]

In[2]:=
D[Piecewise[{{0, x == 0}}, Sin[x]^2/x], x]

Out[2]=
Piecewise[{{0, x == 0}}, (2*Cos[x]*Sin[x])/x - Sin[x]^2/x^2]

Both are incorrect because the derivative at x=0 is equal to 1.

D[Piecewise[{{Sin[x]^2/x, x < 0 || x > 0}}], x]

returns the correct answer.

Sometimes working with generalized functions is the right thing to do, in  
particular when we're dealing with probabilities. For example,  
(UnitStep[x] + UnitStep[x - 1])/2 is the CDF for the random variable xi  
such that P(xi = 0) = P(xi = 1) = 1/2. This expression can be viewed  
either as an ordinary or a generalized function, but to obtain the correct  
expression for the PDF we have to differentiate it as a generalized  
function to get (DiracDelta[x] + DiracDelta[x - 1])/2 (Piecewise wouldn't  
do). Now this PDF can be used to evaluate the moment integrals.

Incidentally, here we encounter another error:

In[3]:=
Integrate[(x - 1/2)^2*(DiracDelta[x] + DiracDelta[x - 1])/2,
   {x, -Infinity, Infinity}]

Out[3]=
0

If we apply Expand to the integrand we get 1/4 as expected.

Maxim Rytin
m.r at inbox.ru


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