Re: Re: Need a better Integrate
- To: mathgroup at smc.vnet.net
- Subject: [mg43042] Re: [mg43019] Re: [mg42976] Need a better Integrate
- From: Dr Bob <drbob at bigfoot.com>
- Date: Sat, 9 Aug 2003 02:57:42 -0400 (EDT)
- References: <200308080426.AAA05622@smc.vnet.net>
- Reply-to: drbob at bigfoot.com
- Sender: owner-wri-mathgroup at wolfram.com
As best I can tell from a few experiments, that method is useful only when
the integrand is of the form D[g, x] g^n, where g is a polynomial in x, in
which case my integration rule works much faster.
integratePower[x_] = {(b_)*(a_)^(n_) :> (b*a^(n + 1))/((n + 1)*D[a, x]) /;
FreeQ[D[a, x]/b, x], (a_)^(n_) :> a^(n + 1)/((n + 1)*D[a, x]) /; FreeQ[D[a,
x], x]}; g = 1 - 3x^2 + x^4;
int = D[g, x] g^5;
Timing[int /. integratePower[x]]
Timing@MyIntegrate[int, x]
{0.*Second, (1/6)*
(1 - 3*x^2 + x^4)^6}
{0.45300000000000007*Second, (1/6)*(1 - 3*x^2 + x^4)^6}
My rule misses the simplification, however, if g is itself the power of a
polynomial.
g = Expand[(1 - x)^3]; int = D[g, x]*g^5; Timing[int /. integratePower[x]]
Timing[MyIntegrate[int, x]]
{0.*Second, (1/6)*(1 - 3*x + 3*x^2 - x^3)^6}
{0.203*Second, (1/6)*(-1 + x)^18}
But using Factor helps:
g = Expand[(1 - x)^3]; int = D[g, x]*g^5; Timing[Factor[int] /.
integratePower[x]]
Timing[MyIntegrate[int, x]]
{0.016*Second, (1/6)*(-1 + x)^18}
{0.172*Second, (1/6)*(-1 + x)^18}
(I Quit and reloaded definitions between tests. Otherwise, the second test
of MyIntegrate goes faster -- but still slower than integratePower[x].)
MyIntegrate goes into a seemingly infinite loop if you give it anything but
a polynomial, so a better definition would start with
MyIntegrate[f_, x_] /; PolynomialQ[f, x] :=
Bobby
On Fri, 8 Aug 2003 00:26:32 -0400 (EDT), Andrzej Kozlowski
<andrzej at platon.c.u-tokyo.ac.jp> wrote:
> This seems quite hopeless except in the case when you are dealing with
> polynomials (as in your example). That is because for general functions
> "constants" can be "disguised" as all kinds of complicated expressions
> involving identities such as Cos[x]^2+Sin[x]^2==1 etc, and there is not
> general procedure for simplifying them, (although FullSimplify is pretty
> good at it). Even for polynomials you can't hope for very much. In order
> to apply "some clever algebra" you need an algebraic criterion of
> simplicity, and LeafCount is not algebraic enough. So given a polynomial
> f, which of the polynomials f+C (where C is a constant) will be the
> simplest? Well, I propose the following very imperfect choice: if there
> is a C such that f+C has discriminant 0 and FullSimplify[f+C] has
> LeafCount less than f, we choose f+C, otherwise we stick with f. To
> implement this we first define Discriminant ((borrowed from Daniel
> Lichtblau)
>
>
> Discriminant[p_?PolynomialQ, x_] := With[{n = Exponent[p, x]},
> Cancel[((-1)^(n(n - 1)/2)Resultant[p, D[p, x], x])/Coefficient[
> p, x, n]^(2n - 1)]]
>
> and now here is a rather clumsy first version of MyIntegrate:
>
> MyIntegrate[f_, x_] := Module[{
> g = Integrate[f, x], ls}, First[Select[ls = FullSimplify[Append[g +
> C /. Solve[Discriminant[g + C, x] == 0, C], g]], (LeafCount[#] ==
> Min[LeafCount /@ ls]) &]]]
>
>
> In the case of your example it behaves as you wanted:
>
>
> MyIntegrate[x*(-5 + x^2)^3, x]
>
>
> (1/8)*(-5 + x^2)^4
>
> Of course , in many cases you will get the same answer as Integrate
> gives, when a "simpler one" may be possible but at least you will never
> get a more complicated one.
>
> Andrzej Kozlowski
> Yokohama, Japan
> http://www.mimuw.edu.pl/~akoz/
> http://platon.c.u-tokyo.ac.jp/andrzej/
>
>
>
> On Thursday, August 7, 2003, at 06:53 AM, Ersek, Ted R wrote:
>
>> A user sent me an email with an interesting problem. See below.
>>
>> In[1]:= deriv = D[(-5 + x^2)^4/8, x]
>>
>> Out[1]= x*(-5 + x^2)^3
>>
>>
>> In[2]:= int = FullSimplify[Integrate[x*(-5 + x^2)^3, x]];
>>
>> Out[2]= (x^2*(-10 + x^2)*(50 - 10*x^2 + x^4))
>>
>>
>> Why doesn't FullSimplify return (-5 + x^2)^4/8 which is simpler?
>> It doesn't because the expression Integrate returned is different from
>> (-5 + x^2)^4/8 by 625/8. Of course both answers are correct because
>> the
>> result of an indefinite integral includes an arbitrary constant.
>> However,
>> Mathematica never includes the constant in the result Integrate returns.
>> In the next line we get the result we expected above.
>>
>> In[3]:= Factor[int + 625/8]
>>
>> Out[3]= (-5+x^2)^4/8
>>
>>
>> I wonder if somebody can figure out how to define a better integrate
>> that
>> would do the following:
>>
>> MyIntegrate[f_,x_]:= FullSimplify[Integrate[f,x]+const]
>> (* Through some clever algebra find the value of (const) that will give
>> the simplest answer. *)
>>
>>
>> We would then get:
>> In[4]:= MyIntegrate[x*(-5 + x^2)^3, x]
>>
>> Out[4]= (-5+x^2)^4/8
>>
>>
>> -------------------
>> Regards,
>> Ted Ersek
>>
>>
>>
>>
>
>
--
majort at cox-internet.com
Bobby R. Treat
- References:
- Re: Need a better Integrate
- From: Andrzej Kozlowski <andrzej@platon.c.u-tokyo.ac.jp>
- Re: Need a better Integrate