mathlib-develop

[Mathlib-develop] Re: Symbolic calculations

[Stefan M. <st...@he...>]

>>Hi Mark and Stefan!
>>
>>1) Why don't we put it on a new class on Functions/Calculus package? I have
>>free access to some "numerical methods" books at the university, so if you
>>need it I'll pass you some hints.
> 
> 
> What I meant was how to handle cases like derivative( sin(x), x) => cos(x) 
> or even derivative( UserFunction(x), x) => OtherFunction(x)
> 
> My idea was that the implementation of sin and UserFunction would contain a 
> method called derivative which would return the correct value.
I aggree to add the two methods to the individual functions.
But what about complex integration/derivative rules, like the chain
rule?
e.g.: d/dx of sin(cos(x))


>>2) Do you mean find the factors of a number (15= 3*5)
>>If true, this page can help you http://www.alpertron.com.ar/ECM.HTM
>>I have tried it with 128 digits on 16 segons!!!
> 
> 
> Actually I meant finding factors of polynomials 
> i.e. x^2 + 2*x + 1 => (x+1)^2
> 
> Though I'll check the link out since the the current implementation of 
> numerical factorization is pretty slow.
Sorry, I've no clue at all. I've never done anything with symbolic
stuff.

Kind regards,
   Stefan.

-- 
------------------------------------
Dr.-Ing. Stefan Mueller
email: St...@he...
------------------------------------

[Mathlib-develop] Re: Symbolic calculations

[Alejandro T. <ate...@ho...>]

Hi Mark and Stefan!!

> > Actually I meant finding factors of polynomials
> > i.e. x^2 + 2*x + 1 => (x+1)^2
> >
> > Though I'll check the link out since the the current implementation of
> > numerical factorization is pretty slow.
> Sorry, I've no clue at all. I've never done anything with symbolic
> stuff.

Me too, but I think we can give us hints to a proper sollution, as we have
done before ;-)

Finding the roots of polynomials, I have read something about the
Tartaglia's triangle, Pascal's or Newton's one, but I don't remeber now.

You can see a plenty tutorial about it at
http://www.krysstal.com/binomial.html

But this works only for binomial power of expressions like (x + y)^z where
x,y are integers and z is a natural number.

If we want general root finding, including complex roots, we should search a
bit more.

Regards,
Alejandro.

Re: [Mathlib-develop] Re: Symbolic calculations

[mark <msp...@ya...>]

On Saturday 15 Feb 2003 10:04 pm, Alejandro Torras wrote:
> Hi Mark and Stefan!!
>
> > > Actually I meant finding factors of polynomials
> > > i.e. x^2 + 2*x + 1 =3D> (x+1)^2
> > >
> > > Though I'll check the link out since the the current implementation=
 of
> > > numerical factorization is pretty slow.
> >
> > Sorry, I've no clue at all. I've never done anything with symbolic
> > stuff.
>
> Me too, but I think we can give us hints to a proper sollution, as we h=
ave
> done before ;-)
>
> Finding the roots of polynomials, I have read something about the
> Tartaglia's triangle, Pascal's or Newton's one, but I don't remeber now=
=2E
>
> You can see a plenty tutorial about it at
> http://www.krysstal.com/binomial.html
>
> But this works only for binomial power of expressions like (x + y)^z wh=
ere
> x,y are integers and z is a natural number.
>
> If we want general root finding, including complex roots, we should sea=
rch
> a bit more.
>
> Regards,
> Alejandro.
>

Well I'm not sure that we need to worry about complex roots to begin with=
 but=20
it does need to be more general than just binomial expansions. Things lik=
e

x^3 + 6x^2 + 11x + 6 =3D (x+1)(x+2)(x+3)

I wasn't able to find anything on google but there is a section on it in=20
volume 2 of The Art Of  Computer Programming. I'll read through that toda=
y.

Best Regards

Mark

Re: [Mathlib-develop] Re: Symbolic calculations

[mark <msp...@ya...>]

Hi

This is an overview of the state of the Symolic section.

When an expression containting a symbolic token is evaluated it returns a=
=20
symbolic expression. there are three types of SymbolicExpression dependin=
g on=20
the type of the operator

SymbolicExpression
=09ArithmaticSymbolicExpression
=09GeometricSymbolicExpression
=09ExponentialSymbolicExpression

When a function that contains either a SymbolicToken or a SymbolicExpress=
ion=20
is evaluated it returns itself.

ArithmaticSymbolicExpression stores an ArrayList of the operands and a nu=
meric=20
multiplier. So X + Y + Z is stored as a single ArithmaticSymbolicExpressi=
on.

GeometricSymbolicExpression stores an ArrayList of the numerator and=20
denominator as well as a numeric multiplier. So it can store expressions =
like=20
(U*V*W)/(X*Y)

ExponentialSymbolicExpression stores the mantissa and exponent.

At the moment the following functions work to varying degrees
Simplify(exp)
Expand(exp)
Derivative(exp, deriveBy)
Integral(exp, integrateBy)
Subst(exp, old, new)

The following doesn't work

Derivative for expressions of the form U^V
Derivatives only work for a small number of functions
The output from Derivative can be messy

Integrals for expressions of the form U*V, U/V, U^V and fn(U)

Factorization still doesn't work at all

Despite this I'd suggest removing the code for the meditor library since =
it's=20
not being used any more.

if there are any queries, comments or suggestions then let me know  =20

Best Regards

Mark

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Re: [Mathlib-develop] Re: Symbolic calculations

[mark <msp...@ya...>]

On Wednesday 12 Feb 2003 6:55 pm, Stefan Mueller wrote:
> >>Hi Mark and Stefan!
> >>
> >>1) Why don't we put it on a new class on Functions/Calculus package? =
I
> >> have free access to some "numerical methods" books at the university=
, so
> >> if you need it I'll pass you some hints.
> >
> > What I meant was how to handle cases like derivative( sin(x), x) =3D>
> > cos(x) or even derivative( UserFunction(x), x) =3D> OtherFunction(x)
> >
> > My idea was that the implementation of sin and UserFunction would con=
tain
> > a method called derivative which would return the correct value.
>
> I aggree to add the two methods to the individual functions.
> But what about complex integration/derivative rules, like the chain
> rule?
> e.g.: d/dx of sin(cos(x))
>

IIRC d/dx f(u) =3D d/dx u * d/du f(u)=20

so d/dx sin(cos(x)) =3D -sin(x) * cos(cos(x))

So differentiating chained functions shouldn't be too hard.

Though we should probably limit derivatives to functions that take 1=20
parameter.

Best Regards

Mark