MathAction #176 Factored Polynomials aren't differentiated correctly

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  - #176 Factored Polynomials aren't differentiated correctly <-- You are here.

axiom

p := -x*y^2+x*y+x^3-x^2

$\label{eq1}-{x \ {y^2}}+{x \ y}+{x^3}-{x^2}$

(1)

Type: Polynomial(Integer)

axiom

)se ou algebra on

D(factor(p),x)

            2         2
   (2)  - (y  - y - 3x  + 2x)

$\label{eq2}-{\left({y^2}- y -{3 \ {x^2}}+{2 \ x}\right)}$

(2)

Type: Factored(Polynomial(Integer))

axiom

D(p,x)

           2         2
   (3)  - y  + y + 3x  - 2x

$\label{eq3}-{y^2}+ y +{3 \ {x^2}}-{2 \ x}$

(3)

Type: Polynomial(Integer)

Note that the factorization is correct. It's the D(.,x) that misses the sign.

Forgot to categorize... --unknown, Mon, 20 Jun 2005 10:18:00 -0500 reply

Category: Axiom Compiler => Axiom Mathematics Severity: normal => serious

Fix --kratt6, Tue, 21 Jun 2005 05:43:03 -0500 reply

Severity: serious => critical Status: open => fix proposed

Fix --kratt6, Tue, 21 Jun 2005 05:45:04 -0500 reply

The mistake is in differentiate$FR which currently reads:

    differentiate(u:%, deriv: R -> R) ==
      ans := deriv(unit u) * ((u exquo (fr := unit(u)::%))::%)
      ans + fr * (_+/[fact.xpnt * deriv(fact.fctr) *
       ((u exquo nilFactor(fact.fctr, 1))::%) for fact in factorList u])

It intends to use the formula

$\frac{d}{dx} f(x)) = \sum_{i=1}^n \frac{f(x)}{f_i(x)}\frac{d}{dx}f_i(x)$

where

$f(x)=\prod_{i=1}^n f_i(x).$

Therefore, the fix is to leave away the 'fr':

    differentiate(u:%, deriv: R -> R) ==
      ans := deriv(unit u) * ((u exquo unit(u)::%)::%)
      ans + (_+/[fact.xpnt * deriv(fact.fctr) *
       ((u exquo nilFactor(fact.fctr, 1))::%) for fact in factorList u])

Martin

applied in patch-41 --kratt6, Tue, 04 Oct 2005 05:51:35 -0500 reply

Status: fix proposed => closed

... --kratt6, Fri, 28 Dec 2007 12:40:55 -0800 reply

Category: Axiom Mathematics => Axiom Library

... --kratt6, Fri, 28 Dec 2007 12:46:13 -0800 reply

unfortunately, TeX output misses a parenthesis. See Issue #95

Subject: Be Bold !!

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