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Reduction to a rational expression

asked 4 years ago

sal gravatar image

updated 4 years ago

I have the the following expression:

tan(mactan(x)+narctan(y))

For what values of n,mN is this expression reducible to a rational expression in R[x,y], where R is a generic ring?

Is there some rule lhs=>rhs that is needed in order to get a rational expression form the expression above?

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Homework ?

And, by the way, what do you call a "generic ring" ?

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 4 years ago )

Yeah, homework. A generic ring is a ring over the rationals. But I don't see why the ring itself is relevant, the answer could be written as a function of some elements of the ring. They don't have to be specified.

sal gravatar imagesal ( 4 years ago )

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answered 4 years ago

Emmanuel Charpentier gravatar image

updated 4 years ago

slelievre gravatar image

Since it's homework, some hints :

  • tan(a+b)=tan(a)+tan(b)1tan(a)tan(b) is a rational expression in tan(a),tan(b).

  • This is true for b=(m1)a, mN. By recurrence, tan(ma) is a rational expression in tan(a). For details, see wikipedia, the French version has formulas more easily transcriptible in Sage...).

  • in C, tan(arctan(x))=x (but the reverse is no true !).

But those rational expressions are fractions, which have a meaning in the usual fields, not in a "generic ring" where division may not even be defined.

FWIW, Sage won't give you a closed-form for tan(mx) (but you can program the expressions given by Wikipedia). However, for defined values of m and n, it will give you the rational expression sought. Example:

sage: x, y = SR.var('x, y')
sage: a = tan(5*arctan(x) + 6*arctan(y))
sage: b = a.trig_expand().trig_expand()
sage: eq = a == b

sage: eq
tan(5*arctan(x) + 6*arctan(y)) == -(2*(3*y^5 - 10*y^3 + 3*y)/(y^6 - 15*y^4 + 15*y^2 - 1) - (x^5 - 10*x^3 + 5*x)/(5*x^4 - 10*x^2 + 1))/(2*(x^5 - 10*x^3 + 5*x)*(3*y^5 - 10*y^3 + 3*y)/((y^6 - 15*y^4 + 15*y^2 - 1)*(5*x^4 - 10*x^2 + 1)) + 1)

sage: eqq = eq.factor()
sage: eqq
tan(5*arctan(x) + 6*arctan(y)) == (x^5*y^6 - 15*x^5*y^4 - 30*x^4*y^5 - 10*x^3*y^6 + 15*x^5*y^2 + 100*x^4*y^3 + 150*x^3*y^4 + 60*x^2*y^5 + 5*x*y^6 - x^5 - 30*x^4*y - 150*x^3*y^2 - 200*x^2*y^3 - 75*x*y^4 - 6*y^5 + 10*x^3 + 60*x^2*y + 75*x*y^2 + 20*y^3 - 5*x - 6*y)/(6*x^5*y^5 + 5*x^4*y^6 - 20*x^5*y^3 - 75*x^4*y^4 - 60*x^3*y^5 - 10*x^2*y^6 + 6*x^5*y + 75*x^4*y^2 + 200*x^3*y^3 + 150*x^2*y^4 + 30*x*y^5 + y^6 - 5*x^4 - 60*x^3*y - 150*x^2*y^2 - 100*x*y^3 - 15*y^4 + 10*x^2 + 30*x*y + 15*y^2 - 1)

From there, using latex(eq) and latex(eqq) or view(eq) and view(eqq) we get

tan(5arctan(x)+6arctan(y))=2(3y510y3+3y)y615y4+15y21x510x3+5x5x410x2+12(x510x3+5x)(3y510y3+3y)(y615y4+15y21)(5x410x2+1)+1

and

tan(5arctan(x)+6arctan(y))=x5y615x5y430x4y510x3y6+15x5y2+100x4y3+150x3y4+60x2y5+5xy6x530x4y150x3y2200x2y375xy46y5+10x3+60x2y+75xy2+20y35x6y6x5y5+5x4y620x5y375x4y460x3y510x2y6+6x5y+75x4y2+200x3y3+150x2y4+30xy5+y65x460x3y150x2y2100xy315y4+10x2+30xy+15y21

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Asked: 4 years ago

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Last updated: Jul 27 '20