Re: [Phys-L] new puzzle

[David Bowman <David_Bowman@georgetowncollege.edu>]

Regarding JSD's commented doubts:

> I don't want to detract from the above, but I'm not sure I agree
> with the following tangent:
>
> > If the formulas are reduced to find real solutions in terms of
> > only real arithmetic then we need to be able to take trig
> > functions of fractions of arc trig functions in order to
> > evaluate the trig function of interest.
>
> Anything you can do with complex numbers you can do with vectors
> in the plane. Each vector has two real components.
>
> In particular, if you want to find a solution for x = cos(θ),
> then a method that solves for y = sin(θ) and x = cos(θ) at the
> same time counts as a valid method IMHO.

All true, but it doesn't help things.

> Consider y to be a temporary variable introduced for
> convenience, allowing geometric insight to be applied to the
> problem. This sticks closer to the original question about
> angles.

When one looks under the hood at complex arithmetic one finds just a lot of real arithmetic.  And taking power law roots of complex numbers, which formally doesn't look like it doesn't involve trig functions, actually does when the underlying real arithmetic is exposed.  For instance, consider the process of taking the cube root of a complex number z, i.e. w = z^(1/3).  This looks like it doesn't involve any trig and inverse trig functions.  But it really does so when one looks under the hood at the actual arithmetical computations.  Suppose, for maximal simplicity, z happens to be in the first quadrant of the complex plane, so its polar angle argument is acute and positive.  Then when taking the principal cube root of z we find the real and imaginary parts of w are actually

Re(w) = cos((1/3)*arccos(x/√(x^2 + y^2)))*(x^2 + y^2)^(1/6) and
Im(w) = sin((1/3)*arcsin(y/√(x^2 + y^2)))*(x^2 + y^2)^(1/6)

Where x is the real part of z and y is the imaginary part of z.

Thus the act of finding the real part of a power law root of a generic complex number necessarily involves the use of the type of both trig and inverse trig functions which are real valued functions of real arguments.

Likewise, consider the standard solution-by-radicals formula for solving a cubic equation of the form
z^3 = 3*m*z + 2*n where m and n are real number coefficients.

If this equation happens to have one real root and 2 complex conjugate roots (i.e. if n^2 > m^3) the formulas for these roots involve just straightforward real square and cube root-finding artihmetic, i.e.

z = (n + √(n^2 - m^3))^(1/3) + (n - √(n^2 - m^3))^(1/3)
for the real root, and

z = (-1/2)*((n + √(n^2 - m^3))^(1/3) + (n - √(n^2 - m^3))^(1/3)) ± i*(√(3)/2)*((n + √(n^2 - m^3))^(1/3) - (n - √(n^2 - m^3))^(1/3))
for the complex conjugate pair of roots.

*But* if the polynomial equation has 3 real roots, namely when n^2 < m^3, (as is the case for finding the separate real & imaginary parts of the 7th roots of unity, so we can find trig functions for angles which are integer multiples of π/7) then the above formulae are still valid, but they suddenly require taking square roots of negative numbers and cube roots of complex numbers.  But, magically, all the imaginary parts of the solutions cancel out in the end (and only at the end) of the calculation.  But unfortunately the cube roots of the complex numbers involve finding trig and inverse trig functions as illustrated above.  And, when the smoke clears, the formulas for those 3 real roots boil down to the real-valued formula

z = 2*√(m)*cos((1/3)*arccos(n/m^(3/2)) + 2*π*k/3)

where k takes on the values of any 3 consecutive integers (for instance -1, 0, 1) to get all 3 distinct real roots for z.
So, formally, the solution of the cubic equation only involves finding the radicals of square and cube roots, but, in practice when all of the roots of the cubic equation are real, those radicals hide the actual state of affairs, which actually require one to be able to take trig and inverse trig functions when the actual underlying real arithmetic is eventually done.  We have a similar situation, in spades, when finding trig functions of integer multiples of π/9  where the corresponding polynomial equation is quartic.

> > we are still assured that they are algebraic numbers because
> > they can each be shown to obey a polynomial equation with
> > integer coefficients
>
> Agreed, yes, they are /algebraic/ by definition.
>
> But /algebraic/ does imply nice,

Whether or not algebraic implies nice probably depends strongly on one's definition of nice.  At best, they may be only a little nicer than transcendental when the degree of the polynomial is at least 5, if only because the equation required to be solved to get to them is easily evaluated, even if those actual solution roots are not.

>                                  and doesn't imply solutions can
> be found using high-school algebra (add, subtract, multiply,
> divide, roots).
>
> Abel and Galois had something to say about this.
>   https://mathworld.wolfram.com/QuinticEquation.html

Yep.

David Bowman