A mathematical issue -- finding planar coordinates

Hi folks.

First of all, I'm sorry to bring up a topic that has actually nothing to do with scales. But I've asked a few people about this and noone seemed capable of giving me the answer. As there are some skilled mathematicians here, I'm hoping that maybe here someone might be able to tell me what I should do with this.

I have the following pair of equations:

x^3 - 3*x*y^2 + 1/2 = 0

y^3 - 3*x^2*y + sqrt(3/4) = 0

And now I'm getting stuck on the silly fact that I have no idea what the necessary steps are for finding the values of "x" and "y" (I mean, real numbers, not complex ones). Seems like some cubic equations but I don't know what to do when they're actually two of them.

Any suggestion is much appreciated. Thanks in advance.

Petr

--- In tuning@yahoogroups.com, Petr Pařízek <petrparizek2000@...> wrote:
> x^3 - 3*x*y^2 + 1/2 = 0
> y^3 - 3*x^2*y + sqrt(3/4) = 0

I don't know a nice method but can do it by searching and refining - there may be a solution near x=0.17364817767, y=-0.98480775301, is that any good?

Steve M.

----- Original Message ----- From: "martinsj013" <martinsj@...>
To: <tuning@yahoogroups.com>
Sent: Sunday, April 15, 2012 2:05 PM
Subject: [tuning] Re: A mathematical issue -- finding planar coordinates

--- In tuning@yahoogroups.com, Petr Pa�Tízek <petrparizek2000@...> wrote:
> x^3 - 3*x*y^2 + 1/2 = 0
> y^3 - 3*x^2*y + sqrt(3/4) = 0

> I don't know a nice method but can do it by searching and refining - there > may be a solution near
> x=0.17364817767, y=-0.98480775301, is that any good?

Thanks, Steve. I was rather more interested in the algebraic expressions leading to the numbers (similarly to saying "log2(3/2)*1200" instead of "~701.955") but the sizes probably do work.

Petr

I wrote:

> Thanks, Steve. I was rather more interested in the algebraic expressions
> leading to the numbers (similarly to saying "log2(3/2)*1200" instead of
> "~701.955") but the sizes probably do work.

Actually, I can't verify it at this moment. If there are 3 real roots for "x" and 3 real roots for "y", then it will probably be necessary to pick the right pair (like x_1 with y_1, x_2 with y_2 and so on).

Petr

--- In tuning@yahoogroups.com, Petr Parízek <petrparizek2000@...> wrote:

> Actually, I can't verify it at this moment. If there are 3 real roots for
> "x" and 3 real roots for "y", then it will probably be necessary to pick the
> right pair (like x_1 with y_1, x_2 with y_2 and so on).

You can solve problems like this using the resultant, Taking the resultant and eliminating y leads to 512*x^9+192*x^6-192*x^3+1 = 0, which you can solve and back-substitute. Rather than doing all that work, make you computer algebra system do it for you. Either way, you get:

y = root of 8*y^3 - 6*y + sqrt(3) = 0
x = 8*y^3 + sqrt(3)*y - 2

or else

y = root of 64*y^6+96*y^4+16*3^(1/2)*y^3+36*y^2+6*3^(1/2)*y+3 = 0
x = 3*y/(16*y^3-sqrt(3))

--- In tuning@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

> You can solve problems like this using the resultant, Taking the resultant and eliminating y leads to 512*x^9+192*x^6-192*x^3+1 = 0, which you can solve and back-substitute.

Actually more "by hand" isn't hard. The above polynomial factors, with one factor 8*x^3-6*x+1 and the other 64*x^6+48*x^4+16*x^3+36*x^2+6*x+1.

--- In tuning@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

> Actually more "by hand" isn't hard. The above polynomial factors, with one factor 8*x^3-6*x+1 and the other 64*x^6+48*x^4+16*x^3+36*x^2+6*x+1.
>

"By hand" you can end up better off than the machine solution I gave if you want to work for it, as if we pick a particular root and set z=2*root, so that z^3-3*z+1=0, then the first factor factors further as (2*x-2+z+z^2)*(-2*x-2+z^2)*(-2*x+z) and the second as -(-4*x^2-4*x+2*x*z+2*x*z^2-2-z)*(4*x^2+2*x*z+z^2)*(4*x^2-4*x+2*x*z^2+4-z-z^2).

Gene,
thanks so-so-so much. I've tried to solve this for a few weeks, actually, and everyone whom I asked responded by giving me approximations to a handful of decimals.
At first, I thought I wouldn't post my question here as it's not related to most topics discussed here. But finally I did.

Petr

--- In tuning@yahoogroups.com, Petr Parízek <petrparizek2000@...> wrote:
> Thanks, Steve. I was rather more interested in the algebraic expressions leading to the numbers (similarly to saying "log2(3/2)*1200" instead of "~701.955") but the sizes probably do work.

No problem, Petr. But I think the algebraic expressions are going to be complicated (and I see Gene has given some help that I couldn't have done). Let us know how it turns out!

A separate question from me: have you found/used the interval
(sqrt(13)-1)/2
in your work?

Steve M.

--- In tuning@yahoogroups.com, "martinsj013" <martinsj@...> wrote:

> No problem, Petr. But I think the algebraic expressions are going to be complicated (and I see Gene has given some help that I couldn't have done). Let us know how it turns out!

I think actually the algebraic expressions are likely to be quite simple if you do it right, which people seldom do. This is because I suspect Petr is only interested in real solutions, which come in terms where you can use Chebychev cube roots or cosines. If Petr would like that, I'll give the real solutions in such simple terms.

Gene wrote:

> I think actually the algebraic expressions are likely to be quite simple if you
> do it right, which people seldom do. This is because I suspect Petr is only
> interested in real solutions, which come in terms where you can use Chebychev
> cube roots or cosines. If Petr would like that, I'll give the real solutions in
> such simple terms.

I'm afraid cosines wouldn't give me the answer I was looking for. As to Chebishev cube roots, I'm not sure what that means. Anyway, I was expecting nested cube and square roots in the results -- right from the moment I first arrived at these equations.

Petr

Steve wrote:

> A separate question from me: have you found/used the interval
> (sqrt(13)-1)/2
> in your work?

Scala evaluates it as ~457.9-something cents -- just checked. Did you mean this one?
No, I haven't thought about that.
Where did you find it?

Petr

--- In tuning@yahoogroups.com, "petrparizek2000" <petrparizek2000@...> wrote:

> I'm afraid cosines wouldn't give me the answer I was looking for.

What are you looking for?

As to Chebishev cube roots, I'm not sure what that means. Anyway, I was expecting nested cube and square roots in the results -- right from the moment I first arrived at these equations.

Cube roots are a bad way to solve most cubic equations, and that goes double for the Casus irreducibilis, which is what you have here.

Gene wrote:

> What are you looking for?

I think you've already given me the answer in the preceding messages.

The reason why I didn't think cosines would give me the right answer was that I was more interested in the expressions leading to the required values; and these would be lost if I used polar coordinates and then converted back to rectangular, for example. Similarly, I'm more interested to know that the factor for a quarter-comma fifth is (5)^(1/4) and only then I start caring about the fact that it's equal to 1.495348-whatever.
To pick an example more similar to the current topic, let's say that we're interested to know the exact values of cosine and sine of 2*pi/7 (i.e. use a different method for finding them than doing closer and closer approximations). Luckyly, these very values are listed on this webpage in the context of complex roots:
http://en.wikipedia.org/wiki/Root_of_unity#Examples

Petr

--- In tuning@yahoogroups.com, "petrparizek2000" <petrparizek2000@...> wrote:
>
> Gene wrote:
>
> > What are you looking for?
>
> I think you've already given me the answer in the preceding messages.
>
> The reason why I didn't think cosines would give me the right answer was that I was more interested in the expressions leading to the required values; and these would be lost if I used polar coordinates and then converted back to rectangular, for example.

Why would a real algebraic function of one real value not give the right answer, but the sum of two complex-valued algebraic functions does. What mystic property does the second solution possess?

Gene wrote:

> Why would a real algebraic function of one real value not give
> the right answer, but the sum of two complex-valued algebraic
> functions does. What mystic property does the second solution possess?

The fact that the two values were part of a single complex number rather than two separate real values was not crucial here. I picked that webpage just because it contained expressions equal to the cosine and sine of 2*pi/7, which I gave as another example. Had they been given as two separate real values, those expressions would stay the same anyway and would be equally meaningful for me.

Petr

--- In tuning@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
> Cube roots are a bad way to solve most cubic equations, and that goes double for the Casus irreducibilis, which is what you have here.

Making use of Gene's hints, I find that the solution can be stated as x=cos(4pi/9). And, well, y^2=(x^2)/3 + 1/(6x). But that might not satisfy Petr!

Steve M.

--- In tuning@yahoogroups.com, "petrparizek2000" <petrparizek2000@...> wrote:
Steve> (sqrt(13)-1)/2
Petr> Where did you find it?

In the 2-d plot of triadic harmonic entropy (3HE) there are "valleys" of lower 3HE; each one seems to link rational triads A:B:C that satisfy an equation like pA+qB+rC=0. The most prominent is A-2B+C=0 containing 3:4:5, 4:5:6 etc, but another is 3A-B-C=0 containing 3:4:5, 10:13:17, 4:5:7, etc. I wanted to find the point on that (latter) curve having its lower interval equal to its upper interval, i.e. C/B=B/A (not constrained to be integers of course) - the answer B/A=x satisfies x^2+x-3=0. At first I thought it was near Mike B's 21:27:35 but I was wrong - but it is near 0-5-10 in 13EDO.

Steve M.