Math Is Fun Forum

Hi ElainaVW

Thanks.

Hi NickMeyers

One question - does it matter if J or S was chosen in the last example?

The one Elaina mentioned above, Mathematica.

For Q1, use the fact that a matrix is non-invertible if its determinant is 0. Do you know how to calculate the determinant of that matrix?

Both approaches need to be made more rigorous, I think. I should note that those equations are all valid only under the circumstance that |x|<1. You need to prove that the sum converges in order to use that.

If you mean the one in post #7, that would be hard, because if you pick like that, the distribution of b over (0,1) is not uniform. You would need to be picking it from (0,1-a) with a non-uniform distribution, I think.

(3,4) is correct. (7,16) seems to be the point that maps to (-3,-8).

I have to look at it a bit more.

Glad we're finally online at the same time. It's a bit late over there, isn't it?

I think we can look at this geometrically. If we plot in R³ which points satisfy the conditions (i.e. x+y+z=1) it's an equilateral triangle with a side of

l = Select[Table[{#1, #2, 1 - #1 - #2} &[RandomReal[], RandomReal[]], {100000}], #[[3]] > 0 &];
l1 = Select[l, #[[1]] < 0.5 && #[[2]] < 0.5 && #[[3]] > 0 && #[[3]] < 0.5 &];
Length[l1]/Length[l] // N

It has to do with the nature of the roots of real polynomials. When you have a polynomial with real coefficients, if it has a complex root, the the conjugate of that root is also a root. So, you have three roots, and all must be complex conjugates of some root. So, what can happen is:
1. We have three real roots, each being its own complex conjugate;
2. We have 2 complex root, which means the third one must be its own complex conjugate, and therefore real.

This logic works for any odd number of roots.

Well, for the first one, you can say it's equivalent to

Let's first look at A. What that notation means is we are looking at the set of all points P for which the line segment PQ has length 1. Do you know what that might be?