But you really want to get the original one and zoom in... There are places of beauty which are not that visible in the scaled down version.
That's really something. I didn't know what to expect, but it wasn't that. Can you talk us through this, in Fisher-Price math? Let me see how far I can get on my own. A complex number, like 2 + 3i, has a real (2) and imaginary (3i) part. The real part is simple enough, and we can pretend the i in the imaginary part is just another variable like x, except that when two i's get multiplied together they change into a -1. You wrote a program to solve an equation, actually a lot of equations. Let's start with one equation. A simple polynomial might have the form ax² + bx + c = something, you didn't say what but I'll guess it's zero. You restricted the coefficients, a, b, and c to the values 1 and -1. Each of the three coefficients can take two values, so we should get 2 × 2 × 2 = 8 equations. 1x² + 1x - 1 = 0 1x² - 1x + 1 = 0 1x² - 1x - 1 = 0 -1x² + 1x + 1 = 0 -1x² + 1x - 1 = 0 -1x² - 1x + 1 = 0 -1x² - 1x - 1 = 0 To find solutions, we ask what values x can take that give true equations. Considering real numbers first, we might be lazy and plot all eight equations on a graph and see where they cross the x-axis. We find four values of x that work, each time yielding zero in two different polynomials. The highest is near x = 1.618, where the fourth and fifth polynomials both equal zero. In other words, so or = (1 ± √(1 - 4(1)(-1)))/2 = (1 ± √5)/2 which, in agreement with the graph, is about -0.618 and 1.618, the latter of which is phi, the golden ratio. I thought that number looked familiar. So we could plot these values on a number line, but it would be two boring dots. So let's consider complex numbers. Now x can have two parts, one a real number like before and the other imaginary. For the first equation, 1x² + 1x + 1 = 0, we need the imaginary part or it's hopeless, because the sum is always positive if x is real. If x is complex, the equation takes the form 1(a + b)² + (a + b) + 1 = 0, and when we solve it we will change any b² to a -1 and hopefully the b coefficient will be zero at the end because we're trying to get to zero and there's no i in zero. So a² + 2ab - 1 + a + b + 1 = 0 a² + 2ab + a + b = 0 a² + a + 2ab + b = 0 a² + a + b(2a + 1) = 0 Maybe now we just assume b is zero so that part drops out, and a(a + 1) = 0 and a = 0 or -1. But those don't satisfy the first equation, so maybe it was a bad assumption and this equation has no complex roots. How about the second one, 1x² + 1x - 1 = 0. It has two real roots, near x = -1.618 and x = 0.618. If x is complex, it takes the form a² + ab + b² + a + b - 1 = 0 a² + ab - 1 + a + b - 1 = 0 a² + ab + a + b - 2 = 0 a² + a + ab + b - 2 = 0 a² + a + b(a + 1) - 2 = 0 b(a + 1) = 2 - a² - a b = (2 - a² - a) / (a + 1) and cheating again with the graphing tool I find that when a = -2 or 1 the b equals zero. So that gives me x = (-2 + 0i) and x = (1 + 0i), but these are both real. Is my Fisher-Price arithmetic off, or did I get off track by assuming you would set the polynomials to zero, or perhaps do things only get interesting in the higher degree polynomials? 1x² + 1x + 1 = 0 1x² - 1x - 1 = -1x² + 1x + 1 2x² - 2x - 2 = 0
Now we turn to the quadratic formula to solve for x: x² - x - 1 = 0 x = (-b ± √(b² - 4ac)) / 2a a² + 2ab + b² + a + b + 1 = 0 a² + a = 0 1(a + b)² + 1(a + b) - 1 = 0
Side comment first: yes, root of the polynomial is something that makes it to be equal to zero, so there is 0 in the right side of every equation. Square equations You're completely right up to the point where you're trying to find complex roots of a square equation. It is done using exactly the same way as for real roots: So if your equation is 1x² + 1x + 1 = 0, then a=1, b=1, c=1 and roots are Since √(-3) = i√3 (because i√3 x i√3 = (i x i) x (√3 x √3) = -1 x 3 = -3), the roots are -1/2 ± (√3/2)i It is possible to find complex roots your way You can find the roots the way you calculate things, but be more careful with imaginary part; let's represent x not as 'a + b' as you do but as 'a + bi', where both a and b are real numbers, and i is, well, square root of -1. Then we will get: a² + 2abi + (bi)² + a + bi + 1 = 0 => a² + 2abi - b² + a + bi + 1 = 0 => (a² - b² + a + 1) + (2ab + b)i = 0 Now look at the number at the left side: it has real part (a² - b² + a + 1) and imaginary part (2ab + b)i. For this number to be zero, both parts should be zero (you cannot make zero out of real number by adding imaginary part to it). So we got that I'm going to look at the imaginary part first. 2ab + b = 0 => (2a + 1)b = 0 Now we have two numbers (2a + 1 and b) such that their product is zero, so one of those numbers is zero. It cannot be b (try to substitute b = 0 -- the equation turns out to a² + a + 1 = 0, which has no real solutions, and a was a real number). Hence Now, when we know a, let's look at the real part. 1/4 - b² - 1/2 + 1 = 0 => b² = 3/4 => b = ±√3/2 We substituted x = a + bi, so we can get our two roots now: -1/2 ± (√3/2)i General situation There is a theorem that the number of (complex) roots a polynomial has is equal to its degree. Well, the whole thing is a bit less straightforward as some roots can repeat themselves, like in x² - 2x + 1 = 0 -- there is only one solution, x = 1, but since the same equation can be represented as (x - 1)(x - 1) = 0, the solution x = 1 is, well, counted twice -- once for each "x - 1". I'm not going deeper here, but what's important to know: if you take random polynomial of degree n, almost always it will have exactly n different complex roots. It is not possible to calculate exact roots for general polynomial of degree over 4 -- not that we don't know how, it's proven theorem that the formula cannot exist. About my picture So, I have equations which look like ..x^24 + ..x^23 + ...and so on... + ..x^2 + ..x + .. = 0, where every '..' is either 1 or -1. I lose nothing assuming that x^24 always has 1 (think about it), so there are 24 coefficients left. It gives me 2^24 = 16777216 equations. I find approximate solutions for each of those, and each has exactly 24 solutions. In this way I'm getting 2^24 x 24 = 402653184 complex numbers. For each such number, if it is a + bi, I put a point with coordinates (a, b). And that's how I've got my picture! x = (-b ± √(b² - 4ac)) / 2a (-b ± √(b² - 4ac)) / 2a = (-1 ± √(1 - 4)) / 2 = (-1 ± √(-3)) / 2 1(a + bi)² + 1(a + bi) + 1 = 0 => a² - b² + a + 1 = 0 and (2ab + b)i = 0 (2ab + b)i = 0 => 2a + 1 = 0 => a = -1/2 a² - b² + a + 1 = 0 =>