Studying Mathematics Thread

[polymath257]

Very nice! Good work, and it shows what I was saying about how understanding helps the memory. It all follows, it's not just meaningless drivel to memorise.

Now do the formula for cubics Just kidding, I didn't see that done until university, and it hurt my poor brain.

Funnily enough, I remember hearing that there is also a formula for power 4 polynomials, but that there cannot be a general formula for power 5 or more. I wonder if my memory is correct there. There's probably some restriction about the kind of formula we're talking about.

Instead, we can use iterative methods and such, for approximate solutions.

Your memory is correct. The allowed operations are addition, subtraction, multiplication, division, and extraction of roots.

The quadratic formula shows it is possible to solve a general quadratic via these operations. And yes, there is a procedure for cubic and a quartic equations

it turns out that for every polynomial there is an object called a group that describes the symmetries between the roots of that polynomial. If that group has a certain property (called solvability) , then the polynomial is solvable via radicals. It turns out that the relevant group is NOT solvable for the general equation of degree 5 or higher.

A specific quintic that cannot be solved via radicals is x^5 -5x+5.

---

Another interesting thing about complex numbers is that you sometimes get extra solutions you wouldn't normally think of.

For example, consider

x^3 = 1

What's the solution?

x = 1

Obviously. But there's also two other solutions, if we allow x to be complex. They fall on the unit circle drawn in the complex plane around the origin, at angles 120 and 240 degrees from the positive real axis. Multiplying those numbers by themselves rotates them around the origin, so 3 * 120 = 360 and 3 * 240 = 720 leaving you back on the positive real axis at 1.

The same thing happens for every polynomial equation of the form x^n =1. The result is a regular n-sided polygon in the complex plane.

---

Very nice! Good work, and it shows what I was saying about how understanding helps the memory. It all follows, it's not just meaningless drivel to memorise.

Now do the formula for cubics Just kidding, I didn't see that done until university, and it hurt my poor brain.

Funnily enough, I remember hearing that there is also a formula for power 4 polynomials, but that there cannot be a general formula for power 5 or more. I wonder if my memory is correct there. There's probably some restriction about the kind of formula we're talking about.

lol, I don't even know what it's supposed to be like (the cubic formula). But I can imagine it being rather tedious to derive. Maybe a few years from now once I get into studying proofs for real.

The formula itself is nasty, but the procedure can be understood.

Given a cubic equation, a x^3 +b x^2 +c x +d=0, first divide by the leading coefficient to get

x^3 +B x^2 +C x +D=0

Then let  y=x-B/3. It turns out that when you do the algebra, this will eliminate the quadratic term, so you have
an equation of the form

y^3 + E y +F=0.

Now comes the trick. Multiplying out shows that (u+v)^3 -(3uv)(u+v) =u^3 + v^3. We try to force our equation into this form.

So, we put y=u+v, giving one more variable to play with and then require that -3uv=E, so v=-E/(3u),

Plugging in gives
0=y^3 +Ey +F=(u+v)^3 -(3uv)(u+v) +F =u^3 +v^3 +F =u^3 +(-E/(3u) )^3 +F.

Expand this and multiply through by 27u^3 gives

27 u^6 +27F u^3 -E^3 =0.

This is now a quadratic in u^3, which can be solved by the quadratic formula. This u^3, which gives u, which gives v, which gives y, which finally gives x.

In finding u from u^3, a cube root has to be taken and it is crucial to remember that there are three complex cube roots, which ultimately gives the three roots to the cubic.

Whew!