Some More Maths

0 is the easiest root i guess xD

Yes indeed, now find the irrational root >:D

I feel like just using Newton’s Method for the irrational one then using that to find the other values, but that seems a bit cheaty.

Since 0 is a root, the 3 other roots can be found by solving x^3 - 3x^2 + 10 = 0 which is a real pain using Cardano’s method, so i guess there’s a trick, right ? Solving cubic equations isn’t interesting in any way if it’s simply done by following the steps without truly understanding what’s going on :p

yeah cardano’s seems to be the only way to do it

and yeah @LogalDeveloper the taiwanese always 不诚实

@Nyxify and @LogalDeveloper I already know the answers :P @Shroomyjoe @Tuong As far as I know, there are three algebraic ways to solve it, Newton’s method is one, Cardano’s method (substitution) is another one, and I solved it a third way. Either one will only get you the irrational roots, the complex roots must be found using a second method. :-)

I know I’m pretty late, but I’ve found a solution that doesn’t use Cardano’s method or Newton’s method. You substitute 1/y for x, then use the fact that cos3x = 4cos^3 x - 3cosx. Then there’s a ridiculous amount of complex shenanigans I have to go through so that inverse cos of a number >1 makes sense.

The final result is absolutely ridiculous, but I’ve checked it with my Mac’s insanely powerful graphing program.

this handwriting

looks interesting but it’s not elaborate enough on what you’re actually doing like when you decide to substitute y for sqrt(-2/5)cos(t) which btw, you should be writing 2i/sqrt(5) cos(t) there

This looks very cool, it’s very clever to use trigonometric functions to do this (‘never seen this before :D), though the whole thing isn’t exactly rigorous, for example

• some people do not appreciate when you take the square root of a negative number :p
• k^2 = -2/5 ⇒ k = sqrt(-2/5) is missing a ±
• e^(-iz) = -i*sqrt(5/2) - sqrt(-7/2) ⇒ -iz = ln( -i*sqrt(5/2) - i*sqrt(7/2) ) I’m not sure if you’re allowed to do that since the exponential function isn’t a bijection of ℂ in ℂ (i think adding a +2nπ where n∈ℤ would fix this, but that’s not too important since it would get merged with the 2nπ you get from getting rid of the cosine)

Anyway, the whole thing is very cool C:

I was a bit rushed to get to bed when I did it, so there may be some questionable bits in the working out.

O wait, if I understand correctly what the k is, you were looking for a number k that satisfies both 10k^3 = -4 and -3k = 3, but I think such a number does not exist :/

@Tuong sort of. I made it work by factoring out -√(-2/5).

I wanted a value of k such that 10k^3 : 3k = -4 : 3. This is because I wanted to replace the x’s with a function in terms of cos theta in which I could factor out a number and change the remaining 4 cos^3 theta - 3cos theta into cos3theta. In this case I found that k = √(-2/5) worked if I just factored out a -√(-2/5) afterwards. The writing on the side with the k’s is just to figure out what number I needed, the cos3theta = 4 cos^3 theta - 3cos theta is the trig identity I used, and everything else in that part of the proof is just algebra.

Actually, just to clarify what I did with an analogy, here’s another example of using the same technique, but with nicer numbers.

x^3 - 3x - 1 = 0

We want k such that k^3 cos^3 ø - 3kcos ø = n(4cos^3 ø - 3cos ø) By equating like terms: k^3 = 4n and -3k = -3n Divide the first by the second: k^3 / (-3k) = 4n / (-3n) So k^2 = 4 k = 2 works (I am aware k = -2 is also a solution, but we do not need it, as k is not an important part of the solution.) Now we can continue the question:

let x = 2cosø (2cosø)^3 - 3(2cosø) - 1 = 0 8cos^3 ø - 6cosø - 1 = 0

Now we can use the fact that they are in the right ratio and factor out a 2:

2(4cos^3 ø - 3cosø) - 1 = 0

Now using the trig identity I mentioned earlier,

2(cos3ø) - 1 = 0 2cos3ø = 1 cos3ø = 1/2

And now we can solve it in terms of ø instead of x. The k does not actually become part of the solution, it’s just used to make the substitution actually useful.

Oh it’s 4n and -3n, not just 4 and -3; I completely get it now :p