# Reformulating exact values

1. Jun 1, 2010

### Zetison

1. The problem statement, all variables and given/known data

Find exact values for z without any trigonometrical components or imaginary numbers
$$x = \frac{79}{60} + \frac{1}{30} \sqrt{6121} cos\left(\frac{1}{3}arccos(z)\right)$$
where
$$z = \frac{473419}{6121\sqrt{6121}}$$

2. Relevant equations

$$cos\left(\frac{1}{3}arccos(z)\right) = \frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}}$$

3. The attempt at a solution

I pretty fast get imaginary numbers from the equation above. But they shall, as I know, cancel one way or the other. So this is what I get:

$$x = \frac{79}{60} + \frac{1}{30} \sqrt{6121} cos\left(\frac{1}{3}arccos(z)\right) = \frac{79}{60} + \frac{1}{30} \sqrt{6121} \frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}}$$

$$x = \frac{79}{60} + \frac{1}{30} \sqrt{6121} \left(\frac{\left({ \frac{473419}{6121\sqrt{6121}} + \sqrt{ -\frac{5207760000}{229333309561}}}^{1/3}}{2} + \frac {1}{2\left({ \frac{473419}{6121\sqrt{6121}} + \sqrt{ -\frac{5207760000}{229333309561}}}\right)^{1/3}}\right)$$

$$x = \frac{79}{60} + \frac{1}{60} \sqrt{6121} \left(\left({ \frac{473419}{6121\sqrt{6121}} + i \sqrt{\frac{5207760000}{229333309561}}}\right)^{1/3} + \frac 1 {\left({ \frac{473419}{6121\sqrt{6121}} + i \sqrt{\frac{5207760000}{229333309561}}}\right)^{1/3}}\right)$$

$$x = \frac{79}{60} + \frac{1}{60} \left({473419 + 600 i \sqrt{14466}}\right)^{1/3} + \frac {6121}{60\left({473419 + 600 i \sqrt{14466}}\right)^{1/3}}$$

Let
$$u = 473419 + 600 i \sqrt{14466}$$

Then I have

$$x = \frac{79}{60} + \frac{1}{60} {u}^{1/3} + \frac {6121}{60{u}^{1/3}}$$

$$x = \frac{79}{60} + \frac{{u}^{2/3} + 6121}{ {60{u}^{1/3}}}$$

$$x = \frac{79}{60} + \frac{{u}^{4/3} + 6121{u}^{2/3}}{ {60u}}$$

But from here, I can not see how I can get rid of the i-s...

Last edited: Jun 2, 2010
2. Jun 1, 2010

### tiny-tim

Hi Zetison!

(btw, to make LaTeX brackets that fit, type "\left(" and "\right)" )

Don't you se where that formula comes from?

If cosy = x, then (x + √(x2 - 1))1/3 = (cosy + isiny)1/3 = cosy/3 + isiny/3, and 1/that = cosy/3 - isiny/3.

So you can either add and divide by 2, or you can just take the real part of one of them.

3. Jun 1, 2010

### Zetison

But that doesn't give me an answer without trigonometrical components and imaginary numbers...

And btw:
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary9.jpg [Broken]

Thank you for LaTeX tip

Last edited by a moderator: May 4, 2017
4. Jun 4, 2010

### tiny-tim

Last edited by a moderator: May 4, 2017
5. Jun 4, 2010

### Mentallic

Last edited by a moderator: Apr 25, 2017
6. Mar 12, 2011

### Zetison

I don't think it is possible to solve this problem, all tho my grandfather say he has done it. (he can't remember how, but I guess he cheated...)

7. Mar 12, 2011

### Mentallic

Why do you say that?

8. Mar 13, 2011

### Zetison

Because I have tried for several years now :) But if you think it can be solved, be my guest ;)

9. Mar 13, 2011

### Mentallic

Are you referring to the exact same question? That is, to answer the given question without complex or trigonometric manipulations?

10. Mar 13, 2011

### Zetison

Yes ;)

11. Mar 13, 2011

### Mentallic

Ahh, well then I'd be wondering why you are so curious in answering this problem without complex or trig. Why spend years on such an endeavour?

12. Mar 13, 2011

### Zetison

Well, it's just for fun actually. My grandfather is claim that he has solved it, but he can't recall how. So I only got he's word for it. But I really want to solve the problem. He claimed that he solved it when he was at my age (21), so I got a pressure to so too ;)

13. Mar 13, 2011

### Mentallic

Fair enough, but if you started trying to solve it years ago, then you must've been solving a slightly different or more general question?

Such as $$x=a\cdot \cos\left(\frac{\cos^{-1}\left(b\cdot a^{-3}\right)}{3}\right)$$

where b has some kind of special relationship to a which I can't quite see - if you take a look at the OP in this thread, $b=473419$ and $a=\sqrt{6121}$

14. Mar 14, 2011

### Zetison

Well, the original problem is to solve for x here:

[PLAIN]http://folk.ntnu.no/jonvegar/images/flaggstangoppgaven.jpg [Broken]

[PLAIN]http://folk.ntnu.no/jonvegar/images/math.gif [Broken]

This will lead to a 3.degree-equation. And the result is a answer with either trigonometrical components or imaginary numbers, which is a solution, but not the solution my grandfather got...

Last edited by a moderator: May 5, 2017
15. Mar 14, 2011

### Mentallic

Oh I see, but what do you mean by:
If your grandfather got a different solution then he can't be right, because there is only one distinct solution to this problem, whichever way it is solved. If he claims he was able to find the exact solution by another method which didn't involve any use of complex or trig then I'd be quite sceptical about it, considering the cubic formula was derived using these methods.

16. Mar 14, 2011

### Zetison

Yes, I know. I'm a bit sceptical my self. But he is a really smart guy and has also study math like me. The problem is that he is no so old that he can't remember anything :P

But if there is no solution without trig or i-s, then we shuld be able to prove it?

17. Mar 15, 2011

### Mentallic

Fermat also claimed he had a proof for Fermat's Last Theorem, but that claim has been disputed and regarded as false amongst the masses :tongue:

That's not to say that he didn't solve it, because I don't know of any existent side-steps you could possibly take in solving such a problem, but it's unlikely in my books.

Sorry I didn't quite catch that, do you mean we should be able to prove/disprove that one can't find a solution without the use of complex numbers or trig?

18. Mar 15, 2011

### Zetison

Yes. My english sucks, I know :P