Reformulating Exact Values without Trigonometric Components or Imaginary Numbers

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Homework Help Overview

The discussion revolves around finding exact values for a variable \( z \) in a mathematical expression that involves trigonometric components and imaginary numbers. The original poster presents a complex equation and expresses difficulty in eliminating the imaginary components while adhering to the requirement of avoiding trigonometric functions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of the original equation and question the validity of the methods that lead to imaginary numbers. Some suggest that the formula used may not yield a solution without trigonometric or complex components. Others reflect on the historical context of the problem and the challenges faced in finding a solution.

Discussion Status

The conversation is ongoing, with participants sharing their thoughts on the feasibility of solving the problem without trigonometric or imaginary components. There is a mix of skepticism regarding the possibility of such a solution, and some participants express a desire to prove or disprove the existence of a solution that meets the stated criteria.

Contextual Notes

Participants note the constraints of the problem, including the requirement to avoid trigonometric components and imaginary numbers, which complicates the solution process. There is also mention of personal motivations tied to family history in mathematics, adding a layer of personal investment to the discussion.

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Homework Statement



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}}

Homework 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}}

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 = <br /> \frac{79}{60} + \frac{1}{30} \sqrt{6121} \left(\frac{\left({ \frac{473419}{6121\sqrt{6121}} + \sqrt{ -\frac{5207760000}{229333309561}}}^{1/3}}{2} + <br /> \frac {1}{2\left({ \frac{473419}{6121\sqrt{6121}} + \sqrt{ -\frac{5207760000}{229333309561}}}\right)^{1/3}}\right)x = <br /> \frac{79}{60} + \frac{1}{60} \sqrt{6121} \left(\left({ \frac{473419}{6121\sqrt{6121}} + i \sqrt{\frac{5207760000}{229333309561}}}\right)^{1/3} + <br /> \frac 1 {\left({ \frac{473419}{6121\sqrt{6121}} + i \sqrt{\frac{5207760000}{229333309561}}}\right)^{1/3}}\right)x = <br /> \frac{79}{60} + \frac{1}{60} \left({473419 + 600 i \sqrt{14466}}\right)^{1/3} + <br /> \frac {6121}{60\left({473419 + 600 i \sqrt{14466}}\right)^{1/3}}

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

Then I have

x = <br /> \frac{79}{60} + \frac{1}{60} {u}^{1/3} + <br /> \frac {6121}{60{u}^{1/3}}

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

x = <br /> \frac{79}{60} + \frac{{u}^{4/3} + 6121{u}^{2/3}}{<br /> {60u}}But from here, I can not see how I can get rid of the i-s...
 
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Zetison said:
… But from here, I can not see how I can get rid of the i-s...

Hi Zetison! :smile:

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

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. :wink:
 
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

Thank you for LaTeX tip :smile:
 
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Zetison said:
And btw:
[PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary9.jpg[/QUOTE]

No, I meant cos(y/3) + isin(y/3) :wink:
 
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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...)
 
Zetison said:
I don't think it is possible to solve this problem

Why do you say that?
 
Because I have tried for several years now :) But if you think it can be solved, be my guest ;)
 
Zetison said:
Because I have tried for several years now :) But if you think it can be solved, be my guest ;)

Are you referring to the exact same question? That is, to answer the given question without complex or trigonometric manipulations?
 
  • #10
Mentallic said:
Are you referring to the exact same question? That is, to answer the given question without complex or trigonometric manipulations?

Yes ;)
 
  • #11
Zetison said:
Yes ;)

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

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

Which lead to:

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

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...
 
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  • #15
Oh I see, but what do you mean by:
Zetison said:
And the result is a answer with either trigonometrical components or imaginary numbers, which is a solution, but not the solution my grandfather got...

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
Mentallic said:
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.

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
Zetison said:
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
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 :-p

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.

Zetison said:
But if there is no solution without trig or i-s, then we shuld be able to prove it?
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
Yes. My english sucks, I know :P
 

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