Reformulating Exact Values without Trigonometric Components or Imaginary Numbers

  • Thread starter Zetison
  • Start date
In summary: 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...?Yes. But I guess that will be really hard...Yes, it has been proved that the general cubic equation with real coefficients has no solutions (in the sense that it can't be solved using the algebraic operations and root extractions that we're familiar with) if it has no real roots.I really don't think this problem can be solved without either trig or i-s. :shrug:
  • #1
Zetison
35
0

Homework Statement



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

Homework Equations



[tex]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}} [/tex]

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:

[tex]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}}[/tex]

[tex]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)[/tex][tex]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)[/tex][tex]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}}[/tex]

Let
[tex]u = 473419 + 600 i \sqrt{14466}[/tex]

Then I have

[tex]x =
\frac{79}{60} + \frac{1}{60} {u}^{1/3} +
\frac {6121}{60{u}^{1/3}}[/tex]

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

[tex]x =
\frac{79}{60} + \frac{{u}^{4/3} + 6121{u}^{2/3}}{
{60u}}[/tex]But from here, I can not see how I can get rid of the i-s...
 
Last edited:
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  • #2
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:
 
  • #3
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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  • #4
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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  • #6
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
Zetison said:
I don't think it is possible to solve this problem

Why do you say that?
 
  • #8
Because I have tried for several years now :) But if you think it can be solved, be my guest ;)
 
  • #9
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 [tex]x=a\cdot \cos\left(\frac{\cos^{-1}\left(b\cdot a^{-3}\right)}{3}\right)[/tex]

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, [itex]b=473419[/itex] and [itex]a=\sqrt{6121}[/itex]
 
  • #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 :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.

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
 

1. What is the purpose of reformulating exact values?

The purpose of reformulating exact values is to simplify complex mathematical expressions and equations into a more manageable form, while still retaining the same value and meaning. This can make calculations and problem-solving easier and more efficient.

2. How is reformulating exact values different from rounding?

Reformulating exact values involves rewriting a number or expression in a different form, while rounding involves approximating a number to a certain number of decimal places. Reformulating exact values retains the same value, while rounding may result in a slightly different value.

3. What are some common techniques used to reformulate exact values?

Some common techniques used to reformulate exact values include factorization, simplification, and substitution. These methods can be used to manipulate expressions and equations in order to make them more concise and easier to work with.

4. How can reformulating exact values be useful in real-world applications?

Reformulating exact values can be useful in real-world applications such as engineering, physics, and finance. It allows for more accurate and efficient calculations and can also help to identify patterns and relationships between different variables.

5. Are there any potential drawbacks to reformulating exact values?

One potential drawback of reformulating exact values is that it may result in a loss of precision. This can be especially problematic in fields where very precise calculations are necessary. Additionally, reformulating exact values may not always be possible or may be more time-consuming than simply using the original expression or equation.

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