How do you solve x=1+ω^x+ω^2x
- Thread starter Darkside
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Solve for what - x or ω?
Dr.Peterson
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I assume you meant x=1+ω^x+ω^(2x) .
Ordinarily I would not expect an equation like this, with the variable both in an exponent and outside, to be solvable algebraically. But the right side should look familiar; you might try multiplying both sides by (1-ω^x) to see what happens. I think it will still be tricky, but there should at least be something you can do.
You might want to tell us what techniques you have learned that you might be expected to use. This is not "beginning algebra"; what course is it?
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x = 1 + w^x + w^(2x)
x(1-w^x) = (1- w^x)[1 + w^x + w^(2x)]...... assuming w^x \(\displaystyle \ne\)1
x(1-w^x) = [1- w^(3x)]...... given w^3 = -1
x(1-w^x) = [1- (-1)^x]
we know:
w = -1 and
w = (-1/2) ± (i/2)√3 = -e^[±i*π/3)
I do not see an algebraic solution for 'x' (other than by observation - same as Denis) ...... blinded by the light....
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Dr.Peterson
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No, it was specified, in the second submission, that ω is a given constant (one of the non-real cube roots of 1), not a variable. So this is not allowed.
No, ω is a cube root of 1, not of -1, and not the real root, though that wasn't stated explicitly. (It's a common convention.) That is, we can take \(\displaystyle \omega=\dfrac{-1 + i \sqrt{3}}{2}\).
But I didn't say I have an actual solution. Things stay tricky even when you start the way I suggested; and Wolfram Alpha gives only numerical solutions (which I don't quite trust anyway - it had trouble understanding what I was asking for). That's why I primarily asked for more information. It may be particularly useful to know what the OP knows about complex logarithms.
This is not a topic I am at all knowledgeable about, but there are three such roots. 1 + 0i is one, but if it is understood that real roots are excluded, there are still two complex roots.
\(\displaystyle \left ( \dfrac{-\ 1 + i\sqrt{3}}{2} \right )^3 = \dfrac{(-\ 1)^2 + 2(-\ 1)(i\sqrt{3}) + (i\sqrt{3})^2}{2^2} * \dfrac{-\ 1 + i\sqrt{3}}{2} =\)
\(\displaystyle \dfrac{1 - 2i\sqrt{3} - 3}{4} * \dfrac{-\ 1 + i\sqrt{3}}{2} = \dfrac{-\ 2 - 2i\sqrt{3}}{4} * \dfrac{-\ 1 + i\sqrt{3}}{2} =\)
\(\displaystyle \dfrac{2 - 2i\sqrt{3} + 2i\sqrt{3} - 2(-\ 1)(3)}{8} = \dfrac{2 + 6}{8} = 1.\) And
\(\displaystyle \left ( \dfrac{-\ 1 - i\sqrt{3}}{2} \right )^3 = \dfrac{(-\ 1)^2 - 2(-\ 1)(i\sqrt{3}) + (i\sqrt{3})^2}{2^2} * \dfrac{-\ 1 - i\sqrt{3}}{2} =\)
\(\displaystyle \dfrac{1 + 2i\sqrt{3} - 3}{4} * \dfrac{-\ 1 - i\sqrt{3}}{2} = \dfrac{-\ 2 + 2i\sqrt{3}}{4} * \dfrac{-\ 1 - i\sqrt{3}}{2} =\)
\(\displaystyle \dfrac{2 + 2i\sqrt{3} - 2i\sqrt{3} - 2(-\ 1)(3)}{8} = \dfrac{2 + 6}{8} = 1.\)
So why do you identify only one case as worthy of consideration?
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Dr.Peterson
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As he said, "ω, ω² are the cube roots of unity", that is, they are those two values you refer to; and the equation, x=1+ω^x+ω^(2x) is symmetric in ω, in the sense that if you replace ω with ω^2, the equation is unchanged because ω^4 = ω. As a result, it doesn't matter which one you call ω.
Thank you.
Dr.Peterson
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Have you followed the discussion and tried doing what I said to do? You can find a rather simple real solution; the complexity I mentioned is largely due to wanting to be sure we have all solutions, especially if x is not required to be real.
I'd still like to know the context of the question. What course is this for, and what techniques have you learned?