Solving [imath]x^7-1=0[/imath]

[Aion]

Hi, I need help solving this equation exactly using only square roots. I'm aware that one solution for this problem is

[math]x_k=e^{\frac{2ki\pi}{7}}, k=0,1,...,6.[/math]
These roots are evenly partitioned as vertices in a regular heptagon in the complex plane. But this doesn't seem to help as I only obtained trigonometric solutions with this approach. I was given a hint by the author to define [imath]t=x+x^{-1}[/imath]. This was my attempt using that hint. Consider the following

[imath]x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0.[/imath] Hence it would be enough to study the cyclotomic polynomial [imath]x^6+x^5+x^4+x^3+x^2+x+1[/imath]

By factoring out [imath]x^3[/imath] from this polynomial, we obtain the following expression

[math]x^6+x^5+x^4+x^3+x^2+x+1=x^3((x^3+x^{-3})+(x^2+x^{-2})+(x+x^{-1})+1)=0.[/math]
Given that [imath]t=x+x^{-1}[/imath], we have [imath]t^2=x^2+x^{-2}+2[/imath] and [imath]t^3=x^3+x^{-3}+3t.[/imath]. Hence

[math](x^3+x^{-3})+(x^2+x^{-2})+(x+x^{-1})+1=t^3+t^2-2t-1=0.[/math]Define [imath]t=y+\alpha[/imath]. Where [imath]\alpha=-1/3[/imath]. Then this polynomial reduces into [imath]y^3-\frac{7y}{3}-\frac{7}{27}=0.[/imath] Given that [imath]\Delta=(p/3)^3+(q/2)^2=-49/108<0[/imath]. The third-degree polynomial has three real roots. Set [imath]-\Delta =\Delta^{'}>0.[/imath]. Then the three roots are given by Cardanos formula, where

[math]\frac{-q}{2}+i\sqrt{\Delta^{'}} =\frac{7}{54}+\frac{7i}{6\sqrt3}=re^{i\phi}[/math]
Then by trigonometry, the angle [imath]\phi=arctan(3\sqrt3)[/imath]. Now it follows that [imath]y_j=\frac{2\sqrt7}{3}cos(\frac{\phi}{3}+\frac{2k\pi}{3})[/imath] for [imath]j=1,2,3[/imath] with corresponding values of [imath]k=0,1,2.[/imath]. Then you could conclude that [imath]t_1=y_1+\alpha, t_2=y_2+\alpha, t_3=y_3+\alpha[/imath]. But we already know that [imath]t=x+x^{-1}[/imath]. By multiplying by [imath]x[/imath], we get the quadratic equation [imath]x^2-tx+1=0[/imath], which by the quadratic formula gives the roots
[math]x=\frac{t\pm \sqrt{t^2-4}}{2}[/math]
By substitution of the three roots [imath]t_1,t_2[/imath] and [imath]t_3[/imath] into this expression, we obtain six possible roots as required. But this method only seemed harder and more complicated than the one above and does not yield any exact roots. The solution is still only an approximation. Can someone please help point me in the right direction. Thank you.

 

[The Highlander]

Hi, I need help solving this equation exactly using only square roots. I'm aware that one solution for this problem is

[math]x_k=e^{\frac{2ki\pi}{7}}, k=0,1,...,6.[/math]
These roots are evenly partitioned as vertices in a regular heptagon in the complex plane. But this doesn't seem to help as I only obtained trigonometric solutions with this approach. I was given a hint by the author to define [imath]t=x+x^{-1}[/imath]. This was my attempt using that hint. Consider the following

[imath]x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0.[/imath] Hence it would be enough to study the cyclotomic polynomial [imath]x^6+x^5+x^4+x^3+x^2+x+1[/imath]

By factoring out [imath]x^3[/imath] from this polynomial, we obtain the following expression

[math]x^6+x^5+x^4+x^3+x^2+x+1=x^3((x^3+x^-3)+(x^2+x^-2)+(x+x^-1)+1)=0.[/math]
Given that [imath]t=x+x^{-1}[/imath], we have [imath]t^2=x^2+x^{-2}+2[/imath] and [imath]t^3=x^3+x^{-3}+3t.[/imath]. Hence

[math](x^3+x^{-3})+(x^2+x^{-2})+(x+x^{-1})+1=t^3+t^2-2t-1=0.[/math]Define [imath]t=y+\alpha[/imath]. Where [imath]\alpha=-1/3[/imath]. Then this polynomial reduces into [imath]y^3-\frac{7y}{3}-\frac{7}{27}=0.[/imath] Given that [imath]\Delta=(p/3)^3+(q/2)^2=-49/108<0[/imath]. The third-degree polynomial has three real roots. Set [imath]-\Delta =\Delta^{'}>0.[/imath]. Then the three roots are given by Cardanos formula, where

[math]\frac{-q}{2}+i\sqrt{\Delta^{'}} =\frac{7}{54}+\frac{7i}{6\sqrt3}=re^{i\phi}[/math]
Then by trigonometry, the angle [imath]\phi=arctan(3\sqrt3)[/imath]. Hence it follows that [imath]y_j=\frac{2\sqrt7}{3}cos(\frac{\phi}{3}+\frac{2k\pi}{3})[/imath] for [imath]j=1,2,3[/imath] with corresponding values of [imath]k=0,1,2.[/imath]. Then you could conclude that [imath]t_1=y_1+\alpha, t_2=y_2+\alpha, t_3=y_3+\alpha[/imath]. But we already know that [imath]t=x+x^{-1}[/imath]. By multiplying by [imath]x[/imath], we get the quadratic equation [imath]x^2-tx+1=0[/imath], which by the quadratic formula gives the roots
[math]x=\frac{t\pm \sqrt{t^2-4}}{2}[/math]
By substitution of the three roots [imath]t_1,t_2[/imath] and [imath]t_3[/imath] into this expression, we obtain six possible roots as required. But this method only seemed harder and more complicated than the one above and does not yield any exact roots. The solution is still only an approximation. Can someone please help point me in the right direction. Thank you.

Please stop posting all this rubbish!

Your stupid April fool 'joke' was bad enough but we can do without you taking up otherwise valuable space with this nonsense!

I only read the first paragraph before it occurred to me to check who the OP was and then it was obvious that it was your usual drivel!

Please go bother people somewhere else.

Thank you.
(& Goodbye?)

 

[Aion]

Please stop posting all this rubbish!

Your stupid April fool 'joke' was bad enough but we can do without you taking up otherwise valuable space with this nonsense!

I only read the first paragraph before it occurred to me to check who the OP was and then it was obvious that it was your usual drivel!

Please go bother people somewhere else.

Thank you.
(& Goodbye?)

What April fool joke? I'm asking a genuine question and posted my attempt. What have I done to offend you?

 

[Dr.Peterson]

What April fool joke? I'm asking a genuine question and posted my attempt. What have I done to offend you?

There was someone recently who wrote several nonsense questions as "jokes", which I think have been deleted. Apparently he is confusing you with that person, somehow. I see nothing wrong with your question ... except that I, too, don't see the benefit of doing all that.

Can you show us the source that made this suggestion, including context, so we can try to figure out what is intended by the hint?

I only read the first paragraph before it occurred to me to check who the OP was and then it was obvious that it was your usual drivel!

What did you find that was "usual drivel"? I see nothing of the sort. My understanding is that, after I reported it, the thread I have in mind ('The predictability theorem as juxtaposed binomial abacus permutation?') was deleted (so I can't confirm who wrote it) and the poster, who had posted nothing real, was banned (so it presumably can't be Aion). Or do you have something else in mind?

 

[blamocur]

The third-degree polynomial has three real roots.

but you follow this with the expression for a complex root --

 

[blamocur]

but you follow this with the expression for a complex root --

Silly me. State on this page:

"When D>0 all three roots are real and distinct. However,according to Cardano's formula, the roots are expressed in terms of cube roots of imaginary quantities. Although in this case both the coefficients and the roots are real, the roots cannot be expressed in terms of the coefficients by means of radicals of real numbers; for this reason, the above case is called irreducible."

 

[The Highlander]

There was someone recently who wrote several nonsense questions as "jokes", which I think have been deleted. Apparently he is confusing you with that person, somehow. I see nothing wrong with your question ... except that I, too, don't see the benefit of doing all that.
What did you find that was "usual drivel"? I see nothing of the sort. My understanding is that, after I reported it, the thread I have in mind ('The predictability theorem as juxtaposed binomial abacus permutation?') was deleted (so I can't confirm who wrote it) and the poster, who had posted nothing real, was banned (so it presumably can't be Aion). Or do you have something else in mind?

No, now that you have confirmed that there is nothing wrong with the post, I don't have anything else in mind.

You are right that I (wrongly) assumed this was another post from the person who posted the one you mention ('The predictability theorem as juxtaposed binomial abacus permutation?').

I was tired (and a little frustrated) after having been irritated by a couple of Agent Smith's posts: here & here). His posts aren't malicious but they can be annoying when they are so often random and off-topic interjections in a thread.

Then @Aion's post appeared and, when I looked at it, it seemed obvious (to me) that:-

\(\displaystyle x^7-1=0\iff x=1\)​

and a quick graphing of the function in Desmos appeared to confirm that...


I'm afraid the idea of there being complex roots for higher degree polynomials is way beyond my ken and so the remainder of the post just looked like gobbledygook (to me) and I presumed, therefore, that it was another prank.

My sincere and abject apologies to @Aion.​

 

[Aion]

There was someone recently who wrote several nonsense questions as "jokes", which I think have been deleted. Apparently he is confusing you with that person, somehow. I see nothing wrong with your question ... except that I, too, don't see the benefit of doing all that.

Can you show us the source that made this suggestion, including context, so we can try to figure out what is intended by the hint?


What did you find that was "usual drivel"? I see nothing of the sort. My understanding is that, after I reported it, the thread I have in mind ('The predictability theorem as juxtaposed binomial abacus permutation?') was deleted (so I can't confirm who wrote it) and the poster, who had posted nothing real, was banned (so it presumably can't be Aion). Or do you have something else in mind?

Sadly I can't give you the source, as it's no longer in the public domain. The problem is from a short Swedish paper written by Torbjörn Tambour 2003, Stockholm University called Elementary polynomials and equations. But to give more context to the question, according to Tambour, the equation [imath]x^n-1=0[/imath] is a type that can be solved with root extraction, i.e. in a formula of the same form as the one from Cardano-Ferrari. The paper doesn't prove this statement as it requires Galois theory, but he gives one example in the case of [imath]x^5=1[/imath] or [imath]x^4+x^3+x^2+x+1=0[/imath] (by dividing away the root [imath]x=1[/imath]). Set [imath]\epsilon =e^{\frac{2i\pi}{5}}[/imath] then the roots are [imath]\epsilon^k[/imath] for [imath]k=1,2,3,4[/imath]. He then proceeded to solve this equation by a systematic use of the relationship between roots and coefficients. The argument starts with the observation that [imath]\epsilon^4+\epsilon^3+\epsilon^2+\epsilon=-1[/imath]. He then defines [imath]\eta_1=\epsilon+\epsilon^4[/imath] and [imath]\eta_2=\epsilon^2+\epsilon^3[/imath]. Then proceeds to show that [imath]\eta_1+\eta_2=-1[/imath] and [imath]\eta_1\eta_2=-1[/imath] since [imath]\epsilon^5=1.[/imath] Therefore [imath]\eta_1[/imath] and [imath]\eta_2[/imath] are the roots of the equation [imath]X^2+X-1=0[/imath]. These are [imath]\frac{-1\pm \sqrt5}{2}[/imath]. Since [imath]\eta_1=\epsilon+\epsilon^4=2cos(\frac{2\pi}{5})>0[/imath], we have [imath]\eta_1=\frac{\sqrt5-1}{2}, \ \eta_2=-\frac{\sqrt5-1}{2}[/imath]. To determine [imath]\epsilon[/imath] we use [imath]\epsilon^4=\epsilon^{-1}[/imath]. This gives the equation

[math]\epsilon + \frac{1}{\epsilon}=\eta_1=\frac{\sqrt5-1}{2}.[/math] Hence

[math]\epsilon^2-\eta_1\epsilon+1=0[/math] Thus

[math](\epsilon^2-\frac{\sqrt5-1}{4})^2=(\frac{\sqrt5-1}{4})^2-1=\frac{-10-2\sqrt5}{16}[/math] And so

[math]\epsilon=\frac{\sqrt5-1}{4}\pm i\frac{\sqrt{10+2\sqrt5}}{4}[/math] The imaginary part of [imath]\epsilon[/imath] is [imath]sin(2\pi/5)>0[/imath], so we have lastly

[math]\epsilon=\frac{\sqrt5-1}{4}+ i\frac{\sqrt{10+2\sqrt5}}{4}[/math] Hence

[math]cos(\frac{2\pi}{5})=\frac{\sqrt5-1}{4}, \ sin(\frac{2\pi}{5})=\frac{\sqrt{10+2\sqrt5}}{4}[/math].

I've tried to generalize this proof and apply a similar method on the [imath]x^7=1[/imath] case. But I find it hard to generalize the method, where does the inspiration for [imath]\eta_1[/imath] and [imath]\eta_2[/imath] come from? It seems necessary to use these resolvents to solve the cyclotomic polynomial. I'm not a Gauss genius and I've not studied higher mathematics. But I've seen Gause's proof about the regular-sided 17-polygon and I think you need to use something similar, like subgroups of order 3 of the Galois group of [imath]Q(\epsilon)[/imath]?, these are elements that are left fixed by them, (whatever that means).

 

[Aion]

Sadly I can't give you the source, as it's no longer in the public domain. The problem is from a short Swedish paper written by Torbjörn Tambour 2003, Stockholm University called Elementary polynomials and equations. But to give more context to the question, according to Tambour, the equation [imath]x^n-1=0[/imath] is a type that can be solved with root extraction, i.e. in a formula of the same form as the one from Cardano-Ferrari. The paper doesn't prove this statement as it requires Galois theory, but he gives one example in the case of [imath]x^5=1[/imath] or [imath]x^4+x^3+x^2+x+1=0[/imath] (by dividing away the root [imath]x=1[/imath]). Set [imath]\epsilon =e^{\frac{2i\pi}{5}}[/imath] then the roots are [imath]\epsilon^k[/imath] for [imath]k=1,2,3,4[/imath]. He then proceeded to solve this equation by a systematic use of the relationship between roots and coefficients. The argument starts with the observation that [imath]\epsilon^4+\epsilon^3+\epsilon^2+\epsilon=-1[/imath]. He then defines [imath]\eta_1=\epsilon+\epsilon^4[/imath] and [imath]\eta_2=\epsilon^2+\epsilon^3[/imath]. Then proceeds to show that [imath]\eta_1+\eta_2=-1[/imath] and [imath]\eta_1\eta_2=-1[/imath] since [imath]\epsilon^5=1.[/imath] Therefore [imath]\eta_1[/imath] and [imath]\eta_2[/imath] are the roots of the equation [imath]X^2+X-1=0[/imath]. These are [imath]\frac{-1\pm \sqrt5}{2}[/imath]. Since [imath]\eta_1=\epsilon+\epsilon^4=2cos(\frac{2\pi}{5})>0[/imath], we have [imath]\eta_1=\frac{\sqrt5-1}{2}, \ \eta_2=-\frac{\sqrt5-1}{2}[/imath]. To determine [imath]\epsilon[/imath] we use [imath]\epsilon^4=\epsilon^{-1}[/imath]. This gives the equation

[math]\epsilon + \frac{1}{\epsilon}=\eta_1=\frac{\sqrt5-1}{2}.[/math] Hence

[math]\epsilon^2-\eta_1\epsilon+1=0[/math] Thus

[math](\epsilon^2-\frac{\sqrt5-1}{4})^2=(\frac{\sqrt5-1}{4})^2-1=\frac{-10-2\sqrt5}{16}[/math] And so

[math]\epsilon=\frac{\sqrt5-1}{4}\pm i\frac{\sqrt{10+2\sqrt5}}{4}[/math] The imaginary part of [imath]\epsilon[/imath] is [imath]sin(2\pi/5)>0[/imath], so we have lastly

[math]\epsilon=\frac{\sqrt5-1}{4}+ i\frac{\sqrt{10+2\sqrt5}}{4}[/math] Hence

[math]cos(\frac{2\pi}{5})=\frac{\sqrt5-1}{4}, \ sin(\frac{2\pi}{5})=\frac{\sqrt{10+2\sqrt5}}{4}[/math].

I've tried to generalize this proof and apply a similar method on the [imath]x^7=1[/imath] case. But I find it hard to generalize the method, where does the inspiration for [imath]\eta_1[/imath] and [imath]\eta_2[/imath] come from? It seems necessary to use these resolvents to solve the cyclotomic polynomial. I'm not a Gauss genius and I've not studied higher mathematics. But I've seen Gause's proof about the regular-sided 17-polygon and I think you need to use something similar, like subgroups of order 3 of the Galois group of [imath]Q(\epsilon)[/imath]?, these are elements that are left fixed by them, (whatever that means).

After thinking more carefully about the problem, I realised I could restate the question in a simpler form, and find the roots of a polynomial [imath]p(x)[/imath], which is satisfied by [imath]cos(\frac{2\pi}{7})[/imath]. I'm not sure if this would help, however.

 

[blamocur]

Sadly I can't give you the source, as it's no longer in the public domain. The problem is from a short Swedish paper written by Torbjörn Tambour 2003, Stockholm University called Elementary polynomials and equations. But to give more context to the question, according to Tambour, the equation [imath]x^n-1=0[/imath] is a type that can be solved with root extraction, i.e. in a formula of the same form as the one from Cardano-Ferrari. The paper doesn't prove this statement as it requires Galois theory, but he gives one example in the case of [imath]x^5=1[/imath] or [imath]x^4+x^3+x^2+x+1=0[/imath] (by dividing away the root [imath]x=1[/imath]). Set [imath]\epsilon =e^{\frac{2i\pi}{5}}[/imath] then the roots are [imath]\epsilon^k[/imath] for [imath]k=1,2,3,4[/imath]. He then proceeded to solve this equation by a systematic use of the relationship between roots and coefficients. The argument starts with the observation that [imath]\epsilon^4+\epsilon^3+\epsilon^2+\epsilon=-1[/imath]. He then defines [imath]\eta_1=\epsilon+\epsilon^4[/imath] and [imath]\eta_2=\epsilon^2+\epsilon^3[/imath]. Then proceeds to show that [imath]\eta_1+\eta_2=-1[/imath] and [imath]\eta_1\eta_2=-1[/imath] since [imath]\epsilon^5=1.[/imath] Therefore [imath]\eta_1[/imath] and [imath]\eta_2[/imath] are the roots of the equation [imath]X^2+X-1=0[/imath]. These are [imath]\frac{-1\pm \sqrt5}{2}[/imath]. Since [imath]\eta_1=\epsilon+\epsilon^4=2cos(\frac{2\pi}{5})>0[/imath], we have [imath]\eta_1=\frac{\sqrt5-1}{2}, \ \eta_2=-\frac{\sqrt5-1}{2}[/imath]. To determine [imath]\epsilon[/imath] we use [imath]\epsilon^4=\epsilon^{-1}[/imath]. This gives the equation

[math]\epsilon + \frac{1}{\epsilon}=\eta_1=\frac{\sqrt5-1}{2}.[/math] Hence

[math]\epsilon^2-\eta_1\epsilon+1=0[/math] Thus

[math](\epsilon^2-\frac{\sqrt5-1}{4})^2=(\frac{\sqrt5-1}{4})^2-1=\frac{-10-2\sqrt5}{16}[/math] And so

[math]\epsilon=\frac{\sqrt5-1}{4}\pm i\frac{\sqrt{10+2\sqrt5}}{4}[/math] The imaginary part of [imath]\epsilon[/imath] is [imath]sin(2\pi/5)>0[/imath], so we have lastly

[math]\epsilon=\frac{\sqrt5-1}{4}+ i\frac{\sqrt{10+2\sqrt5}}{4}[/math] Hence

[math]cos(\frac{2\pi}{5})=\frac{\sqrt5-1}{4}, \ sin(\frac{2\pi}{5})=\frac{\sqrt{10+2\sqrt5}}{4}[/math].

I've tried to generalize this proof and apply a similar method on the [imath]x^7=1[/imath] case. But I find it hard to generalize the method, where does the inspiration for [imath]\eta_1[/imath] and [imath]\eta_2[/imath] come from? It seems necessary to use these resolvents to solve the cyclotomic polynomial. I'm not a Gauss genius and I've not studied higher mathematics. But I've seen Gause's proof about the regular-sided 17-polygon and I think you need to use something similar, like subgroups of order 3 of the Galois group of [imath]Q(\epsilon)[/imath]?, these are elements that are left fixed by them, (whatever that means).

This Wikipedia page mentions constructibility of angles involving Fermat primes (which include 17). The page also shows how trig functions of multiple of [imath]3^\circ[/imath] can be expressed, but 360/7 is not among those angles.

 

[Dr.Peterson]

This Wikipedia page mentions constructibility of angles involving Fermat primes (which include 17). The page also shows how trig functions of multiple of [imath]3^\circ[/imath] can be expressed, but 360/7 is not among those angles.

That was my first thought; but constructibility is linked to square roots, and this will probably involve cube roots, which is a different story. I am not pursuing this any deeper, because I don't have a lot of knowledge in this area.

 

[Aion]

I've been pondered for a long time why Wolfram Alpha does not give me any exact roots. But I don't think the fact that a regular heptagon cannot be constructed with a compass and straightedge implies that the roots of the polynomial [imath]x^7-1=0[/imath] cannot be expressed using a finite sequence of arithmetic operations and square root extractions.

 

[Aion]

That was my first thought; but constructibility is linked to square roots, and this will probably involve cube roots, which is a different story. I am not pursuing this any deeper, because I don't have a lot of knowledge in this area.

Yes, my original interpretation of the question was to determine it exactly using only square roots. I realize now that is probably not possible, you need cube roots as well.

 

[Aion]

I gave up as it was too hard for a beginner and googled the problem. After reading the exact values of cos(2π/7), I understand one solution now. Set [imath]\epsilon=e^{\frac{2i\pi}{7}}[/imath]. Then my original polynomial [imath]p(t)=t^3+t^2-2t-1=0[/imath] must be satisfied by [imath]t_k=\epsilon^k+\epsilon^{-k}=2cos(\frac{2k\pi}{7})[/imath] for [imath]k=1,2, 3[/imath]. Need to sleep now