Can We Solve A Hexic Equation? | Problem 240

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[@SidneiMV]

basically: z⁷ = 1 and z ≠ 1 z⁷ = e^(2kπi) *z = e^(2kπi/7)* when k ≠ 7n (because z ≠ 1) in other words k can be 0, n, 2n, 3n, 4n, 5n, 6n when n is an integer

[@Nobodyman181]

I need an exact value

[@SidneiMV]

@@Nobodyman181 there are infinite of them

[@krabbediem]

Thank you for the video :D

[@aplusbi]

Of course!

[@Nobodyman181]

Thank you very much

[@aplusbi]

You are welcome

[@NadiehFan]

It is obvious for anyone familiar with summing a (finite) geometric series that we have (z⁷ − 1)/(z − 1) = z⁶ + z⁵ + z⁴ + z³ + z² + z + 1 so we are actually dealing with the equation z⁶ + z⁵ + z⁴ + z³ + z² + z + 1 = 0 and it is also clear from this that the solutions of this hexic are the seventh roots of unity except for unity itself, so we have the solutions zₖ = exp(²⁄₇·k·π·i), k = 1..6 There seem to be different names for these types of polynomials and equations. The English Wikipedia speaks of _self-reciprocal_ polynomials and notes that these are also called _palindromic_ polynomials (and equations), since the coefficients of such a polynomial in standard form read from left to right are the same as the coefficients read from right to left. But the common term in _inter alia_ French and German for these types of polynomials and equations is _reciprocal_ polynomials and equations. The designation _reciprocal equation_ was coined by Euler, since replacing the variable of such an equation with its reciprocal yields the same equation. The substitution t = z + 1/z results in the cubic equation t³ + t² − 2t − 1 = 0 Your claim at 6:35 that this cubic can be solved algebraically merits some discussion, because if you would have attempted this you would have discovered that you actually _can't._ This cubic has three distinct _real_ roots which are all irrational but if you attempt to express these roots algebraically you will find that the roots can only be expressed algebraically using cube roots of complex numbers. This defies the whole purpose of solving the equation algebraically because any attempt to extract the cube roots of those complex numbers algebraically will only lead to cubic equations which are essentially equivalent with the cubic equation you were trying to solve in the first place. So, you are back at square one (or cube one, really). This is why early algebraists confronted with this catch-22 referred to this as the _casus irreducibilis_ (the irreducible case). To see this, let's try to solve our cubic algebraically anyway. First we depress the cubic by performing a substitution t = (x − 1)/3 which reduces the equation to x³ − 21x − 7 = 0 Now, the roots of a depressed cubic x³ + px + q = 0 where p and q are real numbers and D = (¹⁄₂q)² + (¹⁄₃p)³ are given by x₁ = u + v x₂ = −¹⁄₂·(u + v) + ¹⁄₂√3·(u − v)·i x₃ = −¹⁄₂·(u + v) − ¹⁄₂√3·(u − v)·i where u = ³√(−¹⁄₂q + √D) v = ³√(−¹⁄₂q − √D) As you can see, if D > 0 then the depressed cubic has _one_ real and _two_ conjugate complex roots. But for our present cubic x³ − 21x − 7 = 0 we have p = −21, q = −7 so D = (⁷⁄₂)² + (−7)³ = −¹³²³⁄₄ which is negative, so u = ³√(−¹⁄₂q + √D) = ³√(⁷⁄₂ + √(−¹³²³⁄₄)) = ³√(⁷⁄₂ + i·²¹⁄₂√3) and v = ³√(−¹⁄₂q − √D) = ³√(⁷⁄₂ − √(−¹³²³⁄₄)) = ³√(⁷⁄₂ − i·²¹⁄₂√3) are complex. Assuming the cube roots of the complex numbers to refer to their _principal_ value the roots of the equation x³ − 21x − 7 = 0 are x₁ = ³√(⁷⁄₂ + i·²¹⁄₂√3) + ³√(⁷⁄₂ − i·²¹⁄₂√3) x₂ = −¹⁄₂·(³√(⁷⁄₂ + i·²¹⁄₂√3) + ³√(⁷⁄₂ − i·²¹⁄₂√3)) + ¹⁄₂√3·(³√(⁷⁄₂ + i·²¹⁄₂√3) − ³√(⁷⁄₂ − i·²¹⁄₂√3))·i x₃ = −¹⁄₂·(³√(⁷⁄₂ + i·²¹⁄₂√3) + ³√(⁷⁄₂ − i·²¹⁄₂√3)) − ¹⁄₂√3·(³√(⁷⁄₂ + i·²¹⁄₂√3) − ³√(⁷⁄₂ − i·²¹⁄₂√3))·i Since ³√(⁷⁄₂ + i·²¹⁄₂√3) and ³√(⁷⁄₂ − i·²¹⁄₂√3) are complex conjugates, their sum must be real and their difference must be purely imaginary, so it is clear that x₁, x₂, x₃ are _all real._ Yes, we have 'solved' the equation x³ − 21x − 7 = 0 but the result is useless. Now, you may be tempted to try to evaluate ³√(⁷⁄₂ + i·²¹⁄₂√3) and ³√(⁷⁄₂ − i·²¹⁄₂√3) algebraically. Since this is the a + bi channel let's say that we have ³√(⁷⁄₂ + i·²¹⁄₂√3) = a + bi Cubing both sides we have ⁷⁄₂ + i·²¹⁄₂√3 = (a³ − 3ab²) + (3a²b − b³)i so we have a³ − 3ab² = ⁷⁄₂ 3a²b − b³ = ²¹⁄₂√3 and since ³√(⁷⁄₂ + i·²¹⁄₂√3) = a + bi implies ³√(⁷⁄₂ − i·²¹⁄₂√3) = a − bi we also have a² + b² = ³√((⁷⁄₂ + i·²¹⁄₂√3)(⁷⁄₂ − i·²¹⁄₂√3)) = ³√(⁴⁹⁄₄ + ¹³²³⁄₄) = ³√(¹³⁷²⁄₄) = ³√343 = 7 So, we have b² = 7 − a² and substituting this in a³ − 3ab² = ⁷⁄₂ we have a³ − 3a(7 − a²) = ⁷⁄₂ a³ − 21a + 3a³ = ⁷⁄₂ 4a³ − 21a = ⁷⁄₂ 8a³ − 42a − 7 = 0 which is another cubic equation in a. Worse, if we substitute a = c/2 we get c³ − 21c − 7 = 0 which is _exactly_ the cubic equation x³ − 21x − 7 = 0 we were trying to solve in the first place, only with a different letter for the variable. So we are back at cube one. Is all hope to solve this equation lost? No, but we _cannot really solve this cubic equation algebraically_ and this is precisely what had the Italian algebraists who worked on cubic equations stumped for the better part of the 16th century, until Vieta came to the rescue. Vieta had been working on identities for the sine and cosine of integer multiples of angles and since, using modern notation of course, we have cos 3φ = 4·cos³φ − 3·cos φ this implies, for example, that 4·cos³ 20° − 3·cos 20° = cos 60° = ¹⁄₂ We can conclude from this that x = cos 20° is an _exact_ root of the cubic equation 4x³ − 3x = ¹⁄₂ Vieta hit upon the idea to use this to convert depressed cubic equations into an equation of the form 4x³ − 3x = c by applying a suitable _scaling_ (substitution) of the variable. To see how this works, again, in a modernized version, let's go back to our stubborn cubic x³ − 21x − 7 = 0 which we first rewrite as x³ − 21x = 7 If we substitute x = r·cos φ we get r³·cos³φ − 21·r·cos φ = 7 What we really want to get at the left hand side is 4·cos³φ − 3·cos φ so we begin by multiplying both sides by 4/r³ which gives 4·cos³φ − (84/r²)·cos φ = 28/r³ Now we want to have 84/r² = 3 so we need to have r² = 28 and so we can choose r = 2√7 which gives us 4·cos³φ − 3·cos φ = 1/(2√7) and because 4·cos³φ − 3·cos φ = cos 3φ and 1/(2√7) = ¹⁄₁₄√7 this gives cos 3φ = ¹⁄₁₄√7 Now, there is a single value of 3φ on the interval [0, π] which satisfies this, which is 3φ = arccos(¹⁄₁₄√7) but since the cosine is a periodic function with a period 2π we have 3φ = arccos(¹⁄₁₄√7) + 2·k·π, k ∈ ℤ and therefore φ = ¹⁄₃·arccos(¹⁄₁₄√7) + ²⁄₃·k·π, k ∈ ℤ. Of course, since the cosine is a periodic function with a period 2π this gives only three different values for cos φ by selecting any three consecutive integers for k. And since we substituted x = r·cos φ and since r = 2√7 we find using k = 0, k = −1, k = 1 that the equation x³ − 21x − 7 = 0 has the solutions x₁ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7)) x₂ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) − ²⁄₃·π) x₃ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) + ²⁄₃·π) And, since we started with a substitution t = (x − 1)/3, we can express the roots of t³ + t² − 2t − 1 = 0 as t₁ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7)) t₂ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) − ²⁄₃·π) t₃ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) + ²⁄₃·π) But this is not the whole story. Remember that we obtained the equation t³ + t² − 2t − 1 = 0 by substituting t = z + 1/z in z⁶ + z⁵ + z⁴ + z³ + z² + z + 1 = 0 which has the roots zₖ = exp(²⁄₇·k·π·i), k = 1..6 Since for z = exp(φi) we have 1/z = exp(−φi) and therefore tₖ = zₖ + 1/zₖ = exp(²⁄₇·k·π·i) + exp(−²⁄₇·k·π·i) = 2·cos(²⁄₇·k·π) the roots of the equation t³ + t² − 2t − 1 = 0 can actually be expressed as t₁ = 2·cos(²⁄₇·π) t₂ = 2·cos(⁴⁄₇·π) t₃ = 2·cos(⁶⁄₇·π) Of course we find only three different values for tₖ = zₖ + 1/zₖ with k = 1, 2, 3 because with k = 4, 5, 6 we again get the same three values t₄ = t₃, t₅ = t₂, t₆ = t₁ since cos(2π − φ) = cos φ. It is not evident at all how these much simpler expressions for the roots of t³ + t² − 2t − 1 = 0 relate to or can be derived from the much more complicated trigonometric expressions for the roots of this equation which we obtained using the standard method for solving a cubic equation with three real roots trigonometrically. For example, it is not clear at all why we should have −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7)) = 2·cos(²⁄₇·π) Needless to say, WolframAlpha is incapable to come up with either type of trigonometric expression for the roots of t³ + t² − 2t − 1 = 0. Neither is it able to simplify −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7)) into 2·cos(²⁄₇·π) although when entered in isolation it _does_ observe that 2·cos(²⁄₇·π) is a zero of x³ + x² − 2x − 1.

[@Obotron7]

This is brilliant. I have never seen a polar substitution used to solve polynomials like that. I'll be sure to try that sometime

[@takyc7883]

incredibly well written explanation. Reading through it actually gave me more insight than the entire video itself! Nice to see some history of things I just learnt in a textbook and cool to have the worked example, as well as equivalent unsatisfying but valid answers. also i remember from school something about the trigonometric and polar form approaches being equivalent but im not sure i remember why. im guessing something to do with eulers formula but honestly my brain is fried already. but this comment was a really good read

[@phill3986]

Stats wasn't my cup of tea either. Multivariable calc was my favorite

[@moeberry8226]

6 roots of unity Syber.

[@aplusbi]

I guess I meant to say 7th roots of unity 😉

[@scottleung9587]

6 of the 7, that is.

[@moeberry8226]

@@scottleung9587 there is no 7 this is really a hexic equation. This is not a polynomial of degree 7.

[@AndDiracisHisProphet]

3.00 how can i stay positive knowing that there is no quintic forumula!? that makes me angry!!! argh!!!

[@aplusbi]

I know. I feel the same way!

[@bobbyheffley4955]

In the first method, you obtained the minimal polynomial for a regular heptagon.

[@stvp68]

European Z and European 7, but American 1? 🙂

[@aplusbi]

yep

[@stvp68]

@@aplusbi you write American Xs too, if I recall correctly.

[@thomaslangbein297]

You are still obsessed with 0^0😂? 0^0=0^(a-a)=0^a/0^a=0/0. And that remains, till the universe will end in zillions of years, undefined.

[@aplusbi]

let a = 0. then 0^0=0^(a-a)=0^a/0^a = 0^0/0^0 = 1 😁

[@thomaslangbein297]

@@aplusbiyou should know that this argument of yours is wrong. If not, terribly sad. I have thought, up to now, that you try to think like a mathematician. You changed my mind with this comment though. It‘s a bit like the flat earthers. I am disappointed. Sorry to say that.

[@NadiehFan]

@@thomaslangbein297 There is an entire Wikipedia article _Zero to the power of zero_ about this where it is shown that for many applications it makes sense to define 0⁰ to be equal to 1 and this is also implemented in many computer algebra systems and programming languages although e.g. WolframAlpha considers 0⁰ to be undefined.

[@surelyred]

@@thomaslangbein297 0^1 = 0^(1+2) / 0^2 = 0/0. So therefore 0^1 is not well-defined?! I can also invent indeterminate forms from nothing.

[@thomaslangbein297]

@@surelyred Shaking my head. 0^1 = 0. 0^2 = 0. 0^3 = 0. As 0/0 is not defined, therefore 0^1 can never be 0^3/0^2. You are claiming 3 = 3 + 0 = 3 + 0/0 therefore addition is not well defined. HAHAHAHA Who can read and think has a clear advantage.