*mistake in the video* at 3:54, the solutions should be -𝜑 and "+1/𝜑" Thanks to Angel Mendez-Rivera and Egill Andersson for pointing it out I just typed up the solutions (all solutions are complex) in radical form on my Twitter: twitter.com/blackpenredpen/status/1261726124758888453/photo/1 1st way: 0:50 2nd way: 6:20
By the way, you could see it just by looking at the complex form, where the real part can be positive twice(apart from 1), but at the same time your first sol had only negative real parts)))
@@blackpenredpen does the video [x^x^x^2017=2017] has a second episode?i thought you said you are going to talk about why the infinite power dont have a solution
@@cillo71 it just took me 10 secs to solve the question as it's just fifth root of unity. I'm glad to see someone else knows this stuff. The truth is that most of his questions, I'm able to solve by writing them in eular form!
Yet another alternative: Substitute u=x^2. Separate the odd and even terms and square the equation, observe you get the exact same equation in u. Then consider the directed graph with the vertices being the 4 complex roots, with edges pointing from x to x^2. Consider the length shortest cycle, you either get x^2=x, x^4=x, x^8=x or x^16=x for some x. From there, solving should be easy.
Blackpenredpen: please pause the video and try for yourself Me: tries it using the quartic formula Blackpenredpen: Of course you can do it using the quartic fomula, but dont do that Me:...
x^5-1 can be factored into (x-1)(x^4+x^3+x^2+x+1), so the solutions are the 5 roots of unity of x^5-1, excluding the solution x=1. ok yep thats the second solution
Isn't there a mistake at 3:54 ? The answers are -𝜑 but +1/𝜑, isn't it ? So, the sign error continues to the second whiteboard. Fortunately Teddy made no mistake ! 🤣🤣
OK, I see your correction in your self-pinned comment, that the solutions to your 1st quadratic are (x + 1/x) = -φ and 1/φ (rather than -1/φ), so your 4 quartic solutions are, 1st 2 the same: x = ½[-φ ± √(φ² - 4)] and last 2 are now: x = ½[1/φ ± √(1/φ² - 4)] where you stop, because you don't want to get into the messy substitution of φ = ½(1+√5) into those. But with φ you can take advantage of the "powers of phi" rules: φⁿ = F[n]φ + F[n-1] In this case, φ² = φ + 1 1/φ = φ - 1 1/φ² = 2 - φ So this, together with recognizing that both your radicals have negative contents, gives: x = ½[-φ ± i√(3 - φ)] x = ½[φ - 1 ± i√(φ + 2)] Fred
@@letstalksciencewithshashwa9527 Copy-paste from a (MacOS) TextEdit file (which I've put together mainly by copy-pasting from online comments that contain them), where I keep special characters. It might also be in Character Viewer, in MacOS. Some special characters can be generated with combinations of Cmd, Option, and Shift. In Windows, the Unicode characters are available with special keystrokes; I'm not familiar with how to do that, but I've seen Windows users mention it. I believe you can find out by searching online. (ASCII characters are 1-byte; so there are 256 of them; Unicode characters are 2-byte; so there are 65536 of them, which includes the ASCII character set.) Fred
Nice! I was made aware by one of my viewers that you made a video of this a while ago. My solution to this problem did not contain your first method which is really cool! 🤩
@@blackpenredpen I am a student but I also came across this equation to find the fifth root of unity and their properties. Your solution is much more simple and easier than mine, thank you.
when i saw the equation, i immediately thought of vectors in the complex plane and roots of unity. you just have to find equally distributed vectors on the unit circle that make the equation work (so the vectors need to form a closed polygon). this immediately gives the fifths roots of unity except for 1 as solutions.
It is the 5th cyclotomical polynomial, the product of all X-ζ, with ζ in the set of the 5th primitive roots of unity. Primitive mean that they generate the multiplicative group of the 5th roots of unity, and when we have a prime number, here 5, any of them generates the group (not 1 of course). More generally, the polynomial 1+X+...+X^p-1 with p prime has all his roots in the set {exp(2iπk/p), k∈[1,...,p-1]}
@@addi.1813 if you multiply the polynomial in the problem by x-1 you get x^5 -1. It's solution are the 5-roots of unity. Since we multiplied by x-1, we erase 1 and the solutions we're looking for are the four 5-roots of unity that are not 1 (these are called primitive in this case).
Addison Think about Euler’s formula. Draw a circle that intersects the x and y axis at 1, i, -1, -i. Then place 1, x, x2, x3, and x4 equidistant around the circle. x = e(i 2Pi/5) is the obvious solution. Then you have all the other solutions when you go around the circle multiple times. Solved it in my head in a few seconds without writing anything down.
This video is very similar to something I just looked into. I read a new article on the recent discovery regarding the repulsiveness of polynomials, by Vesselin Dimitrov. The cyclotomic polynomial, x^4-x^3+x^2-x+1 and the unique properties discussed by Dimitrov might make a good video :)
Did teddy's way instantly, since this was just a sum geometric series. But seriously if we put that phi equation equal to euler's form, that would be one big identity
The first method utilized here is used to help solve (or only reduce if degree>=10) a more general class of polynomials - reciprocal equations. Reciprocal equations are those which can written as (x-x1)(x-1/x1)(x-x2)(x-1/x2)...(x-xn)(x-1/xn) - i.e. those which only have pairs of multiplicative inverse roots. For a related family of first-type even-degree reciprocal polynomials, it can be shown that in general: k^n*a1*x^(2n) + k^(n-1)*a2*x^(2n-1) + ... + k*an*x^(n+1) + a(n+1)*x^(n) + m*an*x^(n-1) + ... + m^(n-1)*a2*x + m^n*a1 = 0 has root set = {x1, m/(k*x1), x2, m/(k*x2), ... ,xn, m/(k*xn)}. The substitution t=k*x + m/x can be used to reduce the above equation to a non-reciprocal polynomial of halved degree (degree n in variable t). EDIT: I wrote a small paper on reciprocal equations as part of my coursework some months back, just sharing a result I discovered here.
Interesantísimo problema, que bello que las matemáticas sean universales, aunque lo expliques en inglés lo entiendo de todos modos, claro es más difícil pero si se puede. Gracias por tus vídeos, haces un gran trabajo.
These are pretty popular bunch of equations called reciprocal equations.... We can see that if k is a root, 1/k will also be a root.... But in this particular equation, the relation with roots of unity was very interesting!
Can't we take 1 on rhs and take x as commmon in lhs diveded both sides by x and take the one on rhs and take x^2 as commmon in lhs and cancel x+1 on both sides and x=-1 the we have x3+1=0. And solve for it can't we??
I remember when my teacher put this exercise on the exam, an easier way to do it is know than x^5-1=(x-1)(x⁴+x³+x²+x+1), then you only have to find the other four roots of x^5-1.
Can someone help me out with this. At 13:04 wont cos(2*pi/5) - i sin(2*pi/5) be a solution, and likewise we can change the sign in all the solutions. Thanks
You can say it's e^(2nπi/5), for n in ±1,±2. Saves a whole lot of writing that way, and verifies that you're getting complex conjugate pairs like you should from a polynomial with real coefficients.
It’s easier to write it as a G.P and it becomes x^5-1=0 ( since x-1 can not be 0) so the solutions are omega, (omega)^2,..., (omega)^4 where omega is the fifth root of unity
2nd solution is elegant, somehow I missed that one. I did it by going (x^2+1)(x^2+x+1)-x^2=0 and then (x^2+x/2+1-x/2)(x^2+x/2+1+x/2)-x^2=0, which gives two quadratic equations.
I tried 🐻's method when you said 'as always plzz Pau.... " but I thought it must be a real solution since you had made video. But trying again I got if the question was x^3 + x^2 + x + 1 = 0 That can be factorised as (x^2+1)(x + 1) Infact every polynomial like that can be factorised when deg is of form: (2^m) - 1
(I haven't watched the video yet) x = 1 isn't a solution, so we can multiply both sides with (x-1) != 0. Now, we have the equation x^5 = 1, which is easy to solve. (using the complex plane for exaple) We only have to exclude 1 as a solution to x^5 = 1 and we are left with 4 solutions to our original equation.
There was a small mistake in the video. You said the solutions to the first quadratic equation, which was with respect to x + 1/x, were -φ and -1/φ. However, the second solution should actually be 1/φ, because φ - 1 = 1/φ, and φ - 1 = [sqrt(5) - 1]/2, which is what you wrote on the board. Other than that, this is a great video.
Not to be too pedantic, but there is an "error" in Teddy's solution. (I put error in quotes because technically speaking it's not a mistake ... it's just kind of inelegant IMHO.) The primitive 5th roots of unity in this case are determined by the equation: x^5 = exp(2π n*i), adding the factor of 2π n to the exponent _after_ the taking the fifth root is superfluous. Taking the fifth root of the above expression gives: x = exp(2π n/5 *i) s.t. n ∈ { 1, 2, 3, 4} as solutions to the original quartic equation. n = 0 is not a member of the solution set because x = 1 does not satisfy the original quartic equation.
@@blackpenredpen No worries. My "n" is your "n+1." Hence when you reach n = 4, you've already begun to wrap around the circle for the second time. Also your n = 4 yields x = exp(2π (1 + 4)/5 *i) = 1 + 0i, but we already stipulated that x =1 is not a solution. 😎
Nice video always! Yes, solving for u = x + 1/x is a neat little trick that works for every "symmetric" equation with ax^4 + bx^3 + cx^2 + bx + a = 0 form. It's quite pretty cool :)
x⁴ + x³ + x² + x + 1 = 0 Multiply by (x-1) x⁵ - 1 = 0 x⁵ = 1 x⁵ = e^(2πni) x = e^(⅖πni) 2π/5, 4π/5, 6π/5, 8π/5 We have to exclude 0, or n=5m, because it doesn't solve the original equation So we have our answers which I'm too lazy to write out lol
Hello! --When you made the perfect square from --1:27-- to --1:58-- you promised that you would subtract the added "2x(1/x)", but you were not specifying that 1/x times x cancels into one basicaly. This was very confusing to me. At first i was thinking you failed to subtruct the entire thing, because the subtraction must had been minus two times ex times one over ex, not only minus two. I have just understood that ex times one over x cancels into one, and then one times minus two counters the added part.-- It has turned out that you say it actualy, i just did not hear it. Nice! Good video
An other way : x^4+x^3+x^2+x+1=0 x^2(x^2+x+1)= -x-1 x^2= -x-1 or x^2+x+1= -x-1 x^2+x+1=0 or x^2+2x+2=0 no solution OR no solution This equations hasnt real solutions
Use geometric progression sum. 1+x+x^2+x^3+x^4 = (x^5 - 1) / (x-1) so x^5 = 1 and x ≠ 1. So, x = all other 4 roots of unity = e^(i 2 π n / 5) where n = { 1,2,3,4 } = cos(t) + i sin(t), where t = ±72°, ±144° = cos(72°) ± i sin(72°) and -cos(36°) ± i sin (36°) where each real (cos) parts are computable using, cos(t) = sin(90-t) and sin(36° ± 18°) = (√5 ± 1)/4 and imaginary (sin) parts using the real (cos) parts, sin(t) = √[1 - cos^2(t)].
Use a tactic I figured out a few years ago: divide the equation by x^2. Then 0 = x^2 + x + 1 + 1/x + 1/x^2 = (x^2 + 2 + 1/x^2) + (x + 1/x) - 1 = (x + 1/x)^2 + (x + 1/x) - 1. Solve the quadratic equation: (x + 1/x) = (-1 +- sqrt(1 + 4))/2 = (-1 + sqrt(5))/2 or (-1 - sqrt(5))/2. Then solve x + 1/x = A for each value: x^2 - Ax + 1 = 0. There's no real solution. Instead, there are two complex-conjugate solution pairs. If x > 0 then x + 1/x is at least 2. If x < 0, then x + 1/x is at most -2. But both values of A are between -2 and 2.
just for curiosity i was trying to resolve a general solution using the second metod and is suprising easy sum for i=0 to n of x^i=0 multiplying both side you get x^(n+1)-1=0 ---> x^(n+1)=1 ---> x^(n+1)=e^(2p*(i+k)) x=e^(2P/n+1*(i+k)) for k=1,2,3,... n the solution n+1 get the solution x=1, but we now is not a solution
I solved it using geometric series. I got: 0=(1-r^5) / (r - 1). We can automatically skip 1 here. Now I just isolate the r and get r= 1^(1/5) and used recursive formula to find all roots. The formula: Wk = (|z|)^n * (cos((φ+2kπ) / n) + i * sin((φ+2kπ) / 2)) Wk+1 = Wk * (cos(2π/n) + i * sin(2π/n)) |z| = √(Real(z)^2 + Imaginary(z)^2) φ is just an angle on a unit circle
This could be used to find exact solutions to sin or cos(-4(pi)/5,-2(pi)/5,2(pi)/5,4(pi)/5). And you can also find the exact solution to sin or cos(even number*(pi)/any integer). eg. find solutions to 2(pi)/9 by first solving x^9=1, and equating real and imaginary parts to the corresponding solutions to x^8+x^7+...+x+1=0. The real difficulty though is finding an alternative way to solving the 8 degree polynomial with exact roots.
Although it is suggested at the beginning of this video that factoring the left hand side of x⁴ + x³ + x² + x + 1 = 0 into two quadratics _directly_ is hard this is in fact not hard at all. This can be done by completing the square twice and then applying the difference of two squares identity like this (x² + 1)² + x³ + x − x² = 0 (x² + 1)² + x(x² + 1) − x² = 0 (x² + 1 + ½x)² − ¼x² − x² = 0 (x² + ½x + 1)² − (½x√5)² = 0 (x² + (½ − ½√5)x + 1)(x² + (½ + ½√5)x + 1) = 0
It can be written as x*(x-1)*(1+x^2) = -1 ------->1st equation 1+x^2 is always positive and x(x-1) has minimum value at x= -1/2 which is -1/4 quickly looking at it as a quad graph you can see that real solutions are not possible for this equation
Nice 2 techniques. But I don't understand why you had to use 2pi +2pi n. Just use 2pi n. When n=0, you get the x=1 solution. The others are the same. Or even better, let n go -2,-1,1,2. Same solutions but simpler trig arguments. Cos(4pi/5)+ or - isin(4pi/5), Cos(2pi/5)+ or - isin(2pi/5).
x⁴ + x³ + x² + x + 1 = 0 Actually it's an easy quartic to solve, if you know the trick. It's a cyclotomic polynomial. Multiply it by (x-1) and you get x⁵ - 1 = 0 x⁵ = 1 the solutions of which are the 5 complex 5th roots of unity. And by multiplying by (x-1), we introduced the root x = 1; so the roots of the original quartic equation are the 4 non-real 5th roots of unity: x = cis(2kπ/5) = cos(2kπ/5) + i sin(2kπ/5), k = {1, 2, 3, 4} = cos(2kπ/5) ± i sin(2kπ/5), k = {1, 2} You could leave it in that form, or try to find purely algebraic, and possibly, numerical forms of those trig quantities. ⅖π = 72º; ⅘π = 144º cos(⅖π) = sin(18º) = 1/(2φ) = 0.309016994374947424... sin(⅖π) = cos(18º) = ½√(φ+2) = 0.951056516295153572... cos(⅘π) = -cos(36º) = -½φ = -0.809016994374947424... sin(⅘π) = sin(36º) = ½√(3-φ) = 0.587785252292473129... where φ is the Golden Ratio = ½(1+√5) I see from the Description that you have 2 ways of doing it. I'm guessing that this was one of them. I know of another, fairly clever algebraic method, which arrives at the algebraic forms of the solutions; I don't recall how it goes, but maybe that is your other method here? Post-view: That method (your 1st one) is even more clever than the one I dimly recall. Thanks for a newer, more elegant solution method for the complex 5th roots of unity! (See my other comment for how to simplify your 1st method results a little.) Fred
I think I noticed a mistake that you did. This was at the time 3:00. When you were plugging in the values to the quadratic formula and solving, you added 1 with 4 instead of minusing it in the equation 1^2-4•1•-1.
Very elegant solution. Can this be generalized to calculate roots of any polynomial of the form 1+x^2+x^3.....x^(n-1) as e^((2pi+2npi)/n)i . It looks like so. Or I am wrong.
wait a minute. you can't devide by X^2 when you are working in IR or even complex numbers. because you add the chance to devide by zero if X=0. or you can say at the very first that your gonna solve the equation in IR* ( real numbers excluding 0) or C*.
Tut, tut, tut, Mr BPRP, you did not specify the allowed domain in the question, so we have to assume it was x∊ℝ. Hence irriducible, end of problem. Then Teddy throws in x∊ℂ! Anyhow it helps I suppose.
If you ask an iit jee aspirant this is a very easy question These equations are called reciprocal equation and are of the form..ax^4(+,-)bx^3(+,-)cx^2(+,-)bx+a
I have a great problem that I would like someone to solve. The problem is that you need to find/evaluate the sum of the first n squared tangents as tan(4°)+tan(8°)+...+tan(88°).
triggered because two of your answers in the mod-arg form aren't given with the argument of -π < θ < π . REEEEEEEE [ just playing great video as always ]
Notice that 1 is not a solution to the equation. Now, multiplying both sides by (x-1), the equation becomes x^5-1=0, and thus the answers are all of the fifth roots of unity except for 1.
