*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!
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
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! 🤩
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.
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.
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
@@shashwatsharma2406 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
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 ! 🤣🤣
@@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.
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.
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!
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
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
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 :)
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.
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 :)
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.
For those familiar with complex representation of vectors, the solutions of this equation can actually be visualized by rephrasing the question as: find all possible vectors v=(x,y) in 2D plane such that, the sum of vectors v_n formed by rotating v by a certain angle theta by n times (n=0, 1, 2, 3, 4): v_n = R(theta)^n v would vanish. Obviously, these vectors must form an angle of integer multiples of 2pi/5 with the x-axis.
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.
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.
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.
The second method is pretty obvious, but the first method is kind of neat. You can also use it to solve ax^8+bx^7+cx^6+dx^5+ex^4+dx^3+cx^2+bx+a=0 for a,b,c,d,e arbitrary complex numbers with a not equal to 0. I strongly advise against working out the general formula, but showing it exists is kind of neat.
(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.
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
An alternative method would have been to add x^2 to both sides, at which point x^4 +2x^2 +1 can be grouped and rewritten as (x^2 +1)^2, followed by grouping x^3 and x together, and rewriting that group as x(x^2 +1). Let h=(x^2+1), leaving you with h^2 +xh=x^2, from here you can complete the square twice or quad formula twice and you end up with the four solutions in a +bi form.
Well, it is really just the value of cos, sin at 2pi/5 that are related to the golden ratio. And this is not too surprising as they are both related to sqrt(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
Use geometric series then nth root of unity can be written using Euler's formula and then just plug in in the values Note: in the end disregard 1 as a solution because that would make denominator zero. There you have your four solutions
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
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.
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.
In fact in this equation is easily to guess value of introduced parameter which makes rhs perfect square without solving resolvent cubic Moreover it this situation works also De Moivre theorem x^4+x^3+x^2+x+1=0 x^4+x^3=-x^2-x-1 x^4+x^3+x^2/4=-3/4x^2-x-1 (x^2+x/2)^2=-3/4x^2-x-1 On RHS we have quadratic and it will be perfect square when its discriminant is equal to zero We have to introduce parameter to force discriminant to be zero (x^2+x/2+y/2)^2=(y-3/4)x^2+(1/2y-1)x+y^2/4-1 (1/2y-1)^2-4(y-3/4)(y^2/4-1)=0 (1/2y-1)^2-(4y-3)(y/2-1)(y/2+1)=0 (1/2y-1)((1/2y-1)-(4y-3)(y/2+1))=0 (1/2y-1)(2y^2-2y+2)=0 (y-2)(y^2-y+1)=0 (x^2+x/2+1)^2=5/4x^2=0 (x^2+x/2+1)^2-(sqrt(5)/2x)^2=0 (x^2+1/2(1+sqrt(5))x+1)(x^2+1/2(1-sqrt(5))x+1)=0 When we have ax^4+bx^3+cx^2+bx+a=0 we will get resolvent cubic partially factored
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
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)].
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
After diving with x^2 and arranging the terms, one can rewrite the equation like: (X+1/2)^2 + (1/X+1/2)^2 = -1/2 Since, sum of squares of two real numbers can't be negative, We can deduct X can't be real.
x^5 - 1 = ( x - 1 ) * (x^4 + x^3 + x^2 + x^1 + 1) so if right side is equal to 0 left must be equal to zero as well. so x^5 must be equal to 1 . this equation have 5 solutions . x = 1 solution comes from (x-1) part so other 4 solutions should be the answer.
I have the symbolic solution worked out. I have noticed that each of the poles have angles that are multiples of the first pole. Pole 0 has a angle of 1.257 radians from 0 or 72 degrees which 1/5 of 360. Pole 1 has an angle 2.513. Pole 2 has an angle of -2.513 which is 3 * the angle of pole 0 - 2*PI. Pole 4 has an angle of -1.257 which is 4 times the angle of pole 0 -2*PI. Also, all the magnitudes are 1. I can see why the pole at 1 was added now. The pole at 1 would have an angle of 0. I need to think more about this using magnitudes and angles. What are Grobner Basis vectors or Dixon Resultants? When I solve for the roots of messy polynomials using Mathematica, it seems there are a lot of common terms that are used over and over again in the solutions for the poles.
If you compare the solutions of both cases, you can find the formulas for sin(72º + 36ºn) and cos(72º + 36ºn). Without this, it's hard to calculate it without resorting to trigonometry.
My approach: Multiply both sides with (x - 1). This causes major cancellation, and when the dust has settled you are left with x⁵ - 1 = 0. Which is easily solved with the set {1, exp(±2/5 iπ), exp(±4/5 iπ)}. The 1 is extraneous (added by the initial multiplication), so it has to be left out of the final answer: x = {exp(±2/5 iπ), exp(±4/5 iπ)}
Could be further simplified reflecting acute angles as follows: x = {exp(±2/5πi), -exp(±π/5 i)}: this is very useful to confirm that product of 2 complex conjugates will produce real numbers.
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
my solution: noticing that these terms are a geometric series i can re write it as (x^5 - 1)/(x - 1) And x doesn't equal one, then we get that x^5 =1 And by using de moivre theorem we can figure out the complex roots of the equation If i'm wrong , please correct me
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. 😎
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
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??
se înmulțește ecuația cu x-1 care este diferit de 0 și obținem x^5-1=0 iar soluțiile sunt rădăcinile de ordinul 5 ale unității ;acestea ,vor fi rădăcinile ecuației date, mai puțin 1 care nu verifică ecuația inițială !!!
Hey, the equation you got is a geometric propgression with common ratio being x. Use the sum of terms of geometric progression formula and simplify :) you will get X^5 = 1 with x not equal to one as you said :)
hey blackpenredpen, there's been a question ive been struggling on and i hope you can help me solve it. The question in full is: 15*(5/4)^(n-1) > 5n+10 and solve for the conditions for n. This question is related to the geometric and arithmetic nth terms, I skipped ahead and made the conditions for both sides. Please have a look at this for me
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.
*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
Hey now few days ago a blue soft pi and now a blue pi shirt .Ithink he knows the channel 3b1b
Just kidding , of course he does.; - )
pls can u respond to my comment as i am a great fan of yours i will be blessed.
@@SupriyoChowdhury5201 Yes!
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
blackpenredpen: As always please pause the video and try this first
Me: *ignores every single time and just watches video*
Lol thanks for being honest : )
Lol
Same with me
OMG SAME HAHAHAHAH I really love his videos tho
There are two types of these problems:
1. Those where answer is 1
2. Those where answer is cos(2π/5)+i sin(2π/5)
those are the two nice answer types.
the much more common answer is just a horrible irratiional mess
Yes, it is nice truth and a pretty pentagon...I have solved it with a geometric approach (adding angles of roots of unity) and it works perfectly
@@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!
how can it be 1 ?
@@namansingla2975 pls teach me
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
So, that generalizes to polynomials of any degree where all of the coefficients are 1.
This was the solution I was thinking of. Seems much easier than the video.
Good but ı say
Exactly, this is one of the _easiest_ quartic equations out there
Yes, this is an example of the famous cyclotomic polynomials. There is even a special name for these!
The first thing I thought of: use geometric series.
I thought of dividing it by x² but that sounds as much better method
That's prety much the second way :)
@@regulus2033 Yes I know
If we want to solve 1+x+x^2.....x^n =0 we are obliged to go to complex number
Geometric series is actually the same as the second method
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!
🤩
It sure is!
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.
that wall is becoming dirtier and dirtier, but you're forgiven ❤️
Lol, thanks!
I don't think the apartment manager will forgive me tho...
Hard working man
That teddy looks similar to mr bean's teddy isn't it
@@blackpenredpen
I'll talk to him...😂
@@blackpenredpen Teddy will clean it up!
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]}
Correct. Though I think BPRP was going for a broader audience, hence the explanation he used in the vis
Okay... I have no idea what this means...
@@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.
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.
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
HOW YOU WROTE PHI IN PHI KEYBOARD !?
@@shashwatsharma2406 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
@@ffggddss Thx
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 ! 🤣🤣
Ah yes!! You are right! Thanks for pointing it out!
🦅 eye
Funny, there is a relation to what I'll be posting soon
Nice! Root of unity?
@@blackpenredpen Yes. They'll make an appearance!
@@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.
@@ProfOmarMath Yay!!
Brilliant ... Smart trick to solve hard quadratic eqn....
Anyway where are u from bro??
Problem caught at 3:54
the values of x + 1/x should be
- (golden_ratio) and 1/(golden_ratio)
No. It is not.
Wait. Ahhh. Got it wrong wjduskske
Wow, actually the most interesting thing about this is the fantastic relation between the golden ratio and sines + cosines you've just shown :O
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.
You watch Mr. Bean???? That show brings so much nostalgia!!!!!! 😭😭
Yes. They got all the episodes on prime videos so I watched them again while doing work. Lol
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!
Cool way to express exp(i*j*2*pi/5) of the form a + bi where 0
Representing 3Blue1Brown!
Blackpenredpen YEAH!!
Blue pen: i wanna join too...
Great stuff bprp. Thank you
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
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
Oooh this is pretty cool!
I would pause the video and try for myself if it wasn't like 4am every time I come here
Method 2:We can write it as (x^5-1)/(x-1)
So the answer will be fifth roots of unity (excluding 1)
Really elegant and interesting
Great method,sir.
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 :)
This could be a nice way to compute cos(pi/5) and sin(pi/5)
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.
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 :)
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
Yes it is, but it's cos(8pi/5) + i sin(8pi/5)
cos(2π/5) - i·sin(2π/5) = cos(8π/5) + i·sin(8π/5), so this is not a new solution. There is no need to change the signs.
Note that trig solutions are periodic, in general.
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.
We can use geometric progression similar to teddy method.
1+x+x^2+x^3+x^4=(x^5-1/x-1)=0
Duude. Loved this video. The solution got me shocked
Sergio Bejar thanks
It would have been nice to see at the end how the four roots obtained in the two ways correspond with each other...
But I loved the video overall!
whats interesting is that the second method can be used for any equation following the similar pattern
Smart solutions, thanks
I divided both sides with x and get
x^3+x^2+x+1 = -1/x
Original equation
x^4+x^3+x^2+x+1 = 0
x^3+x^2+x+1 = -x^4
-1/x =- x^4
1 = x^5
I stucked here
For those familiar with complex representation of vectors, the solutions of this equation can actually be visualized by rephrasing the question as: find all possible vectors v=(x,y) in 2D plane such that, the sum of vectors v_n formed by rotating v by a certain angle theta by n times (n=0, 1, 2, 3, 4): v_n = R(theta)^n v would vanish. Obviously, these vectors must form an angle of integer multiples of 2pi/5 with the x-axis.
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.
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.
Great! I like how you setup two x values in the middle of your presentation!
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.
(-1+sqrt(5))/2 should be positive?
Hyperbolium yes. Positive 1/phi
The second method is pretty obvious, but the first method is kind of neat. You can also use it to solve ax^8+bx^7+cx^6+dx^5+ex^4+dx^3+cx^2+bx+a=0 for a,b,c,d,e arbitrary complex numbers with a not equal to 0. I strongly advise against working out the general formula, but showing it exists is kind of neat.
(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.
Can I mention that high quality 3b1b shirt? ;p
3b1b merch, nicee
Yes and thanks!!
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
Geometric series is much simpler: (x^5-1)/(x-1)=0 if x isn‘t 1. x^5=1.Then take complex roots of 1. x=exp(2*Pi*i/5*n), n=1,2,3,4
An alternative method would have been to add x^2 to both sides, at which point x^4 +2x^2 +1 can be grouped and rewritten as (x^2 +1)^2, followed by grouping x^3 and x together, and rewriting that group as x(x^2 +1). Let h=(x^2+1), leaving you with h^2 +xh=x^2, from here you can complete the square twice or quad formula twice and you end up with the four solutions in a +bi form.
9:47
should be
e^((2/5*pi + 2*pi*n)*i)
I am amazed to discover that the trigonometric functions of 2pi×n/ are related to the golden ratio.
Well, it is really just the value of cos, sin at 2pi/5 that are related to the golden ratio. And this is not too surprising as they are both related to sqrt(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
Basically the vertexes of a pentagon inscribed on a unit circle in the complex plain.
this is awesome
😳😳😳😳😳his t shirt is 3Blue1Brown pi 😍😍😍
That's great to see
Use geometric series then nth root of unity can be written using Euler's formula and then just plug in in the values
Note: in the end disregard 1 as a solution because that would make denominator zero. There you have your four solutions
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
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.
“Its teddy “
I come here just to see these cute actions of him
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.
In fact in this equation is easily to guess value of introduced parameter which makes rhs perfect square without solving resolvent cubic
Moreover it this situation works also De Moivre theorem
x^4+x^3+x^2+x+1=0
x^4+x^3=-x^2-x-1
x^4+x^3+x^2/4=-3/4x^2-x-1
(x^2+x/2)^2=-3/4x^2-x-1
On RHS we have quadratic and it will be perfect square when its discriminant is equal to zero
We have to introduce parameter to force discriminant to be zero
(x^2+x/2+y/2)^2=(y-3/4)x^2+(1/2y-1)x+y^2/4-1
(1/2y-1)^2-4(y-3/4)(y^2/4-1)=0
(1/2y-1)^2-(4y-3)(y/2-1)(y/2+1)=0
(1/2y-1)((1/2y-1)-(4y-3)(y/2+1))=0
(1/2y-1)(2y^2-2y+2)=0
(y-2)(y^2-y+1)=0
(x^2+x/2+1)^2=5/4x^2=0
(x^2+x/2+1)^2-(sqrt(5)/2x)^2=0
(x^2+1/2(1+sqrt(5))x+1)(x^2+1/2(1-sqrt(5))x+1)=0
When we have ax^4+bx^3+cx^2+bx+a=0
we will get resolvent cubic partially factored
He has gone crazy during lock down.
Dolls
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
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)].
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
After diving with x^2 and arranging the terms, one can rewrite the equation like:
(X+1/2)^2 + (1/X+1/2)^2 = -1/2
Since, sum of squares of two real numbers can't be negative, We can deduct X can't be real.
True, but that doesn't actually tell us the solutions.
Teddy you lil genius 🤣🤣
Multiply by (x-1) on both sides then you'll get x^5=1 , then use De moivre theorem to get all 5 roots of the equation.
I considered the polynomial as a geometrical series , so it is equivalent to (1-x^5)/(1-x). That avoids the introduction of x-1 out of the blue.
I learned the symmetric coefficent way while preparing for the MO in my country... So yea this was easy!
I used calculus to insure that there is no real value of x
x^5 - 1 = ( x - 1 ) * (x^4 + x^3 + x^2 + x^1 + 1) so if right side is equal to 0 left must be equal to zero as well. so x^5 must be equal to 1 . this equation have 5 solutions . x = 1 solution comes from (x-1) part so other 4 solutions should be the answer.
and you should never comment before watching the whole video.
I have the symbolic solution worked out.
I have noticed that each of the poles have angles that are multiples of the first pole. Pole 0 has a angle of 1.257 radians from 0 or 72 degrees which 1/5 of 360. Pole 1 has an angle 2.513. Pole 2 has an angle of -2.513 which is 3 * the angle of pole 0 - 2*PI. Pole 4 has an angle of -1.257 which is 4 times the angle of pole 0 -2*PI. Also, all the magnitudes are 1. I can see why the pole at 1 was added now. The pole at 1 would have an angle of 0. I need to think more about this using magnitudes and angles.
What are Grobner Basis vectors or Dixon Resultants? When I solve for the roots of messy polynomials using Mathematica, it seems there are a lot of common terms that are used over and over again in the solutions for the poles.
Multiply both sides with x-1
Plot the answers in the complex plane to show
even spacing on unit circle
If you compare the solutions of both cases, you can find the formulas for sin(72º + 36ºn) and cos(72º + 36ºn). Without this, it's hard to calculate it without resorting to trigonometry.
My approach: Multiply both sides with (x - 1). This causes major cancellation, and when the dust has settled you are left with x⁵ - 1 = 0. Which is easily solved with the set {1, exp(±2/5 iπ), exp(±4/5 iπ)}. The 1 is extraneous (added by the initial multiplication), so it has to be left out of the final answer: x = {exp(±2/5 iπ), exp(±4/5 iπ)}
Could be further simplified reflecting acute angles as follows:
x = {exp(±2/5πi), -exp(±π/5 i)}: this is very useful to confirm that product of 2 complex conjugates will produce real numbers.
Interesting way to solve those sin and cos functions in terms of phi.
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
3:53 shouldn't it be -φ and ϕ
my solution: noticing that these terms are a geometric series i can re write it as (x^5 - 1)/(x - 1)
And x doesn't equal one, then we get that x^5 =1
And by using de moivre theorem we can figure out the complex roots of the equation
If i'm wrong , please correct me
Thanks teddy, I too thought of the fifth roots of unity first.
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.
John Nolen
Oh, the way I did it was 2pi+2pi*n where n is 0,1,2,3,4
And when n is 4, it’s not a solution
@@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. 😎
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
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??
btw this also shows the relationship between ∅ and 36°
se înmulțește ecuația cu x-1 care este diferit de 0 și obținem x^5-1=0 iar soluțiile sunt rădăcinile de ordinul 5 ale unității ;acestea ,vor fi rădăcinile ecuației date, mai puțin 1 care nu verifică ecuația inițială !!!
Hey, the equation you got is a geometric propgression with common ratio being x. Use the sum of terms of geometric progression formula and simplify :) you will get X^5 = 1 with x not equal to one as you said :)
hey blackpenredpen, there's been a question ive been struggling on and i hope you can help me solve it. The question in full is:
15*(5/4)^(n-1) > 5n+10 and solve for the conditions for n. This question is related to the geometric and arithmetic nth terms, I skipped ahead and made the conditions for both sides. Please have a look at this for me
This is the cyclotomic polynomial Phi_5(x) so the roots are the primitive fifth roots of unity and since 5 is prime they are exp(2 pi i k/5) k = 1..4
Every time he says seero an angel gets his wings
was hoping you'd do the method where you eliminate the x^3 term and find the resolvent cubic like some renaissance mathematician :p
phi doesn't want to be at the bottom, phi wants to be at the top: 1/phi = phi - 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.
Geometric series formula can be used