Most equations like this are not solvable.

[@sinecurve9999]

Today I learned: trigonometric functions of rational multiples of pi are algebraic numbers. Neat.

[@Nitin-_-_-_-_-_-_-00971]

Are you a JEE aspirant?

[@thenateman27]

I remember when I first learned this and my mind was blown

[@johannesh7610]

Pretty cool, yeah. Follows directly from cos(qτ) = Re(e^(iqτ)), but e^(iqτ) is a (denominator of q)-th root of unity

[@matthieumoussiegt]

it is not a +1 at the end of the polynomial but a -1 to ensure that it is the correct solution

[@XJWill1]

I noticed that also. He forgot to plug his solution back into the original equation to see if it works! Then maybe he would have caught the error. Guess that was not a good place to stop!

[@davidblauyoutube]

@@XJWill1 So did I. Wolfram confirms this with MinimalPolynomial[2Cos[Pi/11],x].

[@ramzikawa734]

This was driving me insane, “what did I do wrong in my algebra?!?”

[@koenth2359]

Noticed that too. I wondered whether the problem was intended with -1 and just copied wrongly afterwards. Maybe the problem was intended with the +1. You'd get a nastier solve: (t^11+1)/(t+1) = 2t^5.

[@no1ofinterst]

Ah, thanks! I was quickly cheking it on desmos to see if it worked. That explains why all the solutions seemed to be finding "2s" of the equation

[@scp3178]

Nice approach ("Ansatz"), Michael. Thank you. I think there is a little mistake at (9:56): You wrote exp(i*pi)=exp(i*(pi + 2k))

[@infopek3221]

Yep, he definitely forgot to include the pi in the product xd

[@landsgevaer]

Obviously, he meant k itself to be a multiple of pi...

[@hbowman108]

Riemann "solved" the quintic using a generalization of the trigonometric/hyperbolic solution of the cubic. Those functions have a triple "angle" formula that can be used for a cubic solution. Riemann showed that by using doubly periodic analytic functions of a complex variable (elliptic functions), there were identities that could be used to solve any quintic. Of course, then to "compute" the roots, you have to evaluate the elliptic functions, which are simply solutions to a certain- differential equation. Or they have series expansions.

[@dysrhythmia]

I was wondering about that for a long time! Do have a reference for that so i can look into it more?

[@tutordave]

yeah, just when I plugged 2cos(pi/11) into the equation, it gave the result of 2, not 0. I checked my result twice for mistakes. The mistake is that after the t + 1/t substitution, you get t^5 - t^4 + t^3 - t^2 + t + 1 + 1/t - 1/t^2 + 1/t^3 - 1/t^4 + 1/t^5. That's not your result. So I think maybe the mistake is in the original equation. I think it should be x^5 - x^4 - 4x^3 + 3x^2 + 3x - 1 = 0.

[@holyshit922]

x=2cos(π/11) is solution to the equation x^5-x^4-4x^3+3x^2+3x-1

[@ib9rt]

It would really help confused watchers if you would update the comments or description to correct the mistake in the video. The constant term in the original polynomial needs to be −1, not +1. You could also mention the good practice of sketching the function and locating the approximate solution in order to check the result. In this case the approximate solution to the polynomial as presented is around -1.775, which does not correspond to 2*cos(pi/11).

[@johnchessant3012]

It'd also be a nice problem to go in reverse, i.e., find rational polynomials for which 2cos(pi/n) is a root. For example, 2cos(pi/7) is a root of x^3-x^2-2x+1, and 2cos(pi/13) is a root of x^6-x^5-5x^4+4x^3+6x^2-3x-1.

[@timeWaster76]

I would guess 2cos(pi/n) is a root for an infinite number of polynomials

[@TheEternalVortex42]

Of course, so presumably you ask for a polynomial of minimal degree with leading coefficient 1 (I guess this is called a 'minimal polynomial').

[@weltkaiserendzeit2417]

so what's the minimal polynomial of 2cos(pi/n) if this number is algebraic

[@ecoidea100]

I think there is a mistake, the sign of 1 in the expansion is not negative, it is t^5-t^4+t^3-t^2+t+(+1)+1/t...

[@elkincampos3804]

In general if p is prime and 5 divides p-1. Then there is a sum of p-roots, s_p of unity such that s_p is root of polynomial of 5 degree. We can generalize, for all n.

[@cameronspalding9792]

The reason why we can’t solve a general quintic is because of how the permutation group of 5 elements decomposes into simple groups The largest simple group in S5 is A5 which is simple and since it is non communicative you can’t write the general solution to the quintic

[@jopicocco]

Oh yeah i remember algebra 2 (italian università) what a beautiful exam (solvable groups and galois)

[@japanada11]

Exercise: What is the Galois group of the polynomial from the video?

[@soyoltoi]

in radicals

[@TacoDude314]

@@japanada11 Technically the Galois group of the polynomial from the video is S_5 but he actually meant to use the polynomial x^5 - x^4 - 4 x^3 + 3 x^2 + 3 x - 1 which has Galois group C_5. This is pretty easy to verify because the five solutions are 2cos(n pi/11) for n=1,3,5,7,9. The formula for cos(n x) is polynomial in cos(x) (called a Chebyshev polynomial). Therefore Q(2cos(pi/11)) is the splitting field for x^5 - x^4 - 4 x^3 + 3 x^2 + 3 x - 1. Since Q(2cos(pi/11))/Q is a degree five Galois extension, its Galois group is the group of order 5, and is therefore C_5. The discriminants for the splitting fields of these polynomials are 19761 and 14641 respectively, which you can use to search for them in the L-functions and modular forms database (LMFDB).

[@mobatyoutube]

For extra credit, show why this method generates only five unique solutions to the original quintic even though there are eleven solutions to t^11 + 1=0?

[@GiornoYoshikage]

Obviously -1 is not a root. For any other root there's an inverse which is also a root - (e^(2πin/11))^(-1) = e^(2πi(11-n)/11) - and we can make exactly 5 pairs from them. For x=t+1/t there's no difference if we put t of 1/t into here so we have only 5 different values of x

[@f5673-t1h]

Because the original polynomial is a 5th degree. But for something a little more illuminating, because these are roots of unity, so their inverses are also roots of unity, and also solutions. For a given solution t, t+1/t is the corresponding solution to the original polynomial. But this can also be written as (1/t) + 1/(1/t), so it also corresponds to 1/t.

[@cameronspalding9792]

x=t + 1/t The reason why we only have 5 solutions to the original polynomial is that some pairs of solutions to t^11=-1 generate the same value of x

[@neilgerace355]

Because cosine is even?

[@monkerud2108]

Well any polynomial with n real roots, has a derivative with at least n-1 real roots.. constant function cant have roots, its derivative doesn’t either, from there 2nd degree polynomials can only have (max#1st degree roots + 1) and so on leading to -> polynomial of degree N only has maximum N real roots. Pretty trivial proof.

[@seok7960]

The constant should be negative 1 to make solution correct anyway solving quintic that is awesome !!

[@Null_Simplex]

Any quintic equation is as “solvable” as any polynomial of degree 4 or less. The key is realizing that quadratic equations were also unsolvable until roots were invented. Similarly, you can solve any polynomial of degree 5 or higher so long as you are willing to introduce new functions. In the case of quintic equations, one could use the addition of Bring Radicals to solve them. If inventing new functions to solve these feels like cheating, just realize that the square root of 2 is not something you can solve without a computer, since calculating the root of a number is (usually) an infinitely long process.

[@ivannz01]

the image of the reals under the map t\mapsto t +\frac1t is (-\infty, -1] U [+1,+\infty), i.e excludes the open interval (-1, +1). why didn’t we check for solutions to the original poly in x within the excluded range?

[@digxx]

Why is that substitution common?

[@get2113]

Very clever, a nice trick to learn.

[@landsgevaer]

Except that it doesn't generalize at all, as far as I can see... A similar substitution t = x + p/x is much more generally useful when solving a cubic though.

[@get2113]

@@landsgevaer Agree. I only encountered numerous quadratics and one quartic over career. Cubics rarely.

[@karlbindl7883]

Could you possibly do some videos about galois theory...?

[@PunmasterSTP]

Not solvable? More like "nice overall", because your videos are always amazing. Thanks again for making and posting them!

[@TheNameOfJesus]

@1:25 it warms my heart that you wrote 1 4 4 before fixing it to 1 4 6. It gives me hope.

[@robwilkinson5682]

Think there's a mistake here - the expression in t at 5:00 should have constant +1 not -1. Then when cancelling terms we're left with t^11 + 2t^6 +2t^5 +1 in the numerator, which sucks as the t=-1 sol isnt valid because of the t+1 term at the bottom. Not sure where to go from there!

[@txikitofandango]

Just change the original problem to -1 ;-)

[@TheRealSamSpedding]

@@txikitofandango always the best solution

[@rosiefay7283]

4:31 Each step in this process might be easy, but there are an awful lot of them. Would your students be expected to write out so much algebra to solve an exam question? And how useful is this substitution x = t + 1/t anyway? It seems as if it is good pretty much only for this exact question.

[@afuyeas9914]

Asking Maple the question of what are the roots in radicals you get an extremely complicated expression but from my understanding it essentially boils down to adding the fifth roots of a the solutions to an equation of degree 4 because the galois group of the equations which have cos(kpi/11) as roots is the dihedral group of order 10 which has Z5 has a normal subgroup... i think?

[@rotiferphile]

Looking at a graph of the original problem, there is a single real root near x=-2 - definitely not 2cos(pi/11). There is a mistake at 4:48 - the constant term should be +1. ??

[@japanada11]

There's a typo in the problem - the original polynomial should end with a -1, not a +1

[@goodplacetostop2973]

10:01

[@carultch]

Did you make your user name, specifically to make this comment?

[@goodplacetostop2973]

@@carultch Yes

[@holyshit922]

Basic QR for eigenvalues fails to converge many times but here can approximate the roots (Basic means with no shifts and deflations and something like that)

[@ldjq]

задача украдена у Российского и Советского профессора Михаила Абрамовича. Автора этой задачи 1987 года. Она была вступительной задачей в МГУ. Вам Стыдно должно быть. Если вы крадете, указывайте автора. И приезжайте учиться в Россию. Мы вас научим математике) task stolen from Russian and Soviet professor Mikhail Abramovich. The author of this problem is 1987. She was an introductory task at Moscow State University. You should be ashamed. If you steal, indicate the author. And come to study in Russia. We will teach you math

[@yoav613]

Very nice! Are there more common subtitutions like this for solving those equations?

[@landsgevaer]

The substitution t = x + p/x can be used in one of the steps when solving a cubic quite generally. But the fact that t = x + 1/x gets you anywhere here is just a happy accident.

[@srikanthtupurani6316]

The hint may be (x+1) ^3. I am not sure. These problems can sometimes be annoying.

[@surbix1356]

Why do you use t + 1/t for x? I get that you were using substitution for X but why that specific equation?

[@l.p.7585]

I stopped caring about factorization tricks a while back but this was a fun problem! Thanks for showing it.

[@alonamaloh]

So is the Galois group associated with the polynomial solvable? Is there a general procedure to compute the Galois group, see if it's solvable and in that case find the roots exactly? I think I could say I understand Galois theory if I could write a program to do that.

[@sgkhk2397]

X=t+1/t.....here is a problem replacing x with t+1/t you have missed numbers between (-2, 2) because you can't make t+1/t any number between (2, -2) .... So you are considering their is no root between(2, -2)....

[@Reidemeistermoves]

Where does the $t + 1/t$ trick come from? Are there other ones?

[@mulletronuk]

Not just most, almost all!

[@williamandrade-universoele6661]

Michael after replacement and development, the polynomial at "t" has its positive fifth-degree term. If it were negative, it would be worth its simplification using fraction equivalence.

[@8hng]

This is a polynomial, and it has one root at (-1.77312, 0.0)... what does this (x=2 cos(pi/11),,, x=1.9189...) mean then?

[@Kuratius]

There is a solution to arbitrary quintics ( since they can be reduced to Bring Jerrard form) using a series generated by the Lagrange Bürmann theorem. So I would say they are solvable.

[@XJWill1]

What is the solution for x^5 - x + 1 = 0

[@nna7yk]

@@XJWill1 Ohoooo, so you're following Mathologer's videos, too - just like me :D :D

[@shriramtomar3442]

@@nna7yk which video are you talking about?

[@japanada11]

@@XJWill1 One solution is given by x = 3F4 (1/5, 2/5, 3/5, 4/5; 1/2, 3/4, 5/4; 3125/256), where 3F4 is a hypergeometric function (and therefore x can be expressed as a power series)

[@XJWill1]

@@japanada11 An infinite series is not what most people are thinking of when saying that an equation is "solvable". For most applications, people expect a solution to be finite.

[@kennethkan3252]

x=-1

[@karimkamil1787]

Thanks for your instructive vidéos Could you please try this problem from a Moroccan contest Solve in integers m³+n²=9 Thanks

[@josephmartos]

I love this channel

[@mab9316]

Can't understand why you choose that variable change: t+1/t ?

[@landsgevaer]

Because it worked, by accident really. Or, the problem was designed such that it worked. It is not something that generalizes or that you would come up with without that prior knowledge.

[@choiyatlam2552]

I remember this is a crucial step in deducing the cubic formula though.

[@alessandrochen607]

Cannot t be -1?In the final passages we see that the exponent of t is odd, and (-1)^11 is - 1, so why we use complex numbers?

[@fplancke3336]

If you plug t=-1 in the polynomial in t you should see it doesn't work. That's why we have to go through complex numbers, to get other solutions for t = 11th roots of -1 to start from.

[@konstantinospapadimitriou1735]

2*cos(pi/11) is not a solution of this equation. The constant should be -1 not +1...

[@SuperSilver316]

Integral Suggestion Can you do int(x*ln(abs(cos(x)))) from 0 to pi I think I have some ideas of how to get there, but it may need some gentle manipulations to solve it more formally.

[@SuperSilver316]

Nvm I think I got it, but you should do it on your channel!

[@yoav613]

@@SuperSilver316 yes it is very nice. Split it to integral from 0 to pi/2 of xln(cosx) plus integral from pi/2 to pi of xln(-cosx) you can solve each of them by writing cosx= (e^ix+e^(-ix))/2

[@yoav613]

@@SuperSilver316 i am sorry there is simple way to solve it. After you split it to integral from 0 to pi/2 x ln cosx + integral xln(-cosx) from pi/2 to pi you can use the identity -cosx=cos(x+pi). Now make the sub z=pi-x for the second integral and it become integral from 0 to pi/2 of pi lncosx -x lncosx,then the xlncosx cancells and you left with the integral pi lncosx from 0 to pi/2 which michael already solved it is -pi^2ln2/2

[@jarikosonen4079]

If looks like "good luck" with the substitution as it must be to get it done...

[@manucitomx]

Thank you, professor!

[@srikanthtupurani6316]

How did you guess that x=t+1/t works.

[@XJWill1]

That substitution is useful in solving cubic equations, so it makes sense to at least try it.

[@kevinmartin7760]

@@XJWill1 I have to wonder if it is useful in solving higher-degree equations in general, or more so for those that appear in math contests because these tend to be selected so this substitution works?

[@japanada11]

The trick here should work any time the roots of a polynomial are all real and lie in the interval [-2,2] - this is something that can be checked in advance (eg using the intermediate value theorem) if you don't want to dive right in to the substitution.

[@japanada11]

(the reason being that any real number x in the interval [-2,2] can be written as x = e^(r pi i) + e^(-r pi i) for some r; so when you make the substitution x=t+1/t, you end up searching for points on the unit circle, which are typically much easier to find)

[@japanada11]

In fact, even if there's just one real root in [-2,2], the resulting polynomial in t will have a root on the unit circle (but it's still usually hard to find unless all the other roots are on the unit circle as well)

[@Everth97]

Pi -> Pi(1+2k) at the end for the periodicity!

[@void7366]

Quintic polynomials may have a general form solution, just not an algebraic one

[@pardeepgarg2640]

And what that general form??

[@XJWill1]

What is the solution for x^5 - x + 1 = 0

[@japanada11]

@@XJWill1 One solution is given by x = 3F4 (1/5, 2/5, 3/5, 4/5; 1/2, 3/4, 5/4; 3125/256), where 3F4 is a hypergeometric function (and therefore x can be expressed as a power series)

[@XJWill1]

@@japanada11 An infinite series is not usually what people mean when they talk about an exact general solution to an equation. I realize that it was stated "not an algebraic one" so the statement was technically correct, but as a practical matter I do not think it will get the point across to many people.

[@japanada11]

@@XJWill1 Actually, technically speaking, the solution I provided IS algebraic! The definition of "algebraic function" is one that can be expressed as a root of a polynomial in rational functions, and this hypergeometric series qualifies. The only distinction here is whether a solution can be expressed _in radicals_ - which is really not a conceptually significant distinction, aside from the fact that we as a society have come to privilege radicals as being somehow special because they come up often enough in applications to get their own calculator buttons. There is no fundamental difference between radicals and hypergeometric functions (both are defined algebraically as the root of a certain polynomial, and analytically as a power series)

[@theophonchana5025]

#QuinticEquation #polynomial

[@Amoeby]

4:38 how did he get -1? -x^4 gives us -6 and 3x^2 gives +6 (3*2), so they cancel each other. And the only constant left is in original equation but it has +1 not -1.

[@sauzerfenicedinanto]

I Agree, prof Penn probably made a spelling mistake in the initial polynomial; it should have been x^5- x^4 - 4x^3 + 3x^2 + 3x -1 which has as solution 2cos (pi / 11) With the constant +1 would not have come the development with alternating T signs that would have led to the remarkable product that Prof Penn used. .

[@fedeqwerty]

Michael spilled the T on this one

[@dmitryweinstein315]

The solution is not correct as with x=2*cos(pi/11) the left hand side =2, not 0.

[@suttoncoldfield9318]

Just found that myself

[@algebraicoo]

You need to check the solutions because t substitution give you 11 solutions,and the original equation have just 5

[@TheRealSamSpedding]

One will be t = -1, but that was not a solution - we multiplied everything by (t + 1) earlier on so couldn't have been zero as this would break our logical equivalences. The other 10 roots will be 5 sets of complex conjugate pairs, which will only yield five distinct values of x, as the complex conjugation preserves the real part.

[@lexyeevee]

i believe quintics are /solvable/ - at least, wolfram alpha can give exact solutions - just not necessarily in radicals. and arguably you haven't given a solution in radicals here either :)

[@TaladrisKpop]

Yes, quintics are not all solvables in radicals. But we can use other functions

[@NicholasOfAutrecourt]

When a mathematician says something isn't "solvable," they mean exactly solvable. But yes, you can always brute-force an analytical approximation. I'd imagine that's what Wolfram and similar software does.

[@lexyeevee]

@@NicholasOfAutrecourt no, wolfram alpha gives *exact* solutions, just in terms of more esoteric functions than radicals. specifically hypergeometric ones, which i don't really know anything about. we could quibble about the meanings of "exact" and "solution" here, but they do give you a straightforward power series you can evaluate to arbitrary precision, which is arguably what a radical is too

[@landsgevaer]

Yes, if you allow that, then just define f(x) to be one branch of the inverse of this polynomial. Then f(0) is a solution. Voila, solved.

[@TaladrisKpop]

@@landsgevaer Sure, but elliptic functions are not random functions. They have applications in many areas of mathematics.

[@bekhaddaderrar2111]

OMG ! You are a super genius in maths wow! How could you become like this?

[@ldjq]

lol. he completely stole this problem from the Soviet Russian professor Mikhail Abramovich. The author of this problem, which he invented in 1987. And your "mathematician" stole both the problem and the complete solution of the Russian professor)))). I feel sorry for you. you have a very poor education. ((Come to study in Russia! We will teach you

[@NicholasOfAutrecourt]

There's a mistake in there somewhere. 2cos(pi/11) is approximately 1.918, whereas the real root of the given polynomial is more like x = -1.773...

[@BigDBrian]

he wrote the original equation wrong, the constant term should've been a -1. notice that under the substitution of x=t+1/t nothing affects the constant term, yet it mysteriously flips from +1 to -1

[@NicholasOfAutrecourt]

@@BigDBrian So, he knows he made a mistake in the video, but doesn't come back and acknowledge it in the comments section and lets the video stand as it is? Or maybe he simply doesn't read comments? Either way ... odd.

[@random19911004]

Right at the end, you meant "plus 2k * pi"

[@landsgevaer]

Unless he meant k to be a multiple of pi. He didn't actually claim k is an integer... 😉

[@Nekuzir]

Why does desmos think it can solve any quintic?

[@Mephisto707]

Interesting that an irrational value obtained from a trigonometric function is a solution to a polynomial with integer coefficients. I guess this would be impossible with polynomials with degree less than 5?

[@kevinmartin7760]

This is easy, but the argument of the trig function is banal so the trig function gets simplified out. x^2+1 = 0 has solutions e^(±iπ/2) = ±i×sin(π/2) but sin(π/2) is known to be 1 so this simplifies to ±i. Even for cubic polynomials you get trig functions of multiples of π/3 which all have well-known values. The only thing special here is that there are not easily-obtained values for trig functions of multiples of π/11 so the trig function can't be simplified out. I suspect that if your tried punching out cos(π/11) using double-angle, half-angle, angle-sum and angle-difference identities you would eventually just end up back at the original polynomial.

[@japanada11]

Following up on Kevin's excellent answer - if you take any of the polynomials x-1 x^2-x-1 x^3-x^2-2x+1 x^4-x^3-3x^2+2x+1 and perform the x=t+1/t trick, you get something of the exact same form (alternating sum of powers of t) and therefore the same argument gives roots: 2cos(pi/3) 2cos(pi/5) 2cos(pi/7) 2cos(pi/9). The first of these is not that interesting, but even the second is pretty cool (it tells you that cos(pi/5) is half the golden ratio).

[@Mephisto707]

@@japanada11 Interesting! But something still puzzles me. If explicit formulas exist for 3rd and 4rd degree polynomials, doesn't this mean that cos(pi/7) and cos(pi/9) actually CAN be expressed in terms of sums, products and root extractions of real rational numbers? I thought this was impossible (the heptagon and the nonagon being non constructible polygons).

[@japanada11]

@@Mephisto707 yes they can! The issue is that being constructible is more restrictive - it only allows for _square_ roots (which can be computed by intersecting lines and circles or circles and circles). The expressions for cos(pi/7) and cos(pi/9) use cube roots, and cube roots are generally not constructible (cf the impossibility of doubling the cube with straightedge and compass)

[@XJWill1]

@@japanada11 Yes, those nasty cube roots are so much more problematic than square roots! Here is an expression for cos(pi/9) 1/sqrt(1-(1-4/(2+(-4+sqrt(-48))^(1/3)))^2) which does not look too bad, but that 1/3 power hides a lot of complexity.

[@paulu_]

6:52 Does this factorization pattern have a name? Edit: Okay, found it. It’s called a telescoping sum.

[@TheRealSamSpedding]

yes it's linked to geometric series

[@kaptenkrok8123]

quintic equations are very much solvable just doesnt exist a general solution formula...

[@Examinx]

I tried doing this on my own before watching the video and arrived at the same initial steps, but got stuck due to the mistake in the problem (polynomial should have -1 at the end instead of +1). The hard part of the problem seems to be coming up with those initial steps, so I figured I'd give my personal thoughts on how I came up with those initial steps and substitution. The polynomial initially appears to be "almost factorable" when you split the -4x^3 into -x^3-3x^3, and this gives x^5-x^4-x^3-3x^3+3x^2+3x-1=0. I did this because the 4 coefficient appeared to be the sum of the 1 and 3 coefficients. Splitting up that term makes it clear that the polynomial can almost be factored into (x^3-3x)(x^2-x-1)-1=0. Then, the x^3-3x term motivates the substitution x=t+1/t to give t^3+1/t^3 since the substitution is nice (this is a common trick especially when the numbers are nice like in the case of x^3+-3x). Substituting into the entire equation gives (t^3+1/t^3)(t^2-t+1-1/t+1/t^2)-1=0, which also has nice coefficients when expanded and can be solved easily. Hope this helps motivate why this approach is chosen.

[@비기-y8c]

Awesome

[@mikeduffy4450]

t = -1 is an introduced solution. The other 10 solutions are genuine.

[@robinche95]

in fact by doing the same process, you only have 5 double roots which in term only gives the 5 roots of your initial polynomial.

[@bitterlemonboy]

-1.77313

[@Blabla0124]

small typo at the very end: 2k => 2kpi

[@Happy_Abe]

Not just +2k but +2k*pi

[@semjonadlaj259]

+2 k pi

[@sadeghsadeghi7783]

Hello Dear Mr.Penn.You make a mistake in solving this problem.t+1/t>=2 or t+1/t

[@sgkhk2397]

Yes you are absolutely right..... While watching this I also commented the same thing...nice to see I am not the only one who noticed it🙃🙃🙃🙃

[@sgkhk2397]

Where are you form????.... Are you a college student?

[@adandap]

That's only true for real values of t. If t is allowed to be complex then you can get those values. For example, i + 1/i = 0.

[@playgroundgames3667]

One of many videos where he answers the question. I don't know this one

[@abdollahkhorshidi7367]

pascal-khayyam khayyam is iranian

[@Aleko64]

Confermo l'errore, verificato con Geogebra il polinomio deve finire con -1

[@БратецСлавянин]

At the end, must be -1, no +1

[@kaishang6406]

5am, why am I still awake

[@shubhamdawda7288]

Wow

[@A241-d7k]

Things get imaginary faster than u think

[@nikitakipriyanov7260]

No, the solutions of the original equation (correct one, with -1 in the end, not +1) are real.

[@starpawsy]

I go with x=1 and x=-1. If that doesnt work, I give up.

[@RexxSchneider]

Just use a quick sketch to get an idea of where any roots are, then apply Newton-Raphson. Using a spreadsheet simplifies the tedious calculations.

[@starpawsy]

@@RexxSchneider You force me to repeat myself. if it aint +/-1, I give up.

[@RexxSchneider]

@@starpawsy I don't force you to do anything. These comments are viewed by a large audience, not just you. Even if you want to give up easily, there will almost certainly be a lot of others who will want to try a bit more.

[@starpawsy]

@@RexxSchneider Agree completely. And I applaud them for their efforts.

[@erfanmohagheghian707]

No sir unfortunately you did it wrong. that polynomial has a real root around -1.77 which is not equal to 2cos(pi/11). I wonder why you did not bother to check!

[@비기-y8c]

It must have 5 solutions since it is 5th order What about other 4 sols?

[@robinche95]

you can do the same thing for other 11th roots of -1 : cos((2k+1)pi/11), for k =0..4. Fun fact: since it's a polynomial of degre 5, and you have one root, factor it and you can solve the remaining degree 4 polynomial :)

[@비기-y8c]

@@robinche95 thank u :)

[@Vincent-lw9ix]

why -1 is not equal to e^(3ipi), e^(5ipi)……?

[@bregottmannen2706]

yes it is

[@nikhilnagaria2672]

it is

[@carultch]

It is. e^(i*pi*(2*n+1)) = -1, where n is any integer. (2*n-1) is our odd integer generator. Every odd multiple of i*pi in that exponent, will yield -1 as the output of the exponential. The famous example of e^(i*pi) = -1 is the principle case of this rule, where n=0. The function e^(i*theta) is periodic, so there is a repeating pattern of it throughout the domain of theta. Odd multiples of pi for theta, yield -1, and even multiples of pi for theta yield +1. Odd multiples of pi/2 will yield purely imaginary numbers, and odd multiples of pi/4 will yield complex numbers where the real and imaginary components are equal to each other in absolute value.

[@Vincent-lw9ix]

Yes, but why it is e^(ipi) in this question? Maybe the answer should also be 2cos(3pi/11), or 2cos(5pi/11) or any odd multiple of pi over 11. But why is it 2cos(pi/11) here?

[@bregottmannen2706]

@@Vincent-lw9ix cos is periodic so it doesnt matter if you take cos(3pi/11) or cos(3pi/11 + 2pi)

[@MofoMan2000]

hax

[@damianbla4469]

08:12 When we get "t^11 = -1", the most people would do this: t = 11-th root of (-1) t = -1 x = t + 1/t x = (-1) + 1/(-1) x = (-1) + (-1) x = (-1) - 1 x = (-2) But we would skip the other solutions :(

[@damianbla4469]

@@gregorymorse8423 Sorry, I haven't written clearly enough what I think. I just wanted to emphasise that the most people would solve the equation "t^11 = -1" in the "easy" way (by just taking the 11-th root) even though - as you wrote and as Mr Michael Penn said in 07:53 - it brings us to the wrong answer. In addition, this way give us only one (wrong) answer - but Mr Penn showed there are more answers. Thank you for your comment which helped the people better understand my previous comment :)

[@maciuikanikoda7809]

question: which percentage of people per country in 2023 can solve this? and understand the history of it and name the mathematicians like Evarice Gallois? Can Vladimir Putin do that by himself alone? If no, "his" Russia is in deep trouble.

[@maciuikanikoda7809]

dido for other countries

[@barisdemir7896]

mr.penns mistake

[@Anonymous-df8it]

OMG! My comments are always the last. Giving this reply so your comment doesn't get pushed down!

[@mmmmmark9751]

Get annoyed with this channel quite often........many simple procedural errors (crossed plus and minus), calculation errors, spending time detailing simple techniques like foiling and cross-mults......and then flies through complex and difficult stuff (like number theory) with no explanations......