We're no strangers to love You know the rules and so do I A full commitment's what I'm thinking of You wouldn't get this from any other guy I just wanna tell you how I'm feeling Gotta make you understand Never gonna give you up Never gonna let you down Never gonna run around and desert you Never gonna make you cry Never gonna say goodbye Never gonna tell a lie and hurt you We've known each other for so long Your heart's been aching, but You're too shy to say it Inside, we both know what's been going on We know the game and we're gonna play it And if you ask me how I'm feeling Don't tell me you're too blind to see Never gonna give you up Never gonna let you down Never gonna run around and desert you Never gonna make you cry Never gonna say goodbye Never gonna tell a lie and hurt you Never gonna give you up Never gonna let you down Never gonna run around and desert you Never gonna make you cry Never gonna say goodbye Never gonna tell a lie and hurt you (Ooh, give you up) (Ooh, give you up) Never gonna give, never gonna give (Give you up) Never gonna give, never gonna give (Give you up) We've known each other for so long Your heart's been aching, but You're too shy to say it Inside, we both know what's been going on We know the game and we're gonna play it I just wanna tell you how I'm feeling Gotta make you understand Never gonna give you up Never gonna let you down Never gonna run around and desert you Never gonna make you cry Never gonna say goodbye Never gonna tell a lie and hurt you Never gonna give you up Never gonna let you down Never gonna run around and desert you Never gonna make you cry Never gonna say goodbye Never gonna tell a lie and hurt you Never gonna give you up Never gonna let you down Never gonna run around and desert you Never gonna make you cry Never gonna say goodbye Never gonna tell a lie and hurt you

i can never get over how he switches between markers so effortlessly

omg how he cleaned the board with his will at 5:24 that's so nice

This would be a great way to introduce the cyclotomic polynomials.

In the quintic equation, the polynomial can also be factorised as (x³ + 1)(x² + x + 1) = 0 and these two are very simple to solve

The shirt says algebra in arabic 🚬💀☕

Actually you can generalize the method of the quintic equation for all degrees of that kind of equation: x^n+x^(n-1)+...+x+1=0 | *(x-1)/=0 => x^(n+1)-1=0 And now calculate all roots of unity for k=1,...,n (and without k=0 of course).

Beautiful! The solutions to each equation were all the (n+1)'th roots of unity (except 1). A little hard to see the pattern, but the last one revealed it spectacularly.

I really wonder to see how you handle different colour pens in a single hand

Level 1: anyone can understand this Level 2: *MAXIMUM CONFUSION UNLOCKED*

In Taiwan, Asia, we have to complete the above knowledge in three years of high school. Although mathematics is very interesting, the pressure of the exam often makes me breathless

Now for Level 0: 1=0 Try to solve that one

Why it is difficult? n-th equation is : (X^(n+1)-1)/(x-1)=0 So we need to find all roots of X^(n+1)=1, except x=1., or to find all roots of power n of 1. It could be done easily in exponential form on complex plane. cos(2*pi*k/n)+i*sin(2*pi*k/n) where 0

I barely understand this but it's still fun to learn something. I love your videos and they are so well done.

I love how you casually hold a pokeball while solving difficult equations

I solved the last problem in this way.I can't speak English so I only write formulas. ‪ 𝒙⁵ +‪ 𝒙⁴ + 𝒙³ +‪ 𝒙² + 𝒙 + 1 = 0 ‪ 𝒙⁶ - 1 = 0 (‪ 𝒙 ≠ 1) (‪ 𝒙³ + 1)(‪ 𝒙³ - 1) = 0 (‪ 𝒙 + 1)(‪ 𝒙² - ‪𝒙 + 1) (‪ 𝒙 - 1)(‪ 𝒙² + ‪𝒙 + 1) = 0 ∴ ‪ 𝒙 = -1, (1±√3𝒊)/2 , (-1±√3𝒊)/2 , (‪ 𝒙 ≠ 1)

2:48 Although it's obvious, you should briefly mention that x cannot be equals to 0

For the quartic equation after dividing by x^2 my first instinct was to let x= cos@ + isin@, so I could then get x + 1/x to be 2cos@ and x^n + 1/x^n to be 2cosn@. After which I got a hidden quadratic, found a few values of @ and put it back into my original substitute. I do think your way was cleaner since I did need a good calculator to keep things in exact form.

One of the last roots of the last equation is -1,since it is an inverse relationship with an odd degree. The resulting quartic equation can be solved by using the method of solving inverse relations.

In quintic you can also factor (x+1). Or you can solve all five of them by multiplying (x-1) as shown in the last one.

Each polynomial is actually just a geometric series, the reason he multiplied by (x-1) for the Quintic is because of the geometric sum formula. Sum x^r from r=0 to r=n equals (x^(n+1)-1)/(x-1). So infact each and every one of this polynomials could have been multiplied by (x-1) and then solved very easily by getting the nth root of unity. In general a polynomial sum x^r, r=0 to n, i.e. x^n + x^n-1 + ….. + x +1 = 0 can be multiplied by (x-1) to obtain x^n+1 -1 =0 to then get the nth roots of unity

Great video. It's interesting how the quartic was actually the harder than the quintic.

this is so cool omg i noticed how we got some interesting roots of unity in the x^3 + x^2 + x + 1 case thats such a beautiful connection!!!

The solution to the quartic equation was very nice and creative, I love seeing quadratics in something weird

Also, x⁴+x³+x²+x+1 can be factored as (x²+φx+1)(x²+Φx+1); where φ and Φ are the Golden ratios, (1±√5)/2, the solutions to x²-x-1=0

The fact he can hold two markers and so easily switch between the two while also holding a Poké ball in his other hand and explain his process is astounding to me

I like the surprisingly simple solution for the quintic (if trig functions qualify as simple).

Great video as always! I wish you would have added just one thing in the end and drew a circle to show the 6th roots of 1, its a really great visual connection for complex numbers.

if we have these types of maths teachers nobody hates maths 🙏❤ LOVE FROM INDIA

x^6-1=0 can be written (x^3+1).(x^3-1) =0 and then we have all the local methods to solve the 5th deg polynomial. We can apply then Cardan's rule too I guess, he used the Complex-Euler geometry and brought in e, pi! Very interesting 🔥

My first thought was use i for the first ones but you found another way, amazing!

I feel like he made the solutions were wayy more complicated than needed, like i know all of this, but i have never seen anyone solbe it this way

Correction: the 4th level will be a quartic equation, not a quadratic equation (as mentioned in the description)

Hi tnx for the great content you make. I've learned so so much in here Ican't even thank you enough. Can you plz make a Fourier transform marathon as well? or Fourier series marathon? thank you so much

Calculus by substitution. Brilliant... I haven't had to do that for 10 years. Nice work.

I used the last method for the cubic onwards. The set of solutions for the n degree polynomial, call it X_n, are the numbers of the form cis(2×k×pi/(n+1)), k going from 1 to n

You should note that 0 isn’t a root of the equation when you divide by x for transformation to be equivalent though.

For last equation X^5+x^4+x^3+x^2+x+1 X³(x²+x+1)+x²x+1 (X²+x+1)(x³+1)=0 -1,omega,omega²,-omega, -omega² Can we Do like this

For n th order problem the solution will be vertices of n+1 th order regular polygon inscribed in a unit circle, with one vertice being on 1.

I like the way you handle two markers at the same time

That quintic was very elegant. As long as you grasp the Euler Identity, it’s deceptively simple for a very intimidating equation. One might even call it genius

Thank you for all your videos, I try to understand as much as I can but I'm kinda to young for this xD But I can't wait to see all of this in my future classes

The first precious knowledge I earned... From your educative channel... Is... 2:40 - 7:00 Quartic = Quadratic

I like solving x^(1/6)=1 graphically, much easier and more pleasing to eye

Wait, so can we generalize the last method to work for polynomial equations of any order?

Thanks for this good results 😊

perfect and lovely. thanks

Also, x == -1 is a solution to all of the ones that start with an odd power. So divide by (x+1) and you';ve simplified down one order.

For the quintic(if you do by grouping) x⁴(x + 1) + x²(x + 1) + 1(x + 1) = 0 (x + 1)(x⁴ + x² + 1) = 0 x = -1 x² = - 1 ± sqrt(3)i/2 x = ±sqrt(- 1 ± sqrt(3)i/2)

🤔Started to see the pattern with the cubic solution being i, i^2, and i^3; and I was kind of expecting it to end with a general solution. It wasn't quite the general solution but did end up being a general method. The n+1th roots of unity minus 1.

Bro, you made me fall in love with maths again 😁

Try this EXTREME quintic equation 👉 kzfaq.info/get/bejne/fdV3prl908ycgp8.html

Love your arabic for Algebra shirt

Wonderful video! 👍👍

the way he switch the marker is so smooth

الجبر

The last equation can be solved by another way: x⁴(x+1)+x²(x+1)+x+1=0 (x+1) (x⁴+x²+1)=0 x1=-1 let t=x² t²+t+1=0 this is the second equation, that we solved. But solving by using Euler's formula is beautiful

The solution for the 4th eqn is so cool!

In signal processing it is known as a moving average filter. It has a low-pass characteristic, and it is often used for smoothing a noisy signal.

6:18 Caution, 1/𝜑²-4 is negative, so we cannot take the square root so, but you can write i√(4-1/𝜑²) (because 4-1/𝜑² is positive). Same at 6:58 for 𝜑²-4 which is also negative.

E=MC^2

it would be nice to have a drawing diagram of the 6 roots of unity to complete the nice presentation, Thank You :)

You can use the trick at the end generally. Then you get the general solution of x^n + x^(n-1) + ... + x^1 + 1 = 0 is x_j = exp(j* 2π/(n+1) * i) with j = 1,...,n It is also possible to find this solution graphically, if you use the unit circle in the complex plane and divide the full angle in the equal parts and add the unit "vectors".

lost me at level 2

Remember, when dividing by term with an x (such as x^2), you must make sure that none of the solutions makes that process a division by zero.

Well the first step for the quintic would be to check the galois group and see if it is solvable. If not you could use Briggs radical or elliptic functions.

5:24 lets appreciate that cut there

What’s about 1+x+x^2+x^3+x^4+x^5+…=0

N Roots of unity, other than unity...

The method used for que quintic equation can be used for all type from grade 1 to infinite.

Thank you master

Am I missing something on the golden ratio? From Goggle, golden ratio is (1+root (5))/2. The negative of golden ratio is correct. However, the reciprocal of golden ratio seems weird to me.

wait, why not x^3(x^2+x+1)+x^2+x+1 = 0 (x^3+1)(x^2+x+1) = 0 (x+1)(x^2-x+1)(x^2+x+1) = 0 then easy solve by quadratic which is doable without university knowledge

This video is very satisfying to watch

these videos remind me of how much i liked doing algebra and calculus in high school

am i the only 17 yo kid stuck at level 2 or

Let p(x) = xⁿ⁻¹ + ... + x + 1, n > 1. Then (x-1)p(x) = (x-1)(xⁿ⁻¹ + ... + x + 1) = xⁿ - 1 So the solutions to p(x) = 0 are the complex n'th roots of unity, except 1 itself. Confession: I've seen this before. Like, before there was a bprp YT channel; like before there was YT; like before there was an internet/ARPAnet. Addendum: I really thought you were going to work the last one, the quintic, by grouping, because that would have reduced the problem to one(s) that we've already solved; the fervent quest of every mathematician. x⁵ + x⁴ + x³ + x² + x + 1 = (x+1)(x⁴ + x² + 1) which we know gives x = -1 and x² = ½(-1 ± i√3) ... then go back to your existing video about how to find √(a + bi), to find those last 4 roots of the quintic. . . . or x⁵ + x⁴ + x³ + x² + x + 1 = (x³+1)(x² + x + 1) = (x+1)(x² - x + 1)(x² + x + 1) which gives x = -1 and x = ½(±1 ± i√3) Final note: Please take these remarks as they are intended - additional discussion on your excellent video. In your final problem, you build the bridge needed to get to the remaining infinity of problems in this "sequence." I think you're doing wonderful work here, spreading the fun of math to the YT community. Thanks! Fred

I remember solving a quintic equation. This is actually a nice way to solve that equation.

We can use sum of n terms of gp to solve these kinds of polynomial equations and then. We get a equation whose roots we can find by de moivre theorem

For the last one we can also just factorise by x^3 the 3 first therms and then everything goes by itself

Find vertical asymptotes of tan(sinx). Of course if you can do it.

Comments before watching: My immediate guess for the fifth equation in the thumbnail was -1, and I see from calculating it out that that is in fact a solution. I am excitedly anticipating that there will also be some less trivial solutions. After watching: I knew I would be reminded of how to calculate roots of unity by this video. Thanks.

Wow I love the way he is swapping red and black on the board 😍

Very nice!

I love how he changes pens so smoothly

Such a convoluted way to solve a quadratic equation

We need part 2!

It shall be noted that when taking the six roof we should account for plus or minus but since the solutions lie on the unit circle the -e^(pi/3)I and the rest of the other negative values will just correspond to another value with a different angle theta. For example e^ipi=-1 but -e^ipi=1 which corresponds to n=0 but was discarded since it wasn’t a solution of the original equation since we multiplied by an extra factor (x-1). And similarly for e^(ipi/3) corresponds to e^(4ipi/3).

the awesomeness is based on the quantity of terms. Awesome.

loved level 3. cuz i actually understood it

You're basically attempting to write the n-th roots of unity in radical form (which is possible for any root of unity).

Good great awesome outstanding AMASING!

These are really getting awesome as it gets

For each equation with n terms (the highest exponent being n-1) the solutions are just the nth roots of unity, excluding unity itself (aka 1).

Im new to the channel and when i saw the way he switched between different colors by holding two pens in his hands at the same time is just amusing to see haha

When you had x^6-1=0, you could notice that x^6-1 is a difference of squares. Then we have (x^3+1)(x^3-1)=0, factor the sum and difference of cubes, use the zero factor property, and then you have all 5 solutions for x, where we have to exclude x=1 for multiplying by x-1 on both sides.

Level 4 was so cool!

Nice video thanks

I just recently looked into the equation x^n = 1 for any positive integer n. The roots of this equation has some interesting properties many of which you can figure out without actually solving the equation. For example, it is clear that the absolute value is equal to 1. If it were less than one, the value would get smaller and smaller and approach 0 as you multiply up the values and no matter the value of n it will never reach 1 but just have smaller and smaller absolute value as n is higher. If the absolute value is larger than 1 it will spiral out and grow larger and larger and never reach 1 as you increase the powers, so regardless of n it will never be 1. Further, any such equation can be written as (x-r0)(x-r1)(x-r2)... (x-rn) which when multiplied out becomes x^n - (r0 + r1 + r2 + r3... + rn)x^(n-1) + .... and for n > 1 we have this sum equal to 0 so the sum of the roots are always 0. Another interesting result for this particular equation is that if r is a root to the equation so that r^n = 1 then r^j for any integer j is also a root of the same equation this is because (r^j)^n = (r^(nj)) = (r^n)^j = 1^j = 1. So if you go around the unit circle which has all values with absolute value equal to 1 and the first root you encounter (e^((2*pi/n)i)) and we call it r then r^0 = 1 = r0, r^1 = r = r1, r^2 = r2 and so on r^(n-1) = r(n-1) and r^n = 1 = r0 again and it cycles from there since r^(n+j) = r^n r^j = 1 r^j = r^j. Further you can find that if you set L(x) = (1 + x)(1+x^2)(1+x^3)....(1+x^n) then L(r) can be found as follows: x^n - 1 = (x-r0)(x-r1)(x-r2)...(x-rn) is an identity and is valid for all x including x=-1 and plugging that in you get on the left side (-1)^n-1 and on the right side you get n terms of the form (-1 + rj) = (-1)(1 - rj) = (-1)(1 - r^j) so the whole left side is equal to (-1)^nL(r), If n is even (-1)^n = 1 so left hand side is equal to 0 and right hand side is also equal to 0 but when n is odd left side is equal to -1 - 1 = -2 and right hand side is equal to -L(r) so we get L(r) = 2. I. e. (1+r)(1+r^2)(1+r^3)....(1+r^n) = 2 regardless of n so for example the roots of x^241 = 1 is such that the 241 factors together multiply to 2 while the roots of x^2021 is such that the 2021 factors together multiply to precisely 2. and so on. It is easy to verify this for small values of n such as n=3 or n=5. For n = 3 we have r = -1/2 + sqrt(3)i/2 and (1+r)(1+r^2)(1+r^3) = 2. Since the latter factor is also 2 (1 + r^3) = (1 + 1) = 2 we must have that the factors of all roots except r=1 must equal to 1. so (1 + r)(1+r^2) = 1 - we get 1 + r + r^2 + r^3 = 1 + r + r^2 + 1 but the sum of all roots is zero so 1 + r + r^2 = 0 so this is equal to (1 + r + r^2) + 1 = 0 + 1 = 1 as I just stated. I got into these properties when looking at the problem of finding number of subsets of the set (1..2000) which adds up to a multiple of 5. In this problem you look at x^5 = 1 but the findings are more general than that and apply to any prime number n but also to non-prime numbers when you add certain considerations and complications to the general formula for that problem.

u can use the summation of geometric sequence