solving equations but they get increasingly awesome
Solving polynomial equations but they get increasingly more awesome. We will also be solving them with different methods such as using the quadratic formula, factoring by grouping, a special way to complete the square, and also utilizing the 6th root of unity.
Try this EXTREME quintic equation 👉 • EXTREME quintic equati...
0:00 5 levels of a polynomial equation, from good to AWESOME!
0:3 linear equation x+1=0
0:16 quadratic equation x^2+x+1=0
1:36 cubic equation x^3+x^2+x+1=0
2:41 quadratic equation x^4+x^3+x^2+x+1=0
7:01 quintic equation x^5+x^4+x^3+x^2+x+1=0
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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 have to pin this!
@@blackpenredpen you didn't have too
@@stevenwade147 yeah... No, he actually had to
@@stevenwade147 He was actually required by law to pin that... sorry.
@Pooshan HalderChess this is one of the best ways I've ever been rickrolled ngl
i can never get over how he switches between markers so effortlessly
I was thinking the same brooo
He has a tutorial on it. Its really easy actually.
omg how he cleaned the board with his will at 5:24 that's so nice
Thanks 😆
He didnt..its editing
@@pratheekreddy8912 r/woooosh
@@pratheekreddy8912 u didnt get the joke apparently 😂
@@ajkanver9301 Have u heard of sarcasm?
This would be a great way to introduce the cyclotomic polynomials.
I actually forgot about it 😆
Yey, I was thinking the same thing while he was solving the quintic.
Can you please elaborate?
@@createyourownfuture5410 cyclo = cycle = circle, tomic = cut. Cyclotomic polynomials are those that have roots that are evenly spaced on the unit circle in the complex plane (and oriented to line up with the real number 1 as a solution) that no other cyclotomic polynomial of lesser degree has. So the first cyclotomic polynomial is x -1 = 0. The second splits the complex unit circle in two, so that would be x^2 = 1 for x = {-1, 1}, but 1 is already a root, so it's (x^2 - 1) / (x - 1) = 0, or x + 1 = 0. The third cyclotomic polynomial has three roots evenly spaced around the unit circle, but nix x=1, so that's x^2 + x + 1 = 0. The fourth cyclotomic has 4 roots evenly spaced, but x = {-1, 1} are already taken by the second and first cyclotomics, respectively, so it's a polynomial whose roots are {-i, i}, or x^2 + 1 = 0. The p-th cyclotomic polynomial (where p is prime) is always in the form sum (n=0 to p-1) x^n. See if you can prove why. The nth cyclotomic polynomial is always in the form x^n - 1 / (product ( j divides n ) j-th cyclotomic polynomial).
@@chx1618 I don't understand some things but thanks anyways!
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
LOL. i found also (x+1)(x^4+x^2+1)= 0. The 2 equations are correct but what is the connection between the 2 ??? ^^^....
He literally said it in the video that you can solve the same by factorising, but since he already used it earlier in the video, he wanted to use a differe t meathod
@@annoyingbroccoli3939 -- He should *not* have used a different method in the video, because it clashes with the others, and it is needless.
@@vincentdescharmes7897 the relation between both is the self factorization who utilizes the roots e always be products of (x-root).
@@vincentdescharmes7897 All these equations have solutions equal to 6th roots of 1. The different factorizations of the equation are linked to the different factorizations of the cyclic group Z/6Z.
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).
Yes, that's why he showed the method.
Or you can use partial sum of a geometric series formula: ∑[𝑘=0..𝑛] 𝑥ᵏ = (𝑥ⁿ⁺¹ − 1) / (𝑥 − 1) which is equal to iff the numerator is zero but the denominator is not. So, the solutions are all (𝑛+1)ᵗʰ roots of unity except the unit itself. Which is basically the same method.
Yeah i was like why he didnt use it on 4th degree equation, but glad he didnt, had forgotten the symetric method
I found that too. You always learn something in maths.
yes!
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.
1 is literally a root of unity lmao
@@ffc1a28c7 I mean to say that 1 is one of the roots of unity but is excluded from the set of solutions. The solutions are all the roots of unity except 1.
Level 1: anyone can understand this Level 2: *MAXIMUM CONFUSION UNLOCKED*
True tho
I really wonder to see how you handle different colour pens in a single hand
The shirt says algebra in arabic 🚬💀☕
whats with the emojis
@@jay8762 mood
@@aboal-fuad16yearsago55 nice
Yes Arabic
That's icy
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
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.
Now for Level 0: 1=0 Try to solve that one
If 1=0 x⁰=0 x€R
Empty set
no solution
Just call 1 or 0 a variable and you're done.
level zero would be x^0 = 1, so yeah math still works
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.
The cos theta method won't work everytime if the roots are greater than one then the method will fail to work It's a nice trick though If its given in the question that x^2 + y^2 is equal to one then you can use it as y = cos and x = sin
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 like the surprisingly simple solution for the quintic (if trig functions qualify as simple).
I barely understand this but it's still fun to learn something. I love your videos and they are so well done.
bruh
@@thuglife1219 'sup cuh
@@thuglife1219 Sup thug
@@thuglife1219 sup thug
I love how you casually hold a pokeball while solving difficult equations
There's a microphone in it
everything until x^6 is school curriculum tho
Those are not difficult at all.
🤓
@@amirnuriev9092lol no. I mean in some countries maybe, but most stop at x^3
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.
My first thought was use i for the first ones but you found another way, amazing!
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
Thank you so much, I have this question to many mathematics teachers and non was able to give me a sound explanation. Finally you did it for me. Thanks 👍🙏
Also don't forget to exclude x=1 from the solutions set as we multiplied by (x-1) and introduced the extraneous root
You are an absolute legend bro, thanks, I only have one question, for the quintic, when you multiply by (x-1) on both sides, the quintic turns into just (x^5-1)right? Because before, the partial sum for geometric series was (x^5-1)/(x-1) so why did he write x^6-1? Shouldn't it have been x^5-1? Sorry i'm still learning
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!!!
Glad you like it!
@@blackpenredpen i am not that advanced, just wanted to know how you get 1 over 4 in the quadratic part
Great video. It's interesting how the quartic was actually the harder than the quintic.
The quartic can also be solved by the method he used in the quintic, it’s just that he used different methods.
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
this is cheating
["Transcending dimensions" (Jacob's Ladder 🪜)/ 0-5th dimensions corrected & clarified] by C. M. Elmore 2D shape-name or 3D shape-name for determining 4D end-shape and 5D end-shape: (2D 'circle' vs. 3D 'sphere', in this case) 🆗️ Earth IS a 4D 'circle' (2D shape-name); a "quaternion". 🚫 Earth IS NOT a 4D 'sphere' (3D shape-name); a "hypersphere" (lit. 5D/contradictory). 🆗️ 2D 'circle' 🚫 2D 'sphere' contra. 🚫 3D 'circle' contra. 🆗️ 3D 'sphere' 🆗️ 4D 'circle' (2D shape-name) = 4D end-shape "quaternion". 🚫 4D 'sphere' (3D shape-name) = contradictory 🚫 5D 'circle' (2D shape-name) = contradictory 🆗️ 5D 'sphere' (3D shape-name) = 5D end-shape "hypersphere" 2D shape-name for determining 4D end-shape. 3D shape-name for determining 5D end-shape. "Evens and Odds", transcendentally speaking. 0D: point 🔘 1D: length/line 2D: shape-name + L/W "circle/flat earth". 3D: shape-name + L/W + depth "sphere/globe earth". 4D: shape-name + L/W/D + time(flow); current, 'contingent' (not simulated) 4D 'circle' "quaternion" earth. 5D: (H1; hypersphere) Google 'quaternion' stereographic video/images/info to see for yourself. "we're 4D?" -me, "🔫 always have been" -also me.
@@Annihillation how so? There's a reason we have a specific symbol for the golden ratio, it appears in nature and clearly math cares about it. If we are able to do u-substitutions, how is this any different? Just substitute φ for (1+√5)/2
how come
@@Annihillation no its math
2:48 Although it's obvious, you should briefly mention that x cannot be equals to 0
True
lol
I am failing to see how x can not be zero here. Care to share?
@@kruksog it says that the left part = 0, so there's only a few Xs that verify this equation. if u replace X by 0, it is not equal to 0, so the equation is not true, so its impossible. hence, X is not equal to zero
@@kruksog and by the way the number of Xs that verify this equation is equal or less than 4, because if we call the left part P, a polynom, deg(P) = 4.
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).
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.
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
You should note that 0 isn’t a root of the equation when you divide by x for transformation to be equivalent though.
it would be nice to have a drawing diagram of the 6 roots of unity to complete the nice presentation, Thank You :)
Calculus by substitution. Brilliant... I haven't had to do that for 10 years. Nice work.
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)
keep writing formulas, cause you're writing poems
I wonder how did u write that "x"
Correction: the 4th level will be a quartic equation, not a quadratic equation (as mentioned in the description)
He said quartic in the video.
@@stumbling but he said quadratic in the description, which may not be a problem for you since you don't read them, but it still needs to be fixed
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.
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
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 KNOW RIGHT
if we have these types of maths teachers nobody hates maths 🙏❤ LOVE FROM INDIA
The method used for que quintic equation can be used for all type from grade 1 to infinite.
Thanks for this good results 😊
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 🔥
Unlike quadratic, cubic, and quartic polynomials, the general quintic cannot be solved algebraically in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions, as rigorously demonstrated by Abel (Abel's impossibility theorem) and Galois.
I like solving x^(1/6)=1 graphically, much easier and more pleasing to eye
I like the way you handle two markers at the same time
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".
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
How old r u
@Clash Royale Montages true that
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.
the last equation was complicated for no reason
@@RithwikVadul what are you talking about? The level of difficulty is actually pretty much the same. The solution for 1+x+...+x^n = 0 is simply x = e^{2πk/(n+1)}, with k going from 1 to n. That's it. You people were rickrolled.
🤔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.
Wait, so can we generalize the last method to work for polynomial equations of any order?
Yes
@@blackpenredpen So, for 4-th power roots can be e^(i*Pi*N*2/5) where N=1,2,3,4 ? Does that mean that you just skip a proof that 5-th equation hasn't real roots and just solve in complex?
For the last one we can also just factorise by x^3 the 3 first therms and then everything goes by itself
Just asking: Do you plan on doing additional mathematics for O'Levels?
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
Please reply
Should be yes
And x³ + 1 can be factorized as (x + 1)(x² - x + 1)
Excellent, where omega is one of the complex cube root of unity.
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)
The solution for the 4th eqn is so cool!
Bro, you made me fall in love with maths again 😁
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
🐵🤓🐵🤓
Is this taught in some weird way somewhere? As far as I would've thought, until I saw it being mentioned this way multiple times throughout the comment section, referring to an understanding of polar form complex numbers as "grasp[ing] the Euler identity" itself shows a lack of understanding. This video has nothing to do with the Euler identity. It's like watching a random video on geometry and being like "oh yeah Pythagoras, got it."
@@alansmithee419 What are you talking about? e^πi = -1 is *exactly* what this video utilizes
@@fanamatakecick97 The polar form is not defined based on Euler's identity, rather Euler's identity is a single case of the more general polar form for values at unity. Perhaps I should revise my original comparison. It's more like being asked if you know what prime numbers are and replying "yes, I know what 2 is" as if that somehow covers everything. It's an example, yes, but it doesn't define the set.
@@alansmithee419 That’s part of grasping the Euler identity
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.
You use the complex square roots. Anyway, the approach used for the 5th order version can be used for all other orders, and is much simpler.
Wonderful video! 👍👍
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
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.
when he divides by x^2, in the equation, 0 can't be a solution because 0^3 + 0^2 + 0 +1 is different from 0 so 0 can't be a solution.
No. Just ... no.
Love your arabic for Algebra shirt
Thanks.
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).
5:24 lets appreciate that cut there
Haha!
E=MC^2
we don't do that here.
Einstein spotted
physics
applied mathematics, **shudders**
الجبر
in the quintic equation- one solution for x can also be -1, idk how to do the method but u can try out that value it works out as 0
He just presented -1 as e^(i*pi), note that e^(i*pi) = -1
the way he switch the marker is so smooth
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
Sure, but your method is way more intuitive. Nice job!
Trouble is then you're taking square roots of the solutions for t, e.g. sqrt(-1/2+isqrt(3)/2) which is unpleasant to say the least. Would also point out that Euler's formula does not appear in this video. Polar representation of complex numbers is not unique to Euler's formula.
he mentions this at 7:16
@@alansmithee419 Well yeah, but e^(ipi/3) is the cube root of -1 which is -1, in that way you do use Euler's formula.
you can also use the method he used for the quintic equation to solve every equation in this video
What’s about 1+x+x^2+x^3+x^4+x^5+…=0
I don’t think it ever = 0. Oddly enough.
The LHS is 1/(1-x), which we know using the geometric series formula. So 1/(1-x) = 0, which has no solution because if u multiply 1-x on both sides you get 1=0
I tried the method used in level 5 but with a limit. It doesnt work because the variable im letting approach infinity will only allow integers to plug in. Another problem is I falsely showed 1-1+1-1…=0 by accident. Sorry if i explained poor
@@Ninja20704 I agree with u but 1+x+x^2…=1/(1-x) , if |x|
@@schizoframia4874 I mean, if |x|>=1 it doesn't converge, so it's definitely not going to be able to answer an equation.
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
The first precious knowledge I earned... From your educative channel... Is... 2:40 - 7:00 Quartic = Quadratic
N Roots of unity, other than unity...
Yep, with this observation the video would have lasted less than a minute.
Try this EXTREME quintic equation 👉 kzhead.info/sun/etOAo7qEpZhuhaM/bejne.html
Excuse me sir can you solve my doubt if we write x =y as x/y =1ans if we put the value of x =0 then y =0 but if we put this value of x and y in x/y =1 then we get 0/0 =1 how is it possible. Sir it's my humble request to you that please solve my doubt please please please
@@historywallah--princeshukl619 x/y=1 only for y≠0
@@Noname-67 can you send your number please
Makes sense
We need part 2!
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.
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.
you can show that 2/(1+sqrt(5))=(sqrt(5)-1)/2 by multiplying the numerator and denominator by sqrt(5)-1. Additionally, since the golden ratio is a solution to the equation: x^2-x-1=0, that means φ^2-φ-1=0, or φ^2=φ+1 If you divide both sides by φ and simplify you get 1/φ=φ-1 =(1+sqrt(5))/2-1 =(-1+sqrt(5))/2
Just rationalise the denominator after taking the reciprocal and you’ll get it
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
He said he didnt want to use that method because it would be similar the the cubic eqn earlier.
You're basically attempting to write the n-th roots of unity in radical form (which is possible for any root of unity).
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
Yes you are this is basic stuff
Pretty much yea
13 and stuck on 4, I think so
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
Hi Fred! Thanks for your thoughtful comment, just like always! I mentioned that we could do the quintic by grouping but we did that already for the cubic. And I just wanted to make the last one "awesome", that's why I did it that way! : )
@@blackpenredpen Yes, I agree that was a good choice. And as I said, by doing that, you actually answered an infinite number of problems in one solution! ;-) And it's good practice to use that strategy of variety in your "lesson plan" for the video, to maximize the use of screen time. Fred
perfect and lovely. thanks
Somewhat falling into golden ratios got introduced. In algebra class we only solved polynomials by factoring and never went pass polynomials to the 4th degree. How or what class do you take to learn to manipulate the values so well?
Find vertical asymptotes of tan(sinx). Of course if you can do it.
Nice. Using the formula for the sum of a geometric sequence can be used for all of these as X is never equal to 1. Then the solutions are the roots of unity except for 1.
This video is very satisfying to watch
Very nice!
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
Thank you master
Wow I love the way he is swapping red and black on the board 😍
I remember solving a quintic equation. This is actually a nice way to solve that equation.
not all quintic can be solved
@@leeyc0 true but the quintic i solved was solvable
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.
for the last equation, you can consider it to be sum of a GP and apply the formula for sum of a GP to get the resultant equation as (x^6-1)/x-1=0 where x is not equal to 1.
Level 4 was so cool!
For the second one I like the pq-formula (1/2*p +/- √(1/2*p)² - q)
Such a convoluted way to solve a quadratic equation
Fun Fact: (Sum of (x^n ) where n starts from 1,2..) - n will always have (x-1) as it's factor.
I love how he changes pens so smoothly
Funnily enough, I just did an animated 3D graph on Desmos for that exact quadratic, with x as the real part of x, y as the real part of y, z as the imaginary part of y and animating "a" between -1 and 1 as the imaginary part of x. I found the y-intercept at x=-.5,a=sqrt(3)/2 and x=-.5,a=-sqrt(3)/2, as expected.
That pokeball is adorable and OMG I JUST SAW KIRBTY!! :D
I love that Kirby plush I have one that's exactly the same