The roots that Dr Peyam found are on the unit circle. In the complex plane, taking the nth power of such a number means spinning it around the unit circle until its angle is n times as large as it was at first, but not changing the radius (from the origin). So if we take one of these roots -- say, the "positive" one in the first quadrant, and raise it to the 2020 power, it spins around the circle hundreds of times (n × pi/3 radians), then stops in the third quadrant, specifically at -1/2 - i(sqrt(3))/2. To add 1 means moving 1 unit to the right. Do that to our point in the third quadrant, and it reaches 1/2 - i(sqrt(3))/2, which is also reached by spinning the point again (i.e. raising the root to the 2021 power). Thank you Dr Peyam!
Thanks Dr Peyam! I looked for real roots... a sketch graph shows there is only one and it will be a little greater than 1.. put x=y+1, rearrange for y*(1+y)^2020=1. Put y=a/2020, (1+y)^2020 ~ exp(a), a*exp(a) ~ 2020, a ~ W(2020), x~1+W(2020)/2020 ~ 1.00289. This is accurate to about the last decimal place. (W is the Lambert W function)
There is a paper over polynomials the form x^n+-x-1 . Note that if x^n-x^(n-1)-1=0 then (1/x)^n+1/x-1. The paper proves that the unique possible irreducible factor is the form x^2-x+1. Also we know the polynomials the form x^n-x-1 are irreducible and its Galois grup is S_n. That is, the roots this polynomials are the most possible complicated. I think the same over the other factor than x^2-x+1.
Happy New year to you sir. May you bring interesting math videos in coming year and bring smiles on those faces who fears the math most 🎉🎉 again happy new year
Nice video - a cool and unexpected solution! I think that when you said that x^2021 - x^2020 - 1 = 0 implies x²-x+1=0, you meant it the other way round: the solutions to x²-x+1=0 will also be solutions to x^2021 - x^2020 - 1 = 0.
Very neat interpretation of an otherwise rather boring equation. Btw: His t-shirts' frontal imprint "Körper" literally means "body". Pretty hilarious. What a great guy!
@@FromTheMountain Although the back has "ab=ba", whereas if fields are called bodies, the term may include non-commutative division rings such as quaternions. Or maybe that's a dated use of the word.
@@FromTheMountain Let's ruin Peyam's nice tongue-in-cheek joke by explaining it. It's the other way around: Dedekind introduced "Körper" which was (badly, ihmo) translated as "field". The Dutch translation of Dedekind's "Körper" is perfect: "Lichaam" (if wiki is right). And Peyam's joke is that his natural(*) body (i.e. "Körper") sticks in a T-shirt saying "Körper" ("Lichaam") on it's front, and shows the mathematical "Körper"'s axioms on the back. As a native speaker Peyam was most certainly playing this joke intentiously. Unfortunately, this specific joke is dead, now (i.e. became a "Leichnam") Anyway: Happy New Year to all.
Viete's indicates both the sum and product of all the zeros is 1. The sum and product of the two "famous zeros" you found happens to be 1, but we know there is at least one real zero since degree is odd(2021). BlackPenRedPen said is was a little more than 1. So the hidden zeros in your "JUNK" factor probably account for that :) The modulus x^2-x+1 you used was very kewl!! It is a cyclotomic I think. Bigger question I have is, if f(x) congruent g(x) (mod( x^2-x+1)), will f and g have the same zeros or what is the relationship? Also, the subtleties between equality and congruence can be dizzying at times. Happy new year and thanks for the wonderful math experience you create here!
This all came a bit out of the blue for me... is there any reason why we might guess that X^2 - X + 1 divides the original polynomial? Because if you can't motivate that, you might as well just start plugging in random numbers into the equation and hope you find a root by pure luck.
That comes from the difference of cubes formula. There’s methods for know how to factor a difference and sum of cubes. Here they are a³+b³=(a+b)(a²−ab+b²) and a³−b³=(a−b)(a²+ab+b²) The way I remember the signs is SOAP (Same, Opposite, Always Positive). In general, aˣ−bˣ always have a-b as a factor for any positive integer x, and you can find what’s left over using polynomial long division or synthetic division. But if I’m not mistaken, the left over is always like a^(x-1)+a^(x-2)•b+a^(x-3)•b^2+...+a^(x-k)•b^(k-1)+...+a^2•b^(x-3)+a•b^(x-2)+b^(x-1). A good way to see this is true is do the exercise where you take a-b and multiply to polynomials of the above stated form, of larger and larger x, and see how the cancellation works. This is why peyam wrote it as x³−(-1), so he could do the a-b since that holds generally for larger and larger x. The a+b one, however, does not. It becomes clear why when you multiply and see the cancellation as I said.
I understand your concern. x^3+1=(x+1)(x^2-x+1) implies x^3 congruent -1 (mod x^2-x+1) which is useful as a congruence reducing aid. As I understood it, he just showed that original polynomial was congruent to x^2-x+1 which apparently means any zero of x^2-x+1 is a zero of original. Not certain.
New Year, new channel name? :> I like it :) All the best in the New Year everyone! (And let's hope this one doesn't follow in his older brother footsteps)
If x^3 = -1, the original equation can be simplified as x²-x+1=0. So, it is needed that at least 2 roots of x^3 +1 =0 can satisfy x²-x+1=0. Luckily, x²-x+1 is a modulus of x^3 +1. An analytical way is to set (x^3)^673 = a, and it can be determined that a = -1.
"Körper" means "Cuerpo" in español, that in English is "Body" but "Body" is not a mathematical concept in English. Strange. Why you people does not just say "Body" to yours Fields? Anyways nice T-shirt. Happy new year Dr Peyam. Cheeers
X is the Nth root of e, where N is from the equality 2022 = N • ln[N]. An definite approximation would be, X ~ the 2022th root of ( 2022/ln2022 ) I’ll send you the steps through Twitter.
Huh was news to me but we'll explained as usual 🤪👋🇦🇺🐨 happy NY Doc it's 1am NY right now...you still 8.30am another 15 hours to go Go out n get drunk ton tonight ! haha
When you say -1 is not a solution of the original equation, which one are you referring to? The 2021 and 2020 one? Because if you mean the x³+1 one then -1 most definitely is a solution
Hi bro! Couldn´t you have gotten the same result with just about any numbers!!!!!!?????? Let´s say X to the 50 and X to the 49. Please think about it!!!!!!!!!!
Happy new year!! I thought the eq was obvious but no…. 😆
same lol, tried solving it but didn't get close lmao
Happy new year!!! 🥳🥳🥳
😁😁
@@drpeyam happy new year I hope you will be safe from everyone you are a good teacher 😉❤❤❤❤❤
X is just 1 isn't it
So I came back because I noticed the only real solution for that equation is about 1.0029, while the abs value of the sol u got is 1! WOW
Wow so cool!!
Dr. Peyam announces two eagerly-awaited awards for the year 2021:
● Theorem of the year: *-1 ≡ -1*
● Roots of the year: *(1 ± i√3)/2*
The roots that Dr Peyam found are on the unit circle.
In the complex plane, taking the nth power of such a number means spinning it around the unit circle until its angle is n times as large as it was at first, but not changing the radius (from the origin). So if we take one of these roots -- say, the "positive" one in the first quadrant, and raise it to the 2020 power, it spins around the circle hundreds of times (n × pi/3 radians), then stops in the third quadrant, specifically at
-1/2 - i(sqrt(3))/2.
To add 1 means moving 1 unit to the right. Do that to our point in the third quadrant, and it reaches
1/2 - i(sqrt(3))/2,
which is also reached by spinning the point again (i.e. raising the root to the 2021 power).
Thank you Dr Peyam!
Thanks Dr Peyam! I looked for real roots... a sketch graph shows there is only one and it will be a little greater than 1.. put x=y+1, rearrange for y*(1+y)^2020=1. Put y=a/2020, (1+y)^2020 ~ exp(a), a*exp(a) ~ 2020, a ~ W(2020), x~1+W(2020)/2020 ~ 1.00289. This is accurate to about the last decimal place. (W is the Lambert W function)
Thanks so much!!
1+ 1/(167.4) = 1.00597
OMG! Thanks for a stupefying finish to the year, Dr. Wizard.
Dr Wizard, I love that hahaha
Always love the complex roots of unity. Very satisfying.
There is a paper over polynomials the form x^n+-x-1 . Note that if x^n-x^(n-1)-1=0 then (1/x)^n+1/x-1. The paper proves that the unique possible irreducible factor is the form x^2-x+1. Also we know the polynomials the form x^n-x-1 are irreducible and its Galois grup is S_n. That is, the roots this polynomials are the most possible complicated. I think the same over the other factor than x^2-x+1.
I think I need to watch this a couple of more times.
Happy New year to you sir. May you bring interesting math videos in coming year and bring smiles on those faces who fears the math most 🎉🎉 again happy new year
Nice video - a cool and unexpected solution! I think that when you said that x^2021 - x^2020 - 1 = 0 implies x²-x+1=0, you meant it the other way round: the solutions to x²-x+1=0 will also be solutions to x^2021 - x^2020 - 1 = 0.
Right, that’s what I meant
Happy new year to you too! And thanks for the great content!
Very neat interpretation of an otherwise rather boring equation.
Btw: His t-shirts' frontal imprint "Körper" literally means "body". Pretty hilarious.
What a great guy!
It is also the German word for the mathematical concept "field" (note how the back of the shirt has the axioms of a field written on it).
@@FromTheMountain Although the back has "ab=ba", whereas if fields are called bodies, the term may include non-commutative division rings such as quaternions. Or maybe that's a dated use of the word.
@@FromTheMountain Let's ruin Peyam's nice tongue-in-cheek joke by explaining it. It's the other way around: Dedekind introduced "Körper" which was (badly, ihmo) translated as "field". The Dutch translation of Dedekind's "Körper" is perfect: "Lichaam" (if wiki is right). And Peyam's joke is that his natural(*) body (i.e. "Körper") sticks in a T-shirt saying "Körper" ("Lichaam") on it's front, and shows the mathematical "Körper"'s axioms on the back.
As a native speaker Peyam was most certainly playing this joke intentiously.
Unfortunately, this specific joke is dead, now (i.e. became a "Leichnam")
Anyway: Happy New Year to all.
Ah, yes.... some good old complex solutions for this brand new 2022 :) Happy new year, Dr. Peyam and thanks for all the funthematics! :D
A very very happy new year sir,
may the new year open new wonders of the universe especially in math🥰
Viete's indicates both the sum and product of all the zeros is 1. The sum and product of the two "famous zeros" you found happens to be 1, but we know there is at least one real zero since degree is odd(2021). BlackPenRedPen said is was a little more than 1. So the hidden zeros in your "JUNK" factor probably account for that :)
The modulus x^2-x+1 you used was very kewl!! It is a cyclotomic I think. Bigger question I have is, if f(x) congruent g(x) (mod( x^2-x+1)), will f and g have the same zeros or what is the relationship?
Also, the subtleties between equality and congruence can be dizzying at times. Happy new year and thanks for the wonderful math experience you create here!
Happy new year, love your equations
Thank you and all the best in New Year!
Thank you for fantastic equation!
Happy new year😊
awesome way to end the year! thanks Dr. Peyam
Thank you and happy new year!
Happy New Year and thank you for this videoooo
Ingenious!
04:50 "Well .. if you set this equal to zero, this in particular implies x^2 - x + 1 = 0 .. " No. That is in fact backwards.
"-1 =-1 theorem of the year "
seriously the best dialogue i heard in new year till now i laughed for straight 20 min
Happy New Year!
OMG the Yamaha firework sound effect XD
Happy new year in 2022.
Happy new year!
This was so much fun!😀
Happy new year :)
you make me happy, that's pretty much all i came for :)
Dat math-körp though
Happy new year dr peyam 🎉🎉❤️
Using Bolzano's theorem I noticed there must be a real root between 1 and 2!
Indeed!
@@ludo-ge9fb that’s not obvious to some people
Was amazing thank you
2022 is 6*337. 337 is the next prime after 333. That is an interesting property. And 6 is precisely the sum of its digits.
Cool, thanks. Assuming (x^3)^673 = a, it can be determined that a = -1.
Happy new year 🎊
*Happy New Year Boss* ❤💜❤
This all came a bit out of the blue for me... is there any reason why we might guess that X^2 - X + 1 divides the original polynomial? Because if you can't motivate that, you might as well just start plugging in random numbers into the equation and hope you find a root by pure luck.
I use "Guess and check" all the time for root finding. You?
😉
That comes from the difference of cubes formula. There’s methods for know how to factor a difference and sum of cubes. Here they are
a³+b³=(a+b)(a²−ab+b²) and
a³−b³=(a−b)(a²+ab+b²)
The way I remember the signs is SOAP (Same, Opposite, Always Positive). In general, aˣ−bˣ always have a-b as a factor for any positive integer x, and you can find what’s left over using polynomial long division or synthetic division.
But if I’m not mistaken, the left over is always like a^(x-1)+a^(x-2)•b+a^(x-3)•b^2+...+a^(x-k)•b^(k-1)+...+a^2•b^(x-3)+a•b^(x-2)+b^(x-1).
A good way to see this is true is do the exercise where you take a-b and multiply to polynomials of the above stated form, of larger and larger x, and see how the cancellation works. This is why peyam wrote it as x³−(-1), so he could do the a-b since that holds generally for larger and larger x. The a+b one, however, does not. It becomes clear why when you multiply and see the cancellation as I said.
I understand your concern. x^3+1=(x+1)(x^2-x+1) implies x^3 congruent -1 (mod x^2-x+1) which is useful as a congruence reducing aid. As I understood it, he just showed that original polynomial was congruent to x^2-x+1 which apparently means any zero of x^2-x+1 is a zero of original. Not certain.
You can work out the reasoning using complex numbers.
New Year, new channel name? :> I like it :)
All the best in the New Year everyone!
(And let's hope this one doesn't follow in his older brother footsteps)
Thank you!!!
Génial ! 🥳 🇫🇷
Mercii
We could provide an analytic expression for the other roots with the Lagrange inversion theorem, although it would be kinda messy.
So COOL!
That's nice but what about the junk ?
Now, I want to know the solution of x^2022=x^2021+1
Thx sir
Happy New Year! Were those added SFX fireworks or actual fireworks going off outside the room you were filming in?
SFX fireworks hahaha
was that firework at the beginning from a piano? I recognise that exact sound.
Yeah. It was some fsx sound
Correction - real root x= 1+(1/345.15) or so.
Happy new year 🥳🥳🥳🥳🥳🥳
ohh,my brain
Wait a minute, isn't it 2022 but not 2020
Why can't there be some roots in the "JUNK" ?? I mean, the commutativity implies the X² - X + 1 equals 0 OR the junk equals 0... Thanks !
Of course there can! Junk gives you the other 2019 roots
@@drpeyam Thanks for your fast answer :D so, maybe a video about how to find the last 2019 roots for 2022 ? 😉
@@drpeyam "the other 2019 roots" I laughed at this point :D
How do you know that "JUNK" is not equal to zero? If it was, then the implication that x²-x+1 does not hold.
The opposite implication obviously does hold though. But there's (according to desmos and wolframalpha) one real root of approximately 1.0029
Of course, I never said I found all the roots
@@drpeyam ohh okay my bad
I don't understand the step at 2:49 why do we just assume x^3 = -1 ?
This starts from the assumption that x⁴+1=0, in other words trying to find the roots of the polynomial
If x^3 = -1, the original equation can be simplified as x²-x+1=0. So, it is needed that at least 2 roots of x^3 +1 =0 can satisfy x²-x+1=0. Luckily, x²-x+1 is a modulus of x^3 +1. An analytical way is to set (x^3)^673 = a, and it can be determined that a = -1.
Can you make a video of Quotientenraum and Homomorphiesatz ?
Why did u assume x3=-1?
I kinda wanna find every root now
"Körper" means "Cuerpo" in español, that in English is "Body" but "Body" is not a mathematical concept in English. Strange. Why you people does not just say "Body" to yours Fields? Anyways nice T-shirt.
Happy new year Dr Peyam. Cheeers
X=1.0029
A real root is approx x=1+ 1/167.4
Why not 2021 and 2022?
And the other 2019 solutions?!
Stuck in 2019
There are 2021 roots since complex one counts
I want to start 2022 with an integral
Wow, I didn't know you speak German! Frohes Neues!
Dankeschön!!!
Today I leant that X^2021 = -X^2
Phương trình này có vấn đề đó. Cảm ơn nhé.
X is the Nth root of e, where N is from the equality 2022 = N • ln[N]. An definite approximation would be, X ~ the 2022th root of ( 2022/ln2022 )
I’ll send you the steps through Twitter.
Huh was news to me but we'll explained as usual 🤪👋🇦🇺🐨
happy NY Doc it's 1am NY right now...you still 8.30am another 15 hours to go
Go out n get drunk ton tonight ! haha
Thank youuuu, will do 🍷
why does x^3 + 1 = 0 mod x^2 - x + 1? i thought it should equal x + 1
anyway, happy new year :D
なるほど!
My TI-86 calculator gives answer: x= 1.0028972542072
and HP Prime gives answer: x= 1.00289725421
The doctor forgot to check the answer. Plug in exp(i.pi/3) into the x^2021-x^2020-1=0 equation.
Left as an exercise to the viewer
Happy New Year's message, may the Creator bless you (whatever you might believe in) 😊
Why x^2 - x+1=0
When you say -1 is not a solution of the original equation, which one are you referring to? The 2021 and 2020 one? Because if you mean the x³+1 one then -1 most definitely is a solution
The original one
What gives you the right to replace X by minus 1!?
Where did that happen?
modular arithmetic
Awwww..... Where's the -1/12 gone??
Love the shirt
Got lost halfway
you are missing 2019 roots
That was so 2019 😉
many people miss about 2019 year
Hello Dr. Peyam. Nice T - shirt. I want that shirt. Where can I buy?
It’s in my Teespring store, see description
Hi bro! Couldn´t you have gotten the same result with just about any numbers!!!!!!?????? Let´s say X to the 50 and X to the 49. Please think about it!!!!!!!!!!
🌸
What a jank solution.
I did not understand it.....
Wait what
2 sol down 2019 sol to go.
طب ماشي ياعم شكرا
I understood nothing 😁
Happy New Year :)
Happy new year!!
You made this video last year: th-cam.com/video/G4xbwpG2zaE/w-d-xo.html
You could tell us special properties of the number 2022 as well.
Sadly 2022 is not that interesting but there will be a fun 2022 video coming :)
gained
Why are you Guy
I don’t think I had a say in this 😂
Junk lol
Thanks a lot! Happy New year!
Happy new year!