I just did one of my end of year exams for year 12 and this video is the reason I was able to correctly answer probably the hardest question on the whole exam! Thank you so much for your videos, please keep making them. The question required finding the root of a cubic and I always forget how to do polynomial long division but I found 1 was a root by just trying it and then solved the cubic by factoring to (x-1)(a^x+bx+c) then expanding again and solving for a,b and c just like you did in this video
At 1:03 he said synthetic division so i left the video and spent 3h researching synthetic divisipn and everything else i needed to understand it, and here i am again. Coincidentally now it's 4 am and i have school at 7:45 and im probably gonna kms but idk yet.
Simpler Solution - Use n'th root of unity x^5 = 1 = e^(i 360° n) so, x = e^(i n 360° / 5) = e ^ (i n 72°) for n = 0, 1,2, 3,4 leading to 1 (the only real root) and 2 complex conjugate roots with angles ±72° and ±144° So, *the 5 roots are* 1, cos 72° ± i sin 72° and cos 144° ± i sin 144° (Use cos ±72° = sin 18° = (√5-1)/4) and (cos ±144° = -cos 36° = -(√5+1)/4) and similarly the sine counter-parts to express the solution without any un-computed trigonometric functions)
Also simpler because you skipped an essential part of the solution. Why is sin 18° equal to (√5-1)/4 ? To make your solution complete you have to make that part of the solution. And also your statement 1 = e^(i 360° n) is not correct. It has to be 1=e^(i2𝜋*n). The exponent has to be a number and not something with degrees in it.
@henkhu100 can't be bothered to read either the solution or the criticism. The solutions are 1/5 rotations around the complex plain starting at 1. That is what was intended to be communicated
@@governmentis-watching3303 Ok, but that does not mean that what he writes may be wrong. An expression like e^(360 degrees) has no meaning at all. If we write e^(2*pi) we have a real number for the exponent. We don't write e^(2*pi radians) because that does not have any meaning. Just like we can write √ 360 but not √(360 degrees). So my comment was: if you solve a problem use correct formulations and write things that have a menaning, so don't use expressions like e^(some degrees).
You can solve this either algebraically or geometrically. For the algebraic solution, just factor out x - 1 and solve the resulting 4th degree polynomial: x⁵ - 1 = 0 (x - 1) (x⁴ + x³ + x² + x + 1) = 0 etc. For the geometrical solution, just use Euler's formula: xₙ = ⁵√1 ⋅ [cos(n ⋅ 360° / 5) + i ⋅ sin(n ⋅ 360° / 5)] n = 1→5 So, basically: x₁ = cos(72°) + i ⋅ sin(72°) x₂ = cos(144°) + i ⋅ sin(144°) x₃ = cos(216°) + i ⋅ sin(216°) x₄ = cos(288°) + i ⋅ sin(288°) x₅ = cos(360°) + i ⋅ sin(360°)
One of the best videos I have seen. Excellent mastery of Ceva versus Delanges math when you said that it didn't matter because they were symmetric. Ambiguities and dualities were combined at that point in the calculations.
The golden ratio emerges here, sometimes pi, sometimes e^x, etc.... show up. When I look back, it is those surprising or unexpected little nuggets we find that got me hooked on mathematics early on.
An easier way at 2:58 if u divide by x² then u will get a equation like x² + 1/x² + x + 1/x + 1 . Then u can assume x + 1/x as y so new equation will be y² + y - 1 and then u can solve it from there
Please explain how if (x + (1/x)) = Y how did you get y² +y -1 To me it doesn't make sense, its like you combined the x+1/x or something. Waiting for explanation please. 😊
@@AbdullaAlQemzi1 ok so heres detailed explanation take "1/x + x" as y now y² = 1/x²+ x² +2 y² - 2 = 1/x² + x²( Subtract 2 from both sides) so the new eqn becomes y² - y -2 + 1 = 0 y² - y - 1 = 0 if u still dint get it dont hesitate to ask again
@@OgSuperHeroWoah.😮 Thanks bro I understood. If you don't mind can you explain how to move on from your equation to find solutions. Do you use euler theory ? And why not use Z⁵=1 from the beginning and use euler on that?
Exactly! And phi^2 could be replaced by phi+1, to make the roots look a bit simpler... And you don't need to put b=-1/a and replace and multiply by a to get the quadratic equation, because you have the sum S and the product P of a and b, and by theorem of Vieta, a and b ar the solutions to y^2 - Sy + P = 0
9:30 You can not only figure out sin(72) but also the sin and cos value of 72, 144, 216, 288. Those angles which you can raise to the fifth power to rotate back to 1 on the complex plane.
Thanks for that. When he said he wasn't going to use complex/ polar, I was disappointed. Rotations in the complex plane was the first thing I thought of to find all five solutions.
@@mikefochtman7164 Yes, I did not want to point that out, since it was an exercise to solve it this way. Just reflected that real and imaginary parts give us all the other values as well.
I was hoping for five values for x. Thanks for continuing these videos, it's been ages and I just came back to playing with math. You aged a little, but that makes it all the more beautiful ❤ Much love from here! ❤️ ❤
Since phi^2=phi + 1, it follows that phi^2 - 4=phi - 3, so the first two solutions can be presented as [phi+i*(3-phi)^1/2]/2 and [phi-(phi-3)^1/2]/2. For the other two, consider that -1 - 4*phi^2=-1 - 4*phi - 4=-5 - 4*phi. So we can present these solutions as [1+i*(5+4*phi)^1/2]/(2*phi) and [1-i*(5+4*phi)^1/2]/(2*phi).
very nice ! my first thought was to take 1 as e^(i2pi*k) with k any integer, so the fifth root would be e^(i2pi*k/5) giving us the five different solutions of e^0=1, e^(i2pi/5), e^(i4pi/5), e^(i6pi/5), e^(i8pi/5)
I figured as much. In the complex plane, the results of a given power (example: 4) has four results, and all of them are in the complex plane. And if you connect the results' coordinates, you'll notice the bigger the power is, the results will form a circle.
x is 1 the other 4 solutions (since its a power of 5 equation is has 5 solutions) are all equally spread around the complex plane (72 degrees apart) with a size 1, use trigonometry to find out the real and imaginary componants of each
Having 5 solutions, we can take 1, than we take the non arytmetic solution (with the module of a complex and its angle), select a numbrr with 1 as the module and with either 1/5 of 2π, or 2/5 , 3/5 and 4/5 of 2π as the angle Than just take cos and sin and you have your 5 solutions (And sorry for my bad english(
Wow. I can't remember all those tricks to factor the polynomial. I'd go another route: the solution would be any complex number z which modulus is 1 and the 5*arg(z) mod 2pi = 0. Then you may figure the arguments as 0, 2*pi/5, 4*pi/5, 6*pi/5 and 8*pi/5.
What an amazing video bprp! I am only an 8th grader but ever since I saw your calculus videos, I have been engrossed with your channel. Is it ok if you can make a video with a question featuring an improper integral, a derivative, a series and a limit all in one question where it is aimed at Calculus 1 and 2 students. Thank you!
Mmm if you bodice with any of these polynomials of the form x^n-1=0, 1 is always a solution. To get the remaining polynomial you find need to do synthetic division! Notice that x^n-1/x-1 is the sum of a geometric series with common ratio of x upto the n-1th term
An easier way to factor it is to use geometric series. (x^5 - 1) / (x -1) is a geometric series with first term 1, 5 terms, and ratio x. So 1 + x + x² + x³ + x⁴
es una ecuacion de grado 5, por lo tanto, siempre generaria 5 respuestas correctas, las cuales se pueden apreciar en el video, el numero de respuestas siempreesta indicado por el indice de una variable, por ejemplo, x'2=5x, x=5 y x=0
I love how he says "now, this is the part where we have to think a little bit" So,what in the math have we been doing the whole time???? We did all this without thinking??? 😮😂
You know that thinks fun 1 through 10 and then we show the expression upon chemistry. What won through is in standard in chemical makeup, Constance and biological result with identity of 1 whole 2 maximum hole and cube effect in unstable acceleration, as well as I can't remember what is called in division but when it. Divides an acts across the axes and catalysts. I think that's fun. You should do that 1 'cause that 1's fun. Yeah, one is an expression, is it an? Identity, and when one is expressed, it may not have a hole that accepts 10 as a whole of interval. The interval set makes extend to the 10th place before or after whole one equals in holes. It's weird, but it's our universe and it's how it actually works.
2:33 “any polynomial can be factored in terms of linear and quadratic” Wouldn’t that mean any polynomial would have solutions expressible in terms of linear terms and square roots (possible complex roots)? That’s clearly not possible. x^3-2=0 for starters.
Why not use Eulers Identity? e^iπ = -1? Therefore: e^i2kπ = 1, x^5 = e^i2kπ, etc. In the end, it is a circle in the complex plane with roots at: (0°), (72°), (144°), (216°), and (288°).
At the very beginning he said he didn't want to use that method and wanted to show it in the more difficult algebraic way which is juuuust barely possible.
7:07 этот типок лепит детские ошибки: если b=-(1/a), то подставив а из строки выше, мы получим, что b=-(2/(1±sqrt5)). ну и фи получается равно 0, ибо два его корня должны быть равны 0, ибо только 0 в любой степени дает 1. что это за начинающий математик?
To solve the equation \( X^5 = 1 \) where \( X eq 1 \), we are looking for the fifth roots of unity. The solutions to this equation can be expressed in the form: \[ X = e^{2\pi i k / 5} \] where \( k \) is an integer. The fifth roots of unity are: 1. \( k = 0 \): \( X = e^{2\pi i \cdot 0 / 5} = 1 \) 2. \( k = 1 \): \( X = e^{2\pi i / 5} \) 3. \( k = 2 \): \( X = e^{4\pi i / 5} \) 4. \( k = 3 \): \( X = e^{6\pi i / 5} \) 5. \( k = 4 \): \( X = e^{8\pi i / 5} \) Since we want the solutions where \( X eq 1 \), the valid solutions are: 1. \( X = e^{2\pi i / 5} \) 2. \( X = e^{4\pi i / 5} \) 3. \( X = e^{6\pi i / 5} \) 4. \( X = e^{8\pi i / 5} \) In summary, the solutions to \( X^5 = 1 \) with \( X eq 1 \) are: - \( e^{2\pi i / 5} \) - \( e^{4\pi i / 5} \) - \( e^{6\pi i / 5} \) - \( e^{8\pi i / 5} \)
And that was the only real root in the given solution set. What exactly is the problem here? Edit: Hey all, I will also take the opportunity to edit my comment. The original comment used to say that the video was straight-up wrong, but after my reply, this statement has been removed.
@@isavenewspapers8890 I really thought you commented long after I edited it. But if not, then youre right👍 because I did first comment implying/claiming there was a problem.
No. The quartic is the end-of-the-line, when it comes to a closed-form formula in elementary functions as the master key to solve any polynomial of that degree. There are special case quintics where there exists a formula, but Galois proved there can be no such general formula for quintics, or anything beyond.
@@carultch OK, I don't know anything about Galois, but I asked about quartics, not quintics. When I write out the general equation x^4+ax^3+bx^2+cx+f I can use 4 variables, p, q, d and e in the form (x^2+px+d)*(x^2+qx+e) and that gives me de = f p + q = a pq+d+e = b pe+dq = c These reduce to the values in this video where d = 1 and e = 1. That should be solvable, right? It might not be trivial to solve, but there are 4 equations/4 unknowns.
@@flowingafterglow629 Sorry, I misunderstood your question. I had my mind on quintics, since his original video was about quintics. The quartic in general is solvable. Ferrari, who worked with Cardano, the namesake of the cubic formula, derived the general quartic formula, using many of the same principles as deriving the cubic formula. You're method is legit, however you will ultimately need to solve a quartic anyway, to solve that system of four equations. So it doesn't really make any progress.
Given: sqrt(2*x^2) - sqrt(2*x) = 3 Extract all coefficients from square root terms, and shuffle to the left: sqrt(2)*sqrt(x^2) - sqrt(2)*sqrt(x) - 3 = 0 Divide thru by sqrt(2): sqrt(x^2) - sqrt(x) - 3/sqrt(2) = 0 Reverse the nesting order of the first term: sqrt(x^2) = sqrt(x)^2 Let u = sqrt(x), and rewrite in the u-world. u^2 - u - 3/sqrt(2) = 0 Now we can solve for u with a quadratic formula: u = (1 +/- sqrt(1 - 4*-3/sqrt(2)))/2 Simplify: u = 1/2 +/- sqrt(1 + 6*sqrt(2))/2 Since u = sqrt(x), this means u can only be a positive number, by definition of the square root function. So we can rule out the lower solution to the quadratic formula as an extraneous solution. Upon evaluating it, you'll see that u is approximately -1.04 and +2.04. Only the larger solution for u, is the one that applies to the original equation. Thus: u = 1/2 + sqrt(1 + 6*sqrt(2))/2 Square it, to find the corresponding x-value: x = (1/2 + sqrt(1 + 6*sqrt(2))/2)^2 Expand and simplify, and you get: x = 1/2 + 3*sqrt(2)/2 + 1/2*sqrt(1 + 6*sqrt(2))
Try my "extreme quintic equation" next: x^5-5x+3=0
th-cam.com/video/GoGsVLnf8Rk/w-d-xo.htmlsi=WumfxJqZYwNfBWe_
Use demoiver's theorem 😊
Co-efficient (ab+2) should be with X^2.
Doesn’t ANY number above one increased when ^5? And ANY number under one decrease?
Hi, would you like doing number theory problems and proofs of its theorems?
@@Jordan-gt6gd yes
2:35
His face when he double thinks about what he said is gold.
“Tf did I just say”
"Never mind it's right "@@mbapum6363
Everyone says that x⁵ = 0 is easier than this. But i think that x⁵ = x⁵ is much easier
How about x⁵=x⁵+1? Instead of everything, it's just nothing. Just as simple.
That's a violation.😢
It's not a quintic though
it's not quintic.
Quintic is ax⁵+bx⁴+cx³+dx²+ex+f=0 for non zero a.
@@YarinGD you win!
I just did one of my end of year exams for year 12 and this video is the reason I was able to correctly answer probably the hardest question on the whole exam! Thank you so much for your videos, please keep making them. The question required finding the root of a cubic and I always forget how to do polynomial long division but I found 1 was a root by just trying it and then solved the cubic by factoring to (x-1)(a^x+bx+c) then expanding again and solving for a,b and c just like you did in this video
At 1:03 he said synthetic division so i left the video and spent 3h researching synthetic divisipn and everything else i needed to understand it, and here i am again. Coincidentally now it's 4 am and i have school at 7:45 and im probably gonna kms but idk yet.
Curiosity killed the human
I can’t believe im seeing this now cuz synthetic division is literally the most recent thing i learned
@@M7m7at But satisfaction brought it back
Simpler Solution - Use n'th root of unity
x^5 = 1 = e^(i 360° n) so, x = e^(i n 360° / 5) = e ^ (i n 72°) for n = 0, 1,2, 3,4 leading to 1 (the only real root) and 2 complex conjugate roots with angles ±72° and ±144°
So, *the 5 roots are* 1, cos 72° ± i sin 72° and cos 144° ± i sin 144°
(Use cos ±72° = sin 18° = (√5-1)/4) and (cos ±144° = -cos 36° = -(√5+1)/4) and similarly the sine counter-parts to express the solution without any un-computed trigonometric functions)
Also simpler because you skipped an essential part of the solution. Why is sin 18° equal to (√5-1)/4 ? To make your solution complete you have to make that part of the solution.
And also your statement 1 = e^(i 360° n) is not correct. It has to be 1=e^(i2𝜋*n). The exponent has to be a number and not something with degrees in it.
@henkhu100 can't be bothered to read either the solution or the criticism. The solutions are 1/5 rotations around the complex plain starting at 1. That is what was intended to be communicated
@@governmentis-watching3303 Ok, but that does not mean that what he writes may be wrong. An expression like e^(360 degrees) has no meaning at all. If we write e^(2*pi) we have a real number for the exponent. We don't write e^(2*pi radians) because that does not have any meaning. Just like we can write √ 360 but not √(360 degrees).
So my comment was: if you solve a problem use correct formulations and write things that have a menaning, so don't use expressions like e^(some degrees).
I think he said no polar form at the start
Does anyone else miss the pokeball mic? Maybe a special return of the pokeball episode before the end of the year?
4:26 THAT X² IN THAT STEP STAYED AS "X" FOREVER RAAAAAAAA
Unsatisfactory 😭 once you see it, you can't unsee it
I saw it immediately because he actually mentioned it but didn't fix it
Yes, I spotted that, too, and wanted it to be corrected as BPRP did correcting a previously missed one at 4:24.
8:46 You can use ψ (psi), it's the reverse symmetrical of the golden ratio φ (phi)
You can solve this either algebraically or geometrically.
For the algebraic solution, just factor out x - 1 and solve the resulting 4th degree polynomial:
x⁵ - 1 = 0
(x - 1) (x⁴ + x³ + x² + x + 1) = 0
etc.
For the geometrical solution, just use Euler's formula:
xₙ = ⁵√1 ⋅ [cos(n ⋅ 360° / 5) + i ⋅ sin(n ⋅ 360° / 5)]
n = 1→5
So, basically:
x₁ = cos(72°) + i ⋅ sin(72°)
x₂ = cos(144°) + i ⋅ sin(144°)
x₃ = cos(216°) + i ⋅ sin(216°)
x₄ = cos(288°) + i ⋅ sin(288°)
x₅ = cos(360°) + i ⋅ sin(360°)
Love the shirt in the thumbnail. For those who do not know, it means algebra in Arabic.
Well,algebra came from middle east soo,you are completely right
I love the connection between phi and this quintic/pentagons
Is it just me that saw this as a tiktok search thing? TH-cam has done this with other shorts
@@Chessbutmostlyrandomstuff Nice profile picture :)
@@Chessbutmostlyrandomstuff
I think it's like some random words just get that search thing, it's a bit odd
@@Chessbutmostlyrandomstuff holy hell
@@isavenewspapers8890 ty
One of the best videos I have seen. Excellent mastery of Ceva versus Delanges math when you said that it didn't matter because they were symmetric. Ambiguities and dualities were combined at that point in the calculations.
Hence why I like mutlivalued functions. x=⁵√1, write out the 5 values by rotating by 2π/5 each time from 1, and we're done.
Como se usa esse texto azul?
@@KrybiaAs far as I noticed, those are added automatically by TH-cam
Ok but he said he wouldn’t use polar forms
@@Krybia I don't. I think TH-cam adds it automatically. Which text was blue?
@@xinpingdonohoe3978 the smol 5
I love seeing these kind of videos and knowing how to solve them
The golden ratio emerges here, sometimes pi, sometimes e^x, etc.... show up. When I look back, it is those surprising or unexpected little nuggets we find that got me hooked on mathematics early on.
An easier way at 2:58 if u divide by x² then u will get a equation like x² + 1/x² + x + 1/x + 1 . Then u can assume x + 1/x as y so new equation will be y² + y - 1 and then u can solve it from there
Please explain how if (x + (1/x)) = Y how did you get y² +y -1
To me it doesn't make sense, its like you combined the x+1/x or something. Waiting for explanation please. 😊
I checked using gbt it cannot do it.
@@AbdullaAlQemzi1 ok so heres detailed explanation take "1/x + x" as y now
y² = 1/x²+ x² +2
y² - 2 = 1/x² + x²( Subtract 2 from both sides)
so the new eqn becomes
y² - y -2 + 1 = 0
y² - y - 1 = 0
if u still dint get it dont hesitate to ask again
@@OgSuperHeroWoah.😮 Thanks bro I understood. If you don't mind can you explain how to move on from your equation to find solutions. Do you use euler theory ? And why not use Z⁵=1 from the beginning and use euler on that?
we can also divide by x^2 and reduce the equation into quadratic by substituitng x+1/x as t
He has done that before in previous videos on the exact same quartic equation.
Only by assuming x0.
@@alanclarke4646 it is not assumed because we can very easily verify from the original equation that x cannot be 0
1:10 I am surprised this is my first time seeing this type of method
"-1/phi" could just be called "1-phi".
Exactly! And phi^2 could be replaced by phi+1, to make the roots look a bit simpler...
And you don't need to put b=-1/a and replace and multiply by a to get the quadratic equation, because you have the sum S and the product P of a and b, and by theorem of Vieta, a and b ar the solutions to y^2 - Sy + P = 0
Phi low so phi.
9:30 You can not only figure out sin(72) but also the sin and cos value of 72, 144, 216, 288. Those angles which you can raise to the fifth power to rotate back to 1 on the complex plane.
Thanks for that. When he said he wasn't going to use complex/ polar, I was disappointed. Rotations in the complex plane was the first thing I thought of to find all five solutions.
@@mikefochtman7164 Yes, I did not want to point that out, since it was an exercise to solve it this way. Just reflected that real and imaginary parts give us all the other values as well.
9:24 You mean sine of 72 DEGREES.
Granted 72 radians is a ridiculous statement
Yeah 2π/5
Basically this describes all rotations that are k * 2 pi / 5 for k element of {1,2,3,4,5} expressed as a complex number
So satisfying! I was never good at this kind of math, so it's really interesting to see how you go about solving these equations.
IS THE FISRT TIME I NOTICED YOU TYPED ON YOU T SHIRT AL GEBRA ARABIC "I AM VERY PROUD"
I was hoping for five values for x.
Thanks for continuing these videos, it's been ages and I just came back to playing with math.
You aged a little, but that makes it all the more beautiful ❤
Much love from here! ❤️ ❤
+-
You got five values, the second and third boxes are pairs of complex conjugates
Since phi^2=phi + 1, it follows that phi^2 - 4=phi - 3, so the first two solutions can be presented as [phi+i*(3-phi)^1/2]/2 and [phi-(phi-3)^1/2]/2.
For the other two, consider that -1 - 4*phi^2=-1 - 4*phi - 4=-5 - 4*phi. So we can present these solutions as [1+i*(5+4*phi)^1/2]/(2*phi) and [1-i*(5+4*phi)^1/2]/(2*phi).
Wouldn't x^5 = 0 be the easiest? 😂😂
We want the simplest, not the freeest
@@Musterkartoffel😂😂
That would be the most trivial
Unless you're willing to extrapolate and include x⁵=x⁵ I guess.
!No
You could have used φ² = φ + 1, 1/φ = φ - 1, 1/φ² = 2 - φ for simplification!
=> x = 1 || x = ( - φ ± i √(3 - φ) ) / 2 || x = ( φ - 1 ± i √(φ + 2) ) / 2
I think, you could even reduce this to simple roots and numbers, as a homework.
Can we not use nth root of unity and just get all the roots in their respective Euler's form? 😅
OMG I LOVE THAT PHI POPPING OUT AS RELATES TO THE COMPLEX CIRCLE
very nice ! my first thought was to take 1 as e^(i2pi*k) with k any integer, so the fifth root would be e^(i2pi*k/5) giving us the five different solutions of e^0=1, e^(i2pi/5), e^(i4pi/5), e^(i6pi/5), e^(i8pi/5)
Wow! That was really enlightening.
So, the simplest quintic equation involves a whole bunch of factoring, a non-linear system of equations, and the golden ratio. Pretty cool
1-sqrt(5)/2 is usually the conjugate of phi, or phi bar.
Fu bar’s little brother.
So you have derived a closed form of cos 2n pi/5, sin 2n pi/5 n 0..4 . Thats nice
I figured as much. In the complex plane, the results of a given power (example: 4) has four results, and all of them are in the complex plane. And if you connect the results' coordinates, you'll notice the bigger the power is, the results will form a circle.
x is 1
the other 4 solutions (since its a power of 5 equation is has 5 solutions) are all equally spread around the complex plane (72 degrees apart) with a size 1, use trigonometry to find out the real and imaginary componants of each
@@md-sl1io he explicitly said he wasn't doing that because it was too easy
The marker switches are just as impressive
Oh wow I didnt even notice I was early becuase I was so invested! Amazing video
i like your channel
curiosity and perseverance with which you bring the decision to the end
my like and follow you
Brilliant! Now integrate the regular factorial and super factorial of x (sf(x))
since b = -1/a shouldn't b be equal to -2/(1 + sqrt(5))?
BTW, for those who don't know the word " الجبر ", it means " Algebra "
As an Arab person, I feel very happy to see you using this word ❤
Kfo
Question. After you do polynomial division of x^5/(x-1), you get coefficients 111110. What does this mean, and how does it translate to x^4+...+1?
The 0 is garbage, unless used for the other side! For the rest: All coefficients are 1: 1 x^4 + 1 x³ + 1 x² + 1 x + 1 (= 0) || x -1 = 0
1*x^4 + 1*x^3 + 1*x^2 + 1*x^1 + 1*x^0 + 0/(x - 1)
1:03 I know how to divide polynomials the long way but I don't recall ever being taught this way (which you call "synthetic division").
Actually that’s the Ruffini’s rule, and works only for divisions by first degree polynomials
I'm going to use the "come on man" theorem which states that the more 'erm acshually' an argument is in math, the less validity it has.
Having 5 solutions, we can take 1, than we take the non arytmetic solution (with the module of a complex and its angle), select a numbrr with 1 as the module and with either 1/5 of 2π, or 2/5 , 3/5 and 4/5 of 2π as the angle
Than just take cos and sin and you have your 5 solutions
(And sorry for my bad english(
Cu stilul tău de a prezenta problemele vei reuși să îndepărtezi mulți vizitatori de matematica adevărată.
Wow. I can't remember all those tricks to factor the polynomial. I'd go another route: the solution would be any complex number z which modulus is 1 and the 5*arg(z) mod 2pi = 0. Then you may figure the arguments as 0, 2*pi/5, 4*pi/5, 6*pi/5 and 8*pi/5.
What an amazing video bprp! I am only an 8th grader but ever since I saw your calculus videos, I have been engrossed with your channel. Is it ok if you can make a video with a question featuring an improper integral, a derivative, a series and a limit all in one question where it is aimed at Calculus 1 and 2 students. Thank you!
But for a-1/a where a belongs to (-♾️,-2]U[2, ♾️) then how a-1/a = 1
Good job!
nice T-shirt BTW in tha thumbnail "الجبر"
Mmm if you bodice with any of these polynomials of the form x^n-1=0, 1 is always a solution. To get the remaining polynomial you find need to do synthetic division! Notice that x^n-1/x-1 is the sum of a geometric series with common ratio of x upto the n-1th term
Made a lot of typos but I hope people can understand
I think u can use (ln). Like
5ln(x)=ln(1)
Ln(x)=ln(1)/5
X=1
3:59 there: is abx(2) or abx?
isn't x⁵=0 simpler ?
@@charlievane 😂
Thats not simpler,thats free
x⁵=1 is still the simplest one that still gives a challenge, the definition of 'simple' and 'challenge' is left as an exercise to the reader.
@@Musterkartoffel How is it free if x can't be any value?
@@anglaismoyen0?
7:11 you have a mistake, u found b incorrectly, remember that b=-1/a => b=2/-1-√5 and 2/1+√5
Is it coincidence that I did this for fun the other day while waiting between lessons
How about e^(i*0.4*pi)? That works too
This is roughly 0.309 + 0.951*i, but I was just thinking 72 degrees in the complex plane.
Is there actually a practical operation where this sort of thing is necessary?
An easier way to factor it is to use geometric series. (x^5 - 1) / (x -1) is a geometric series with first term 1, 5 terms, and ratio x. So 1 + x + x² + x³ + x⁴
I notice a lot of asian people (myself included) write it out like this. Just kinda uniform then everywhere and I love that I could follow it
Isnt it b = -2/(1+√5) ?
Same thing.
The video was (1-√5)/2
@@nurrohmadi7852 Same thing.
@@isavenewspapers8890 you right
es una ecuacion de grado 5, por lo tanto, siempre generaria 5 respuestas correctas, las cuales se pueden apreciar en el video, el numero de respuestas siempreesta indicado por el indice de una variable, por ejemplo, x'2=5x, x=5 y x=0
Like ur t-shirt ❤
I love how he says "now, this is the part where we have to think a little bit"
So,what in the math have we been doing the whole time???? We did all this without thinking??? 😮😂
That statement probably means critical thinking. You don't need critical thinking to apply a formula.
ax *bx = abx? shouldn’t it be abx^2? 4:03
nevermind
Yes :)
You know that thinks fun 1 through 10 and then we show the expression upon chemistry. What won through is in standard in chemical makeup, Constance and biological result with identity of 1 whole 2 maximum hole and cube effect in unstable acceleration, as well as I can't remember what is called in division but when it. Divides an acts across the axes and catalysts. I think that's fun. You should do that 1 'cause that 1's fun. Yeah, one is an expression, is it an? Identity, and when one is expressed, it may not have a hole that accepts 10 as a whole of interval. The interval set makes extend to the 10th place before or after whole one equals in holes. It's weird, but it's our universe and it's how it actually works.
So good!
Simple. The roots are the complex numbers with magnitude 1 and arguments of 0, 2π/5, 4π/5, 6π/5, and 8π/5.
Needs the final step: calculate the real and imaginary parts of each answer!
JUST learnt this in class today lol
Shouldn't there be an i somewhere in the 4 complex solutions?
Taking log both sides
log(x⁵)=log1
5 log(x) = 0
log(x) = 0
x = e⁰
x = 1
2:33 “any polynomial can be factored in terms of linear and quadratic”
Wouldn’t that mean any polynomial would have solutions expressible in terms of linear terms and square roots (possible complex roots)? That’s clearly not possible. x^3-2=0 for starters.
Love how golden ratio appeared
Why not draw a circle and divide into 2pi/5 sectors?
usually it s easier to go into the complex as z^5=1 and result right as you said
Why not use Eulers Identity? e^iπ = -1? Therefore: e^i2kπ = 1, x^5 = e^i2kπ, etc.
In the end, it is a circle in the complex plane with roots at: (0°), (72°), (144°), (216°), and (288°).
At the very beginning he said he didn't want to use that method and wanted to show it in the more difficult algebraic way which is juuuust barely possible.
x^5-1 = x^5 - x^4 + x^4 - x^3 + x^3 - x^2 + x^2 - x + x - 1
=(x-1)(x^4 + x^3 + x^2 + x + 1)
Great! But why don't use that:
x^n=1
x=e^(2πk/n), k € N
7:07 этот типок лепит детские ошибки: если b=-(1/a), то подставив а из строки выше, мы получим, что b=-(2/(1±sqrt5)).
ну и фи получается равно 0, ибо два его корня должны быть равны 0, ибо только 0 в любой степени дает 1.
что это за начинающий математик?
4:20 (ab+2)x. x should have been raised to the power of 2
I wasn't expecting the Golden Ratio to pop out of this at some point.
De moivres theorem and symmetry around the unit circle for an elegant approach?
To solve the equation \( X^5 = 1 \) where \( X
eq 1 \), we are looking for the fifth roots of unity. The solutions to this equation can be expressed in the form:
\[
X = e^{2\pi i k / 5}
\]
where \( k \) is an integer. The fifth roots of unity are:
1. \( k = 0 \): \( X = e^{2\pi i \cdot 0 / 5} = 1 \)
2. \( k = 1 \): \( X = e^{2\pi i / 5} \)
3. \( k = 2 \): \( X = e^{4\pi i / 5} \)
4. \( k = 3 \): \( X = e^{6\pi i / 5} \)
5. \( k = 4 \): \( X = e^{8\pi i / 5} \)
Since we want the solutions where \( X
eq 1 \), the valid solutions are:
1. \( X = e^{2\pi i / 5} \)
2. \( X = e^{4\pi i / 5} \)
3. \( X = e^{6\pi i / 5} \)
4. \( X = e^{8\pi i / 5} \)
In summary, the solutions to \( X^5 = 1 \) with \( X
eq 1 \) are:
- \( e^{2\pi i / 5} \)
- \( e^{4\pi i / 5} \)
- \( e^{6\pi i / 5} \)
- \( e^{8\pi i / 5} \)
Why not just use the fact that the 5th roots of unity are separated by an angle of 2pi/5 on the unit circle?
Last four roots are complex. ANY number >1 will increase when raised to 5th power. And ANY number
And that was the only real root in the given solution set. What exactly is the problem here?
Edit: Hey all, I will also take the opportunity to edit my comment. The original comment used to say that the video was straight-up wrong, but after my reply, this statement has been removed.
@@isavenewspapers8890i didnt say there was a problem, just that 1 is the only real root and the others are complex.
@@ibperson7765 You know TH-cam shows when your comments are edited, right?
@@isavenewspapers8890 I really thought you commented long after I edited it. But if not, then youre right👍 because I did first comment implying/claiming there was a problem.
@@ibperson7765 I see.
Does that approach for solving the quartic work for all quartics? That was sweet.
No. The quartic is the end-of-the-line, when it comes to a closed-form formula in elementary functions as the master key to solve any polynomial of that degree. There are special case quintics where there exists a formula, but Galois proved there can be no such general formula for quintics, or anything beyond.
@@carultch OK, I don't know anything about Galois, but I asked about quartics, not quintics. When I write out the general equation
x^4+ax^3+bx^2+cx+f
I can use 4 variables, p, q, d and e in the form
(x^2+px+d)*(x^2+qx+e)
and that gives me
de = f
p + q = a
pq+d+e = b
pe+dq = c
These reduce to the values in this video where d = 1 and e = 1.
That should be solvable, right? It might not be trivial to solve, but there are 4 equations/4 unknowns.
@@flowingafterglow629 Sorry, I misunderstood your question. I had my mind on quintics, since his original video was about quintics.
The quartic in general is solvable. Ferrari, who worked with Cardano, the namesake of the cubic formula, derived the general quartic formula, using many of the same principles as deriving the cubic formula.
You're method is legit, however you will ultimately need to solve a quartic anyway, to solve that system of four equations. So it doesn't really make any progress.
Can you make a video explaining how to solve this?
√(2x²) - √(2x) = 3
Given:
sqrt(2*x^2) - sqrt(2*x) = 3
Extract all coefficients from square root terms, and shuffle to the left:
sqrt(2)*sqrt(x^2) - sqrt(2)*sqrt(x) - 3 = 0
Divide thru by sqrt(2):
sqrt(x^2) - sqrt(x) - 3/sqrt(2) = 0
Reverse the nesting order of the first term:
sqrt(x^2) = sqrt(x)^2
Let u = sqrt(x), and rewrite in the u-world.
u^2 - u - 3/sqrt(2) = 0
Now we can solve for u with a quadratic formula:
u = (1 +/- sqrt(1 - 4*-3/sqrt(2)))/2
Simplify:
u = 1/2 +/- sqrt(1 + 6*sqrt(2))/2
Since u = sqrt(x), this means u can only be a positive number, by definition of the square root function. So we can rule out the lower solution to the quadratic formula as an extraneous solution. Upon evaluating it, you'll see that u is approximately -1.04 and +2.04. Only the larger solution for u, is the one that applies to the original equation. Thus:
u = 1/2 + sqrt(1 + 6*sqrt(2))/2
Square it, to find the corresponding x-value:
x = (1/2 + sqrt(1 + 6*sqrt(2))/2)^2
Expand and simplify, and you get:
x = 1/2 + 3*sqrt(2)/2 + 1/2*sqrt(1 + 6*sqrt(2))
I’m trying to think of a practical application of this
If that’s the simplest one, I’d hate to see what a difficult one looks like
X⁵=X⁰
5=0
🗣️🔥🙏🏻
This is meth not math 😭
@obz1357 THE ONE PIECE!
THE ONE PIECE IS REEEEEEEEEEEEEEEEEEEEEEAL!
Mathmathecally wrong
How about you stay on anime plz don't disturb us
Mathmathecally wrong
How about you stay on anime plz don't disturb us
@@bolt2839 I am an imo aspirant bruv ofc it's a joke 💀
When I saw this I immediately thought "roots of unity". But yeah this is a creative way. (Also why not just use the quintic formula? oh wait... :)
Also, why not just use the geometric series formula for x^4+x^3+x^2+x+1? 🤔
@@sudoer-Ht Because this is not a geometric series? A series goes on forever, up to infinity.
@@bjornfeuerbacher5514 series can be finite. Search for geometric series sum formula
@@bjornfeuerbacher5514 No, the term "series" does not mean it goes on forever. The two terms "infinite series" do.
@@tfg601 A series is, by definition, an infinite sum.
At 7:06 it shouldve been b=2/(1-sqrt(5))
-2 but yeah
Yeah I thought so too. He also forgot to add the ^2 on (ab+2)x even after mentioning it. Kind of sloppy stuff
If a = 1/2 + 1/2 sqrt(5) and a + b = 1, then it is obvious that b should be 1/2 - 1/2 sqrt(5), since the 1/2s become 1 and the sqrts cancel out.
Now try doing it using the quartic formula (and use up the entire marker to finish) 😂.
Co-efficient (ab+2) should be with X^2.
Nice one 👍