What do I think about this question? I looked at a more general problem years ago. Suppose x^n=1, then there are n equally spaced roots on a unit circle centered on the origin of the complex plane, and x_0=1 is the "first" of those roots. 1) The sum of all n roots is zero. Why? 2) The sum of x_j^k, 0
You are absolutely the best answer!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! x^5-1=(x-1)(x^4+x^3+x^2+x^1+x^0)=0 Then, with unit circle,..............................................................................................
The right side of the equation is a simple geometric progression that can be rewritten as : (x^5 - 1)/(x - 1) . Setting the above equal to 0 and multiplying both sides by (x - 1), we have : x^5 - 1 = 0 -> x^5 = 1 , which means the solutions are the 5th roots of unity with some restrictions : e^(2*pi*i*k/5) for 1 ≤ k ≤ 4 . k = 0 is not a solution and neither is k = 5 , for obvious reasons. Since there are four (complex) solutions and the equation is a quartic, they’re the only solutions as well.
As this is a cyclotomic polynomial, this was the (much easier!!) solution they were expecting. As pointed out, the answer is that four of the five complex roots of unity (two pairs of complex conjugates) are the solutions sought.
The steps shown for the solution are clear and easy to follow. What’s not shown, and is left a complete mystery, is how you should know to take those steps. What is it about the original equation that tells you to divide by X squared and then organize X + 1/X into a single variable? That’s the part I’d like to hear explained more fully. The more obvious approach is to do as several previous commenters have suggested: multiply the equation by X-1, giving X^5 - 1 = 0, so that the solutions for X are the complex fifth roots of unity. But then you get solutions like X = cos (72 degrees) + i•sin(72 degrees), and the mystery becomes, how do you resolve that into a numerical expression not involving trig functions? Anyone who can explain either of those mysteries, please post a reply!
Are you ready for this? Suppose we have x^n=1, which has n roots, x_j=cos(2jπ/n)+i*sin(2jπ/n), j=0,1,...,n-1, uniformly distributed around a unit circle centered on the origin of the complex plane. Thus, x_0=1, and when n is even there is also a real root x_{n/2}=-1. So what? I'm not a mathematician, and what follows would surely be better explained by one of them guys. Let x_j=e^{2ijπ/n},j=0,1,...,n-1. For any 0
To the host of this video. Just an observation. You have white spots on your nails. This indicates a zinc deficiency. Sol: zinc supplements and Google high zinc foods. You're welcome.
Is this a general way of solving degree 4 polynomials or does it only apply to this specific problem? If it's the latter (which I'm betting it is), then this seems like trivia: either you know the specific trick for this specific equation (and can crank through the necessary algebra), or you don't and you are SOL. Why would knowing random trivia be a good screen for uni admissions???
OMG. I gave up when I hit that question on my admissions test to janitorial school. Which is why I became an engineer because they give you all the answers.
I got through multivariable calculus in college without ever hearing of cardano's formula. Plus, Google tells me it's for cubic equations, not 4th degree equations.
I do'nt know where do yo are or what do yo are learnt! In my Land we know (Euler Like) thar 1 Is the 2kπ+1 imaginary exponential of e, so if U do not ask the precisión of a square root ( my hand) it Is solved as all the 5th roots of 1, except of itselft.
What do I think about this question? I looked at a more general problem years ago. Suppose x^n=1, then there are n equally spaced roots on a unit circle centered on the origin of the complex plane, and x_0=1 is the "first" of those roots. 1) The sum of all n roots is zero. Why? 2) The sum of x_j^k, 0
If we multiply given equation by x-1, we get x⁵=1, or x=exp(2×pi×k×i/5), k=1,2,3,4
Wundervoll! Danke.
cos(72°k)+isin(72°k), k=1, 2, 3, 4.
You are absolutely the best answer!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
x^5-1=(x-1)(x^4+x^3+x^2+x^1+x^0)=0
Then, with unit circle,..............................................................................................
@chien-lihhwang9763 Thank You!
Great! See my more general comment for x^n=1.
x=e^(2n/5)πi ,n∋Z
@@joegillian6781 Check Your answer.
The right side of the equation is a simple geometric progression that can be rewritten as : (x^5 - 1)/(x - 1) .
Setting the above equal to 0 and multiplying both sides by (x - 1), we have : x^5 - 1 = 0 -> x^5 = 1 , which means the solutions are the 5th roots of unity with some restrictions :
e^(2*pi*i*k/5) for 1 ≤ k ≤ 4 .
k = 0 is not a solution and neither is k = 5 , for obvious reasons.
Since there are four (complex) solutions and the equation is a quartic, they’re the only solutions as well.
As this is a cyclotomic polynomial, this was the (much easier!!) solution they were expecting. As pointed out, the answer is that four of the five complex roots of unity (two pairs of complex conjugates) are the solutions sought.
That's my approach by default when I notice a geometric progression.
Both solutions are correct but using the n roots of 1 seems far easier to me.
Drat! Not a single real solution. I fear I'll have to watch it because my first guesses at possible complex one turned up dry.
For x3 and x4 solutions the square root is: 5/8 + sr5/8 (not 5/8 - sr5/8)
The steps shown for the solution are clear and easy to follow. What’s not shown, and is left a complete mystery, is how you should know to take those steps. What is it about the original equation that tells you to divide by X squared and then organize X + 1/X into a single variable? That’s the part I’d like to hear explained more fully.
The more obvious approach is to do as several previous commenters have suggested: multiply the equation by X-1, giving X^5 - 1 = 0, so that the solutions for X are the complex fifth roots of unity. But then you get solutions like X = cos (72 degrees) + i•sin(72 degrees), and the mystery becomes, how do you resolve that into a numerical expression not involving trig functions?
Anyone who can explain either of those mysteries, please post a reply!
Are you ready for this? Suppose we have x^n=1, which has n roots, x_j=cos(2jπ/n)+i*sin(2jπ/n), j=0,1,...,n-1, uniformly distributed around a unit circle centered on the origin of the complex plane. Thus, x_0=1, and when n is even there is also a real root x_{n/2}=-1. So what? I'm not a mathematician, and what follows would surely be better explained by one of them guys. Let x_j=e^{2ijπ/n},j=0,1,...,n-1. For any 0
To the host of this video.
Just an observation. You have white spots on your nails. This indicates a zinc deficiency. Sol: zinc supplements and Google high zinc foods. You're welcome.
This is derived from x⁵ - 1 = 0, which I just watched.
Has anyone else noticed that the same problems keep popping up around the same time.
Is this a general way of solving degree 4 polynomials or does it only apply to this specific problem?
If it's the latter (which I'm betting it is), then this seems like trivia: either you know the specific trick for this specific equation (and can crank through the necessary algebra), or you don't and you are SOL.
Why would knowing random trivia be a good screen for uni admissions???
Because it is designed to keep people like me a thousand miles away from the place.
It’s the same problem as x^5 - 1 = 0…. Just a bit detailed diff
Could we have used synthetic division?
{x^4+x^4 ➖ }+{x^3+x^3 ➖}+{x^2+x^2 ➖ }+{x+x ➖}+{1+1 ➖ }={x^8+x^6+x^4+x^2+2}=2x^20 2x^2^4^6^8 1x^2^2^2^3^3^2^3 1x^1^1^1^1^3^1^1^2 x^1^3^1^2 x^3^2 (x ➖ 3x+2).
🎉🎉 very good question @MATHS TUTORIAL ( BL SAHU)
Admissions question to what program?
Sorry to say that the host is not going to be admitted!😮😂
OMG. I gave up when I hit that question on my admissions test to janitorial school. Which is why I became an engineer because they give you all the answers.
X=-1
X=-1 -1+-1+1= 1❤❤ = 0 ??😅😂
you need cardano formula for 4th degree equation. first class highschool level problem. easy.
And where the complex solutions are in first class high school?
I got through multivariable calculus in college without ever hearing of cardano's formula. Plus, Google tells me it's for cubic equations, not 4th degree equations.
Use, for example, L. FERRARI's formula and J. CARDANO's.😢
It's not that horrible!!😂
I do'nt know where do yo are or what do yo are learnt!
In my Land we know (Euler Like) thar 1 Is the 2kπ+1 imaginary exponential of e, so if U do not ask the precisión of a square root ( my hand) it Is solved as all the 5th roots of 1, except of itselft.