Evaluating a Quintic Polynomial from a Quadratic
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- Опубликовано: 8 сен 2024
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There's an easier way to present the second method. You can perform a polynomial division and say x^5 -5x = (x^2 - x -1)(x^3 + x^2 +2x +3) + 3(the remainder!). Now the first term is zero as x^2-x-1= 0 is given. So remainder=3 is the answer!
You are repeating my solution
Given:
x² = x + 1
To find:
x⁵ - 5x
x² = x + 1
Multiplying both sides by x (as x ≠ 0):
x³ = x² + x = 2x + 1
Multiplying both sides by x²:
x⁵ = (2x + 1)(x²)
= (2x + 1)(x + 1)
= 2x² + 3x + 1
= 2(x + 1) + 3x + 1
= 2x + 2 + 3x + 1
= 5x + 3
Now,
x⁵ - 5x = 5x + 3 - 5x
= 3.
Yes, did the same!
How did you get fifth power on your keyboard? I could only have 1-4th power
@@ultrabingus961 GBoard has all the way to 9.
...but what's x??????
@@marioluigi9599 we're asked to find x⁵ - 5x we don't have to find x.
Ayyyy 1st method is Illegal
Simple rule for powers of phi:
Phi^n=F(n)times phi plus F(n-1), it works for all positive and negative integers of n. F(x) is the Fibbonaci sequence, F(x)=F(x-1)+F(x-2), where F(0)=0 and F(1)=1
Crucially here, the formula φⁿ=Fₙφ+Fₙ₋₁ also works if you replace φ by the other root of x²-x-1=0, namely -φ⁻¹, since in the proof by induction, we only use that φ satisfies that quadratic equation.
The 2nd method is beautiful!
Glad you think so!
x^2=1+x
square and multiply by x both sides
x^5=x+2x^2+x^3
replace x^2 with 1+x and x^3 with x+x^2
x^5=x+2+2x+x+x^2
replace x^2 with 1+x resulting in:
x^5=5x+3
therefore
x^5-5x=3
Really fun problem
Im glad I dont have to do these eq anymore.Well done to the solution solver...
Nice video & solutions as always.
As others have pointed out, in the 2nd method we can use the formula φⁿ = Fₙφ+Fₙ₋₁ to simplify φ⁵ (where Fₙ is the nth term of the Fibonacci sequence, starting F₀=0, F₁=1; the formula is easily proved by induction).
So φ⁵ = F₅φ+F₄= 5φ+3, φ⁵-5φ=3.
Note that the formula for φⁿ, and hence this last equation, both remain true if we replace φ by the other root of x²-x-1=0, namely -φ⁻¹, since in the proof by induction, we only use that φ satisfies that quadratic equation.
Regarding the 1st method, note that φⁿ can be written in radical form as
φⁿ = (Fₙ₊₁+Fₙ₋₁ + Fₙ√5)/2 (easy to check from the above formula for φⁿ by substituting φ=(1+√5)/2) and using Fₙ₊₁=Fₙ₋₁ + Fₙ).
So φ⁵ = (8+3 + 5√5)/2=(11+5√5)/2, which nicely explains where the coefficients come from.
The corresponding result for the other root of x²-x-1=0, namely -φ⁻¹, is (-φ⁻¹)⁵ = (8+3 - 5√5)/2=(11-5√5)/2, and can be justified by noting that the radical expressions in calculations involving this other root are the conjugate surds of those for φ (i.e. replace √5 by -√5).
As variant to the 1st method, the binomial theorem could, of course, have been used to evaluate [(1+√5)/2]⁵.
Np. Thank you for the comment!
I had a "golden" opportunity to solve this problem easily. The first equation is the equation of the golden ratio, so x = phi. A property of the golden ratio phi, is that for some power n, phi^n = phi^(n-1) + phi^(n-2). So let's keep applying these recurrences: x^5 = x^4 + x^3 = 2*x^3 + x^2 = 3*x^2 + 2*x. The first equation tells us that x^2 = x+1. Plug that in and we get x^5 = 5*x + 3. Therefore, x^5 - 5*x = 3.
This time I solved it using both of the two methods. Very nice problem.
Thank you and take care.
This is the first time on your channel that I have no Idea where to start solving this....
Square the given equation and then multiply x to both sides, we have x^5 - 2x^4 + x^3 - x = 0, then x^5 - 5x -2x^2(x^2-x) - x^3 + 4x = 0, then x^5 - 5x - x(x^2 - x) - 3x^2 + 4x = 0. then x^5 - 5x - 3(x^2 - x) = 0, then we get the result 3.
You can easily prove that if x^2 = x + 1, then x^n = f(n)*x + f(n-1) where f(n) is the nth Fibonacci number, given by f(0) = 0, f(1)=1, and f(n) = f(n-1)+f(n-2) for n>=2. Since f(5)=5 and f(5-1)=3, you get the result
Thus got me wondering... if given the quintic x^5 - 5x - 3 =0, is it possible to somehow reverse this to get the roots phi, -1/phi?
You can see from the comment by Karthik Ananthanarayanan that we can factorise x⁵-x-3 as (x²-x-1)(x³+x²+2x+3). The first factor gives the roots φ and -φ⁻¹, the second factor will give other roots.
Last method was just a magical one. Thanks
Most welcome 😊
thank you so much sir!!!
It's not necessary to evaluate x^4. Just calculate x^5 as x^3 * x^2, i.e. (2x + 1)(x + 1).
SyberMath, c'est le Dieu des Maths...
Ouah! C'est un énorme compliment! 😲
If x^2 - x - 1 = 0, then x^n = F(n)·x + F(n - 1), where F(n) is the nth Fibonacci number. Hence, x^5 = F(5)·x + F(4), so x^5 - 5·x = (F(5) - 5)·x + F(4). Notice that F(5) = 5, so x^5 - 5·x = F(4). F(4) = 3.
let us remember that phi is the solution of the golden equation and that (phi)^2 = Phi + 1
Let us go ahead and substitute x by phi in te equation to resolve x^5 - 5x;
we have: phi^5 - 5phi = phi(phi^4-5) = phi((phi+1)^2 - 5) = phi(phi^2 + 2phi + 1- 5) = phi(phi + 1 + 2phi + 1 - 5) = phi( 3phi -3) = 3phi^2 - 3phi = 3(phi + 1) - 3phi , which gives us by identification x = 3 for the solution.
Took me 3 tries to get 3 (using the polynomial method) - oh well, like they always say, third time's the charm!
That's right!
Great video!
Thanks!
@@SyberMath I may use this problem in my classroom.
Beautiful
Thank you
Excellent
My way to solve
We know x²-x=1 and let x⁵-5x=u
The first step is division:
X(x⁴-5)/x(x-1)=u
Simplify (x⁴-5)/(x-1)=u=((x²)²-5)/(x-1)
We know x²=x+1
So ((x+1)²-5)/(x-1)=u
As result (x²+2x-4)/(x-1)=u
Re-substituting x²=x+1
(3x-3)/(x-1)=u So u=3🧐🥳
Thank you so much 😀
Nice problem
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Your voice isn’t fine, i think you are sick. So i hope you fast healing
Thank you!
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Can you add Turkish subtitles to your videos, Turks are watching you too
Use RUclips subtitle and click on parameter and choose your language
Now that I'm actually in the workforce, I really hate that people have to do this kind of stuff in school. It's all so pointless.
No, it is good mathematical training for the mind.
Exactly! Thank you for the feedback. The math you need to perform work functions is so minimal that it's not even fun to study
@@forcelifeforce If I'm not going to become a math teacher then it's useless. I dont agree that it's good for the mind. Most people hate this stuff. It's usually not good to do things you hate.
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