A Vieta Problem

This problem requires a basic application of Vieta's formula for quadratics and a little factoring.

[@Alucs.E]

Worth noting that the solution is the 16th Fibonnaci number! The value for b will end up being minus the 17th Fibonnaci number. That fact that one of the roots is phi allows us to derive this directly. Since phi^n=F_n*phi+F_n-1, where F_n is the nth Fibonnaci number (and the same is true for the other root, -1/phi) the degree 17 polynomial can be transformed into the equations: a(phi*F_17+F_16)+b(phi*F_16+F_15)+1=0, and a(-1/phi*F_17+F_16) + b(-1/phi*F_16+F_15)+1=0, Subtracting the equations from one another, we get a*F_17+b*F_16=0 Substituting this back into either original equation, we obtain a*F_16+b*F_15+1=0. With a bit of algebra this gives us a= F_16/(F_15*F_17-(F_16)^2)) And b= -a*F_17/F_16 You can prove by induction that (F_n+1)*(F_n-1)-F_n^2=(-1)^n, which means the denominator of a is 1. So, a=F_16=987 and b=-F_17=-1597! This method also gives us the general value for the coefficients of polynomials of any degree. If x^2-x-1 is a factor of ax^(n+1)+bx^n+1, then a=F_n*(-1)^n and b= -(F_n+1)*(-1)^n. P.S.: the original proof had a couple mistakes, which were helpfully pointed out. I edited the comment to account for those

[@Grecks75]

I don't think this way to a solution is logically consistent. I see two problems here: 1) The first equation displayed (which you get from plugging in the known root phi) is not correct. It should instead read: a(phi*F_17+F_16)+b(phi*F_16 + F_15)+1=0. (Note the added "+ F_15" in the b term.) This is probably just a typo and thus is a minor point. 2) This is where the logic breaks: If we assume a, b to be real or complex numbers, there is no way you can separate the first equation into two. There is no way you can get *both* a and b from a *single* equation! What you need is *two independent* equations. You can get the second equation from plugging in the second known root, which is (1 - phi) = -1/phi. Then you can solve the resulting system of two independent linear equations for both a and b. Interestingly, you will indeed arrive at the stated solution.

[@Alucs.E]

@@slavinojunepri7648 Okay I took a moment to think about it. The separation I make only works if a and b are rational numbers. In this case they are in fact integers, but that can't just be assumed. The proper way to get to the separated equations is to realize that having two roots allows us to write both a(F_17*phi+F_16)+b(F_16*phi + F_15)+1=0, and a(-F_17*1/phi+F_16)+b(-F_16*1/phi + F_15)+1=0 Subtracting them from one another, and dividing by (phi-1/phi) you get (a*F_17+b*F_16)=0. Substituting this into the original equation gives us back a*F_16+b*F_15+1=0. So you can indeed make the separation, but it needs a bit more justification!

[@Alucs.E]

@@Grecks75 You're totally right on both accounts. I justify why you can actually make the separation in a comment below. Effectively, it's because you can cancel out the integer terms by substracting the identical equation you get with -1/phi instead of phi.

[@bobross7473]

It’s the 17th Fibonacci number, remember the first number in the sequence is 0

[@sammyjskj]

What I love about your videos, is not only how you explain stuff, but your intro. I ABSOLUTELY love it. I hope you never change it

[@kevinbush4300]

Yes... absolutely. The best teachers tell you what they are going to teach you; then they teach it to you; then they tell you what they just taught you.

[@MrJPI]

BTW: What is the music in the intro?

[@PrimeNewtons]

It's my personal music

[@raghvendrasingh1289]

Excellent problem Another solution x^2= x+1 squaring both sides and putting x^2 = x+1 x^4 = 3x+2 Similarly x^8 = 21 x+13 x^16 = 987 x + 610 x^17 = 987 x^2 +610x = 1597 x + 987 If we put these values in P(x) = a x^17 +bx^16+1 1597a+987b = 0 987a+ 610b+1 = 0 Hence a= 987 , b= - 1597

[@graf_paper]

I really enjoy this solution, thanks for sharing

[@Alucs.E]

As someone pointed out to the solution I gave above, you need a bit more to justify why you can divide the condition into those two. Since the relation only holds for particular values of x you can't just separate them by degree!

[@bobross7473]

I don’t understand how the steps follow, I had a much longer method but it worked still

[@raghvendrasingh1289]

@@bobross7473 I used factor theorem by putting reminder = 0 when P(x) is divided by x^2 - x - 1

[@MrJPI]

That was actually incredible based on how difficult the question initially seemed. :-)

[@555amry]

I absolutely LOVE this question, reason why is that you can relate the roots of x^2 - x - 1 to the Fibonacci sequence and find a solution that way. Amazing video, I’m glad I stumbled across this today

[@Noonecc]

The little thing that makes my day better everyday your channel and you And also you quote NEVER STOP LEARNING THOSE WHO STOPS LEARNING THEY STOPS LIVING

[@heythere9380]

Great video as always. You are master not only of the problem and its solution but also of that chalkboard: not a speck of chalk dust on a black shirt! When I was teaching, man, I started the day with a clean shirt and pants and finished the day looking like I had been rolling in chalk!

[@jay_13875]

The roots of x²-x-1 are the golden ratio φ and -φ⁻¹. Plugging the roots into the other polynomial gives two equations: a*φ^(17) + b*φ^(16) = -1 -a*φ^(-17) + b*φ^(-16) = -1 We have the matrix A [φ^(17), φ^(16);-φ^(-17), φ^(-16)] and we're looking for [a,b] = A⁻¹*[-1;-1]. The determinant of A is φ+1/φ= sqrt(5). Using the formula for the inverse of a 2x2 matrix, a is 1/sqrt(5) * (-φ^(-16)+φ^(16)). This can be further simplified using the identity φ^(n)-φ^(-n) = sqrt(5) * Fₙ (where n is an even integer and Fₙ is the n-th Fibonacci number) Therefore a is sqrt(5)*F₁₆/sqrt(5) = F₁₆ = 987.

[@JoshuaMay-e2n]

An alternative, probably not as clever, method is factor directly: a x^17 + b x^16 + 1 = (x^2 - x - 1) * (A_15 x^15 + A_14 x^14 + ... + A_1 X + A_0) Equating powers you get a = A_15 b = A_14 - A_15 0 = A_13 - A_14 - A_15 ... 0 = A_0 - A_1 - A_2 0 = -A_0 - A_1 1 = -A_0 Then straight-forward algebra gives you an (alternating sign) Fibonacci sequence for the A_n's, giving A_15 =a = Fib_16. I'm not sure which way is less easy to mess up the algebra on, though.

[@narangfamily7668]

Simply lovely

[@AZ-by5ps]

Excellent..!!!

[@akiya9216]

pretty neat solution :)

[@hervesergegbeto3352]

C'est merveilleux !

[@ManojkantSamal]

Incredible!

[@Father.Nature--1]

Diese Lösung ist send wunderbar.

[@michaelz2270]

A little trick to shorten the work: Multiplying the equation r_i^2 = r_i + 1 through by r_i^(k-2) gives r_i^k = r_i^(k-1) + r_i^(k-2). Subtracting this for i = 1 and i =2 gives r_1^k - r_2^k = (r_1^(k-1) - r_2^(k-1)) + (r_1^(k-2) - r_2^(k-2)). So the difference of kth powers satisfies the recursion of the Fibonacci sequence and also starts at 0 and 5^(1/2) for k = 0 and 1 respectively. So the difference of kth powers is is 5^(1/2) F_(16), where F(n) is the nth Fibonacci number, and the answer (r_2^(16) - r_1^(16))/(r_2 - r_1) = 5^(1/2) F_(16) / 5^(1/2) F_(1) = F_(16) = 987.

[@andrasferencz7948]

Brilliant!

[@vasileiossophianosd.1840]

You could use φ¹⁶=F_16*φ + F_15 and Φ¹⁶=F_16*Φ + F_15. Since te roots of x²-x-1 are φ,Φ

[@dan-florinchereches4892]

I was thinking along the following line by rewriting the equality: r1^2=R1+1 r2^2=R2+1 Summing up r1^2+r2^2=(R1+R2)+1=1+1=2 If we note r1+r2=S1=1 r1^2+r2^2=S2=2 Then multiplying each equation by the respective root r1^3=r1^2+R1 r2^3=r2^2+R2 Adding up equations S3=S2+S1=1+2=3 And this can be generalised as the next term of the fibonacci series S3=f4 and so on. Then knowing R1 and R2 are roots of the big polinomial ar1^17+br2^16+1=0 ar2^17+br2^16+1=0 Adding up equations a *s17+ b* s16+2=0 a* f18+ b* f17+2 =0 This gives me a lot of solutions. I guess this is why I suck at Olympiad problems If R1>R2 then let D1=r1-r2=√5 r1^2=r1+1 r2^2=R2+1 Subtracting D2=D1=√5 D3=D2+D1=2√5 So in general Dn=Fn√5 where Fn is nth fibonacci number. So a * F18+b*F17+2=0 (from sum) a * F17√5+b*F16√5=0 (from difference) => b=-a*F17/F16 => a(f18-f17^2/F16)+2=0 a=2*F16/(F17^2-F18*F16) I don't think writing Fn=1/√5(r1^n-r2^n) is going to help.working on denominator F18=F17+F16 => F17^2-F17F16-F16^2=F17(F17-F16)-F16^2=(because F17=F16+F15) F17F15-F16^2=F16F15+F15^2-F16^2=(factoring F16 and using identity of fib series)F15^2-F16F14 Looks like F17^2-F18F16=F15^2-F16F14=...=F3^2-F4F2=2^2-3*1=1 =>a=2*F16

[@bestmathuae]

Polynomials used to be my favourite and the easiest........................................................... but then I watched this video ⚠💀

[@VarunPatwal]

Gr8 👍

[@XX-sf1eh]

Awesome 😄

[@Notking444.]

Can you make lecture videos of differentiation and log and all. By the I like your effort ❤❤❤

[@abulfazmehdizada]

To determine \( a \) such that \( x^2 - x - 1 \) is a factor of the polynomial \( ax^{17} + bx^{16} + 1 \), we can use the fact that if \( x^2 - x - 1 \) is a factor, then the roots of this polynomial must also satisfy \( ax^{17} + bx^{16} + 1 = 0 \). The roots of \( x^2 - x - 1 = 0 \) are given by the quadratic formula: \[ x = \frac{1 \pm \sqrt{5}}{2} \] Let's denote the roots as: \[ \phi = \frac{1 + \sqrt{5}}{2} \quad \text{(the golden ratio)}, \quad \text{and} \quad \psi = \frac{1 - \sqrt{5}}{2} \] Now, we need to evaluate \( ax^{17} + bx^{16} + 1 \) at both \( \phi \) and \( \psi \) and set these equal to zero. Using the Fibonacci sequence properties, we know: \[ \phi^n = \frac{\phi^{n+1} - \phi^{n-1}}{\sqrt{5}} \] Specifically, we can derive: - \( \phi^0 = 1 \) - \( \phi^1 = \phi \) - \( \phi^2 = \phi + 1 \) - \( \phi^3 = 2\phi + 1 \) - \( \phi^4 = 3\phi + 2 \) - Continuing this pattern, we find that: - \( \phi^n = F_n \phi + F_{n-1} \) where \( F_n \) is the \( n \)-th Fibonacci number. Using this relation, we can compute \( \phi^{16} \) and \( \phi^{17} \). \[ \phi^{16} = F_{16} \phi + F_{15} \] \[ \phi^{17} = F_{17} \phi + F_{16} \] The Fibonacci numbers relevant to this are: - \( F_{15} = 610 \) - \( F_{16} = 987 \) - \( F_{17} = 1597 \) Thus: \[ \phi^{16} = 987\phi + 610 \] \[ \phi^{17} = 1597\phi + 987 \] Now substituting into the polynomial: \[ a(1597\phi + 987) + b(987\phi + 610) + 1 = 0 \] \[ (1597a + 987b)\phi + (987a + 610b + 1) = 0 \] For this to hold for all \( x \), both coefficients must equal zero: 1. \( 1597a + 987b = 0 \) 2. \( 987a + 610b + 1 = 0 \) From the first equation, we get: \[ b = -\frac{1597}{987}a \] Substituting \( b \) into the second equation: \[ 987a + 610\left(-\frac{1597}{987}a ight) + 1 = 0 \] \[ 987a - \frac{610 \cdot 1597}{987}a + 1 = 0 \] \[ \left(987 - \frac{610 \cdot 1597}{987} ight)a + 1 = 0 \] Multiplying through by \( 987 \): \[ (987^2 - 610 \cdot 1597)a + 987 = 0 \] Now we calculate \( 987^2 \) and \( 610 \cdot 1597 \): - \( 987^2 = 974169 \) - \( 610 \cdot 1597 = 974170 \) Thus: \[ (974169 - 974170)a + 987 = 0 \] \[ -a + 987 = 0 \quad \Rightarrow \quad a = 987 \] Therefore, the value of \( a \) is: \[ \boxed{987} \]

[@wildfire_]

i feel like every video, it takes longer to get into the video.

[@HridikKanjilal4444-k5f]

That day I have sent you that problem but you did not even look at that, I had sent it in the gmail account given in the previous video of yours

[@MBSall]

Can we find the value of b by isolating a this time

[@rohangt1]

I have a question and would like to know the response -: If x > y, then -: Is x^y > y^x true or false or both?

[@carultch]

Not enough information. Assuming positive real values of x&y, and given x > y. I haven't explored negatives or non-reals. If y>e, then y^x is always greater than x^y, when x>y. If y1, then there are two special case points where x^y = y^x, and between which y^x > x^y, for all real values of x&y. For instance, 2^4 = 4^2, and between the two special cases, the base dominates the exponent, such as like 2^3 < 3^2. While outside these points, the exponent dominates, e.g. 2^5 > 5^2. If y=e, there's only one point where x^y can equal y^x, which is when they both equal e. Everywhere else, the exponent dominates and y^x is greater. If y

[@rohangt1]

@carultch Looks like this problem can be a great idea for next video if it's solvable for all possible cases. I don't know if this question comes/can come in any of the exams/competitive exams or not. But if it comes/can come, then solution to this problem can help.

[@ironfbody]

Using the traditional alpha and beta for the quadratic roots would have meant a lot less writing.

[@camilotiznado5325]

should it be r1+r2 = -1?

[@SamuelDonald-pr2uu]

No it’s 1. The formula is: -b/a. The negative sign will be affected and that gives plus.

[@Football.avengers.]

Can you find the integral of x^[ln (x)]

[@mooshroommoon]

Can you solve x to the tetration of itself?

[@PrimeNewtons]

That's going to be tough

[@mooshroommoon]

@ figured it out

[@mooshroommoon]

@ x^^x = y x = the yth root of y

[@ThePayner11]

Is this related to Lucas Numbers?

[@xsimox13]

à is the 16 th Fibonacci number 987

[@anestismoutafidis4575]

x^2-x-1=0 φ^2-φ-1=0; x1=αx^17; x2=bx^16 [-0,1•φ^17 +(φ/10 - 4,5×10^-4)•φ^16+1=0] α=-0,1; b=(φ/10 - 4,5×10^-4); -p=-1; q=18,88^4; [α= - 1/10]

[@pillegraknel4308]

You could have been using that "not so nice" number (root of the x²-x-1) symbolically with the same result don't you? :D You just swap the r1 and r2 with phi, done.

[@pieterschadron3644]

I like you're presentations but I think the mistake at 3 minutes 34 seconds in this video is disturbing. You claim the sum of the roots r1 and r2 is 1. This must be negative 1.

[@PrimeNewtons]

You need to state why you say it was a mistake. Where did you get -1?

[@pieterschadron3644]

@@PrimeNewtons x^2+p*x+q=0 can be solved by finding r1 and r2; r1 + r2=p and r1*r2=q, in your calculation p=-1 and q=-1.

[@DontKnow-6969]

​@@pieterschadron3644r1+r2=-p bro check vieta's theorem

[@robertpearce8394]

​​​@@pieterschadron3644 (x-r1)*(x-r2)=(x^2-x(r1+r2)+r1r2) r1+r2=1 (-p in the general equation) r1*r2=-1

[@camilotiznado5325]

agree. the multiplication of the roots should be the free number (r1*r2=-1), and the sum should be the coeficient of "x", in this case r1+r2=-1