the nth term formula of the Fibonacci sequence from a quadratic equation

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How a simple quadratic equation solves the nth term formula for the Fibonacci sequence (aka Binet's Formula). We will use the quadratic formula to solve x^2-x-1=0 and then use the quadratic equation itself to find the Binet's Formula! This is a very lovely math trick that might help you with math competition problems or math olympiad problems when you have to solve recurrence relations. It's definitely cool enough to impress your math teachers and math friends! #blackpenredpen #Fibonacci #quadratic
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[@blackpenredpen]

Learn more math by visiting Brilliant 👉 brilliant.org/blackpenredpen/ (20% off with this link!)

[@anilkumarsharma8901]

Artificially intelligent calculators banva do phir sara desh main supply karo muft main Sabhi tarah ki economy apkey naam Chalney lagegi

[@rosieroti4063]

Follow up question. Please solve this: What is the value of 1/1 + 1/2 + 1/3 + 1/5 + 1/8 + 1/13 + ..... The denominators are in Fibonacci sequence. Would love to see a video on that.

[@AnonimityAssured]

Ah, the wonderful properties of phi! Here's another interesting set: 0φ² = φ⁰ − 1 1φ² = φ¹ + 1 1φ² = φ² + 0 2φ² = φ³ + 1 3φ² = φ⁴ + 1 5φ² = φ⁵ + 2 8φ² = φ⁶ + 3 13φ² = φ⁷ + 5 21φ² = φ⁸ + 8 This can be extended in either direction. The underlying formula is: Fₙφ² = φⁿ + Fₙ₋₂, where Fₙ is the nth Fibonacci number. (I wish KZbin formatting options were better.)

[@angelmendez-rivera351]

In fact, φ^n = F(n)·φ + F(n - 1) for all n, where F is the Fibonacci sequence.

[@jakobr_]

Yes, this is trivially equivalent to the pattern BPRP explored in this very video

[@AnonimityAssured]

@@jakobr_ _My_ pattern is _not_ equivalent to the one presented in the video, as it has a fixed phi squared term. Angel's pattern, on the other hand, is essentially that derived in the video.

[@chrisg3030]

Here's a property better revealed by the caret symbol ^ rather than a superscript: (φ^1)+1 = φ^(1+1). I call it the bracket shift law.

[@jakobr_]

@@AnonimityAssured Oh, I see that now. My mistake.

[@pseudo_goose]

I didn't recognize it until the end, but that polynomial is the characteristic polynomial of the Fibonacci recurrence! Just learned about this method of solving recurrences in combinatorics this semester, and this is a great example of that method. Wow

[@angelmendez-rivera351]

This is not unique to the Fibonnaci sequence. In fact, all sequences satisfying a recursive relation A(M)·S(N + M) + A(M - 1)·S(N + M - 1) + ••• + A(1)·S(N + 1) + A(0)·S(N) = 0 are given by S(N) = C(0)·[x(0)]^N + C(1)·[x(1)]^N + ••• + C(M - 2)·[x(M - 2)]^N + C(M - 1)·[x(M - 1)]^N, where x(0), ..., x(M - 1) are all the roots, assuming they are distinct, of the polynomial A(M)·x^M + A(M - 1)·x^(M - 1) + ••• + A(1)·x + A(0).

[@Dreamprism]

I was vaguely aware of this explicit formula before but had never proven it. Turns out it was less complicated than I was expecting. Very cool.

[@hydropage2855]

He proved it years back. Look for it. He set it up as a difference equation (not a differential equation but very similar to solve). It was quite the mess he made

[@tubbo2780]

The sequence can also be represented as a system of linear equations, and, when put into matrix form, you can derive the golden ratio by finding the eigenvalues of the transformation matrix.

[@phiarchitect]

The Golden Ratio is at a pivot point between the arithmetic and geometric world. And the value that is the bridge is *one,* the unit. *x^2 = x + 1*

[@dudewaldo4]

Can you be precise in what you mean by a pivot point?

[@phiarchitect]

@@dudewaldo4 maybe _inflection_ point is the better term. It's just poetry, after all.

[@phiarchitect]

@@dudewaldo4 but the simplest way to say it is in the geometric proportion: *a / b = b / (a + b)* or *1 / φ = φ / (1 + φ)*

[@AnonimityAssured]

I know what you mean. Both φ and its negative reciprocal, -1/φ, can form sequences that are simultaneously Fibonacci-type and geometric. Here are a few terms of one of those sequences in two equivalent forms, suitably spaced to keep equivalent terms approximately aligned: 0φ + 1, 1φ + 0, 1φ + 1, 2φ + 1, 3φ + 2, 5φ + 3, 8φ + 5, ... φ^0, φ^1, φ^2, φ^3, φ^4, φ^5, φ^6, ... This is actually just what BPRP was showing us, although he didn't explicitly mention the Fibonacci-type/geometric crossover. Note that each term is both the sum of the previous two terms (Fibonacci-type) and φ times the previous term (geometric). Each term in the sequence can be multiplied by any constant value, so infinitely many distinct sequences are possible, albeit differing only in scale.

[@phiarchitect]

@@AnonimityAssured What is most amazing is that nature uses this as an ordering principle.

[@tobybartels8426]

Whoa, you had to bring out the _blue_ pen for this one!

[@blackpenredpen]

Yup 😆

[@jessetrevena4338]

Love you videos! Best math channel on KZbin.

[@mathisnotforthefaintofheart]

Interestingly, your quadratic equation turns out to be the Characteristic Equation of a 2 by 2 matrix who powers represent successive Fibonacci numbers. From diagonalization you can then find the formula to find the nth term directly by "plugging in". I sometimes use this as an application at the end of my Linear class (time permitting). This is typically well received by the students.

[@10-year-oldcalculus19]

Wow! I’ve been trying to find a formula for the n-th number of the Fibonacci sequence, this turned out to be very helpful!

[@20icosahedron20]

2:51 Nice solving😂

[@bruhyou197]

0:27 lmao

[@agabe_8989]

this is truly mesmerizing, thanks m8

[@EE-ho1iz]

I took a loooong look at the alternate proof he showed starting at 8:00, and I think I get the gist of it. Here's the recursive formula for Fibonacci sequence: F_n = F_(n-1) + F_(n-2) for any n ≥ 2 He assumed F_(n-1) and F_(n-2) to be in the exponential form a*r^(n) for r ≠ 0 and coming up with the F_n = a_1 * r_1^n + a_2 * r_2^n He then solve for r by turning the equation into a quadratic but excluding the coefficient for later. Once he had r, he subbed it back to the F_n = a_1 * r_1^n + a_2 * r_2^n. Now, he plugs in some value of n to figure out the a_1 and a_2 coefficient he excluded in the previous step. He plugs n=0 to find out that a_1 is actually equal to negative a_2. Then, he plug in n = 1 to find out that a_1 = 1. Since a_1 is equal to one, a_2 is equal to negative 1. Having all the pieces of the puzzle, he just plug everything back in and it yields the explitcit formula. Kinda neat! And that is also a good morning mental exercise Honestly, I still have two question though: Why should F_n-1 and F_(n-2) be exponential? Anyhow, thanks for reading! Edit: one question (clearly I can't count😅😅😅. Maybe the second question was "what is my second question?")

[@chessematics]

Fₙ₋₁ and Fₙ₋₂ are NOT exponential in the primary sense. They are the differences of two exponentials, because Fₙ is itself one such difference of two exponentials. Now with this new proof we know that those two expos are φⁿ and 1/φⁿ. It's because we are adding two of the previous terms to get the next term, which is much like the Discrete Derivative of an exponential function. Take, for example, 3ⁿ. ∆3ⁿ = 2•3ⁿ = 3ⁿ+3ⁿ It's like that. Instead of adding the same thing, we add something a bit less. And all of the above is loose, non-rigorous talk.

[@tmsniper9229]

@@chessematics for a more rigorous proof, we can notice that it's a linear equation, so the set of it's solutions is a vector subspace with a dimension of 2(the proof of it is not hard using matrices) because it's a recursive relation between Fn and 2 previous terms so it's enough to find 2 vectors solutions linearly inpdependant and in that way you form a base for that vector subspace, in that base you can express all solution by linear combination between the 2

[@chessematics]

@@tmsniper9229 sir, as a 10th grader, I'm not allowed to touch Linear Algebra yet, no matter how ambitious I might be. But I am very much looking forward to finish Multivariable calculus and then maybe Linear Algebra after the Board exams of 10th, while waiting for the results.

[@paulanthony312]

Ah yes, the good old "second order linear homogeneous recurrence relation with constant coefficients" or solhrrcc for short, if you'd like. I have a proof in my discrete mathematics textbook somewhere, but suffice it to say, this method using quadratics to find an explicit formula is only effective when dealing with a recurrence relation that references the previous two terms in the sequence AND has a nonzero coefficient in front of the term two back from the nth term.

[@tmsniper9229]

@@chessematics well, good luck on your studies fam, keep going and i know you've got what it takes

[@iaroslav3249]

He just casually extended the Fibonacci sequence to any number.

[@Sugarman96]

This is an interesting approach. The one that I'm familiar with is using the Z transform on the recursive formula, that one works very well.

[@netcat22]

Wonderfully elegant method! Great video!

[@gillesphilippedeboissay109]

I used to solve it by solving Fn+1=Fn+Fn-1 directly r^2=r+1 r1=(1+sqrt(5))/2 r2=(1-sqrt(5))/2 Fn=A * r1^n + B * r2^n Then find A an B with F0=F1=1

[@salmonsushi47]

Its so much easy and it also makes more sense.

[@massimhb8674]

This is incredibly elegant, because the only solution I knew of was linear homogeneous recurrence and whatnot. I wonder if you could generalize this line of thinking to other types of series?

[@gillesbarre8527]

very nice demonstration thank you

[@edwardromana]

What a wonderful presentation. !!! :) Thank You so much.

[@JohnDoe-lt4kl]

This is supercool ! Now please provide an even easier proof for the closed form formula for the Bell numbers. which count the number of partitions of a set with n elements.

[@andrewkarsten5268]

Which one are you referring to? Do you mean Σs_(n,k) where s_(n,k) a stirling number of second order?

[@JohnDoe-lt4kl]

@@andrewkarsten5268 Yes, OEIS sequence A000110.

[@chrisg3030]

In 𝔁⁵ - 𝔁³ - 𝔁² - 𝔁¹ - 𝔁¹ - 𝔁⁰ = 0, 𝔁 solves as -1, φ, -1/φ. Does it work for any other Fibonacci segment of exponents?

[@jahidalic54s]

Whao, what a great result ! I'll use this method to create a computer program to print Fibonacci series. Thank you !!!!

[@cook1416]

If you want a really interesting project using this formula, try using non-integer values of n and plotting those values on the complex plane, you’ll be sure to love it!

[@crichtonsaurus.]

I love how he’s holding a pokeball every time he uploads these videos and solves a math problem.

[@cy_rkv]

It's a microphone

[@dr.rahulgupta7573]

Excellent presentation. Vow !!

[@PritamShaw]

I liked just after seeing the back of the tshirt lol

[@mathsenhancersclass]

Nice video thanks

[@EducarePakInd]

nice video sir thanks for sharing this video . Wao, what a great result ! I'll use this method to create a computer program to print Fibonacci series. Thank you !

[@chocomint-tw]

Pretty perfect!

[@harryguanous7198]

I've never been on this video this early, this looks very nice

[@SakretteAmamiya]

It's so brilliant!! It can be applied to any Fibbonacci-like terms, right? like, for a term starts with F1=1, F2=3 then we use x^2 = 3x + 1 to solve Fn

[@yoyoezzijr]

This is super cool

[@scottleung9587]

Very cool!

[@blackpenredpen]

thanks!

[@kquat7899]

Another gem.

[@Ndiedddd]

That derivation is really clever!

[@Animalkingdom.x]

Great sir

[@kianushmaleki]

"It is super super cool" I totally agree. Lovely video.

[@claudebalzano7031]

J'adore !!!

[@squidy7771]

That's really cool

[@leofoxpro2841]

I was literally thinking about this yesterday, And he made the video about it😲😲

[@ntth74]

I read this on Wikipedia and have no idea what they are talking about. But your explanation is very easy to understand. Thank you very much!

[@yoyotir1360]

I dealt with this equation for my final oral evaluation of highschool: The general question was the way to find the number of ways to go up a series of n stairs steps, n being a positive integer. So to find the number of ways we use a mathematical induction (not sure that's how we say it since I study mathematics in french) We want to know the number of ways to climb a staircase with n steps (n being a positive integer) knowing that we can climb one or two steps at a time. For a staircase with 0 steps There is one way to climb, for 1 step there is one way to climb a staircase, 2 ways to climb a staircase with 2 steps either by climbing 2 steps at a time or by climbing one step then a second step, for a staircase with 3 steps there are 3 ways to climb either to climb 2 steps then a step or a step then 2 steps or a step then another step then another step, the more the steps the more the number of ways increases and the longer the calculations. We define the sequence Un giving the number of ways to climb a staircase with n steps, u0=1 and u1=1 Find the number of ways to climb a staircase with n+1 steps: We can either climb a single step, so there will remain n steps to climb and the number of ways to climb n steps is one Or we can climb 2 steps and there will then remain n-1 steps to climb, the number of ways to climb n-1 steps is Un-1 So the number of ways to climb n+1 steps is therefore Un+Un-1 so Un+1=Un+Un-1 The Fibonacci sequence noted Fn is the sequence such that f0=0 and f1=1 and Fn+1=Fn+Fn-1 (Un) has the same initial terms as Fn+1 and Un has the same recurrence formula as Fn+1 so Un is the fibonacci sequence except for one term so we can express Un by Un=Fn+1 then we admit that Un is a geometric sequence Un being defined by a double recurrence we put r²=r+1 its characteristic equation which is the equation that you deal with in your video

[@jet4794]

Can you solve the following? Given any triangle where you know Side a, Area A, Angle alpha. What is the perimeter of that triangle? e.g.: A=15 cm^2 a=10 cm alpha=75 degree

[@biancanistor8606]

Brilliant for sure!

[@rygerety8384]

A message from today's sponsor:

[@Eduardo-tq5sk]

You a genius indeed you my Doctor!

[@star_ms]

Wow...that's elegant...

[@eri4108]

ooh that's cool. I've once tried to solve the fibonacci sequence with generating functions, which also leads me to the phi but i don' know why. that's why!!

[@isolatedpotato5757]

Is it possible if you can convert (9i)^1/3 from rectangular to polar?

[@landsgevaer]

Since the second term becomes exponentially negligible, this also shows that the terms of ((1+sqrt(5))/2)^n / sqrt(5) quickly approach integer values for large n, despite it looking rather irrational. Even more interesting, so do the powers of phi itself.

[@PlayNewApp]

Awesome, my friend. 😍💕 Thanks for sharing. 👍 Have a nice weekend. 🤗✨

[@diskritis2076]

this should be taught at high schools!

[@AdoNir]

Beautiful. I solved this one by diagonalizing the matrix 1 1 1 0 And putting the n-th power.

[@hrperformance]

That's pretty cool I must say

[@WorthlessWinner]

Numberphile did a good video doing this in reverse deriving phi from this quadratic

[@procerpat9223]

pure magic

[@_wetmath_]

what if you put complex n into the formula, what will Fn be?

[@ClickBeetleTV]

The legendary third pen!

[@bruno_343]

is there a formula for the finite sum of the Fibonacci sequence?

[@Leonard_Chan]

Recursive formula can be done by linear differcing equation

[@josephpentony4804]

This is such a beautiful relationship.

[@vishalpatel9159]

This question is ine Pathfinder mathematics Indian edition of PMI chapter example 08

[@christianorlandosilvaforer3451]

what is most surprising for me is that u can get natural numbers trought irracional numbers... damn

[@leolesnjakovic8725]

VERY INFORMATION VIDEO!!! TANK YU!! (soryr for bad egluish)

[@cmilkau]

I think if you just write φ^(n+2) = φ^(n+1) + φ^n the connection to the Fibonacci numbers is really obvious.

[@anshumanagrawal346]

Right

[@manjugagare802]

Q.are there any two numbers whose both addition and multiplication is 1?

[@timothybohdan7415]

No. Let x be the number. Then x^2=1 and x+x=1. The solution to the first is x=+/-1. The solution to the second is x=1/2. Thus, no number x exists.

[@The1RandomFool]

This is relatively simple compared to the generating function method using series that I did to find the formula.

[@rshawty]

BAHAHAHA I noticed this fact alone like a month ago and now you make a video on it ! 😚

[@johnnolen8338]

Sounds like an accomplishment to be proud of, not something to upset 😡 you because somebody else figured it out too. Nothing in mathematics is secret.

[@rshawty]

@@johnnolen8338 noooo i’m happy !

[@rshawty]

@@johnnolen8338 I find the fact funny actually so i’m happy, first because i’m right finally and also because bprp make a video on it

[@johnnolen8338]

@@rshawty Then I am confused by your emoji. 😕

[@rshawty]

@@johnnolen8338 convinced now ?

[@ChizuruNoru]

Does anyone know where the equation x^2 - x - 1 = 0 come from?

[@alikhamraev2809]

Omg. You finally shaved that goatee. Thank God. My prayers were heard. Now it's 10 times more enjoyable to watch your vids

[@xorgate667]

Warning jumpscare at 2:50

[@rahimmazouz574]

رائع

[@adiaphoros6842]

If only all recursive relations can be solved this way.

[@MichaelPennMath]

They can with a bit of work....

[@lelouch1722]

LINEAR recursive relations can. When it's non linear, well... good luck.

[@paulanthony312]

In my limited experience, it's usually either this method (second order linear homogeneous with constant coefficients), or just guessing an explicit formula and then proving with induction..

[@DetectiveAndrey]

@@MichaelPennMath A collab between you and blackpenredpen would be cool. : )

[@thomasolson7447]

@@MichaelPennMath show me Chebyshev Polynomials then. It's recursive. It's just a variable. x^2-2x*X+1. I'll get you started. x^2 = 2*X*x-1 x^3 = (2*X*x-1)*x = 2*X*x^2-x = 4*X^2*x-x -2*X x^4 = (4*X^2*x-2*X-x)*x = 8*X^3*x-4*X*x -4*X^2+1 x^5=16*X^4*x-12*X^2*x+x -8*X^3+4*X x^6=32*X^5*x-32*X^3*x+6*X*x -16*X^4+12*X^2-1 In my experience with these, the x and non x can be separated into two different things. *Oh, I was looking at the wrong one. I got it figured out now. I know what's up. I'll leave it here for you guys.*

[@antoinetarant6025]

The most beautiful is, this equation with sqrt of 5 et power of n always give a Natural number.

[@vijaykulhari_IITB]

Shridharachharya formula is the best🤘🤘

[@georgeb8893]

And the Lucas sequence, 2,1,3,4,7,11,18,..., L0=2, L1=1, L(n+2)=L(n+1)+L(n), n>=0, has an even simpler formula: with phi=Golden Ratio, L(n)=phi^n+(-1/phi)^n, n>=0.

[@thomasolson7447]

What's the lucas sequence for x^2-2*x*X+1?

[@angelmendez-rivera351]

The formula is simpler as far as coefficients are concerned, but its sequential properties are not as nice, nor is the sequence conceptually as fundamental to the theory.

[@georgeb8893]

@@angelmendez-rivera351i think I also like the Fibonacci sequence more, but what I like about the Lucas sequence is that, since the other root -1/phi has magnitude less than 1, the formula shows that simply powers of the golden ratio become asymptotic to it. ie limit as n approaches infinity of phi^n-L(n)=0.

[@universogeometrado859]

challenge d^2/dx^2 of (e^sin(x))/i

[@noahvale2627]

Cool beans for sure.

[@blackpenredpen]

Thanks!

[@holyshit922]

With generating function we can solve recursive equation from scratch

[@FunkyPhilMusic]

Why are you keeping the ball in your hand?

[@mrinalrabha1759]

What about tribonacci sequence ?

[@mrinalrabha1759]

I couldnt solve thats why asking ...

[@paulxx4829]

sin(54°) = (1+sqrt(5)) / 4?

[@sethdon1100]

Yup.

[@hghghghghghghghghgh]

Haha, matrix exponentiation go brrrr

[@botiromondavlatov3299]

👍👍👍👍

[@ashutoshsahoo351]

DIntegrate(((1/Cos(x))*Sqrt(1+Sin(x)^2/Cos(x)^4)),x,0,Pi/3) please solve kardo

[@alibekturashev6251]

Okey I have a problem for you. The circle of x^2 + y^2 = R^2 and parabola y = R - a*x^2 are given (a > 0). Find all values of the parameter a such that the parabola and the circle have only one common point

[@spaghettiking653]

Thanks, this is a nice one :) First, if the curves are touching, then that means that their equations are both satisfied at once, and so are their derivatives, so: x^2+y^2 = R^2 (circle equation) y = R-ax^2 (parabola equation) If both of the equations are true (at the single point where they touch), then we can substitute y = R-ax^2 into the circle equation: x^2 + (R-ax^2) ^2 = R^2, expanding gives x^2 + R^2 -2aRx^2 + a^2 x^4 = R^2 →a^2 x^4 + x^2 -2aRx^2 = 0 →a^2 x^4 + (1 - 2aR) x^2 = 0 This is a quadratic equation (in x^2), so we can solve by factoring or quadratic formula, I'll just tell you right here the answers are x^2 = 0 or x^2 = (2aR-1)/a^2. Now, the derivatives are also equal. If we have x^2+y^2 = R^2, then 2x + 2y (dy/dx) = 0, dy/dx = -x/y. We know that y = R-ax^2, so dy/dx of the circle is -x/(R-ax^2). The parabola's derivative is dy/dx = -2ax; now we set them equal to each other: -x/(R-ax^2) = -2ax, 1/(R-ax^2) = 2a, →2a(R-ax^2) = 1, →-2a^2 x^2 + 2aR -1 = 0 Now you can see why we solved for x^2 :) Plug in the x^2 expression into this equation: -2a^2 ((2aR-1)/a^2) + 2aR -1 = 0 -2(2aR-1) + 2aR -1 = 0 -4aR+2 + 2aR -1 = 0 -2aR + 1 = 0 1 = 2aR Thus a = 1/2R ! Alternatively, if we use the solution x^2 = 0, we get -2a^2(0) + 2aR -1 = 0, 2aR = 1, a = 1/2R once again :)

[@spaghettiking653]

Also, if you let R < 0, then a < 0 still works, so you can extend the possibilities for a to all real numbers if you like.

[@alibekturashev6251]

@@spaghettiking653 no, this problem came from my physics problem where R>0😅. I needed to find minimal velocity of a stone which was thrown from the top of a hemisphere such that it won't touch it again

[@spaghettiking653]

@@alibekturashev6251 Ah, makes sense :L. What was the velocity you got in the end?

[@alibekturashev6251]

@@spaghettiking653 v >= scrt(gR)

[@vitalsbat2310]

next, differentiate the fibonacci sequence

[@LuisFernando21Gta]

I use z-transform to this problem

[@maxvangulik1988]

Could somebody use this formula to make it continuous?

[@maxvangulik1988]

Like the gamma function but not an integral

[@BRYDN_NATHAN]

. thank you for the stories my guess a negative phi is imaginary like square root a negative one. .

[@l.lawliet1551]

Tell me the location of fibonacci.

[@cmilkau]

I wonder how this was discovered. Maybe someone "encoded" the pair (F(n-1),F(n-2)) as a polynomial F(n-1)x + F(n-2).

[@iAzazelHD]

kzbin.info/www/bejne/d2bXc6muebJsjJY On this video exponents are n+1. Here are just n. I am confused

[@EternalLoveAnkh]

I love this proof. RJ

[@aashsyed1277]

But how do we know that x²-x-1 is the correct quadratic?

[@dqrk0]

if u dive into the theory of recursive sequences, u will find about characteristic polynomials of a sequence. for example, for the fibonacci sequence recursive formula is a_(n+2) = a_(n+1) + a(n), so its polynomial is x² - x - 1

[@MarioFanGamer659]

The idea is that you with the sequence f_(n) + f_(n+1) = f_(n+2), the ratio of the current and next number approach a certain value the higher n becomes. At infinity, the values are equal so you can say that f_(n+2) / f_(n+1) = f_(n+1) / f_(n). With a = f_(n), b = f_(n+1) and c = f_(n+2), you get the equations c/b = b/a and c = a + b which in turn can be written as (a+b)/b = b/a which can be rewritten as 1/(b/a) + 1 = b/a. If you replace b/a by Φ, you get 1/Φ + 1 = Φ and multiplied by Φ, the equation turns into Φ² = Φ + 1. Move all the numbers to the left side and you get 0 = Φ² - Φ - 1.

[@angelmendez-rivera351]

We know it because the Fibonacci sequence is defined by the recursion F(n + 2) = F(n + 1) + F(n), which can be rewritten as F(n + 2) - F(n + 1) - F(n) = 0. Now, notice that x^2 - x - 1 = 0 implies x^(n + 2) - x^(n + 1) - x^n = 0, and so F(n) = x^n solves the equation if x solves x^2 - x - 1 = 0.

[@aashsyed1277]

@@angelmendez-rivera351 yes I know but can this method which is used in the video be extended to solve an+1=b*an+c*a n-1 ???

[@angelmendez-rivera351]

@@aashsyed1277 Your notation is incomprehensible, so I am not sure of what you are asking me.

[@paulchapman8023]

I wonder if there is a contiguous real function of x that gives the Fibonacci numbers at integer values of x, but also real values at non-integer values of x. In theory, just taking the absolute value of the function, i.e. f(x) = |([(1+sqrt5)/2]^x - ([(1-sqrt5)/2]^x)/sqrt5| should do the trick.

[@chrisrybak4961]

Nice thought, but the Problem is that phi2 is negative, so non-integer x will generate complex results…

[@paulchapman8023]

@@chrisrybak4961 Absolute values are always real and positive, and every complex number has an absolute value. (When expressed in polar coordinates, i.e. re^(i*theta), r is the absolute value.)

[@assassin01620]

@@paulchapman8023 Is the function f(x) = [ (φ^x) - cos(πx)*(φ^(-x)) ] / sqrt(5) what you are looking for?

[@chrisrybak4961]

@@paulchapman8023 Ah, ok, fair enough!

[@paulchapman8023]

@@assassin01620 Yes, that checks out. Thank you.

[@SmallsProudctions]

Nerds rock 🤓

[@arequina]

To phi or not to phi. That is the question.