Fibonacci Variants

[@worldnotworld]

Thank you. It's fascinating that variations of the Fibonacci series always seem have the F-series "within" them; it has indeed an elemental quality. The first variant, the product variant, looks profoundly different, though, in terms of behavior, with a limit of zero for a, b1... but what if one is >1 and that other

[@DrBarker]

This sounds like a fun idea to explore - do let me know if you find anything interesting!

[@danielmilyutin9914]

I usually solve unhomogenous equatioon as it is done with ODE. I believe it is called variation of constant method. We search unhomohenous soution in form: u_n= A_n*F_n+B_n*F_{n-1} Putting all that stuff into equation will lead to lots of cancellations. And equations for A_n, B_n will look like simple integration equations (have to): A_{n} = A_{n-1} + someG_{n} B_{n} = B_{n-1} + someH_{n}

[@sabotagedgamerz]

I’ve studied other Fibonacci style sequences in the form U_n+2 = a*U_n+1 + b*U_n. Where a and b are constants. I found their respective “golden ratios” and their nth term formulas. It was really fun.

[@dr.rahulgupta7573]

Excellent presentation 👌

[@pmmeurcatpics]

Nice video! It was very satisfying to see the proof to come together at the end

[@user-jc2lz6jb2e]

7:38 the sum at the end looks like a convolution, so maybe it would be useful to use generating functions. It looks like the product of 1/(1-x) with the generating function for the Fibonacci numbers (shifted by x or 1/x)

[@holyshit922]

I prefer generating functions (ordinary or exponential) to solve this equation Z transform also can be useful Lets try another sequence u_{0}=1 u_{n+1}=sum(u_{k}F(n-k),k=0..n) where F(n) is shifted Fibonacci sequence with F(0)=1, F(1)=1, F(n+2)=F(n+1)+F(n) Generating function for sum of Fibonacci sequence is S(x)=x/((1-x)(1-x-x^2)) The first sequence may be problematic for generating functions

[@yuyu-mm8pk]

Im Japanese High school student but your video is amazing! Subscrived.

[@MrRyanroberson1]

here's an idea: are there any simple sequences where the closed formula is something like... the product of fibonacci numbers? or is such a product always reducible to some other fibonacci number?

[@MrRyanroberson1]

also: with F0 = 0, 13:12, it does quite work. notice for n=1: a F(-1) + b F(0) + F(3) - 2 = a + 0 + 2 - 2 = a. for n=2: a F(0) + b F(1) + F(4) - 3 = 0 + b + 3 - 3 = b. This also means you can extend it to n=0 with a F(-2) + b F(-1) + F(2) - 1 = b-a, which works with the definition as U(1) + U(0) + 0 = b-a + a = b = U(2). The magic of fibonacci is you basically never DON'T get something that works once you've found a formula

[@aytunch]

Great video Doc, thanks.

[@hypebeastuchiha9229]

I like your channel. Any plans for more videos on statistics?

[@DrBarker]

Thank you! I haven't covered much statistics lately, but I do like to keep a mix of topics, so will probably do more stats videos at some point in the future.

[@peterhall6656]

nice

[@Earendil_Maths]

:D