We can also use the generating function for the Fibonacci numbers which is x + x² + 2x³ + 3x⁴ + 5x⁵ + ... = Σ Fn x^n (n = 1...∞) = x / ( 1 - x - x²). Here we get S = x + x² + 2x³ + 3x⁴ + 5x⁵ + ... for x = 1/2 and so 1/2 / (1 - 1/2 - 1/4) = 2. So the sum you are looking for at the end of the video would be (1 / λ) / ( 1 - 1/λ - 1/λ²) - 1 = λ / (λ² - λ - 1).
you're obliged to show the series converges first. AFAIK that wants binets formula, which is the natural tool for just bludgeoning the problem into submission anyhow; you express it explicitly as the sum of two geometric series and push around some annoying irrational numbers for a while. for viewers who don't understand the motivation for the steps in the vid: in the expression for S you first segregate the first term from the rest, then for the series part you use F(n) = F(n-1) + F(n-2). this recurrence doesn't apply F(1) (or F(0), which is absent from this problem), which is why we treat that term separately. the series (serises?) with F(n-1) and F(n-2) you reindex each one to restore the subscript to F(n), and with very little clean-up you wind up with S = 1/2 + S/2 +S/4 just as in the vid.
@@theupson ah yes you're totally correct. I forgot to show that it converges first, but this can easily be shown using the limit comparison test. (As the Fibonacci numbers approach are of order ((1+√5)/2)^n which is less than 2^n, and so the sum can be bounded above by a convergent geometric sum.
We can also use the generating function for the Fibonacci numbers which is
x + x² + 2x³ + 3x⁴ + 5x⁵ + ... = Σ Fn x^n (n = 1...∞) = x / ( 1 - x - x²).
Here we get S = x + x² + 2x³ + 3x⁴ + 5x⁵ + ... for x = 1/2 and so 1/2 / (1 - 1/2 - 1/4) = 2.
So the sum you are looking for at the end of the video would be (1 / λ) / ( 1 - 1/λ - 1/λ²) - 1 = λ / (λ² - λ - 1).
Feliz Natal
Merry christmas!
You too🎉
@OliverGoodman-todd and a happy new year!
you're obliged to show the series converges first. AFAIK that wants binets formula, which is the natural tool for just bludgeoning the problem into submission anyhow; you express it explicitly as the sum of two geometric series and push around some annoying irrational numbers for a while.
for viewers who don't understand the motivation for the steps in the vid: in the expression for S you first segregate the first term from the rest, then for the series part you use F(n) = F(n-1) + F(n-2). this recurrence doesn't apply F(1) (or F(0), which is absent from this problem), which is why we treat that term separately. the series (serises?) with F(n-1) and F(n-2) you reindex each one to restore the subscript to F(n), and with very little clean-up you wind up with S = 1/2 + S/2 +S/4 just as in the vid.
@@theupson ah yes you're totally correct. I forgot to show that it converges first, but this can easily be shown using the limit comparison test. (As the Fibonacci numbers approach are of order ((1+√5)/2)^n which is less than 2^n, and so the sum can be bounded above by a convergent geometric sum.
Se eu fosse o produtor do vídeo, daria importância às suas queixas.
pi minutes long
@qqqalo haha what a pleasant coincidence!
attentive like Sherlock Holmes ❤😊