Lucas meets Fibonacci

Calculus Level 3

The n n th Lucas number is defined such that L n = L n 1 + L n 2 L_n = L_{n-1} + L_{n-2} , where L 1 = 1 L_1 = 1 and L 2 = 3. L_2 = 3.
Similarly, the n n th Fibonacci number is defined such that F n = F n 1 + F n 2 F_n = F_{n-1} + F_{n-2} , where F 1 = 1 F_1 = 1 and F 2 = 1. F_2 = 1.
Let Q n Q_n be the ratio L n F n . \frac{L_n}{F_n}. With simple calculation we find that Q 1 = 1 1 = 1 , Q 2 = 3 1 = 3 , . . . Q_1 = \frac{1}{1} = 1,\ Q_2 = \frac{3}{1} = 3, ...

Find lim n Q n . \lim_{n\to\infty} Q_n.


The answer is 2.236068.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Incredible Mind
Feb 10, 2015

quite simple if u know

Ln = F(n-1)+F(n+1)

hence

Qn= F(n-1)+F(n+1) /Fn

Qn=2F(n-1) /Fn + 1

Qn=2(1/golden ration) + 1 .at inf

hence ANS = sqrt5

Lamer Zhang
Feb 11, 2019

用特征根法 由于两个数列的递推式相同,故其特征根相同。 由 f ( n ) = f ( n 1 ) + f ( n 2 ) f(n)=f(n-1)+f(n-2) 可得特征方程 x 2 = x + 1 x^2=x+1 ,解得 x 1 = 1 5 2 , x 2 = 1 + 5 2 x_1=\frac{1-\sqrt5}{2},x_2=\frac{1+\sqrt5}{2} 。 可设 L ( n ) = a x 1 n + b x 2 n F ( n ) = c x 1 n + d x 2 n L(n)=a \cdot x_1^n+b \cdot x_2^n \\F(n)=c \cdot x_1^n+d \cdot x_2^n

L ( 1 ) = 1 , L ( 2 ) = 3 , F ( 1 ) = 1 , F ( 2 ) = 1 L(1)=1,L(2)=3,F(1)=1,F(2)=1 代入,解得

c = 1 5 , d = 1 5 a = 1 , b = 1 c=-\frac{1}{\sqrt5},d=\frac{1}{\sqrt5} \\ a=1,b=1

X = x 1 n , Y = x 2 n , t = Y / X X=x_1^n,Y=x_2^n,t=Y/X

Q n = a X + b Y c X + d Y = a + b Y / X c + d Y / X = a + b t c + d t Q_n=\frac{a\cdot X+b\cdot Y}{c\cdot X+d\cdot Y}= \frac{a+b\cdot Y/X}{c+d\cdot Y/X}=\frac{a+bt}{c+dt} n n\rightarrow \infty 时, t t\rightarrow \infty 可用 L H o s p i t a l s r u l e L'Hospital's\;rule ,则 lim n + Q n = lim t + a + b t c + d t = b d = 5 \lim_{n \to +\infty}Q_n=\lim_{t \to +\infty}\frac{a+bt}{c+dt}=\frac{b}{d}=\sqrt5

thus the answer is 5 \sqrt5

Brock Brown
Mar 3, 2015

Python:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
lucas_mem = {1:1, 2:3}
def lucas(n):
    if n in lucas_mem:
        return lucas_mem[n]
    lucas_mem[n] = lucas(n-1) + lucas(n-2)
    return lucas_mem[n]
fib_mem = {1:1, 2:1}
def fib(n):
    if n in fib_mem:
        return fib_mem[n]
    fib_mem[n] = fib(n-1) + fib(n-2)
    return fib_mem[n]
infinity = 1000
# build tables to reduce recursion
for i in xrange(1, infinity):
    lucas(i)
    fib(i)
print lucas(infinity) / float(fib(infinity))

what a method of brutal force!

lamer zhang - 2 years, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...