Fibonnaci序列黃金比例在R

Question

我嘗試編程一個斐波納契序列，斐波那契數除以其前一個前導的除法後，停止斐波那契序列，使黃金比接近於0.001的差值。但是，下面的代碼似乎沒有工作，不知道爲什麼。提前致謝！

GoldenRatio=(1+sqrt(5))/2 
i=3 
fib=c(1,1) 
while(fib[i-1]/fib[i-2]-GoldenRatio>0.001){ 
fib[i] <- fib[i-1]+fib[i-2] 
i=i+1} 
print(fib) 
length(fib) 

Answer 1

試試這個：

GoldenRatio=(1+sqrt(5))/2 
# 1.618034 
i=3 
fib=c(1,1) 
while(abs(fib[i-1]/fib[i-2]-GoldenRatio)>0.001){ 
    fib[i] <- fib[i-1]+fib[i-2] 
    i=i+1 
} 
print(fib) 
# [1] 1 1 2 3 5 8 13 21 34 55 
length(fib) 
# [1] 10 
print(i) 
# 11 
print(fib[i-1]/fib[i-2]) 
# [1] 1.617647 
print(abs(fib[i-1]/fib[i-2] - GoldenRatio)) 
# [1] 0.0003869299 

而且，我們可以從下面看，是相當快的達到收斂：

fib=c(1,1) 
ratio <- c() 
for (i in 3:20) { 
    ratio <- c(ratio, fib[i-1]/fib[i-2]) 
    fib[i] <- fib[i-1]+fib[i-2] 
    i=i+1 
} 
plot(ratio, pch=19,col='red') 
lines(ratio, pch=19,col='red') 
abline(h=GoldenRatio, col='blue') 
legend('topright', legend=c('f(i)/f(i+1)', 'GoldenRatio'), col=c('red', 'blue'), lwd=2) 

Answer 2

比較的絕對差

GoldenRatio = (1+sqrt(5))/2 
i = 3 
fib = c(1,1) 
while (abs(fib[i-1]/fib[i-2] - GoldenRatio) > 0.001) { 
    fib[i] <- fib[i-1] + fib[i-2] 
    i = i+1 
    } 
print(fib) 
length(fib)