How Many Interations Does It Take Before Successive Iterations Do Not Change More Than 1E-6?

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This is Fibonacci's sequence and f3 is each element in the sequence divided by the one before it. This is supposed to demonstrate the Golden Ratio. However, I need my code to tell me how many times elements there are before each successive element is no greater than 10^-6 before it. I'm pretty sure I need to use a while loop, but I still can't get it right....
%fib seq using Binet Eq
a= 1:10;
b= sqrt(5);
x = (1-b)/2;
y = (1+b)/2;
f = (y.^a - x.^a)./b;
fb=reshape(f,[],5)
%ratio
c=a+1;
f2 = (y.^c - x.^c)./b;
f3=f2./f;

回答(1 个)

Santhana Raj
Santhana Raj 2017-4-13
The equation of b, x and y can be outside the loop and same as your definitions.
k=1;
while true
k=k+1;
f(k) = (y.^k - x.^k)./b;
c=k+1;
f2(k) = (y.^c - x.^c)./b;
f3(k)=f2(k)/f(k);
if(abs(f3(k)-f3(k-1))<10e-6)
break;
end
end
Hope it helps.

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