Which is the smallest natural number n to which it applies a(n)>10?

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I have recursive sequence a(i)= a(i-1) + a(i-2)^(-1), where is a(1) and a(2) equals 1. So I have to find smallest natural number n to which it applies a(n)>10.
This is my code so far:
a = zeros(1,15);
a(1) = 1;
a(2) = 1;
for i = 3:15
a(i)= a(i-1) + a(i-2)^(-1);
end
Any help?

Accepted Answer

David Hill
David Hill on 18 Nov 2021
a(1)=1;a(2)=1;
for k=3:1000 %pick something to overshoot or use while loop
a(k)=a(k-1)+1/a(k-2);
end
n=find(a>10,1);%48!

More Answers (1)

John D'Errico
John D'Errico on 18 Nov 2021
Edited: John D'Errico on 18 Nov 2021
The obvious answer is brute force, thus...
a_i = 1;
a_iplus1 = 1;
plot(a_i,a_iplus1,'o')
hold on
iter = 2;
imax = 1e6;
while (a_iplus1 < 10) && (iter < imax)
iter = iter + 1;
[a_i,a_iplus1] = deal(a_iplus1,a_iplus1 + 1/a_i);
plot(a_i,a_iplus1,'o')
end
grid on
xlabel a_i
ylabel a_iplus1
a_iplus1
a_iplus1 = 10.0922
iter
iter = 48
So when iter = 48, a finally grows larger than 10.
Note that I wrote the code so that no arrays are grown. This is important, since the loop might have gone on forever. This is why I put an upper limit on the loop. The trick using deal to advance the terms is well, just a cute trick.
Does a general analytical solution to this nonlinear difference equation exist? Possibly. Such nonlinear difference equations tend to have "interesting" behaviour. As soon as you dive into the domain of the nonlinear, things can go straight to well, you know where. The plot however, suggests a simple asymptotic behavior, one that makes sense when you look at the expression. Questions now come up, like is there a limiting value? Or will a(i) grow forever, unbounded?

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