making an Xn sequence
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I'm trying to code the function Xn=111-(1130/Xn-1)+3000/(Xn-1*Xn-2)
I don't know how to make this sequence in MatLab
I think I can figure out the rest of the problem, I'm just not sure how to do this part particularly
2 个评论
KSSV
2020-9-11
It is pretty simple and straight forward...what have you attempted?
Walter Roberson
2020-9-11
If n is at least 3 then 
would be written X(n-2) and 
would be written X(n-1) and 
would be written X(n)
would be written X(n-2) and would be written X(n-1) and would be written X(n)回答（1 个）
BOB MATHEW SYJI
2020-9-13
Hope this helps. x1 and x2 are the 1st and 2nd terms of the sequence respectively. n is the nth term you want.
x1=%the first number of the sequence
x2=%the second number of the sequence
n=%the nth element you want
x=[x1 x2];
for i=3:n
x(i)=111-(1130/x(i-1))+(3000/(x(i-1)*x(i-2)));
end
disp(x(n));
2 个评论
John D'Errico
2020-9-13
编辑：John D'Errico
2020-9-13
I would suggest you learn why and how to preallocate vectors when they will be grown. Else your code will be exceedingly slow, leaving you to soon post an anguished question of your own at some time - "Why is this code so slow?"
In this case, the code you wrote will create a vector of length n, but your code grows that vector in length one step at a time, forcing MATLAB to reallocate memory at every iteration. It also forces MATLAB to copy the entire vector over at every iteration. If n is at all large, this creates code that will run in O(n^2) time. It is an unnecesarily inefficient coding style, avoidable by a simple line at the beginning like
x = zeros(1,n);
BOB MATHEW SYJI
2020-9-13
Thank you so much for pointing this out
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2020-9-11
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