Thanks for the quick response, but what this loop seems to do is tell me the entirety of the series. I was only looking for a response that returns the value at the 10, 100 and 1000th place mark on the series.
Iterating over an Array Using a “for” Loop.
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This is a very basic question and I would appreciate any help. I've written code to calculate the Fibonacci sequence using a “for” loop. I want to display the elements of the sequence whose indices are specified in the array “N”. The problem is that all displayed values are the same as the value associated with the first element of “N”.
N=[10 100 1000];
first=1;
second=1;
for i=1:(N-2) %The index has to have two terms removed because it starts with 1 and 1 already
next=first+second; %The current term in the series is a summation of the previous two terms
first=second; %Each term must by iterated upwards by an index of one
second=next; %The term that previously was second is now referred to as next
end
for i=1:length(N)
disp([N(i) next])
end
Thanks
2 件のコメント
Maryam Omar Maabreh
2023 年 2 月 5 日
移動済み: DGM
2023 年 2 月 6 日
for i=start : step : last+1
% ...
end
Where step is the increment to i.
In your case, your looking for
for i=1 : 10 : 10001
i = i*10;
% ...
end
採用された回答
Image Analyst
2016 年 4 月 17 日
編集済み: Image Analyst
2016 年 4 月 17 日
Donald, regarding your new "Answer", you can do it like this:
N=[10 100 1000];
for n = 1 : length(N)
first=1;
second=1;
for k = 3 : N(n) %The index has to have two terms removed because it starts with 1 and 1 already
next = first+second; %The current term in the series is a summation of the previous two terms
first = second; %Each term must by iterated upwards by an index of one
second = next; %The term that previously was second is now referred to as next
end
% Print to command window
next
end
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その他の回答 (1 件)
MathWorks Support Team
2020 年 9 月 2 日
移動済み: DGM
2023 年 2 月 6 日
Use a “for” loop to calculate the elements of the Fibonacci sequence for the maximum value in “N”. Then, use another “for” loop to display the values by indexing into the calculated sequence.
N = [10 100 1000];
f(1) = 1;
f(2) = 1;
for i = 3:max(N)
f(i) = f(i-1) + f(i-2);
end
for i = 1:numel(N)
disp(f(N(i)))
end
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