Finding zero without interp1.
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a=[8000];
for x=2:25
a(x)=[(0.97*a(end))-((250*(x-1)/x))];
end
z=1:25;
plot(z,a)
table(z', a')
I have two questions
- How to find the value of z when a is zero? (Without using interp1) Can I use fzero? Or a For loop & If-Else?
- How do I end the table when Var2=0? (The last row in the table should end with 24.8579 and 0. I do not want to include negatives.)
3 个评论
Walter Roberson
2022-6-5
or to decouple the storage location from the x value... like I show in my Answer
Matt J
2022-6-5
I would argue that that is a choice of interpoaltion...
回答（1 个）
Walter Roberson
2022-6-4
0 个投票
You will need a loop, unless you can solve the recurrence relationship. (MATLAB does not offer any tools for recurrence relationships.)
4 个评论
Capami
2022-6-4
Walter Roberson
2022-6-4
x = 2;
while a(end) > 0
a(end+1)=[(0.97*a(end))-((250*(x-1)/x))];
x = x + 1;
end
This will resolve down to an integer.
Walter Roberson
2022-6-4
x = 2;
for dx = 10.^-(0:5)
while a(end) > 0
a(end+1)=[(0.97*a(end))-((250*(x-1)/x))];
x = x + dx;
end
x = x - dx;
a(end) = [];
end
At the end of this, a(end) will be the last positive a before the zero crossing, and x will be the last x before the crossing. The accuracy will be 1e-5 in this code.
Walter Roberson
2022-6-5
One modification you might want to make here is to keep an x history so that later you can plot(xhist, a) since the a array entries are no longer equidistant in x space.
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