Finding zero without interp1.

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

1. How to find the value of z when a is zero? (Without using interp1) Can I use fzero? Or a For loop & If-Else?
2. 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.)

[Matt J]

How to find the value of z when a is zero? (Without using interp1)

The sequence a(k) is defined only for integer k. Unless a(z)=0 for z also an integer, you will have no choice but to define some kind of interpolation between integer points.

[Walter Roberson]

or to decouple the storage location from the x value... like I show in my Answer

[Matt J]

I would argue that that is a choice of interpoaltion...

Answer 1

[Walter Roberson]

You will need a loop, unless you can solve the recurrence relationship. (MATLAB does not offer any tools for recurrence relationships.)

[Capami]

Could you please provide the code for that?

[Walter Roberson]

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]

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]

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.