Vector where each value is X times larger than the last

I'm trying to recreate some log scale plots reported in this paper, the method they use calls for a vector of bin edges such that each one is 3.1% larger than the previous one.

I can do this quite easily using a while loop:

s1 = 1; % minimum bin value
s2 = 1000; % maximum bin value we would like
m = 1.031; % scaling factor, each bin will be m*the last one in size
xi = s1; % starting point
while 1
    xi(end+1) = xi(end)*m; % new bin is the last one*m
    if xi(end)>s2 % when we exceed value s2, stop
        break
    end
end

This code obviously has some serious drawbacks, but it is also just very ugly.

I was wondering if someone could think of a more elegant and efficient approach?

[Stephen23]

For those interested to know what that loop produces:

s1 = 1; % minimum bin value
s2 = 1000; % maximum bin value we would like
m = 1.031; % scaling factor, each bin will be m*the last one in size
xi = s1; % starting point
while 1
    xi(end+1) = xi(end)*m; % new bin is the last one*m
    if xi(end)>s2 % when we exceed value s2, stop
        break
    end
end
display(xi)

xi = 1×228

1.0000 1.0310 1.0630 1.0959 1.1299 1.1649 1.2010 1.2383 1.2766 1.3162 1.3570 1.3991 1.4425 1.4872 1.5333 1.5808 1.6298 1.6803 1.7324 1.7861 1.8415 1.8986 1.9574 2.0181 2.0807 2.1452 2.2117 2.2803 2.3509 2.4238

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Just for fun, without CUMPROD:

s1 * m.^(0:ceil(log(s2/s1)/log(m)))

ans = 1×228

1.0000 1.0310 1.0630 1.0959 1.1299 1.1649 1.2010 1.2383 1.2766 1.3162 1.3570 1.3991 1.4425 1.4872 1.5333 1.5808 1.6298 1.6803 1.7324 1.7861 1.8415 1.8986 1.9574 2.0181 2.0807 2.1452 2.2117 2.2803 2.3509 2.4238

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[Neuroesc]

I will use cumprod because I prefer to use in-built functions where possible, but thank you for this answer, it helped me a lot to understand why this works. I'm sure it will also be very useful to people who don't want to or can't use cumprod.

Answer 1

[Steven Lord]

Use cumprod.

x = [2 3*ones(1, 4)]

x = 1×5

2 3 3 3 3

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y = cumprod(x)

y = 1×5

2 6 18 54 162

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If you don't know how many elements you want y to have, just what you want the largest value to be, solve the equation (multiplication factor)^n = (upper bound) [or M^n = U] by taking the log of both sides and dividing.

U = 2000;
M = 3;
n = ceil(log(U)./log(M));
x = [1 M*ones(1, n)];
y = cumprod(x)

y = 1×8

1 3 9 27 81 243 729 2187

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Note that since I used ceil the largest value in y is actually greater than U. If I wanted to stop just before the largest value in y would be greater than U I'd have used floor instead.

[Neuroesc]

This is fantastic, thank you.