索引超出矩阵尺寸弹出错误

Question

I am trying to do an approximation of 1/e

I keep getting an error saying index exceeds matrix dimensions. Why?

n = 0;
eqn = 0;

while abs(eqn - (1/exp(1))) > .0001
n = n+1;
eqn = (1 - (1/n))^n;

end

nsav = n;
appr = eqn;
builtin = 1/exp(1);
fprintf ('The built in value is %.4f\n' , builtin)
fprintf ('The approximation is %.4f\n', appr)
fprintf ('The value of n required for such accuracy is %d\n', nsav)

Answer 1

EDIT

This answer doesn't really solve the user's problem. But since what the user wanted was a loop to approximate 1/e, here are some possibilities:

One quick approach:

tic
N = 1e6;
N1 = 1;
N2 = N1 + N - 1;
err_exp = 1e-7;
ctrl = 1;
while ctrl
  n = N1:N2; eqn = (1 - (1 ./ n)) .^ n;
  inds = find(abs(eqn - 1/exp(1)) < err_exp, 1, 'first');
  if ~isempty(inds)
    ctrl = 0;
  else
    N1 = N1 + N;
    N2 = N1 + N - 1;
  end
end
fprintf('n must be at least %d\n', N1 + inds - 1)
toc
% Elapsed time is 0.609669 seconds.

A better approach is the dichotomic search:

tic
N = 1e6;
a = 1 / exp(1);
err_exp = 1e-15;
ctrl = 1;
n = 1;
% 1º Step: find starting values for n1 and n2
while ctrl
  eqn = (1 - (1 / n)) ^ n;
  ctrl = abs(eqn - a) > err_exp;
  if ctrl
    n1 = n;
    n = 2 * n;
  end
end
n2 = n;

% 2º step: search dichotomically between the extremes
ctrl = 1;
while ctrl
  if n2 - n1 < N
    n = n1:n2; eqn = (1 - (1 ./ n)) .^ n;
    n = find(abs(eqn - a) < err_exp, 1, 'first');
    eqn = eqn(n);
    n = n1 + n - 1;
    ctrl = 0;
  else
    n = floor((n1 + n2 + 1) / 2);
    eqn = (1 - (1 / n)) ^ n;
    chck = abs(eqn - a) > err_exp;
    if chck
      n1 = n;
    else
      n2 = n;
    end
  end
end
toc
% Elapsed time is 0.231897 seconds.

[abs(eqn - a), err_exp]
[n1 n n2]

Just for the fun of it.

Answer 2

Type whos and make sure abs and exp aren't listed. If they are listed as variables, clear them with:

clear abs whos

Then make sure there is no where else above this code that sets abs and whos as variables.

For the record, it seems there is some strange version-specific popup error that happens when you do builtin=exp(1) . David pointed this out. Unfortunately, I can't reproduce it with R2013b 64-bit.

Answer 3

OK so if I run this code

a=exp(1)

clear

builtin=exp(1)

I get

a =
    2.7183
builtin =
    2.7183

and this error

However if I run this code

a=exp(1)

clear

builtin=exp(1)

clear

a=exp(1)

I get no error!

This is R2012b on Mac OSX.