MATLAB: Find n for inverse of e approximation

inverse

Hi! I have to find an 'n' that approximates 1/exp(1) correctly to 0.0001, and then diplay which n that was.

% write code to compute the value of the inverse to within the tolerance:
inv = 1 / exp(1);
n = 1;
guess = (1-(1/n))^n; 
while 0.0001 > abs(inv-guess)
    n = n+1;
    guess = (1-(1/n))^n;
end
% store the value of n needed to reach this accuracy:
accurate_n = n

But at the end,

accurate_n = 1 and I'm not sure why it is not treating it like a normal counter.

Anything helps!

Best Answer

According to what I understand, you are trying to get the value of ‘accurate_n’ which gives you a tolerance of ‘0.0001’. The problem is not in your approach but your condition in the while loop

while 0.0001 < abs(inv-guess)

Will give the expected answer which is ‘1840’

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MATLAB: When I assign x(n)=0.5 and x(n-1)=0.055 they are gathered into one vector? as x=[0.5 0.055]

You might as well use rand() for the initial guesses, unless you have physical knowledge that would suggest other values.

You are going to need to change your code. You have

f(x(n))=((-(100-DOT(i)))/(100))+(1/(x(n)-60))*(x(n)*exp(-Time(i)/x1)-60*exp(-Time(i)/60))
for i=1:43
    while abs((f(x1)))>0.001
            x(n+1)=x(n)-((x(n)-x(n-1)))*(f(x(n))/(f(x(n))-f(x(n-1))))
            x0=x1;
            x1=x2;
            n=n+1;
    end
end

which tries to define a formula for f(x(n)) based upon general n and i values. The only way to define a formula is with the Symbolic Toolbox, and symbolic formulas need to be define in terms of variables rather than array elements.

Adjusted code:

x1 = nan;      %we do not know what this variable is for or how it should be initialized

x2 = nan;      %we do not know what this variable is for or how it should be initialized
for i=1:43
  f = @(X, X1) ((-(100-DOT(i)))/(100))+(1/(X-60))*(X*exp(-Time(i)/X1)-60*exp(-Time(i)/60));
      while abs((f(x(n), x1)))>0.001
              x(n+1) = x(n)-((x(n) - x(n-1))) * (f(x(n), x1) / (f(x(n),x1) - f(x(n-1),x1)));
              x0=x1;
              x1=x2;
              n=n+1;
      end
  end

This adjusted code is unlikely to work, as I see errors in your formula, but it will at least pass syntax checking.

Best Answer

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