Find n for inverse of e approximation

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Victoria Gonzalez Canalle
回答： Koushik Vemula 2020-3-3
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!
  2 个评论
Adam
Adam 2020-2-19
编辑：Adam 2020-2-19
while 0.0001 > abs(inv-guess)
That effectively says, while the difference is very small carry on with the loop. I would imagine you want a < instead. Or, as I normally would, reverse the order of the operands to make it easier to understand!
In your case I'm guessing the condition fails first time so the loop is never executed.
These things are trivial to see using the debugger though.
Jacob Wood
Jacob Wood 2020-2-19
It looks like the only values n is able to take on is 1,2,3,4,...
Perhaps a bisection search, where you iteratively refine the value of n, would be a more appropriate solution?
This link might be helpful: https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/10RootFinding/bisection/matlab.html

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Koushik Vemula
Koushik Vemula 2020-3-3
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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