math equation using function
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function will recive input Value a and it caluclate its approximate value to e. e=1/e~~(1-1/n)^n. e value that the difference between the approximation and the actual value of e is smaller than err.function should return the corresponding value of n.
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Image Analyst
2019년 11월 24일
I don't know what that means. Let's say that a=8, and e=2.718281828. Do you simply want to return an ouput = (exp(1) - a)??? And do you just want to do a while loop, incrementing the value of n, until (1-1/n)^n is less than 1/e ???
n=0;
e1 = 1 / exp(1)
while difference > e1
difference = abs(e1 - (1-1/n)^n);
n = n + 1;
end
n = n - 1; % Undo last addition before the break.
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Image Analyst
2019년 11월 26일
Well, only Walter here has the Mind Reading Toolbox, not me, so let's see what code you have so far, if you still want/need help. What modifications did you make to the snippet I gave you?
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