math equation using function

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.

Answer 1

[Image Analyst]

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.

[Roger Nadal]

e is euler number and i/p value should be err. function should calculate an approximation of e using formula such that the difference between the approximation and the actual value of e is smaller than err. The function will then return the corresponding value of n.

[Image Analyst]

It seems like they want you to compute e by summing a series. See if you can construct a while loop that accumulates a sum and breaks out of the loop when the error gets small enough. I think I've given you a good start above. Just construct the sum of the series.

[Roger Nadal]

Function should return the value n and how it will calculate the difference between approximation and the actual value of e is smaller than err?

[Image Analyst]

Correct. Do a for loop over, say a million iterations and break when the error is less. Since it seems to be homework, here's a hint

theSum = 0;
correctValue = 1 / exp(1); % or exp(1) depending on how you're constructing the series.
minAllowableError = whatever you want.
for n = 1 : 1000000
    newTerm = ... compute the nth term.
    theSum = theSum + newTerm;
    theErrorSoFar = theSum - correctValue
    if theErrorSoFar < minAllowableError
        % The error is now small enough and we can bail out now.
        break.
    end
end
fprintf('We did %d iterations and got an error of %f.\n', n, theErrorSoFar)

[Roger Nadal]

Not getting the proper answer in function it should receive input value of err and calculate approx value to e and difference of approx and actual e is small than err

[Image Analyst]

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?