# Why does this function not work for decimals?

5 views (last 30 days)
TheSaint on 18 Feb 2024
Moved: Dyuman Joshi on 18 Feb 2024
x = input("Please enter the value of x (in radians): ");
approximatearctan = 0;
n = 0;
arctanactual = atan(x);
while (abs(approximatearctan - atan(x))>0.00001)
approximatearctan = approximatearctan + (-1)^n * x^(2*n +1) / factorial(2*n+1);
n = n + 1;
end
fprintf('The actual value for arctan(x) to eight decimal places given an input of %0.1f is %0.8f. \n', x, arctanactual)
fprintf('The approximate value of arctan(x) to eight decimal places given an input of %0.1f is %0.8f. \n', x, approximatearctan)
fprintf('The number of terms required to reach a five decimal place agreement between the approximate and actual values of arctan(x) is %0.0f \n', n)
I have this code, and it works fine for any number above one. However, I need it to work for numbers smaller than 1, e.g. 0.7. Whenever I put in a number that is <1, the program just gets stuck in an endless running state. Any input is appreciated.
Dyuman Joshi on 18 Feb 2024
Moved: Dyuman Joshi on 18 Feb 2024
The formula you have used is incorrect. There is no factorial in the formula.
Refer to this webpage for expansion of arc tan for different values - https://proofwiki.org/wiki/Power_Series_Expansion_for_Real_Arctangent_Function

Sulaymon Eshkabilov on 18 Feb 2024
Here is the corrected answer (Note abs(x)<=1):
% x = input("Please enter the value of x (in radians): "); % Here "input" prompt does
% not work, but it works in a MATLAB desktop:
% E.g.:
x = pi/4;
approximatearctan = 0;
n=0;
arctanactual = atan(x);
approximatearctan = 0;
while abs(approximatearctan - arctanactual)>1e-8
approximatearctan= approximatearctan + ((-1)^n * x^(2*n + 1)) / (2*n + 1);
n = n+1;
end
fprintf('The actual value for arctan(x) at x = %0.1f is %0.8f \n', x, arctanactual)
The actual value for arctan(x) at x = 0.8 is 0.66577375
fprintf('The approximate value of arctan(x): %0.1f is %0.8f \n', x, approximatearctan)
The approximate value of arctan(x): 0.8 is 0.66577376
fprintf('The number of terms required: %d \n',n)
The number of terms required: 29
fprintf('The difference is %1.8f \n', abs(approximatearctan - arctanactual))
The difference is 0.00000001

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