Taylor series using For loop to approximate Sin(x).

function y = SIN(x)
    %SIN This function takes the value and processes the approximate sin
    %value of that input
    %   The value of sin is approximately calculated using Taylor Series
    %   from the input value x.
   Sum = 0;
    T = 1E-12; %defining tolerance.
    for n = 0:29
        an = ((-1)^n)*((x^((2*n)+1))/factorial((2*n)+1));
        Sum = Sum + an;
        if abs(an) < T || n==29
            break
        elseif abs(an) == T
            disp 'More Iterations are needed to reach the specified tolerance.';
        end
    end
    y = Sum;    
end

the answer i am getting for lets say SIN(-3) = 4.500

however, the expected answer from using the built-in function sin(-3) = -0.1411.

how can i get the expected answer? I am very confused. THIS HAS TO BE DONE USING FOR LOOPS. How can i fix this code?

Answer 1

[Walter Roberson]

SIN(-3)

More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.
More Iterations are needed to reach the specified tolerance.

ans = -0.1411

function y = SIN(x)
    %UNTITLED Summary of this function goes here
    %   Detailed explanation goes here
    n = 0; Sum = 0; an=1;
    T = 1E-12; %defining tolerance.
    for n = 0:30
        an = ((-1)^n)*((x^((2*n)+1))/factorial((2*n)+1));
        Sum = Sum + an;
        if abs(an) < T || n==30
            break
        else
            disp 'More Iterations are needed to reach the specified tolerance.'
        end
    end
    y = Sum;    
end

You had the wrong starting point for n, and you had the wrong test for breaking.

[Humza Khan]

Thank you so much!