Write a script file that will compute the sine of an angle using the Taylor series formula:

Write a script file that will compute the sine of an angle using the Taylor series formula:

The program will prompt the user to input the angle in degrees, and the number of terms in the series. Use the program to calculate sin(150 degrees) using 5 and 9 terms.

Note: Can anyone help mw with this? So far this is what I have and I am not sure if it is even correct.

disp("Input the angle in degrees (x) and the number of terms (n)")
x = input('x: ');
n = input('n: ');
sum = 0;
deg = x*180/pi;
for k = 0:n
    y = ((((-1)^k)*deg^(2*k+1)))/factorial(2*k+1);
    sum = sum + y;
end;
fprintf('sin(%3.2f) = %1.2fln',x,sum)

Answer 1

[Mathieu NOE]

hello

please don't use sum or any other native matlab function names for variables names in your code; this can shadow the native function and create trouble in code execution (and unpredictable results).

beside that , your conversion from degrees to radian was wrong (as x is supposed to be in rad in the taylor formula)

also no need for for loop as your code can be easily vectorized.

code updated :

disp("Input the angle in degrees (d) and the number of terms (n)")
d = input('d: ');
n = input('n: ');

x = d*pi/180;

% out = 0;
% for k = 0:n
%     y = ((((-1)^k)*x^(2*k+1)))/factorial(2*k+1);
%     out = out + y;
% end;

k = 0:n;
y = ((((-1).^k).*x.^(2*k+1)))./factorial(2*k+1);
out = sum(y);

fprintf('sin taylor (%3.2f) = %1.4f\r',d,out) % taylor serie output
fprintf('sin (%3.2f) = %1.4f\r',d,sin(x)) % sin function output
fprintf('error percentage (%3.2f) = %1.4f\r',d,error_percent) % error (%) output

[John Doe]

There's an error here:

Array indices must be positive integers or logical values.

Error in Three1 (line 12)
out = sum(y);

[Mathieu NOE]

I have no error on ly side (R2020b)

>> sin_taylor

Input the angle in degrees (d) and the number of terms (n)

d: 54

n: 3

sin taylor (54.00) = 0.8090

sin (54.00) = 0.8090

error percentage (54.00) = 0.0003

[John Doe]

Oh okay, noted, but there is also another error. The error_percentage was never declared or identified in the code.

[Mathieu NOE]

sorry

probably error of copy paste - that line disappeared by mistake

correction :

disp("Input the angle in degrees (d) and the number of terms (n)")
d = input('d: ');
n = input('n: ');
x = d*pi/180;
% out = 0;
% for k = 0:n
%     y = ((((-1)^k)*x^(2*k+1)))/factorial(2*k+1);
%     out = out + y;
% end;
k = 0:n;
y = ((((-1).^k).*x.^(2*k+1)))./factorial(2*k+1);
out = sum(y);
error_percent = 100*abs((out-sin(x))./(sin(x)+eps)); % added eps to avoid zero division for case sin(x) = 0
fprintf('sin taylor (%3.2f) = %1.4f\r',d,out) % taylor serie output
fprintf('sin (%3.2f) = %1.4f\r',d,sin(x)) % sin function output
fprintf('error percentage (%3.2f) = %1.4f\r',d,error_percent) % error (%) output

Answer 2

[Chunru]

disp("Input the angle in degrees (x) and the number of terms (n)")

Input the angle in degrees (x) and the number of terms (n)

%x = input('x: '); 
x = 45;
%n = input('n: ');
n = 100;

sum = 0;
xrad = x * pi/180;

for k = 0:n
    y = (-1)^k * xrad^(2*k+1) /factorial(2*k+1);
    sum = sum + y;
end
fprintf('sin(%.2f) = %.6f\n', x, sum)

sin(45.00) = 0.707107

Answer 3

[Matt J]

Easier:

n=10;
x=pi/7;

k=0:n;

T=sum( ((-1).^k .* x.^(2*k+1))./factorial(2*k+1)   )

T = 0.4339

approxError=T-sin(x)

approxError = -5.5511e-17