MATLAB cos(x) Taylor Series

I'm working on a Taylor Series expansion for the cox(x) function. A value is returned for all non-negative integer values of n, but no matter how many terms I have the program provide the sum always approaches ans=1, reaching it by around n=5. The code I have right now is this:

function cosx = mycos1(x,n)
%Evaluate and sum the first n terms of the cosx Taylor Series.

if n~=round(n)
    disp('Error: n must be an integer value.')
    cosx = 0;
    return
end

if n<0
    disp('Error: n must be a non-negative value.')
    cosx = 0;
    return
end

cosx=1;
i=0;

for i=0:n+1
    i = n+1;
    addterm = ((-1)^i)*(x^(2*i))/factorial(2*i);
    cosx = cosx + addterm;
end
end

Can someone find my error? Everything works other than the sum always approaching a value of 1. Thanks!

[Ruben Montoya]

El script es erroneo, no se puede poner cosx=1 porque suma los calculos a uno y si el resultado del desarrollo en 0, el resultado que muestra el programa es 0+1=1. Por ejemplo para el coseno de pi/2.

Answer 1

[Nicolas Vayas]

The i=0 and i=n+1 in lines 14 and 16, respectively, are unnecessary (such as Matias suggests, but add another unnecessary if statement): the for loop runs each iteration assigning integer values between your boundaries. Furthermore, the Taylor series expansion of cos(x), has the leading term of '1'. This means that the approximation of the cos(x) for any x using 1 term is equal to 1. When you ran mycos1(x, 1), the return should always be 1. To fix this error, add n = n-1 before your first if statement. This should end up like:

function cosx = mycos1(x,n)
%Evaluate and sum the first n terms of the cosx Taylor Series.
n = n-1; 
if n~=round(n)
    disp('Error: n must be an integer value.')
    cosx = 0;
    return
end
if n<0
    disp('Error: n must be a non-negative value.')
    cosx = 0;
    return
end
cosx=1;
% i=0;
for i=1:n
% i = n+1;
    addterm = ((-1)^i)*(x^(2*i))/factorial(2*i);
    cosx = cosx + addterm;
end
end

Answer 2

[Walter Roberson]

Why are you looping over "i" and then in the loop setting "i = n+1" ?

Answer 3

[Matias Campos]

 function cosx = mycos1(x,n)
%Evaluate and sum the first n terms of the cosx Taylor Series.
if n~=round(n)
    disp('Error: n must be an integer value.')
    cosx = 0;
    return
end
if n<0
    disp('Error: n must be a non-negative value.')
    cosx = 0;
    return
end

 if n == 1
    cosx = 1;
    return
 end

 cosx=1;
 for i =1:n
    addterm = ((-1)^i)*(x^(2*i))/factorial(2*(i));
    cosx = cosx + addterm;
 end
 end

[Matias Campos]

This should work, you had a slight error on the last for loop.

Answer 4

[Fawaz Hjouj]

What happen if we try

mycos1(100,100)

[Walter Roberson]

You can get a big number.

taylor() gets increasingly inaccurate as you get away from the point that the taylor series was approximated around. The order 100 term is going to be proportional to x^100/100! and for x near 100 that is going to be about 1E+42