Taylor series sum and factorial

Hi,

What is the problem with this code:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [S] = taylorTerm1(x,n)
x=1.1;
n=(1:1:5);
p=0;
for i=1:n
    z = (1:2i+1:1);
    iTerm = (-1)^(i)*(x)^2i+1/z;
    S = p + iTerm;   
end
fprintf('The Sum of the Series is %6.4f.\n',S);

I am trying to find the sum of series (-1)^n * (x^2n+1)/(2n+1)! and the factorial is confusing me.

Thanks

Answer 1

[Geoff Hayes]

Jack - part of the problem may have to do with the equation. You state that you are trying to _find the sum of series (-1)^n * (x^2n+1)/(2n+1)! _. Do you really mean this to be

 (-1)^n * (x^(2*n+1))/(2*n+1)!

where the exponent of x is 2*n+1? The line of code that is written as

 (x)^2i+1/z;

means that the exponent of x is 2*i (I think it more likely that the exponent is considered as 0.0000 + 2.0000i - see below for details), and added to that result is 1/z when what you want instead is x to the power of 2*i+1 and that result divided by z.

The other problem, like you suggest, has to do with the factorial. Your code for this term is z and it is initialized as

 z = (1:2i+1:1);

which is a vector of integers from 1 to 1 with a step-size of 2*i+1. So it will always be just one. Use the factorial function instead which will sum the product of the integers from 1,2,...,2*k+1.

This changes your code to

 function [S] = taylorTerm1(x,n)

 S = 0;

 for k=1:n
    kTerm = (-1)^(k)*(x^(2*k*+1))/factorial(2*k+1);
    S = S + kTerm;   
 end

Note the differences. I've removed the initialization of

 n=(1:1:5);

since this overwrites the input parameter with a vector. This will cause confusion for the line

 for i=1:n

and will probably cause the iterations to terminate early.

I've replaced the i with k since MATLAB uses i and j to represent imaginary numbers. In your code, where there was 2i, this was probably being considered as the imaginary number 0.0000 + 2.0000i and not the desired 2*i.

The kterm has been reworked to something more like what you want, and the summation term was changed so that we were always summing S with the values from the previous iterations and the current iteration.

Try the above and see what happens!

[Jack]

Hi,

Thank you for responding so quick. Using the code you suggested the program doesn't recognise the factorial expression which has lead me to try other means. Any other suggestions?

Thanks.

[Geoff Hayes]

Jack - factorial is a standard function within MATLAB so you should have this functionality. What is the error message if you try

 factorial(4)

Also, in the Command Window, type

 ver
 which factorial -all

What are the results? (Please paste them in a comment.)

----------------

As for an alternative, yes, you could replace the factorial(2*k+1) with

 prod(1:1:(2*k+1))

Here, prod returns the product of the elements in the array. For example, if

 1:1:(2*k+1) = 1 2 3 4 5 ... 2*k 2*k+1

then

 prod(1:1:(2*k+1)) = 1*2*3*4*5 ... *(2*k)*(2*k+1) = factorial(2*k+1)

Try the above and see what happens!

Answer 2

[Roger Stafford]

Here is a method that doesn't require the 'factorial' function. See if you can figure out why this is equivalent to the formula you are using.

   x = 1.1;
   n = 10; % <-- Choose the number of terms to add
   s = 1;
   for k = 2*n:-2:2
     s = 1 - x^2/k/(k+1)*s;
   end
   s = s*x;

You can compare your results to sin(x). Setting n = 10 is adequate for x values this size, but larger values of x would require larger values of n to compare successfully with sin(x).