Re: Evaluating a series

[Przemek Klosowski]

On 08/09/2012 03:22 AM, address@hidden wrote:
> Please assist me to do this: Evaluate the series
>
> 1-1/3+1/5-1/7+1/9-...+1/1001

This series looks like sum_{k=0}^{500} (-1)^k/(2k+1) . When you write it 
out in Octave:

        N=0:500; sum(<appropriate expression on the vector N>)

you should get something close to 0.786. Pay attention to the operators 
acting on matrices---you can't just write (-1)^N/(2*N+1) because of the 
way the exponentiation operator is defined for matrices in Octave, so 
you have to use the element-by-element versions of the operators.

Actually, this problem of evaluating functions on vector arguments pops 
in Octave so often that there are facilities to do it automatically:

* vectorize() makes sure a simple expression such as a^b works for 
vector arguments.

* inline() makes it easy to construct a function from the expression 
created by vectorize().

Therefore, the octave expression

inline(vectorize("(-1)^k/(2*k+1)"))(N)

calculates the elements of the series you want to sum.