On Sep 13, 2010, at 5:24 AM, tuggi wrote:
hello,
can i calculate a sum to infinity in R.
i want to do something like this:
\sum_{i=0}^\infty
\frac{2^{-d-1}}{\Gamma(\frac{d-1}{2})}\left(\frac{\Gamma(2d-3)(2-
d)_{i}\Gamma(i+1,-z/2)2^{i+1}}{\Gamma(d-1)(4-2d)_{i}i!}\right)+
\\
\sum_{i=0}^\inft
hello,
can i calculate a sum to infinity in R.
i want to do something like this:
\sum_{i=0}^\infty
\frac{2^{-d-1}}{\Gamma(\frac{d-1}{2})}\left(\frac{\Gamma(2d-3)(2-d)_{i}\Gamma(i+1,-z/2)2^{i+1}}{\Gamma(d-1)(4-2d)_{i}i!}\right)+
\\
\sum_{i=0}^\infty
\frac{2^{-d-1}}{\Gamma(\frac{d-1}{2})}\left(\fr
Dear Roslina,
For $m \rightarrow \infty$ that sum is exp(1)-1:
> options(digits=20)
> exp(1)-1
[1] 1.718281828459045
> m<-20
> sum(1/factorial(1:20))
[1] 1.718281828459045
HTH,
Jorge
On Wed, Mar 25, 2009 at 8:43 PM, Roslina Zakaria wrote:
>
> Hi r-users,
>
> How do we evaluate the summation
Well, sum of 1/m! is e--does that answer your question?
Generally, I guess you have to decide how much error you're comfortable
with; then using an error approximation formula, you can back out the M at
which you can stop the sum. Then you can write a for loop that ends at M.
Hope that helps.
O
Hi r-users,
How do we evaluate the summation of (1/m!) from 0 to infinity (for example).
Any help is very much appreciated.
Thank you.
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