On Sun, Jul 22, 2012 at 05:34:08PM -0700, bilelsan wrote:
> Hi,
> Thanks a lot for answer. It is what I mean.
> But the code does not seem to work (
Hi.
I am sorry for a late reply. I was on vacations one week.
Can you specify the problem?
I get
e1 <- 2
e2 <- 3
k <- 3
f <- 0
for (r in 1:k) {
f <- f + 1/factorial(r) * sum(e1^(r:0)*e2^(0:r))
}
f
[1] 25.33333
If this the correct answer?
The code may be put to a function.
getF <- function(e1, e2, k)
{
f <- 0
for (r in 1:k) {
f <- f + 1/factorial(r) * sum(e1^(r:0)*e2^(0:r))
}
f
}
getF(2, 3, 3)
[1] 25.33333
Hope this helps.
Petr Savicky.
>
> Le Jul 19, 2012 ?? 8:52 AM, Petr Savicky [via R] a ??crit :
>
> > On Wed, Jul 18, 2012 at 06:02:27PM -0700, bilelsan wrote:
> > > Leave the Taylor expansion aside, how is it possible to compute with [R]:
> > > f(e) = e1 + e2 #for r = 1
> > > + 1/2!*e1^2 + 1/2!*e2^2 + 1/2!*e1*e2 #for r = 2, excluding e2*e1
> > > + 1/3!*e1^3 + 1/3!*e1^2*e2 + 1/3!*e2^2*e1 + 1/3!*e2^3 #for r = 3,
> > > excluding
> > > e2*e1^2 and e1*e2^2
> > > + ... #for r = k
> > > In other words, I am trying to figure out how to compute all the possible
> > > combinations as exposed above.
> >
> > Hi.
> >
> > For a general r, do you mean the following sum of products?
> >
> > 1/r! (e1^r + e1^(r-1) e2 + ... e1 e2^(r-1) + e2^r)
> >
> > If this is correct, then try
> >
> > f <- 0
> > for (r in 1:k) {
> > f <- f + 1/factorial(r) * sum(e1^(r:0)*e2^(0:r))
> > }
> >
> > Hope this helps.
> >
> > Petr Savicky.
> >
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