Martin Doumen - 2 months ago 13x

R Question

I'm trying to figure out how to integrate the following function in R:

`item.fill.rate <- function(x, lt, ib, S){`

1-((((1/(factorial(S)))*((x*lt*ib)^S)))/

(sum(((1/(factorial(0:S)))*((x*lt*ib)^(0:S))))))}

Where x is a variable and lt, ib and S are input parameters

Based on a previous topic on here, I tried the following:

`int.func <- function(lt, ib, S){`

item.fill.rate <- function(x){

1-((((1/(factorial(S)))*((x*lt*ib)^S)))/(sum(((1/(factorial(0:S)))*((x*lt*ib)^(0:S))))))

}

return(item.fill.rate)

}

integrate(int.func(0.25, 1, 1), lower=0.25, upper=0.75)$value

When applying this, I get the following error:

`> integrate(int.func(0.25, 1, 1), lower=0.25, upper=0.75)$value`

[1] 0.4947184

Warning messages:

1: In (x * lt * ib)^(0:S) :

longer object length is not a multiple of shorter object length

2: In (1/(factorial(0:S))) * ((x * lt * ib)^(0:S)) :

longer object length is not a multiple of shorter object length

I evaluated the length of those objects, but that did not give me any indication where the error must be.

I tried to be as specific as possible, so hopefully someone is able to help me out with this!

Answer

The `sum`

function is notorious for returning single items when a longer vector was expected, so integrand functions that have a call to `sum`

generally need to be "vectorized" so they deliver the expected results (a vector of the the same length as a provided "x"-vector) for `integrate`

to succeed. The Vectorize function is a wrapper for `sapply`

and is quite handy for this process. You can set the parameters in the call to `integrate`

. (At the moment I think you may be integrating a constant over a domain of length 1/2.)

```
item.fill.rate <- function(x,lt, ib, S){
1-((((1/(factorial(S)))*((x*lt*ib)^S)))/(sum(((1/(factorial(0:S)))*((x*lt*ib)^(0:S))))))
}
vint <- Vectorize(item.fill.rate)
integrate(vint, S=1, lt=0.25, ib= 1, lower=0.25, upper=0.75)$value
#[1] 0.4449025
```

Source (Stackoverflow)

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