Find the second derivative of a log likelihood function - r

Question

I'm interested in finding the values of the second derivatives of the log-likelihood function for logistic regression with respect to all of my m predictor variables.
Essentially I want to make a vector of m ∂2L/∂βj2 values where j goes from 1 to m.
I believe the second derivative should be -Σi=1n xij2(exiβ)/((1+exiβ)2) and I am trying to code it in R. I did something dumb when trying to code it and was wondering if there was some sort of sapply function I could use to do it more easily.
Here's the code I tried (I know the sum in the for loop doesn't really do anything, so I wasn't sure how to sum those values).
for (j in 1:m)
{
for (i in 1:n)
{
d2.l[j] <- -1*(sum((x.center[i,j]^2)*(exp(logit[i])/((1 + exp(logit[i])^2)))))
}
}
And logit is just a vector consisting of Xβ if that's not clear.

Answer 1

I'm hazy on the maths (and it's hard to read latex) but purely on the programming side, if logit is a vector with indices i=1,...,n and x.center is a nxm matrix:
for (j in 1:m)
dt.l[j] <- -sum( x.center[,j]^2 * exp(logit)/(1+exp(logit))^2 )
where the sum sums over i.
If you want to do it "vector-ish", you can take advantage of the fact that if you do matrix * vector (your x.center * exp(logit)/...) this happens column-wise in R which suits your equation:
-colSums(x.center^2 * exp(logit)/(1+exp(logit))^2)
For what it's worth, although the latter is "slicker", I will often use the explicit loop (as with the first example), purely for readability. Or else when I come back in a month's time I get very confused about my is and js and what is being summed over when.