- #1
rsq_a
- 103
- 1
I'm trying to find the general form for the nth derivative of
<br /> f(x) = \frac{1}{x^m \log x}<br />
where m can be anything (set m = 1 for instance). For ease, you can take m to be integral.
It sounds surprisingly simple, but the most I've been able to say is
<br /> f^{(n)}(x) = (-1)^n x^{-(m+n)} \sum_{k=0}^n a_{k, n} [\log(x)]^{-k}<br />
where the coefficients satisfy
<br /> a_{k,n} = [m + (n-1)] a_{k, n-1} + (k-1) a_{k-1, n-1}<br />
for 0 < k < n, and with a_{0, n} = (m + n - 1)!/(m-1)! and a_{n,n} = n!
Unfortunately, I was hoping to get a general form for the coefficients. Does anyone know a trick?
<br /> f(x) = \frac{1}{x^m \log x}<br />
where m can be anything (set m = 1 for instance). For ease, you can take m to be integral.
It sounds surprisingly simple, but the most I've been able to say is
<br /> f^{(n)}(x) = (-1)^n x^{-(m+n)} \sum_{k=0}^n a_{k, n} [\log(x)]^{-k}<br />
where the coefficients satisfy
<br /> a_{k,n} = [m + (n-1)] a_{k, n-1} + (k-1) a_{k-1, n-1}<br />
for 0 < k < n, and with a_{0, n} = (m + n - 1)!/(m-1)! and a_{n,n} = n!
Unfortunately, I was hoping to get a general form for the coefficients. Does anyone know a trick?
