3 ms·
It starts off with a pretty major error. I'(t)=\int_0^1 \partial/(\partial t)((x^t - 1)/(ln x))dx = \int_0^1 x^t dx=1/(t+1), when it is actually equal to \int_
by impossiblefork 10mo ago
It starts off with a pretty major error.
I'(t)=\int_0^1 \partial/(\partial t)((x^t - 1)/(ln x))dx = \int_0^1 x^t dx=1/(t+1), when it is actually equal to \int_0^1 x^{t-1}/ln(x)dx.
These two are definitely not always equal to each other.
- owalt 10mo agoNo, it is correct. The integral is with respect to x, and the ordinary/partial derivatives are with respect to t. Written out fully, the derivative computation is d/dt (x^t - 1)/ln(x) = d/dt [exp(ln(x)t) - 1]/ln(x) = ln(x)exp(ln(x)t)/ln(x) = exp(ln(x)t) = x^t. Edit: d/dt exp(ln(x)t) = ln(x)exp(ln(x)t) by the chain rule, while d/dt (1/ln(x)) = 0 since the expression is constant with respect to t. There are convergence considerations that were not discussed in the blog post, but the computations seem to be correct.
- impossiblefork 10mo agoAh, yes. I don't understand how I differentiated with respect to x instead of t, but...