User:Dfeuer/Derivative of P-Norm wrt P

Theorem
For each $p \in \R_{\ge 1}$, let $\ell^p$ denote the $p$-sequence space.

Let $\mathbf x$ be a sequence of complex numbers.

If $\mathbf x$ is a finite sequence, extend it with $0$s to an infinite sequence.

Suppose that for some non-degenerate open interval $U$, $p \in U \implies \mathbf x \in \ell_p$.

Let $f:U \to \R$ be defined by:
 * $f(p) = \left\Vert{ \mathbf x }\right\Vert_p$

where $\left\Vert{ \mathbf x }\right\Vert_p$ is the $p$-norm of $\mathbf x$, so
 * $f(p) = \left({\sum_{i=1}^\infty |x_i|^p}\right)^{1/p}$

Then $D_p f(p) = $

Proof
For convenience, let $s_i = \left| {x_i} \right|$ for each $i$.

By the general definition of exponentiation,
 * $\displaystyle f(p) = \exp \left({\frac 1 p \ln \sum_{i=1}^\infty s_i^p}\right)$

So

Now $D_p \sum_{i=1}^\infty s_i^p = D_p \sum_{i=1}^\infty e^{p \ln s_i} = \sum_{i=1}^\infty s_i^p \ln s_i$. Thus