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:

$\map f p = \norm {\mathbf x}_p$

where $\norm {\mathbf x}_p$ is the $p$-norm of $\mathbf x$, so

$\ds \map f p = \paren {\sum_{i \mathop = 1}^\infty \size {x_i}^p}^{1/p}$

Then $D_p \map f p = $

Proof

For convenience, let $s_i = \size {x_i}$ for each $i$.

By the general definition of exponentiation:

$\ds \map f p = \map \exp {\frac 1 p \ln \sum_{i \mathop = 1}^\infty s_i^p}$

So

\(\ds D_p \map f p\)

\(=\)

\(\ds \map f p \map {D_p} {\frac 1 p \ln \sum_{i \mathop = 1}^\infty s_i^p }\)

\(\ds \)

\(=\)

\(\ds \map f p \paren {\paren {D_p \frac 1 p} \ln \sum_{i \mathop = 1}^\infty s_i^p + \frac 1 p D_p \ln \sum_{i \mathop = 1}^\infty s_i^p}\)

\(\ds \)

\(=\)

\(\ds \map f p \paren {\frac {-1} {p^2} \ln \sum_{i \mathop = 1}^\infty s_i^p + \frac 1 p \frac 1 {\sum_{i \mathop = 1}^\infty s_i^p } D_p \sum_{i \mathop = 1}^\infty s_i^p}\)

Now:

$\ds D_p \sum_{i \mathop = 1}^\infty s_i^p = D_p \sum_{i \mathop = 1}^\infty e^{p \ln s_i} = \sum_{i \mathop = 1}^\infty s_i^p \ln s_i$

Thus:

\(\ds D_p \map f p\)

\(=\)

\(\ds \map f p \paren {\frac {-1} {p^2} \ln \sum_{i \mathop = 1}^\infty s_i^p + \frac 1 p \frac 1 {\sum_{i \mathop = 1}^\infty s_i^p} \sum_{i \mathop = 1}^\infty s_i^p \ln s_i}\)