How smooth is the generalized power mean function?

Question

Consider the function of power mean of $1$ and $e$ (so that the computation is simpler). That is, define $\displaystyle f(x)=(\frac{1+e^x}{2})^{1/x}$ when $x\neq 0$ and $f(0)=\sqrt{e}$.

It's not difficult to use L'Hopital rule with std. exponential trick to verify the function is continuous.

The question is, do we know how smooth (in which $C^k$) is this function (and a closed form formula of $f^{(n)}(0)$ if possible)?

The only way I can think of is trying to compute $\displaystyle \frac{f^{(n)}(h)-f^{(n)}(0)}{h}$ to verify if $f^{(n+1)}(0)$ exists, but the formula becomes extremely complicated and it's practically impossible to compute that limit even for $n=1$.

Answer 1

I will stick to $x\in \mathbb R.$ We can write

$$\tag 1 f(x) = \exp [(1/x)\ln ((e^x+1)/2)]$$

for $x\ne 0.$ Now $\ln ((e^x+1)/2)$ is real analytic on $\mathbb R.$ Why? Because $(e^x+1)/2$ is real analytic and positive on $\mathbb R,$ $\ln x$ is real analytic on $(0,\infty),$ and compositions of real analytic functions are real analytic. Furthermore, $\ln ((e^x+1)/2)$ equals $0$ when $x=0.$ It follows that $(1/x)\ln ((e^x+1)/2)$ is real analytic on $\mathbb R,$ once we define $f(0)$ to be the value you found, namely $e^{1/2}.$ Again using "compositions of real analytic functions are real analytic", we see from $(1)$ that $f$ is real analytic on $\mathbb R.$ (Since real analytic functions are $C^\infty,$ we have $f\in C^\infty(\mathbb R).$)

Sorry, I don't have any insight towards a nice formula for $f^{(n)}(0).$

Answer 2

The complex valued function $h(z)=(\frac{1+e^z}{2})^{1/z}$ is analytic in $\mathbb{C}/\{0\}$, and the singularity at $z=0$ is removable because the limit $\lim_{z\to 0}h(z)$ exists (it is equal to $\sqrt{e}$). As a result, the function (overloading the notation for $h$) $$h(z)= \left\{\begin{array}{lr} (\frac{1+e^z}{2})^{1/z} & \text{for } z\neq 0\\ \sqrt{e} & \text{for } z=0\\ \end{array}\right. $$ is an entire function (analytic in $\mathbb{C}$), and in particular, has derivatives of all orders at $z=0$. It follows that the real valued function $f(x)$, which is a restriction $h(z)$ to real valued $z$, also has derivatives of all orders at $x=0$, and so $f$ is infinitely smooth. As for a closed formula for $f^{n}(0)$, inspection of the first values (using Mathematica) shows that: $$f'(0)=\frac{\sqrt{e}}{8},\ \ f''(0)=\frac{\sqrt{e}}{64},\ \ f^{(3)}(0)=\frac{-15\sqrt{e}}{512},\ \ f^{(4)}(0)=\frac{-63\sqrt{e}}{4096},\ \ f^{(5)}(0)=\frac{3619\sqrt{e}}{98304}$$ and while the first four derivatives may raise some hope of having some structure (the denominators are $2^{3n}$), the fifth derivative is yet quite different.

Answer 3

$f$ is analytic near the origin

Write $f(x)=\exp\left(\frac{1}{x}\ln\left(\frac{1+e^x}{2}\right)\right)$ and let $g(x)=\frac{1}{x}\ln\left(\frac{1+e^x}{2}\right)$, so that $f(x)=\exp(g(x))$.

Now, consider that

$$\frac{d}{dx}\,\ln\left(\frac{1+e^x}{2}\right)=\frac{e^x}{1+e^x}.$$

The function $h(z)=\frac{\exp(z)}{1+\exp(z)}$ is holomorphic on the open unit disk $D$, and hence has a family of antiderivatives on the unit disk. Let $H(z)$ be the antiderivative that agrees with $\ln\left(\frac{1+e^x}{2}\right)$ on $(-1,1)$. $H$ is holomorphic and indeed analytic throughout $D$:

$$H(z)=\frac{z}2+\frac{z^2}{8}-\frac{z^4}{192}+\dots$$

Then $g(z)=\frac{1}{z}\cdot H(z)$ is also analytic throughout $D$, with

$$g(z)=\frac{1}2+\frac{z}{8}-\frac{z^3}{192}+\dots$$

Finally, $f$ is analytic throughout as the composition of $f\circ g$ of two analytic functions.

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A sort of formula for $f^{(n)}(x)$

We have that

\begin{align} f'(x)&=\exp(g(x))\cdot g'(x)\\ f''(x)&=\exp(g(x))\cdot \Big({g'(x)}^2+g''(x)\big) \end{align}

and more generally, via a combination of Faà di Bruno's formula and Bell polynomials, we get that

$$f^{(n)}(x)=\exp(g(x))\cdot B_n\big(g'(x),g''(x),\dots,g^{(n)}(x)\big),$$

where $B_n$ is the $n$-th complete exponential Bell polynomial.

For each $k\geq 0$, let $f_k=\lim_{x\to0}f^{(k)}(x)$ and $g_k=\lim_{x\to0}g^{(k)}(x)$. By the first part, $f_k$ exists and equals $f^{(k)}(0)$ for all $k$, and similarly for $g_k$.
By inspection $g_0=1/2$, so that $f_k=\sqrt{e}\cdot B_k(g_1,g_2,\dots,g_k)$. Now, the $B_n$'s satisfy a recursion:

$$B_{k+1}(g_1,g_2,\dots,g_{k+1})=\sum_{i=0}^k\binom{k}{i}B_ {k-i}(g_1,\dots,g_ {k-i})g_ {i+1},$$

where $B_0=1$. Hence:

$$f_{k+1}=\sum_{i=0}^k\binom{k}{i}f_ {k-i}g_ {i+1},$$

where $f_0=e^{g_0}=\sqrt{e}$. Thus, the calculation of the $f_k$'s boils down to calculating the $g_k$'s.

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Calculating the $g_k$'s

I might add this section later.