When $t\mapsto (1+|t|^q)^{1/q}$ is real analytic in a small nbd of $0$?

Question

Here $t$ be a real number. It is easy to check that for $q=2$, it is real analytic. More generally for even $q$ the function is real analytic. I guess that this function is not real analytic for any positive real number $q$ other than even integer, but I can't prove it.

Answer 1

If $q > 0$ is not an even integer, $f(t) := 1 + |t|^q$ is not smooth (infinitely times differentiable) around $0$. Notice that $g(x) = x^{1/q}$ is smooth and maps $(0,\infty)$ bijectively into itself. Thus, $t\mapsto g(f(t)) = (1 + |t|^q)^{1/q}$ is also not smooth and can not be real analytic.

Answer 2

By the Taylor expansion of $f(x) = (1 + x)^{1 \over q}$ near $x = 0$, one has $$(1 + |t|^q)^{1 \over q} = 1 + {1 \over q}|t|^q + O(|t|^{2q})$$ If $(1 + |t|^q)^{1 \over q}$ were real analytic, the terms of its Taylor expansion at the origin of order $< 2q$ would then have to be $1 + {1 \over q}|t|^q$. This has to be a polynomial. Since $|t|^q$ is a polynomial only when $q$ is an even integer, the result follows.