gpt pro 5.2 attempt on Math-5705

FORMAL RESTATEMENT

The problem asks:

Given an entire function $f:\mathbb{C}\to\mathbb{C}$, does there exist a path (continuous curve) $L:[0,\infty)\to\mathbb{C}$ with $|L(t)|\to\infty$ as $t\to\infty$ such that for every integer $n\ge 0$, \[ \left|\frac{f(L(t))}{L(t)^n}\right|\to\infty\quad\text{as }t\to\infty? \] Equivalently: along $L$, $|f(z)|$ eventually dominates $|z|^n$ for every fixed $n$.

As written, $f$ may be a polynomial. If $f$ is a polynomial of degree $d$, then for $n>d$ we have $|f(z)/z^n|\to 0$ along every path to infinity, so the answer is no. The note in the problem statement that Boas proved the first part strongly suggests the intended hypothesis is that $f$ is transcendental entire.

I treat:

• the literal statement (all entire functions): false (counterexample below);
• the corrected statement (transcendental entire): not proved here.

QUICK LITERATURE/CONTEXT CHECK

The problem statement records that Boas (unpublished) proved the existence of such a path in the transcendental case. The additional quantitative questions (length bounds in terms of $M(r)$ and faster-than-$M(r)^\varepsilon$ growth along some path) remain.

ATTACK PLAN

• Disproof-track for literal statement: exhibit a polynomial $f$.
• Proof-track for corrected statement: attempt to build a path through successive "large value" regions of $f(z)/z^n$ as $n\to\infty$.

WORK

Counterexample to the literal statement. Take $f(z)=1$, which is entire. Consider $n=1$. Along any path $L(t)$ with $|L(t)|\to\infty$, \[ \left|\frac{f(L(t))}{L(t)^1}\right| = \frac{1}{|L(t)|}\to 0 \quad\text{as }t\to\infty. \] In particular, it does not tend to $\infty$. Therefore no such path $L$ can exist for $f(z)=1$.

Fast reality check. The same obstruction holds for any polynomial: if $f$ has degree $d$, then for $n>d$, $|f(z)/z^n|\to 0$ along all paths to infinity.

VERIFICATION

• The counterexample uses only the definition of the limit: for $f\equiv 1$ and $n=1$, the expression is exactly $1/|z|$.
• Quantifiers: the claim being refuted is "for every entire $f$ there exists $L$ such that for every $n$ the limit is $\infty$". Providing one $f$ with no such $L$ suffices.

FINAL

FULL SOLUTION

COUNTEREXAMPLE/DISPROOF

Claim (literal reading). "For every entire function $f$ there exists a path $L$ to infinity such that for every $n$, $|f(z)/z^n|\to\infty$ along $L$."

Disproof. Let $f(z)=1$. For $n=1$ and any path to infinity, $|f(z)/z|=1/|z|\to 0$, not $\infty$. Hence the claim is false. $\square$