f(n) = (log n)^2023 Write a correct asymptotic upper bound for each of the above.

To find a correct asymptotic upper bound for the function $f(n) = (\log n)^{2023}$, we will analyze the growth of this function as $n$ approaches infinity.

Step 1: Break Down the Problem

1. Understand the function $f(n) = (\log n)^{2023}$.
2. Determine an expression that serves as an asymptotic upper bound for $f(n)$.

Step 2: Relevant Concepts

• The notation $O(g(n))$ is used to denote an asymptotic upper bound. This means that $f(n)$ grows at most as quickly as $g(n)$ for large values of $n$.
• We need to find a suitable $g(n)$ for $(\log n)^{2023}$.

Step 3: Analysis and Detail

1. The logarithmic function $\log n$ grows slower than any polynomial function of $n$ as $n$ increases.
2. Thus, the function $(\log n)^{2023}$ will grow slower than any polynomial function of $n$. For instance, we can consider functions like $n^\epsilon$ for any small $\epsilon > 0$.
3. We can see that as $n \to \infty$: $(\log n)^{2023} = o(n^\epsilon)$ for any $\epsilon > 0$.

Step 4: Verify and Summarize

• Therefore, we can say that $f(n)$ is bounded above by $n^{\frac{1}{2}}$, $n^{\frac{1}{3}}$, and so forth, since logarithmic growth is significantly slower than polynomial growth.
• A suitable asymptotic upper bound could be $O((\log n)^{2024})$ as it captures the growth of $f(n)$ without exceeding it for sufficiently large $n$.

Final Answer

The correct asymptotic upper bound for $f(n) = (\log n)^{2023}$ is: $f(n) = O((\log n)^{2024})$