求解Big Theta符号问题：(n²+5)^10是否属于Θ(n²⁰)？
Great question—let's unpack this using the formal definition of Big Theta notation, since that's the key to resolving this clearly.
First, a quick recap for context: A function $f(n)$ belongs to $\Theta(g(n))$ if there exist positive constants $c_1$, $c_2$, and $n_0$ such that for all $n \geq n_0$, the following inequality holds:
$$c_1 \cdot g(n) \leq f(n) \leq c_2 \cdot g(n)$$
Applying this to $f(n) = (n^2 + 5)^{10}$ and $g(n) = n^{20}$:
Step 1: Prove the lower bound ($f(n) \geq c_1 \cdot n^{20}$)
For any $n \geq 1$, $n^2 + 5$ is clearly greater than $n^2$ (since we're adding a positive value to $n^2$). Raising both sides to the 10th power (which preserves inequalities for positive numbers) gives:
$$(n^2 + 5)^{10} \geq (n2){10} = n^{20}$$
Here, we can use $c_1 = 1$ and $n_0 = 1$—this works for every integer $n \geq 1$.
Step 2: Prove the upper bound ($f(n) \leq c_2 \cdot n^{20}$)
Again, for $n \geq 1$, $n^2 \geq 1$, so $5 \leq 5n^2$. Substituting this in, we get:
$$n^2 + 5 \leq n^2 + 5n^2 = 6n^2$$
Raising both sides to the 10th power:
$$(n^2 + 5)^{10} \leq (6n2){10} = 6^{10} \cdot n^{20}$$
Here, $c_2 = 6^{10}$ (a fixed constant) and $n_0 = 1$ work for all $n \geq 1$.
What about "all cases"?
If we're talking about every possible positive real number (not just integers representing input size), the bounds don't hold for very small $n$ (like $n = 0.1$, where $(0.01 +5)^{10}$ is roughly $5^{10}$ while $n^{20}$ is an extremely tiny number). But in the standard context of algorithm analysis—where Big Theta is almost exclusively used—we only care about the asymptotic behavior as $n$ grows to infinity. For this purpose, "all cases" refers to all sufficiently large $n$, which aligns perfectly with the definition of Big Theta.
To sum up: In the standard asymptotic complexity context (where $n$ is a positive integer representing input size), $(n^2 +5)^{10}$ is indeed in $\Theta(n^{20})$, and this holds for all $n \geq 1$ as you observed.
内容的提问来源于stack exchange，提问作者Mitul Cristiano
