Prove or disprove that 𝑓(𝑛) ∈ Θ(𝑔(𝑛)) where 𝑓(𝑛) = 5𝑛2 + 2𝑛 log 𝑛 and 𝑔(𝑛) = 𝑛2
.

To prove or disprove that \( f(n) = 5n^2 + 2n \log n \in \Theta(g(n)) \) where \( g(n) = n^2 \), we need to apply the formal definition of \(\Theta\)-notation.

### Definition of \(\Theta(g(n))\):

A function \( f(n) \) is said to be in \(\Theta(g(n))\) if there are positive constants \( c_1, c_2, \) and \( n_0 \) such that for all \( n \geq n_0 \):
\[
c_1 \cdot g(n) \leq f(n) \leq c_2 \cdot g(n)
\]
In other words, \( f(n) \) grows asymptotically at the same rate as \( g(n) \), bounded both above and below by constant multiples of \( g(n) \).

### Step 1: Compare the growth rates of \( f(n) \) and \( g(n) \).

We have:

- \( f(n) = 5n^2 + 2n \log n \)
- \( g(n) = n^2 \)

We need to check whether \( f(n) \) is asymptotically proportional to \( g(n) = n^2 \). To do this, let's analyze both terms in \( f(n) \):

- The term \( 5n^2 \) grows like \( n^2 \), which is the same as \( g(n) = n^2 \).
- The term \( 2n \log n \) grows slower than \( n^2 \), since \( \log n \) grows slower than any polynomial.

Thus, the dominant term in \( f(n) \) as \( n \) becomes large is \( 5n^2 \), and the term \( 2n \log n \) becomes negligible compared to \( n^2 \).

### Step 2: Prove the upper bound \( f(n) \leq c_2 \cdot g(n) \).

For large \( n \), \( 5n^2 \) dominates, so:

\[
f(n) = 5n^2 + 2n \log n \leq 5n^2 + n^2 \quad (\text{since } n \log n \text{ grows slower than } n^2)
\]
\[
f(n) \leq 6n^2
\]
Thus, we can choose \( c_2 = 6 \), and there exists an \( n_0 \) such that for all \( n \geq n_0 \), \( f(n) \leq 6n^2 \).

### Step 3: Prove the lower bound \( f(n) \geq c_1 \cdot g(n) \).

For large \( n \), the \( 5n^2 \) term dominates \( 2n \log n \), so:
\[
f(n) = 5n^2 + 2n \log n \geq 5n^2
\]
Thus, we can choose \( c_1 = 5 \), and for sufficiently large \( n \), we have \( f(n) \geq 5n^2 \).

### Conclusion:

Since we have shown that:
\[
5n^2 \leq f(n) \leq 6n^2
\]
for sufficiently large \( n \), it follows that \( f(n) \in \Theta(n^2) \), meaning:
\[
f(n) \in \Theta(g(n))
\]

### Result:

The claim is **proven**: \( f(n) = 5n^2 + 2n \log n \in \Theta(n^2) \).

---

Would you like further clarification on this, or details on specific steps? Here are some related questions:

1. What is the difference between \(\Theta\)-notation and \(O\)-notation?
2. How do we analyze the growth of logarithmic terms compared to polynomials?
3. Why is the constant factor ignored in asymptotic analysis?
4. How do we find the precise constants \(c_1\) and \(c_2\) in asymptotic proofs?
5. How does the growth rate of \(n \log n\) compare to other common functions?

**Tip:** In asymptotic analysis, always focus on the dominant term of a function to determine its growth rate.

Prove or disprove that 𝑓(𝑛) ∈ Θ(𝑔(𝑛)) where 𝑓(𝑛) = 5𝑛^2 + 2𝑛 log 𝑛 and 𝑔(𝑛) = 𝑛^2.

Explore the proof of whether the function f(n) = 5n^2 + 2n log n is in Θ(n^2). Learn about asymptotic analysis and the application of Big Theta notation in determining the growth of functions. Suitable for undergraduate students studying algorithms or advanced mathematics.

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