If $f(n) = \theta(g(n))$ and $g(n)=\theta(h(n))$ then $h(n)=\theta(f(n))$

So, if we are saying that $f(n)$ has the property of:

$\exists \ c_1, c_2, n_0 \ \ \ s.t. \  \ \ c_1 \ \cdot f(n) \le g(n) \le c_2 \ \cdot f(n)$

Which implies that $g(n)$ is a function that both lower bounds $f(n)$ and upper bounds $f(n)$

Therefore, if $g(n)=\theta(h(n))$ then therefore:

$\exists \ d_1, d_2, n_0 \ \ \ s.t. \ \ \ d_1 \cdot \ g(n) \le h(n) \le d_2 \cdot g(n)$

So, h(n) is both the upper bound and lower bound of *g(n).*

We can form the final inequality, which implies:

So the **=** really means that the notation on the left hand side belongs into the set of functions in the theta formula, such that for the final one, h(n) is both lower bounding and upper bounding f(n)

$\exists \ e_1, e_2, n_0 > 0 \ s.t \ \forall n \ge n_0 \ | \ e_1 \cdot h(n) \le f(n) \ge e_2 \cdot h(n)$

If we expand that inequality by plugging in

I am confused exactly how to prove this?

You're working on proving that if $ f(n) = \theta(g(n)) $ and $ g(n) = \theta(h(n)) $, then $ f(n) = \theta(h(n)) $. This is a transitivity property of Big-$\Theta$ notation. Let’s break down the proof clearly step by step.

### Step-by-Step Proof:

1. **Start with the definition of Big-$\Theta$ for $ f(n) = \theta(g(n)) $**:
   By definition, we know that:
   \[
   \exists \ c_1, c_2, n_0 \ \ \text{such that} \ \ \forall n \ge n_0, \ \ c_1 \cdot g(n) \leq f(n) \leq c_2 \cdot g(n)
   \]
   This means that $ g(n) $ bounds $ f(n) $ from both below and above for all $ n \ge n_0 $.

2. **Now, consider $ g(n) = \theta(h(n)) $**:
   Similarly, by the definition of Big-$\Theta$, we know:
   \[
   \exists \ d_1, d_2, n_1 \ \ \text{such that} \ \ \forall n \ge n_1, \ \ d_1 \cdot h(n) \leq g(n) \leq d_2 \cdot h(n)
   \]
   This means that $ h(n) $ bounds $ g(n) $ from both below and above for all $ n \ge n_1 $.

3. **Combine the inequalities**:
   To prove $ f(n) = \theta(h(n)) $, we need to show that $ h(n) $ bounds $ f(n) $ from both below and above. This can be done by combining the two inequalities.

   - From $ f(n) = \theta(g(n)) $, we know:
     \[
     c_1 \cdot g(n) \leq f(n) \leq c_2 \cdot g(n)
     \]

   - From $ g(n) = \theta(h(n)) $, we know:
     \[
     d_1 \cdot h(n) \leq g(n) \leq d_2 \cdot h(n)
     \]

4. **Substitute bounds for $ g(n) $ into the inequality for $ f(n) $**:
   By substituting the bounds on $ g(n) $ from the second inequality into the first inequality, we get:
   \[
   c_1 \cdot (d_1 \cdot h(n)) \leq f(n) \leq c_2 \cdot (d_2 \cdot h(n))
   \]
   Simplifying:
   \[
   (c_1 \cdot d_1) \cdot h(n) \leq f(n) \leq (c_2 \cdot d_2) \cdot h(n)
   \]

5. **Conclusion**:
   The final inequality shows that there exist constants $ e_1 = c_1 \cdot d_1 $ and $ e_2 = c_2 \cdot d_2 $ such that:
   \[
   e_1 \cdot h(n) \leq f(n) \leq e_2 \cdot h(n)
   \]
   This is exactly the definition of $ f(n) = \theta(h(n)) $.

### Conclusion:
Thus, we've shown that if $ f(n) = \theta(g(n)) $ and $ g(n) = \theta(h(n)) $, then $ f(n) = \theta(h(n)) $. This proves the transitivity property of Big-$\Theta$ notation.

---

Would you like any further details or clarifications?

### Here are 5 related questions:
1. How does Big-$O$ notation relate to Big-$\Theta$?
2. Can you prove a similar result for Big-$O$ or Big-$\Omega$ notation?
3. What are some examples of functions where $ f(n) = \theta(g(n)) $?
4. How does the transitivity property help in algorithm analysis?
5. What are the limitations of Big-$\Theta$ notation in asymptotic analysis?

### Tip:
When proving relationships with asymptotic notations, always carefully track how constants combine during substitutions to maintain the correct bounds.

If $f(n) = \theta(g(n))$ and $g(n)=\theta(h(n))$ then $h(n)=\theta(f(n))$. I am confused exactly how to prove this?

Learn how to prove the transitivity property of Big-Theta notation. This problem demonstrates how $f(n) = θ(g(n))$ and $g(n) = θ(h(n))$ imply $f(n) = θ(h(n))$, with a detailed step-by-step breakdown, formulas, and application to algorithm analysis.

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