Math Problem Statement

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?

Solution

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

Step-by-Step Proof:

  1. Start with the definition of Big-Θ\Theta for f(n)=θ(g(n))f(n) = \theta(g(n)): By definition, we know that:  c1,c2,n0  such that  nn0,  c1g(n)f(n)c2g(n)\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)g(n) bounds f(n)f(n) from both below and above for all nn0n \ge n_0.

  2. Now, consider g(n)=θ(h(n))g(n) = \theta(h(n)): Similarly, by the definition of Big-Θ\Theta, we know:  d1,d2,n1  such that  nn1,  d1h(n)g(n)d2h(n)\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)h(n) bounds g(n)g(n) from both below and above for all nn1n \ge n_1.

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

    • From f(n)=θ(g(n))f(n) = \theta(g(n)), we know: c1g(n)f(n)c2g(n)c_1 \cdot g(n) \leq f(n) \leq c_2 \cdot g(n)

    • From g(n)=θ(h(n))g(n) = \theta(h(n)), we know: d1h(n)g(n)d2h(n)d_1 \cdot h(n) \leq g(n) \leq d_2 \cdot h(n)

  4. Substitute bounds for g(n)g(n) into the inequality for f(n)f(n): By substituting the bounds on g(n)g(n) from the second inequality into the first inequality, we get: c1(d1h(n))f(n)c2(d2h(n))c_1 \cdot (d_1 \cdot h(n)) \leq f(n) \leq c_2 \cdot (d_2 \cdot h(n)) Simplifying: (c1d1)h(n)f(n)(c2d2)h(n)(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 e1=c1d1e_1 = c_1 \cdot d_1 and e2=c2d2e_2 = c_2 \cdot d_2 such that: e1h(n)f(n)e2h(n)e_1 \cdot h(n) \leq f(n) \leq e_2 \cdot h(n) This is exactly the definition of f(n)=θ(h(n))f(n) = \theta(h(n)).

Conclusion:

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


Would you like any further details or clarifications?

Here are 5 related questions:

  1. How does Big-OO notation relate to Big-Θ\Theta?
  2. Can you prove a similar result for Big-OO or Big-Ω\Omega notation?
  3. What are some examples of functions where f(n)=θ(g(n))f(n) = \theta(g(n))?
  4. How does the transitivity property help in algorithm analysis?
  5. What are the limitations of Big-Θ\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.

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Math Problem Analysis

Mathematical Concepts

Asymptotic Notation
Big-Theta Notation
Algorithm Analysis

Formulas

$f(n) = \theta(g(n))$: $\exists c_1, c_2, n_0 \ s.t. \ c_1 \cdot g(n) \le f(n) \le c_2 \cdot g(n)$
$g(n) = \theta(h(n))$: $\exists d_1, d_2, n_0 \ s.t. \ d_1 \cdot h(n) \le g(n) \le d_2 \cdot h(n)$
Final result: $\exists e_1, e_2, n_0 \ s.t. \ e_1 \cdot h(n) \le f(n) \le e_2 \cdot h(n)$

Theorems

Transitivity of Big-Theta Notation

Suitable Grade Level

College-Level (Undergraduate Computer Science or Mathematics)