Proving $(n+a)^b=\Theta(n^b)$ Using Polynomials

[evinda]

Hi! (Wave)

I want to show that $\forall a,b$ with $b>0$:

$$(n+a)^b=\Theta(n^b)$$

According to my notes, it is like that:

• $\forall n \geq n_0=\lceil |a| \rceil$
  
  $$(1) \Rightarrow (n+a)^b \leq (2n)^b=2^b n^b=c n^b \Rightarrow (n+a)^b=O(n^b), \forall n \geq \lceil |a| \rceil \text{ and } c=2^b$$
  $$$$
• $\forall n \geq n_0=2 \lceil |a| \rceil$, it holds that:
  
  $$n+a \geq n-|a| \geq n-\frac{n}{2}=\frac{n}{2} \Rightarrow (n+a)^b \geq \left ( \frac{n}{2}\right)^b=\left(\frac{1}{2} \right)^b n^b=c n^b \Rightarrow (n+a)^b=\Omega(n^b) \forall n \geq 2 \lceil |a| \rceil \text{ and } c=\left[ \frac{1}{2}\right]^b $$
  

$$(1)+(2) \Rightarrow (n+a)^b=\Theta(n^b), \forall n \geq 2 \lceil |a| \rceil$$(Sweating)

Couldn't we also show that $(n+a)^b=\Theta(n^b)$ like that? (Thinking)

$$(n+a)^b=\sum_{k=0}^b n^k y^{b-k}= \binom{b}{0} a^b+ \binom{b}{1} n a^{b-1}+ \dots + \binom{b}{b}n^b$$

So, we have a polynomial with respect to $n$ and the dominant term is $n^b$.

Therefore, $(n+a)^b=\Theta(n^b)$.

(Thinking) (Blush)

 

[mmwave]

Hi! Great question and thank you for sharing your thoughts on this topic. Your approach is definitely valid and shows that $(n+a)^b$ can be bounded by a polynomial function of $n$ with the dominant term being $n^b$. However, it is important to note that in order to prove that $(n+a)^b=\Theta(n^b)$, we need to show that the function is both $O(n^b)$ and $\Omega(n^b)$. This ensures that the function is both upper and lower bounded by $n^b$, which is the definition of $\Theta$ notation.

In your first approach, you have shown that $(n+a)^b=O(n^b)$ by choosing a constant $c$ and showing that the function is bounded by $cn^b$. However, you have not shown that it is also $\Omega(n^b)$, which is necessary for proving $\Theta$ notation. In your second approach, you have correctly shown that $(n+a)^b=\Theta(n^b)$ by using the binomial theorem and showing that the dominant term is $n^b$. This approach is a more direct way of proving $\Theta$ notation.

Overall, both approaches are valid and it is great to see that you are thinking about different ways to prove this statement. Keep up the good work! (Thumbs up)