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

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In summary, the conversation discusses the proof that $(n+a)^b=\Theta(n^b)$ for $b>0$. The first approach uses the fact that $(n+a)^b$ can be bounded by a polynomial function of $n$ with the dominant term being $n^b$. However, the second approach uses the binomial theorem to directly show that $(n+a)^b=\Theta(n^b)$. Both approaches are valid, but the second one is more direct in proving the statement.
  • #1
evinda
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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)
 
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  • #2


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)
 

What is the purpose of proving $(n+a)^b=\Theta(n^b)$ using polynomials?

The purpose of proving this statement is to show the relationship between two functions, $(n+a)^b$ and $n^b$, and to demonstrate that they have similar growth rates. This can be useful in analyzing algorithms and determining their time complexity.

What does it mean for two functions to have the same growth rate?

Functions with the same growth rate means that they have a similar rate of increase as the input size grows. In other words, as the input size increases, the two functions will approach infinity at a similar rate.

Why is it necessary to use polynomials in this proof?

Polynomials are used because they are a commonly used type of function and their behavior is well understood. By showing that two polynomials have similar growth rates, it can be generalized to other types of functions.

What is the significance of using $\Theta$ notation in this proof?

The $\Theta$ notation is used to denote the asymptotic behavior of a function. In this proof, it is used to show that the two functions, $(n+a)^b$ and $n^b$, have the same growth rate as $n$ approaches infinity.

What are the steps involved in proving $(n+a)^b=\Theta(n^b)$ using polynomials?

The steps involved include: 1) Expanding both polynomials using the binomial theorem, 2) Comparing the leading terms of the expanded polynomials, 3) Using algebraic manipulation to show that the leading terms are equal, and 4) Using the definition of $\Theta$ notation to show that the remaining terms are negligible in comparison to the leading terms.

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