What is the dominant term of a function and its complexity?

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Discussion Overview

The discussion revolves around determining the dominant term of the function $g(n)=10 \cdot \log (n^{30}+30)+2$ and its complexity in terms of Big Theta notation. Participants explore the implications of different interpretations of "dominant term" and how it relates to asymptotic notation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the dominant term can be identified as $10\cdot \log(n^{30})$ due to the neglect of the constant $30$ in comparison to $n^{30}$.
  • Others argue that the dominant term, taken literally, is $10\cdot \log(n^{30}+30)$, which grows more rapidly than any other term in the expression.
  • A later reply questions whether $g(n)$ can be considered equal to $\Theta(\text{dominant term})$, suggesting that it might not be straightforward.
  • Some participants clarify that $g(n)$ is asymptotically equal to $10\cdot \log(n^{30}+30)$, which leads to the conclusion that it is also equal to $\Theta(\log(n))$.
  • There is a discussion about how to phrase the dominant term and its implications for expressing the complexity of $g(n)$.

Areas of Agreement / Disagreement

Participants express differing views on the definition of the "dominant term" and its implications for the complexity of the function. While some agree on the asymptotic equivalence to $\Theta(\log(n))$, the interpretation of the dominant term remains contested.

Contextual Notes

Participants note that the definition of "dominant term" can vary, which affects the interpretation of the complexity of $g(n)$. There are also unresolved aspects regarding the mathematical steps taken to arrive at conclusions about asymptotic notation.

Who May Find This Useful

This discussion may be useful for students and professionals interested in algorithm analysis, particularly those exploring complexities and asymptotic behavior of logarithmic functions.

evinda
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Hey again! (Smile)

I want to find the complexity of $g(n)=10 \cdot \log (n^{30}+30)+2 $.

We will find that $g(n)=\Theta(\log n)$, right? (Thinking)

What can I say at the beginning? Which is the dominant term of the function? (Thinking)
 
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evinda said:
Hey again! (Smile)

I want to find the complexity of $g(n)=10 \cdot \log (n^{30}+30)+2 $.

We will find that $g(n)=\Theta(\log n)$, right? (Thinking)

What can I say at the beginning? Which is the dominant term of the function? (Thinking)

Hi again! (Happy)

Since $30$ can be neglected with respect to $n^{30}$, I'd say that the dominant term is $10\cdot \log(n^{30})$.
 
I like Serena said:
Hi again! (Happy)

Since $30$ can be neglected with respect to $n^{30}$, I'd say that the dominant term is $10\cdot \log(n^{30})$.

A ok! (Nod) This is equal to $10 \cdot 30 \cdot \log n$, right?
 
evinda said:
A ok! (Nod) This is equal to $10 \cdot 30 \cdot \log n$, right?

Right! (Mmm)
 
I like Serena said:
Right! (Mmm)

So, could we say that the dominant term is equal to $\log n$ ? (Thinking)
 
evinda said:
So, could we say that the dominant term is equal to $\log n$ ? (Thinking)

I guess it depends a bit how the phrase "dominant term" is defined exactly. (Wondering)

Taken literally, it's the term that grows more rapidly than any other in the expression.
As such your dominant term would be $10\cdot \log(n^{30}+30)$. (Nerd)
 
I like Serena said:
I guess it depends a bit how the phrase "dominant term" is defined exactly. (Wondering)

Taken literally, it's the term that grows more rapidly than any other in the expression.
As such your dominant term would be $10\cdot \log(n^{30}+30)$. (Nerd)

So, in this case $g(n)$ isn't equal to $\Theta( \text{ dominant term})$, right? (Thinking)

What else could I write at the beginning, to say that $g(n)=\Theta(\log n)$ ? :confused:
 
evinda said:
So, in this case $g(n)$ isn't equal to $\Theta( \text{ dominant term})$, right? (Thinking)

But it is equal to that! (Wait)

It's just that $\Theta(10\cdot \log(n^{30}+30)) = \Theta(\log(n))$. (Nod)
What else could I write at the beginning, to say that $g(n)=\Theta(\log n)$ ? :confused:

Well, you could say that the dominant term is $10\cdot \log(n^{30}+30)$, so $g(n)$ is asymptotically equal to $10\cdot \log(n^{30}+30)$.

And since the dominant term of $n^{30}+30$ is $n^{30}$, it follows that $g(n)$ is asymptotically equal to $10\cdot \log(n^{30})$, which is the same as $300\log(n)$.

In turn, that is asymptotically equal to $\log(n)$.

So:
$$g(n) = \Theta(10\cdot \log(n^{30}+30)) = \Theta(10\cdot \log(n^{30}))
=\Theta(300 \log(n))=\Theta(\log(n))$$
(Mmm)
 
I like Serena said:
But it is equal to that! (Wait)

It's just that $\Theta(10\cdot \log(n^{30}+30)) = \Theta(\log(n))$. (Nod)

Well, you could say that the dominant term is $10\cdot \log(n^{30}+30)$, so $g(n)$ is asymptotically equal to $10\cdot \log(n^{30}+30)$.

And since the dominant term of $n^{30}+30$ is $n^{30}$, it follows that $g(n)$ is asymptotically equal to $10\cdot \log(n^{30})$, which is the same as $300\log(n)$.

In turn, that is asymptotically equal to $\log(n)$.

So:
$$g(n) = \Theta(10\cdot \log(n^{30}+30)) = \Theta(10\cdot \log(n^{30}))
=\Theta(300 \log(n))=\Theta(\log(n))$$
(Mmm)

I understand! (Nod) Thank you very much! (Smile)
 

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