Question

Show directly that $f(n)=n^2+3n^3\in \mathrm{\Theta }\left(n^3\right)$. That is, use the definitions of $O$ and $\mathrm{\Omega }$ to show that $f(n)$ is in both $O(n^{3)}\text{ and }\mathrm{\Omega }(n^{3)}$

Short answer

The function $f(n)=n^2+3n^3$ belongs to both $O(n^3)$ and $\mathrm{\Omega }(n^3)$. Therefore, it is in $\mathrm{\Theta }(n^3)$.

Step-by-step solution

Proving $f(n)\in O(n^3)$

Define $f(n)=n^2+3n^3$. To show $f(n)\in O(n^3)$, we need to find positive constants $c$ and $n_0$ such that $0\le f(n)\le c\ast n^3$ for all $n\ge n_0$. Let's select $c=4$ and $n_0=1$. For $n\ge n_0$, we have $0\le n^2+3n^3\le n^3+3n^3=4n^3$. Therefore, $f(n)=n^2+3n^3\in O(n^3)$.

Proving $f(n)\in \mathrm{\Omega }(n^3)$

To show $f(n)\in \mathrm{\Omega }(n^3)$, we need to find positive constants $c$ and $n_0$ such that $0\le c\ast n^3\le f(n)$ for all $n\ge n_0$. Let's select $c=1$ and $n_0=1$. For $n\ge n_0$, we have $0\le n^3\le n^2+3n^3=f(n)$. Therefore, $f(n)=n^2+3n^3\in \mathrm{\Omega }(n^3)$.

Conclusion

Since we have shown that $f(n)=n^2+3n^3$ is a member of both $O(n^3)$ and $\mathrm{\Omega }(n^3)$, we can conclude that $f(n)=n^2+3n^3\in \mathrm{\Theta }(n^3)$.

Key concepts

Big O Notation

When we talk about "Big O Notation," we're discussing how to describe the upper limit of the growth rate of a function. It helps us understand the worst-case scenario regarding how a function behaves as its input size grows. In mathematical terms, if a function $f(n)$ is said to be $O(n^k)$, this means there are constants $c$ and $n_0$ such that for every input larger than $n_0$, $f(n)$ will not grow faster than $c\cdot n^k$.
To put it simply:

• "Big O" gives us a high limit prediction.
• It tells us that our function's growth won't exceed a certain threshold.
• It's used to measure algorithm efficiency, particularly in computing.

For example, in the exercise provided, we consider $f(n)=n^2+3n^3$. By selecting suitable constants like $c=4$ and $n_0=1$, we showed that $f(n)$ fits within the boundaries of $O(n^3)$.

Big Theta Notation

"Big Theta Notation" serves as a balanced view between the upper and the lower bounds of a function's growth. It captures the tightest bound of an algorithm, which means it precisely describes the asymptotic behavior as the size of the input approaches infinity.
In more straightforward terms, a function $f(n)$ is in $\mathrm{\Theta }(g(n))$ if it is both $O(g(n))$ and $\mathrm{\Omega }(g(n))$. Here's what this means:

• "Big Theta" provides the exact growth rate "in-between" these two bounds.
• Think of it as being sandwiched firmly by both
"Big O" and "Big Omega."
• This notation shows the efficiency as input size grows towards infinity.

In our example, $f(n)=n^2+3n^3$ was shown to be both $O(n^3)$ and $\mathrm{\Omega }(n^3)$, hence $f(n)$ is in $\mathrm{\Theta }(n^3)$. This implies that both a strong upper and lower bound exist, confirming its exact growth rate behavior.

Big Omega Notation

"Big Omega Notation" helps determine the lower limit of a function's growth rate. In practical terms, it provides the best-case predictability of how a function behaves as the size of its input expands.
For a function $f(n)$ to be considered $\mathrm{\Omega }(g(n))$, it implies that there are constants $c$ and $n_0$ such that for input sizes greater than $n_0$, $f(n)$ will not drop below $c\cdot g(n)$.
Here's a way to break it down:

• "Big Omega" is like seeing the glass half-full. It tells us about the least growth observed.
• This perspective handles efficiency insights under the best scenarios.
• It sets a benchmark on how low the function can genuinely go.

In the case of the given function $f(n)=n^2+3n^3$, by setting constants such as $c=1$ and $n_0=1$, we established that $f(n)$ remains above the limit of $\mathrm{\Omega }(n^3)$. This confirmed its existence as a lower bound in the growth spectrum.