True or false? On the educational website, the text would appear as follows:---\[ \left\lfloor \frac{n}{5} \right\rfloor + 6n + 3 \text{ is } \Theta(n) \]In this equation, $\left\lfloor \frac{n}{5} \right\rfloor$ represents the floor function applied to $ \frac{n}{5} $, which means rounding down $ \frac{n}{5} $ to the nearest integer. The term $6n + 3$ is a linear function of $n$. When combined, the expression $\left\lfloor \frac{n}{5} \right\rfloor + 6n + 3$ has a growth rate that is linearly proportional to $n$, and thus it falls within the asymptotic notation $\Theta(n)$.The $\Theta(n)$ notation indicates that the function grows linearly with respect to $n$, meaning that for large values of $n$, the expression is bound both above and below by linear functions of $n$, up to constant factors.--- ### Analyzing Arithmetic Series in Terms of Big Theta NotationThe following expression represents an arithmetic series:\[ 3 + 6 + 9 + \ldots + 3n \]The sum of this series can be expressed in Big Theta notation as:\[ \Theta(n^3) \]### ExplanationIn this series, the common difference between consecutive terms is 3. To understand the total sum's complexity, we explore the sum of the first $ n $ terms of this arithmetic series.Given an arithmetic series:\[ a + (a+d) + (a+2d) + \ldots + [a + (n-1)d] \]where:- $ a = 3 $ (the first term)- $ d = 3 $ (common difference)- $ n $ is the number of termsThe sum $ S_n $ of the first $ n $ terms of an arithmetic series is given by the formula:\[ S_n = \frac{n}{2} [2a + (n-1)d] \]Substitute $ a $ and $ d $ into the formula:\[ S_n = \frac{n}{2} [2(3) + (n-1)(3)] \]\[ S_n = \frac{n}{2} [6 + 3n - 3] \]\[ S_n = \frac{n}{2} [3n + 3] \]\[ S_n = \frac{n}{2} \cdot 3(n + 1) \]\[ S_n = \frac{3n^2 + 3n}{2} \]In Big Theta notation, higher-order terms and constants are significant for representing the growth rate. For large values of $ n $, $ S_n $ is dominated by the $ n^3 $ term:\[ S_n = \Theta(n^3) \]Thus, the series $ 3 + 6 + 9 + \ldots + 3n $ can be expressed as $ \Theta(n^3) $ in Big Theta notation, indicating that the complexity or growth rate of the sum is cubic.

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Description: True or false? On the educational website, the text would appear as follows:---\[ \left\lfloor \frac{n}{5} \right\rfloor + 6n + 3 \text{ is } \Theta(n) \]In this equation, $\left\lfloor \frac{n}{5} \right\rfloor$ represents the floor function appli

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+6n +3 is Θ (n)

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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True or false?

On the educational website, the text would appear as follows:

---

\[ \left\lfloor \frac{n}{5} \right\rfloor + 6n + 3 \text{ is } \Theta(n) \]

In this equation, $\left\lfloor \frac{n}{5} \right\rfloor$ represents the floor function applied to $ \frac{n}{5} $, which means rounding down $ \frac{n}{5} $ to the nearest integer. The term $6n + 3$ is a linear function of $n$. When combined, the expression $\left\lfloor \frac{n}{5} \right\rfloor + 6n + 3$ has a growth rate that is linearly proportional to $n$, and thus it falls within the asymptotic notation $\Theta(n)$.

The $\Theta(n)$ notation indicates that the function grows linearly with respect to $n$, meaning that for large values of $n$, the expression is bound both above and below by linear functions of $n$, up to constant factors.

---

### Analyzing Arithmetic Series in Terms of Big Theta Notation

The following expression represents an arithmetic series:

\[ 3 + 6 + 9 + \ldots + 3n \]

The sum of this series can be expressed in Big Theta notation as:

\[ \Theta(n^3) \]

### Explanation

In this series, the common difference between consecutive terms is 3. To understand the total sum's complexity, we explore the sum of the first $ n $ terms of this arithmetic series.

Given an arithmetic series:

\[ a + (a+d) + (a+2d) + \ldots + [a + (n-1)d] \]

where:
- $ a = 3 $ (the first term)
- $ d = 3 $ (common difference)
- $ n $ is the number of terms

The sum $ S_n $ of the first $ n $ terms of an arithmetic series is given by the formula:

\[ S_n = \frac{n}{2} [2a + (n-1)d] \]

Substitute $ a $ and $ d $ into the formula:

\[ S_n = \frac{n}{2} [2(3) + (n-1)(3)] \]
\[ S_n = \frac{n}{2} [6 + 3n - 3] \]
\[ S_n = \frac{n}{2} [3n + 3] \]
\[ S_n = \frac{n}{2} \cdot 3(n + 1) \]
\[ S_n = \frac{3n^2 + 3n}{2} \]

In Big Theta notation, higher-order terms and constants are significant for representing the growth rate. For large values of $ n $, $ S_n $ is dominated by the $ n^3 $ term:

\[ S_n = \Theta(n^3) \]

Thus, the series $ 3 + 6 + 9 + \ldots + 3n $ can be expressed as $ \Theta(n^3) $ in Big Theta notation, indicating that the complexity or growth rate of the sum is cubic.

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