+6n +3 is Θ (n)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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![On the educational website, the text would appear as follows:
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\[ \left\lfloor \frac{n}{5} \right\rfloor + 6n + 3 \text{ is } \Theta(n) \]
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.
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Transcribed Image Text: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.
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![### 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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb431a66f-4058-4e70-a20a-31b3452e77d6%2F2d0f3fa9-2b3f-47f5-98bd-a5be5b647e60%2Fmed9ie9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### 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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