Problem 5 - University Calculus: Early Transcendentals

Question

Without sigma notation. Then evaluate them. $$\sum_{k=1}^{3}(-1)^{k+1} \sin \frac{\pi}{k}$$

Step-by-Step Solution

Answer

The evaluated sum is \(-1 + \frac{\sqrt{3}}{2}\).

Step 1: Understanding the Problem

The given problem involves evaluating a finite series expressed using sigma notation, represented by \( \sum_{k=1}^{3}(-1)^{k+1} \sin \frac{\pi}{k} \). The goal is to expand this sum without using sigma notation and then evaluate each term individually.

Step 2: Substituting Values for k

Expand the series by substituting each integer value from 1 to 3 for \( k \):1. When \( k = 1 \), the term is \((-1)^{1+1} \sin \frac{\pi}{1}\).2. When \( k = 2 \), the term is \((-1)^{2+1} \sin \frac{\pi}{2}\).3. When \( k = 3 \), the term is \((-1)^{3+1} \sin \frac{\pi}{3}\).

Step 3: Simplifying Each Term

Now substitute and simplify:1. For \( k = 1 \), \((-1)^{2} = 1\), so the term is \(1 \cdot \sin \pi = 0\) because \( \sin \pi = 0 \).2. For \( k = 2 \), \((-1)^{3} = -1\), so the term is \(-1 \cdot \sin \frac{\pi}{2} = -1\) because \( \sin \frac{\pi}{2} = 1 \).3. For \( k = 3 \), \((-1)^{4} = 1\), so the term is \(1 \cdot \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\) because \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \).

Step 4: Adding the Terms

Add the simplified values from each term:\[ 0 - 1 + \frac{\sqrt{3}}{2} = -1 + \frac{\sqrt{3}}{2} \].

Step 5: Final Evaluation

The final evaluated sum of the series is \(-1 + \frac{\sqrt{3}}{2}\). This is a simplified form, and since it includes \( \sqrt{3} \), it is left in this form as a precise mathematical expression.

Key Concepts

Finite Series

A finite series is a sum of a sequence of numbers or expressions, which terminates after a certain number of terms. It's called "finite" because it doesn't go on indefinitely. In our example, we evaluate a series with three terms. This kind of series is helpful in many areas of mathematics and applied sciences.

When working with finite series, it's essential to determine the number of terms you need to include in the sum. Each term in a finite series is calculated individually, then all terms are summed up. This method allows you to precisely evaluate the expression, as we've done by considering each value of \( k \) from 1 to 3 and simplifying each corresponding term.

Key points to remember about finite series:

• They have a fixed number of terms.
• Each term is computed separately before being added to the others.
• They are often expressed in compact forms, like with sigma notation, for convenience.
• Perfect for situations where you need exact answers with no approximations.

Sigma Notation

Sigma notation, represented by the Greek letter \( \Sigma \), is a way to write and simplify the addition of a series of terms. It's a compact and efficient method for representing long sums, especially those with a regular pattern.

Using sigma notation, you can eloquently express sums without writing out every term. In our exercise, \( \sum_{k=1}^{3}(-1)^{k+1} \, \sin \frac{\pi}{k} \) indicates the sum of terms from \( k=1 \) to \( k=3 \). The variable \( k \) is the index of summation, guiding how the expression changes for each term.

When interpreting sigma notation, follow these steps:

• Identify the range: The lower and upper bounds indicate which terms to include.
• Understand the pattern: Determine how the expression changes with each step in the sequence.
• Substitute and simplify: Replace the index of summation with each integer in the range and simplify those terms individually.

In this exercise, sigma notation helps us describe a pattern but doesn't save work until we interpret and calculate each sum term-by-term.

Trigonometric Functions

Trigonometric functions, such as sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)), are fundamental in calculus and many other areas of mathematics. They relate angles to ratios of sides in right triangles and extend to the unit circle for broader applications.

In our solution, we specifically deal with \( \sin \) at different angles: \( \pi \), \( \frac{\pi}{2} \), and \( \frac{\pi}{3} \). These angles are significant because they correspond to standard angles in trigonometry, often yielding exact values:

• \( \sin \pi = 0 \)
• \( \sin \frac{\pi}{2} = 1 \)
• \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \)

These standard values make calculations straightforward and are useful for simplifying expressions in calculus. Understanding these functions and their properties is crucial, especially when they appear in series or other mathematical constructs.