Problem 8 - Calculus

Question

Find the value of the indicated sum. \(\sum_{k=-1}^{6} k \sin (k \pi / 2)\)

Step-by-Step Solution

Answer

The value of the sum is 4.

Step 1: Set Up the Sum

We need to calculate the sum \( \sum_{k=-1}^{6} k \sin \left( \frac{k \pi}{2} \right) \). This involves evaluating the expression \( k \sin \left( \frac{k \pi}{2} \right) \) for each integer value of \( k \) from -1 to 6, and then adding all these values together.

Step 2: Evaluate Terms Individually

Let's examine each term separately:- For \( k = -1 \), \( k \sin \left( \frac{-\pi}{2} \right) = -1(-1) = 1 \).- For \( k = 0 \), \( k \sin (0) = 0 \cdot 0 = 0 \).- For \( k = 1 \), \( k \sin \left( \frac{\pi}{2} \right) = 1 \cdot 1 = 1 \).- For \( k = 2 \), \( k \sin (\pi) = 2 \cdot 0 = 0 \).- For \( k = 3 \), \( k \sin \left( \frac{3\pi}{2} \right) = 3(-1) = -3 \).- For \( k = 4 \), \( k \sin (2\pi) = 4 \cdot 0 = 0 \).- For \( k = 5 \), \( k \sin \left( \frac{5\pi}{2} \right) = 5 \cdot 1 = 5 \).- For \( k = 6 \), \( k \sin (3\pi) = 6 \cdot 0 = 0 \).

Step 3: Calculate the Sum

Add the results from each evaluated term:\( 1 + 0 + 1 + 0 - 3 + 0 + 5 + 0 = 4 \).

Key Concepts

Trigonometric Functions

Trigonometric functions, like sine and cosine, represent ratios of a triangle's sides corresponding to specific angles. These functions are periodic, meaning they repeat their values in regular intervals. For example, the sine function, denoted as \( \sin(x) \), has a period of \( 2\pi \). This means its values repeat every \( 2\pi \) radians.In the exercise, the sine function is evaluated at different angles derived from integer values of \( k \), like \( \frac{k\pi}{2} \). Depending on the value of \( k \), the angle could be zero, \( \frac{\pi}{2} \), \( \pi \), \( \frac{3\pi}{2} \), or \( 2\pi \), each producing specific outcomes. For example:- \( \sin(0) = 0 \)- \( \sin(\frac{\pi}{2}) = 1 \)- \( \sin(\pi) = 0 \)- \( \sin(\frac{3\pi}{2}) = -1 \)- \( \sin(2\pi) = 0 \)Understanding these standard values is essential for evaluating expressions involving sine at fractional multiples of \( \pi \). Recognizing this periodicity helps simplify complex summations involving trigonometric functions.

Evaluating Series

A series is the sum of the terms of a sequence. Calculus often involves evaluating such expressions to find their total value. The given exercise requires finding the sum of products involving integer multiples of the sine function.When evaluating a series, it is important to:- Determine the starting and ending indices, which are \( k = -1 \) and \( k = 6 \) in this case.- Calculate each individual term based on these indices. Here, each term is calculated as \( k \sin(\frac{k\pi}{2}) \).- Add up the contributions of each term to find the total value of the series.This method of evaluation requires careful attention to the periodic nature of trigonometric functions. It ensures that only those terms that produce non-zero sine values significantly contribute to the final sum. This understanding simplifies processing, as seen with the sine values of zero not contributing to the overall sum.

Step-by-Step Solution

A step-by-step solution can greatly aid in understanding complex mathematical exercises. It breaks down the process into manageable parts, making it easier to follow.1. **Initial Setup**: Establish the expression to be evaluated. We are considering the sum \( \sum_{k=-1}^{6} k \sin(\frac{k\pi}{2}) \).2. **Individual Evaluations**: The expression \( k \sin(\frac{k\pi}{2}) \) is evaluated for each \( k \) from -1 to 6. Here's how it's done: - Assess the sine value at multiples of \( \pi/2 \) for each \( k \). - Multiply by \( k \) to find each term of the series.3. **Summation**: Once the values for each \( k \) are calculated, they are summed: \( 1 + 0 + 1 + 0 - 3 + 0 + 5 + 0 = 4 \).Following a detailed, logical progression of steps clarifies the solution process. It identifies where each value comes from and makes it easier to replicate similar solutions in other exercises.