Problem 19 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Express each of the sums in closed form. Wherever possible, give a numerical approximation of the sum, rounded off to 3 decimal places. $$ \frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\cdots-\frac{3}{2^{6}} $$

Step-by-Step Solution

Answer

The sum of the series in closed form is \(15/4\) or \(3.75\), and the numerical approximation of this value to three decimal places is \(3.750\).

Step 1 - Identify the series

The given series is a geometric series because each term is obtained by multiplying the previous term by common ratio. In this case, observe that each term is obtained by multiplying the previous term by \(-\frac{1}{2}\). So, the common ratio \(r = -\frac{1}{2}\). The first term \(a = \frac{3}{2}\).

Step 2 - Apply the formula for geometric series

The formula for sum of a geometric series is \[S = \frac{a(1-r^n)}{1-r}\] where \(n\) is the number of terms, \(a\) is the first term, and \(r\) is the common ratio. In this case, \(n=6\), \(a=\frac{3}{2}\), and \(r=-\frac{1}{2}\). Plug these values into the formula and calculate the exact value of S.

Step 3 - Calculation of the sum

Now, substitute these values into the formula: \(S = \frac{\frac{3}{2}(1 - (-\frac{1}{2})^{6})}{1 - (-\frac{1}{2})}\). After simplifying, we get: S = \(15/4\) or \(3.75\).

Step 4 - Numerical approximation of the sum

Round off the value of the sum to three decimal places. We get as a result \(3.750\).

Key Concepts

Common Ratio

In a geometric series, the common ratio is a key element that determines how each term in the series relates to the one before it. It is simply the factor by which you multiply one term to get the next. For example, if you have a series like \(3, 6, 12, 24, \ldots\), the common ratio is \(2\) because each term is twice the previous one.
In our exercise, the series is \(\frac{3}{2} - \frac{3}{4} + \frac{3}{8} - \cdots - \frac{3}{2^{6}}\). To find the common ratio here, we observe how each term is derived from the previous one. In this series, each term is created by multiplying the previous term by \(-\frac{1}{2}\). This alternating sign (due to the negative in the common ratio) is what causes the sign change in each term of our series.
If you know the common ratio in a geometric series, you can predict all the terms given the initial term \(a\), using the formula:

• Second term: \(a \times r\)
• Third term: \(a \times r^2\)

This understanding of common ratio is essential to finding the sum of the series.

Sum Formula

The sum of a geometric series can be calculated using a specific formula that makes the process much simpler. Instead of adding up each term individually, the formula provides a neat solution.
The formula to find the sum \(S\) of the first \(n\) terms of a geometric series is: \[S = \frac{a(1-r^n)}{1-r}\] where

• \(a\) is the first term,
• \(r\) is the common ratio,
• and \(n\) is the number of terms to sum.

In our example, with \(a = \frac{3}{2}\), \(r = -\frac{1}{2}\), and \(n = 6\), we plug these values into the formula to find the sum. After inserting and simplifying, the sum \(S\) is calculated as \(\frac{15}{4}\) or \(3.75\).
This formula is powerful because it compacts potentially lengthy calculations into one formulaic expression, which is especially handy with many terms involved or when dealing with infinite series.

Numerical Approximation

Numerical approximation is all about getting as close as possible to the exact number, often by rounding or truncating decimals to a specific number of places. It's a practical approach in many fields. In this exercise, after using the sum formula to calculate the exact value of the series, we get \(3.75\). To give a numerical approximation, we then round this number to three decimal places, resulting in \(3.750\).
This process of rounding can also be useful in real-world applications where exact precision isn't necessary or possible due to constraints like measurement limitations or computational resources. Numerical approximation helps manage long decimal expansions in a way that’s easier to work with and communicate, especially in engineering and scientific calculations.