Problem 2 - Active Calculus

Question

For the following alternating series, \(\sum_{n=1}^{\infty} a_{n}=0.45-\frac{(0.45)^{3}}{3 !}+\frac{(0.45)^{5}}{5 !}-\frac{(0.45)^{7}}{7 !}+\ldots\) how many terms do you have to compute in order for your approximation (your partial sum) to be within 0.0000001 from the convergent value of that series?

Step-by-Step Solution

Answer

Four terms are needed.

Step 1: Identify the Series

The given series is an alternating series: \[ a_n = 0.45 - \frac{(0.45)^3}{3!} + \frac{(0.45)^5}{5!} - \frac{(0.45)^7}{7!} + \ldots \].Each term of the series can be written as \[(-1)^{n+1}\frac{(0.45)^{2n-1}}{(2n-1)!}\] for \(n = 1, 2, 3, \ldots \).

Step 2: Determine the Error Bound

For an alternating series, the error in the partial sum when using \(N\) terms is less than or equal to the absolute value of the next term:\[ |S - S_N| \leq |a_{N+1}| \]. We want this error to be less than 0.0000001.

Step 3: Find the General Term

The general term of the series is \[ a_{N+1} = \frac{(0.45)^{2(N+1)-1}}{(2(N+1)-1)!} \]. We need to find \( N \) such that \[ \frac{(0.45)^{2(N+1)-1}}{(2(N+1)-1)!} < 0.0000001 \].

Step 4: Compute the Terms Iteratively

Calculate \( a_{N+1} \) for various values of \( N \) until the term is less than 0.0000001. Start with small values of \( N \) and incrementally increase it. For example: 1. For \( N = 1 \), \( a_2 = \frac{(0.45)^3}{3!} = 0.0150375 \leftrightarrow 2.5 \times 10^{-3} e 0.0000001 \)2. For \( N = 2 \), \( a_3 = \frac{(0.45)^5}{5!} \approx 0.000213 \leftrightarrow 2.13 \times 10^{-4} e 0.0000001 \)3. For \( N = 3 \), \( a_4 = \frac{(0.45)^7}{7!} \approx 0.0000017 \leftrightarrow 1.7 \times 10^{-6} < 0.0000001 \) Therefore, \( N = 4 \) is where the term is sufficiently small.

Step 5: Conclusion

We compute until we achieve \( N=4 \) terms to ensure the partial sum is within 0.0000001 from the convergent value.

Key Concepts

Alternating series

An alternating series is a series whose terms alternate in sign. This means the terms switch between positive and negative. For example, consider the series \(a_n = 0.45 - \frac{(0.45)^3}{3!} + \frac{(0.45)^5}{5!} - \frac{(0.45)^7}{7!} + \ldots\).
This given series is also an alternating series. Each subsequent term flips sign from positive to negative or vice versa. Alternating series often converge under specific conditions.

Error bound

The error bound in an alternating series tells us how far off our partial sum is from the actual sum. In alternating series, the error bound is very useful. It states that the error after using N terms is less than or equal to the absolute value of the next term \(|S - S_N| \leq |a_{N+1}|\).
For our problem, we require the error bound to be less than 0.0000001. By examining the next unused term, we can determine if additional terms are needed for the approximation to meet the desired accuracy.

Convergence criteria

The convergence criteria for an alternating series are key to understanding when the series will converge. For the series to converge:

• The absolute value of the terms must decrease monotonically (i.e., each term |a_{n+1}| must be smaller than |a_n|).
• The limit of the terms must be zero (\(\lim_{n \to \infty} a_n = 0\)).

Notice that for the series we have, both criteria are satisfied. The terms involve factorials in the denominator, which grow very fast, ensuring that terms become smaller and approach zero eventually.

Partial sum

The partial sum of an alternating series is the sum of a finite number of its terms. For instance, the partial sum of the first N terms is denoted as \(S_N\).
Calculation of the partial sum is straightforward. You simply add and subtract the series terms in the specified order. For our series, if you sum until a certain number of terms such that the error is within the acceptable bound, you have the partial sum close to the series' true value.

General term calculation

General term calculation allows us to inspect each term's contribution to understand the series better. In our example, the general term is given by \((-1)^{n+1}\frac{(0.45)^{2n-1}}{(2n-1)!}\) for \(n = 1, 2, 3, \ldots \).
To find the number of terms needed to achieve a specific accuracy, we use the error bound formula:

\(\frac{(0.45)^{2(N+1)-1}}{(2(N+1)-1)!} < 0.0000001\)
By calculating the size of each term iteratively, we can determine when the next term becomes smaller than 0.0000001, indicating sufficient precision in our partial sum. Following this procedure, it turns out that at least four terms are needed for our example.