Truncation Error Calculation For Infinite Series S₃
In the realm of mathematics, particularly when dealing with infinite series, understanding truncation error is crucial. This article delves into the concept of truncation error within the context of a given infinite series, providing a comprehensive explanation and a step-by-step approach to solving the problem. We will explore the significance of truncation error, its calculation, and its implications in approximating the sum of an infinite series. This discussion is essential for students, educators, and anyone interested in numerical analysis and the behavior of infinite series. Let's embark on this mathematical journey to unravel the intricacies of truncation error.
Introduction to Infinite Series and Truncation Error
In the captivating realm of mathematics, infinite series hold a place of profound significance. An infinite series is, in essence, the sum of an infinite sequence of terms. These series can exhibit a wide range of behaviors – some converge to a finite value, while others diverge, soaring off to infinity. The study of infinite series is a cornerstone of calculus and mathematical analysis, finding applications in various fields, from physics to computer science.
When we set out to compute the sum of an infinite series, we often encounter a practical challenge. Since we cannot literally add up an infinite number of terms, we resort to approximations. This is where the concept of truncation error comes into play. Truncation error arises when we approximate an infinite series by summing only a finite number of terms. In essence, we "truncate" the series, cutting it off after a certain point. The truncation error then quantifies the difference between the true sum of the infinite series and our approximation, which is the sum of the finite number of terms we've included.
To put it more formally, let's say we have an infinite series represented as:
∑(n=1 to ∞) a_n = a_1 + a_2 + a_3 + ...
Here, a_n denotes the nth term of the series. The true sum of this infinite series, which we'll call S, is the limit of the partial sums as the number of terms approaches infinity:
S = lim (N→∞) ∑(n=1 to N) a_n
However, in practice, we can only compute a partial sum, S_N, which includes the first N terms:
S_N = ∑(n=1 to N) a_n = a_1 + a_2 + ... + a_N
The truncation error, denoted as E_N, is then the absolute difference between the true sum S and the partial sum S_N:
E_N = |S - S_N|
The magnitude of the truncation error provides us with valuable insight into the accuracy of our approximation. A smaller truncation error signifies a more accurate approximation, while a larger truncation error suggests that our approximation may not be reliable.
Understanding truncation error is paramount in various applications. For instance, in numerical analysis, where we approximate solutions to mathematical problems using computers, truncation error can impact the accuracy of our results. Similarly, in physics, when we model physical phenomena using infinite series, truncation error can affect the precision of our predictions.
Therefore, comprehending truncation error is not merely an academic exercise; it has profound implications for the practical application of mathematics in diverse fields. In the subsequent sections of this article, we will delve deeper into how to calculate truncation error, explore techniques for minimizing it, and examine its significance in the context of the given problem.
Problem Statement and Series Analysis
The heart of our discussion lies in the following mathematical problem: Given that the infinite series ∑(n=1 to ∞) 1/(4*(2^n)) equals 0.25, our mission is to determine the truncation error for S₃. In simpler terms, we are asked to find the difference between the true sum of the infinite series and the sum of its first three terms.
To embark on this quest, let's first dissect the given infinite series. The series is represented as:
∑(n=1 to ∞) 1/(4*(2^n))
This notation tells us that we are summing terms of the form 1/(4*(2^n)), where n takes on integer values starting from 1 and extending to infinity. The first few terms of this series are:
- For n = 1: 1/(4*(2^1)) = 1/(4*2) = 1/8
- For n = 2: 1/(4*(2^2)) = 1/(4*4) = 1/16
- For n = 3: 1/(4*(2^3)) = 1/(4*8) = 1/32
- For n = 4: 1/(4*(2^4)) = 1/(4*16) = 1/64
- And so on...
Thus, the infinite series can be written as:
1/8 + 1/16 + 1/32 + 1/64 + ...
Now, a crucial observation is that this series is a geometric series. A geometric series is a special type of series where each term is obtained by multiplying the previous term by a constant factor, known as the common ratio. In our case, the common ratio (r) is 1/2. To see this, notice that:
(1/16) / (1/8) = 1/2
(1/32) / (1/16) = 1/2
(1/64) / (1/32) = 1/2
And so forth. The common ratio remains constant at 1/2.
The fact that we are dealing with a geometric series is highly significant because we possess a well-established formula for the sum of an infinite geometric series. The sum (S) of an infinite geometric series is given by:
S = a / (1 - r)
where 'a' is the first term of the series and 'r' is the common ratio. However, this formula is valid only when the absolute value of the common ratio is less than 1 (|r| < 1). In our case, |1/2| < 1, so the formula is indeed applicable.
In our problem, we are given that the sum of the infinite series is 0.25. Let's verify this using the formula. The first term (a) is 1/8, and the common ratio (r) is 1/2. Plugging these values into the formula, we get:
S = (1/8) / (1 - 1/2) = (1/8) / (1/2) = (1/8) * 2 = 1/4 = 0.25
This confirms that the given sum of the infinite series is indeed 0.25.
Now, let's define S₃. S₃ represents the sum of the first three terms of the series:
S₃ = 1/8 + 1/16 + 1/32
To calculate S₃, we simply add these fractions:
S₃ = (4/32) + (2/32) + (1/32) = 7/32
In decimal form, S₃ is approximately 0.21875.
With the true sum (S) and the partial sum S₃ in hand, we are now poised to calculate the truncation error for S₃. In the next section, we will delve into the calculation and interpretation of this error.
Calculating the Truncation Error for S₃
Having established the true sum of the infinite series (S = 0.25) and calculated the sum of the first three terms (S₃ = 7/32 ≈ 0.21875), we are now ready to determine the truncation error for S₃. Recall that the truncation error (E₃) is defined as the absolute difference between the true sum and the partial sum:
E₃ = |S - S₃|
Substituting the values we have, we get:
E₃ = |0.25 - 0.21875|
Now, we perform the subtraction:
E₃ = |0.03125|
Since the absolute value of a positive number is the number itself, we have:
E₃ = 0.03125
Therefore, the truncation error for S₃ is 0.03125. This value represents the magnitude of the error we incur when we approximate the sum of the infinite series by summing only its first three terms. It quantifies the amount by which our approximation deviates from the true sum.
To express this truncation error in a more accessible way, we can round it to three decimal places, which gives us 0.031. This rounded value is very close to one of the options provided in the problem statement.
Now, let's delve a bit deeper into the interpretation of this truncation error. A truncation error of 0.03125 signifies that our approximation, S₃, is within 0.03125 units of the true sum of the infinite series. In other words, if we were to use S₃ as an estimate for the sum of the infinite series, we would be off by at most 0.03125. This provides us with a measure of the accuracy of our approximation.
It's worth noting that the truncation error is influenced by the number of terms we include in our partial sum. Generally, as we include more terms, the truncation error decreases, and our approximation becomes more accurate. This is because we are capturing a larger portion of the infinite series's contribution to the sum. Conversely, if we include fewer terms, the truncation error tends to be larger, indicating a less accurate approximation.
In the context of our problem, the truncation error for S₃ is relatively small, suggesting that summing the first three terms provides a reasonable approximation of the true sum. However, the acceptability of the truncation error depends on the specific application and the level of accuracy required. In some cases, an error of 0.03125 might be tolerable, while in others, a higher level of precision might be necessary, necessitating the inclusion of more terms in the partial sum.
Moreover, the rate at which the truncation error decreases as we add more terms is closely tied to the convergence properties of the infinite series. Series that converge rapidly tend to have smaller truncation errors for a given number of terms, while series that converge slowly might require a larger number of terms to achieve a comparable level of accuracy.
In the upcoming section, we will discuss the significance of this truncation error in the context of the given problem and compare our calculated value with the provided options.
Solution and Answer
In the preceding sections, we have meticulously calculated the truncation error for S₃, arriving at a value of 0.03125. Now, let's connect this result back to the original problem statement and the options provided.
The problem presented us with the infinite series ∑(n=1 to ∞) 1/(4*(2^n)) and stated that its sum is 0.25. Our task was to determine the truncation error for S₃, where S₃ represents the sum of the first three terms of the series. We computed S₃ to be 7/32, which is approximately 0.21875. Subsequently, we calculated the truncation error (E₃) as the absolute difference between the true sum (0.25) and S₃ (0.21875), obtaining E₃ = 0.03125.
The problem statement also provided us with a set of options, and our goal is to identify the option that matches our calculated truncation error. The options were:
A. 0.016
B. 0.031
C. 0.219
D. 0.234
Comparing our calculated truncation error (0.03125) with the options, we observe that option B, 0.031, is the closest match. The slight discrepancy between 0.03125 and 0.031 arises due to rounding. We rounded 0.03125 to three decimal places to obtain 0.031.
Therefore, the correct answer to the problem is option B: 0.031. This signifies that the truncation error incurred when approximating the sum of the infinite series by summing its first three terms is approximately 0.031.
It's crucial to emphasize that understanding the concept of truncation error is paramount in this problem. Truncation error quantifies the error introduced when we approximate an infinite series by considering only a finite number of terms. In practical applications, we often resort to such approximations because summing an infinite number of terms is not feasible. However, it's essential to be aware of the potential error introduced by this approximation and to take steps to minimize it if necessary.
In the context of this problem, the truncation error of 0.031 suggests that summing the first three terms provides a reasonably accurate approximation of the true sum of the infinite series. However, if a higher level of accuracy were required, we would need to include more terms in our partial sum, thereby reducing the truncation error.
In conclusion, we have successfully navigated through the problem, calculated the truncation error for S₃, and identified the correct answer. This exercise underscores the importance of understanding infinite series, truncation error, and their applications in mathematical problem-solving.
Conclusion
In this comprehensive exploration, we have delved into the concept of truncation error within the context of an infinite series. We tackled the problem of determining the truncation error for S₃ in the series ∑(n=1 to ∞) 1/(4*(2^n)), where S₃ represents the sum of the first three terms. Through a step-by-step approach, we calculated the truncation error to be approximately 0.031, corresponding to option B in the problem statement.
Our journey began with an introduction to infinite series and the fundamental concept of truncation error. We established that truncation error arises when we approximate an infinite series by summing only a finite number of terms. This error quantifies the difference between the true sum of the infinite series and our approximation.
We then meticulously analyzed the given infinite series, recognizing it as a geometric series with a common ratio of 1/2. This allowed us to leverage the formula for the sum of an infinite geometric series to verify the given sum of 0.25. We also calculated S₃, the sum of the first three terms, as 7/32, which is approximately 0.21875.
With the true sum and S₃ in hand, we proceeded to calculate the truncation error using the formula E₃ = |S - S₃|. This yielded a truncation error of 0.03125, which we rounded to 0.031 for comparison with the options.
Throughout this process, we emphasized the significance of understanding truncation error in approximating infinite series. Truncation error serves as a crucial measure of the accuracy of our approximation. A smaller truncation error indicates a more accurate approximation, while a larger truncation error suggests a greater deviation from the true sum.
Moreover, we highlighted the factors that influence truncation error, such as the number of terms included in the partial sum and the convergence properties of the infinite series. Including more terms generally reduces the truncation error, while series that converge rapidly tend to have smaller truncation errors for a given number of terms.
The insights gained from this exploration have broader implications in various fields. In numerical analysis, truncation error is a critical consideration when approximating solutions to mathematical problems using computers. In physics, where infinite series are often used to model physical phenomena, understanding truncation error is essential for ensuring the accuracy of predictions.
In conclusion, the concept of truncation error is a cornerstone of working with infinite series. It allows us to quantify the error introduced by approximation and make informed decisions about the level of accuracy required for a given application. By mastering this concept, we equip ourselves with a valuable tool for tackling mathematical problems in diverse domains.
