Question

Find the minimum value of \(N\) such that the remainder estimate \(\int_{N+1}^{\infty} f

Short answer

The minimum value of \(N\) is 1000.

Step-by-step solution

Understand the Problem

The problem requires finding the minimum value of \(N\) such that the partial sum \(\sum_{n=1}^{N} \frac{1}{1+n^2}\) approximates the infinite series \(\sum_{n=1}^{\infty} \frac{1}{1+n^2}\) with an error less than \(10^{-3}\). We use the remainder estimate \(\int_{N+1}^{\infty} f < R_N < \int_N^{\infty} f\) for comparison.

Establish the Inequality

For the series \(a_n = \frac{1}{1+n^2}\), define \(f(x) = \frac{1}{1+x^2}\). We need to solve the inequality \(\int_{N}^{\infty} \frac{1}{1+x^2} \, dx < 10^{-3}\) for \(N\).

Calculate the Integral

The integral \(\int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C\). Thus, \(\int_{N}^{\infty} \frac{1}{1+x^2} \, dx = \lim_{b \to \infty} \left[\tan^{-1}(b) - \tan^{-1}(N)\right] = \left[\frac{\pi}{2} - \tan^{-1}(N)\right]\).

Solve the Inequality

Set up the inequality \(\frac{\pi}{2} - \tan^{-1}(N) < 10^{-3}\). This gives \(\tan^{-1}(N) > \frac{\pi}{2} - 10^{-3}\) or \(N > \tan\left(\frac{\pi}{2} - 10^{-3}\right)\).

Find the Minimum N

Compute \(N\) such that \(N = \tan\left(\frac{\pi}{2} - 10^{-3}\right)\). Approximating numerically, \(N \approx 999.998\). Since \(N\) must be an integer, the smallest \(N\) is 1000.

Key concepts

Partial Sum

In the realm of infinite series, understanding the concept of a partial sum is essential. A partial sum, in essence, is the sum of a finite number of terms from an infinite series. Let's break it down to basics.

When we have an infinite series, say, \( \sum_{n=1}^{\infty} a_{n} \), a partial sum is represented as \( S_{N} = \sum_{n=1}^{N} a_{n} \). This means we are adding up the terms from 1 to \( N \) in that series. It's like watching just a few episodes of a show instead of the entire series.

Partial sums help us approximate infinite series, especially when trying to gauge convergence or determine if the series has a sum that tends toward a particular value. In practical applications, we cannot sum infinitely many terms; hence, we rely on partial sums to estimate these infinite sums.

Remainder Estimate

The remainder estimate plays a vital role when dealing with infinite series and partial sums. It's like the gap or difference between the actual infinite sum and the partial sum you've calculated.

In simpler terms, consider you are painting a room, but it's taking longer than expected. The remainder is what still needs painting. In infinite series, this 'remainder' tells you how much your partial sum is off from the total sum of the series.

• In our exercise, the remainder \( R_N \) is the difference between the actual sum \( \sum_{n=1}^{\infty} a_{n} \) and the partial sum \( \sum_{n=1}^{N} a_{n} \).
• The aim is to make this remainder small, within a certain error margin.
• The inequality \( \int_{N+1}^{\infty} f < R_N < \int_N^{\infty} f \) helps to estimate this remainder more precisely.

Understanding the remainder estimate is key to knowing how good your approximation is.

Integral Calculus

Integral calculus is the superhero that helps us solve many mathematical problems, especially those involving infinite series. It's all about accumulation and finding areas under curves.

In our exercise, we use an integral to estimate the remainder, specifically, we compute \( \int_{N}^{\infty} \frac{1}{1+x^2} \, dx \). This integral gives us an approximation of how much more we need to sum to get close to the total infinite series' sum.

• Integral calculus allows us to calculate continuous sums and is widely useful for finding exact areas under curves.
• Here, the integral \( \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C \) reveals how the function behaves as we approach infinity: it converges to \( \frac{\pi}{2} \).

For estimating remainders or approximating series, integral calculus provides an elegant and accurate approach.

Approximation Error

When approximating infinite series, understanding and managing approximation error is crucial. It measures how much our computed partial sum \( S_N \) deviates from the true infinite sum.

If you were building a bridge, the approximation error would be the small gaps between the planks that could affect the overall structure. Similarly, in mathematics, we want this error to be as small as possible, ensuring high accuracy.

• In this task, our goal was to keep this error \( <10^{-3} \), meaning the approximation should be incredibly close to the actual sum.
• The smaller the error, the more precise your approximation.
• Understanding this concept helps when you decide how many terms of the series you need to consider to achieve a desired level of accuracy.

Learning to estimate and control approximation error is a fundamental skill in both theoretical and applied mathematics.