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 very large, and check calculations again to approximate this value effectively using underlying estimates.

Step-by-step solution

Understanding the Problem

We need to find the minimum value of \(N\) such that the series \(\sum_{n=1}^{N} a_{n}\) estimates \(\sum_{n=1}^{\infty} a_{n}\) accurately to within the given error of \(10^{-4}\). The function \(a_n = \frac{1}{n^{1.1}}\) is a decreasing positive function, and the series resembles a p-series with \(p = 1.1\), which converges.

Applying the Remainder Estimate for Series

As given, \(R_N\) defines the remainder after partial sum \( \sum_{n=1}^{N} a_{n} \). We use the integral test remainder estimate: \[\int_{N+1}^{\infty} \frac{1}{x^{1.1}} \, dx < R_N < \int_{N}^{\infty} \frac{1}{x^{1.1}} \, dx.\]

Calculate the Integral

We evaluate the integral \(\int \frac{1}{x^{1.1}} \, dx\) to get the general solution: \[\int \frac{1}{x^{1.1}} \, dx = \left[-\frac{x^{-0.1}}{-0.1}\right] = 10 x^{-0.1} + C. \]

Evaluate the Improper Integrals

Calculate the improper integrals used in the remainder estimates. - For \(\int_{N}^{\infty} \frac{1}{x^{1.1}} \, dx\), use the formula: \[\int_{N}^{\infty} \frac{1}{x^{1.1}} \, dx = \lim_{b \to \infty} 10b^{-0.1} - 10N^{-0.1} = 0 - 10N^{-0.1} = -10N^{-0.1}. \]Therefore, the remainder estimate is \(R_N < 10N^{-0.1}.\)

Set up the Inequality

We want the error to be less than \(10^{-4}\). Hence, we set up the inequality:\[10N^{-0.1} < 10^{-4}.\]

Solve the Inequality

Simplify and solve the inequality:\[N^{-0.1} < 10^{-5},\]which implies \[\frac{1}{N^{0.1}} < 10^{-5},\]squaring both sides converts to\[N^{0.1} > 10^{5}.\]Thus, solve for \(N\) by raising both sides to the power of 10:\[N > 10^{50}.\]

Calculate the Minimum Integer \(N\)

The minimum integer value for \(N\) satisfying \(N > 10^{50}\) is approximately \(N = 10^{50}\) because that expression exceeds \(10^{5}\). However, recheck the applied critical value considering constraints and rounding necessary number due to process of approximation. Applying logs may simplify this depending on calculation resolution.

Key concepts

p-series

A p-series is a specific type of infinite series expressed in the form \[ \sum_{n=1}^{\infty} \frac{1}{n^p} \]where \(p\) is a positive constant.When the exponent \(p\) is greater than 1, the series converges, meaning it approaches a finite value.However, if \(p\) is less than or equal to 1, the series diverges, implying it grows without bounds or oscillates.For example, the series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \) with \(p = 1.1\) is a convergent p-series.This is because it satisfies the requirement that \(p > 1\). Understanding p-series helps in determining whether a series has a defined sum or not.
In practical terms, knowing a series is convergent can assist in estimating the sum to a desired accuracy, a common task in calculus problems.

remainder estimate

The remainder estimate in series is a tool used to measure the error in approximating an infinite series with its partial sum.The Integral Test is often employed to provide bounds for the remainder, \(R_N\), after summing the first \(N\) terms. For a function \(a_n = f(n)\) that decreases and resembles \(\frac{1}{n^p}\), we use:

• \(\int_{N+1}^{\infty} f(x) \, dx\) as the lower bound, and
• \(\int_{N}^{\infty} f(x) \, dx\) as the upper bound of \(R_N\).

The goal is to ensure the actual sum is within a specific error margin. For example, in estimating \( \sum_{n=1}^{\infty} a_n \) with partial sum \( \sum_{n=1}^{N} a_n \), if the error threshold is \(10^{-4}\), remainder estimates help pinpoint the smallest \(N\) for accurate estimation without excessive calculations.
This ensures efficiency and precision in complex mathematical scenarios.

improper integral

Improper integrals are used when evaluating an integral over an infinite interval or at points of discontinuity (limits or undefined points within). For instance, the integral \(\int_{N}^{\infty} \frac{1}{x^{1.1}} \, dx \) is considered improper due to the infinite upper limit. Evaluating improper integrals involves taking limits to calculate how the integral behaves as the bounds approach infinity.So, for the given example, we write it as\[ \lim_{b \to \infty} \int_{N}^{b} \frac{1}{x^{1.1}} \, dx \].As shown in the solution, the integration process yields:\[ -10N^{-0.1} \].These integrals are crucial for analyzing convergent series, where the integral helps verify the series sum converges and helps gauge the remainder, crucial in obtaining bounds in applied problems.

inequality solving

Inequality solving involves finding the values of variables that satisfy a given inequality condition. In the context of estimating series sums, we often arrive at an inequality dictating the remainder's size must be less than a desired error margin.For example, starting with:\[ 10N^{-0.1} < 10^{-4} \], we aim to determine the smallest \(N\) satisfying this inequality.The steps involve:

• Converting to \(N^{-0.1} < 10^{-5}\), and
• Further manipulation to \(N^{0.1} > 10^{5}\).

Finally, solving for \(N\) approaches finding the smallest integer greater than \(10^{50}\). This process underscores the importance of algebraic manipulation in interpreting analytic outcomes, especially where precision and error control in mathematical estimation are essential.