Use the estimate Rn s f(1) dt to find a bound for the remainder Ry = E an - E an , where a, = f (n). n=1 00 170. E
Use the estimate Rn s f(1) dt to find a bound for the remainder Ry = E an - E an , where a, = f (n). n=1 00 170. E
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can this problem get more broken down from what bartleby already did? I dont under stand how it cam out to 5x10^-7.
![**Chapter 5.3, Problem 170E**
**Textbook Problem**
Use the estimate \( R_N \leq \int_{N}^{\infty} f(t) \, dt \) to find a bound for the remainder \( R_N \).
The remainder \( R_N = \sum_{n=1}^{\infty} a_n - \sum_{n=1}^{N} a_n \), where \( a_n = f(n) \).
\[ 170. \quad \sum_{n=1}^{100} \frac{1}{n^3} \]
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**Explanation:**
This problem involves finding a bound for the remainder of a series using an integral estimate. It provides the formula for the remainder of an infinite series after the first \( N \) terms. The goal is to apply the given estimate to derive a bound for this remainder. Specifically, the series in question involves terms \( \frac{1}{n^3} \) summed from \( n = 1 \) to \( n = 100 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F962f5ce6-0254-4add-b8db-fdb68e5fb759%2F057e49b7-c84a-4bf2-8a1c-94c3d283cfb6%2Fdoseunp_processed.png&w=3840&q=75)
Transcribed Image Text:**Chapter 5.3, Problem 170E**
**Textbook Problem**
Use the estimate \( R_N \leq \int_{N}^{\infty} f(t) \, dt \) to find a bound for the remainder \( R_N \).
The remainder \( R_N = \sum_{n=1}^{\infty} a_n - \sum_{n=1}^{N} a_n \), where \( a_n = f(n) \).
\[ 170. \quad \sum_{n=1}^{100} \frac{1}{n^3} \]
---
**Explanation:**
This problem involves finding a bound for the remainder of a series using an integral estimate. It provides the formula for the remainder of an infinite series after the first \( N \) terms. The goal is to apply the given estimate to derive a bound for this remainder. Specifically, the series in question involves terms \( \frac{1}{n^3} \) summed from \( n = 1 \) to \( n = 100 \).
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