Domanda del libro di testo
17. Even-odd decompositions
b. If f(x) = f_E(x) + f_O(x) is the sum of an even function f_E(x) and an odd function f_O(x), then show that
f_E(x) = (f(x)+f(-x))/2 and f_O(x) = (f(x)-f(-x))/2
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Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 7, Problema 7.AAE.5
Find the limits in Exercises 1–6.
5. lim(n→∞) (1/(n+1) + 1/(n+2) + ... + 1/(2n))
Guida verificata passo dopo passo1
Recognize that the expression is a sum of terms of the form \(\frac{1}{k}\) where \(k\) runs from \(n+1\) to \$2n$. This can be written as \(\sum_{k=n+1}^{2n} \frac{1}{k}\).
Recall that the harmonic series \(H_m = \sum_{k=1}^m \frac{1}{k}\) and use it to rewrite the sum as \(H_{2n} - H_n\).
Use the approximation for large \(n\): \(H_n \approx \ln(n) + \gamma\), where \(\gamma\) is the Euler-Mascheroni constant, to express \(H_{2n} - H_n\) as \(\ln(2n) + \gamma - (\ln(n) + \gamma)\).
Simplify the expression by canceling out \(\gamma\) and combining logarithms: \(\ln(2n) - \ln(n) = \ln\left(\frac{2n}{n}\right) = \ln(2)\).
Conclude that the limit as \(n \to \infty\) of the sum is \(\ln(2)\), since the approximation becomes exact in the limit.
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Limit of a Sequence
The limit of a sequence describes the value that the terms of the sequence approach as the index goes to infinity. Understanding how to evaluate limits helps determine the behavior of sequences and series for very large indices.
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8:22
Introduction to Sequences
Harmonic Series and Partial Sums
The harmonic series is the sum of reciprocals of natural numbers. Partial sums of the harmonic series, such as sums from 1 to n, grow without bound but slowly. Recognizing partial sums helps analyze expressions involving sums of terms like 1/k.
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06:45Intro to Series: Partial Sums
Integral Test and Approximation of Sums by Integrals
The integral test relates sums of sequences to definite integrals, allowing approximation of sums by integrals for large n. This technique is useful to estimate sums like 1/(n+1) + ... + 1/(2n) by comparing them to integrals of 1/x.
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07:25Integral Test
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