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Chapter 9: Problem 18

Verify that the series $$ \sum_{k=1}^{\infty} \frac{\cos k x}{k^{2}} $$ converges uniformly on all of \(\mathbb{R}\).

Short Answer

Expert verified
The series converges uniformly on \( \mathbb{R} \) by the Weierstrass M-test.

Step by step solution

01

Understanding Uniform Convergence

A series of functions \( \sum_{k=1}^{\infty} f_k(x) \) converges uniformly on a set \( E \) if for every \( \epsilon > 0 \), there exists a \( N \) such that for all \( n > N \) and for all \( x \in E \), the inequality \( |S_n(x) - S(x)| < \epsilon \) holds, where \( S_n(x) = \sum_{k=1}^{n} f_k(x) \) is the partial sum and \( S(x) = \lim_{n \to \infty} S_n(x) \).
02

Apply Weierstrass M-test

To verify uniform convergence, we apply the Weierstrass M-test, which states that if there is a sequence \( M_k \) such that \( |f_k(x)| \leq M_k \) for all \( x \in E \) and \( \sum_{k=1}^{\infty} M_k \) converges, then \( \sum_{k=1}^{\infty} f_k(x) \) converges uniformly on \( E \).
03

Determine \( M_k \)

In the series \( \sum_{k=1}^{\infty} \frac{\cos k x}{k^2} \), the function \( f_k(x) = \frac{\cos k x}{k^2} \) satisfies \( |\cos k x| \leq 1 \) for all \( x \). Therefore, we can set \( M_k = \frac{1}{k^2} \).
04

Check the convergence of \( \sum_{k=1}^{\infty} \frac{1}{k^2} \)

The series \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) is known to converge as it is a p-series with \( p = 2 > 1 \). Thus, \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) converges.
05

Conclude Uniform Convergence

Since \( |f_k(x)| \leq M_k \) and the series \( \sum_{k=1}^{\infty} M_k \) converges, by the Weierstrass M-test, the series \( \sum_{k=1}^{\infty} \frac{\cos k x}{k^2} \) converges uniformly on \( \mathbb{R} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weierstrass M-test
The Weierstrass M-test is a powerful tool when it comes to determining the uniform convergence of a series of functions. Simply put, it takes away some of the heavy-lifting work in function series. Imagine a series \( \sum_{k=1}^{\infty} f_k(x) \), where each \( f_k(x) \) is a function dependent on \( x \). The M-test says that if we can find a sequence of numbers \( M_k \) such that:
  • \( |f_k(x)| \leq M_k \) for every \( x \) in the set \( E \)
  • \( \sum_{k=1}^{\infty} M_k \) converges
Then, the original series \( \sum_{k=1}^{\infty} f_k(x) \) also converges uniformly on \( E \).
This test is particularly useful because it transforms a possibly tricky function problem into a number series problem which is typically easier to handle.
series convergence
Understanding series convergence is integral in many areas of mathematics. A series \( \sum_{k=1}^{\infty} a_k \) converges if its partial sums approach a finite limit as \( n \to \infty \). In simpler terms, imagine adding up the terms of the series one by one, forever. If you get closer and closer to a particular number, then the series converges. If not, it diverges.
There are different types of convergence:
  • Absolute Convergence: Every series \( \sum_{k=1}^{\infty} a_k \) converges absolutely if the series made by taking the absolute values \( \sum_{k=1}^{\infty} |a_k| \) converges.
  • Conditional Convergence: A series converges conditionally if it converges but does not converge absolutely.
Recognizing and correctly categorizing the type of convergence is essential for correctly solving series-related problems.
p-series
A p-series is one special type of series that has the form \( \sum_{k=1}^{\infty} \frac{1}{k^{p}} \). These series are called p-series because they are defined by the exponent \( p \). The convergence of a p-series depends on the value of \( p \):
  • If \( p > 1 \), the series converges.
  • If \( p \leq 1 \), the series diverges.
This simple rule makes p-series quite popular when testing for convergence in different situations. For instance, the series \( \sum_{k=1}^{\infty} \frac{1}{k^{2}} \) is a p-series with \( p = 2 \) and notably, it converges because \( 2 > 1 \). Understanding and recognizing p-series can help solve many problems about series convergence quickly and efficiently.

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Most popular questions from this chapter

Thus far we know that \(h\) is differentiable, has a bounded derivative, \(h^{\prime}=0\) on a dense set and \(h^{\prime}\) is positive and discontinuous on another dense set. Show that \(h^{\prime}\) is not Riemann integrable.

If \(f_{n} \rightarrow f\) pointwise at every real number, then prove that $$ \\{x: f(x)>\alpha\\}=\bigcup_{m=1}^{\infty} \bigcup_{r=1}^{\infty} \bigcap_{n=r}^{\infty}\left\\{x: f_{n}(x) \geq \alpha+1 / m\right\\} . $$

Show that the logarithm function can be expressed as the pointwise limit of a sequence of "simpler" functions, $$ \ln x=\lim _{n \rightarrow \infty} n(\sqrt[n]{x}-1) $$ for every point in its domain. If the answer to our three questions for this particular limit is affirmative, what can you say about the continuity of the logarithm function? What would be its derivative? What would be \(\int_{1}^{2} \ln x d x ?\)

Give an example of a sequence of continuous functions \(\left\\{f_{n}\right\\}\) on the interval \([0, \infty)\) that is monotonic decreasing and converges pointwise to a continuous function \(f\) on \([0, \infty)\) but for which the convergence is not uniform. Why does this not contradict Theorem \(9.24 ?\)

Give an example of a function \(F\) that is differentiable on \(\mathbb{R}\) such that the sequence $$ f_{n}(x)=n(F(x+1 / n)-F(x)) $$ converges pointwise but not uniformly to \(F^{\prime}\).
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