Problem 2 Let \(S=\sum_{1}^{\infty}(-1)^{j... [FREE SOLUTION] (2024)

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Chapter 4: Problem 2

Let $S=\sum_{1}^{\infty}(-1)^{j} a_{j}$ be a convergent series, and assumethat all $a_{j} \geq 0$ and $$ a_{1} \geq a_{2} \geq \cdots \geq a_{n} \geq \cdots $$ Prove that $$ \left|S-\sum_{1}^{n}(-1)^{j} a_{j}\right| \leq a_{n+1} $$

Short Answer

Expert verified

$ \left|S - \sum_{1}^{n}(-1)^{j} a_{j} \right| \leq a_{n+1} $

Step by step solution

Understand the Series

The given series is \[ S = \transformed_sum_{1}^{\infty}(-1)^{j} a_{j} \] where each term $ a_{j} $ is non-negative and the sequence is non-increasing.

Partial Sum Representation

Consider the partial sum of the first $ n $ terms: \[ S_n = \transformed_sum_{1}^{n}(-1)^{j} a_{j} \]

Remainder of the Series

The remainder of the series after the partial sum is: \[ R_n = S - S_n = \sum_{j=n+1}^{\infty}(-1)^{j} a_{j} \]

Alternating Series Remainder Theorem

By the Alternating Series Remainder Theorem, the error (remainder) for a convergent alternating series is less than or equal to the first omitted term: \[ |R_n| \leq a_{n+1} \]

Absolute Value of the Difference

Finally, we link the remainder to the absolute value of the difference: \[ \left|S - \sum_{j=1}^{n}(-1)^{j} a_{j}\right| = |R_n| \leq a_{n+1} \]

Key Concepts

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

Convergence of Series

In mathematics, particularly in series analysis, the term convergence refers to the behavior of a series as the number of terms increases. Simply put, a series is said to converge if its partial sums approach a specific value or limit. For example, consider the series $S = \sum_{j=1}^{\infty} (-1)^j a_j $. This series converges if, as we add more and more terms, the sum gets closer to a certain number. A key factor here is that all terms $a_j $ are non-negative and decrease monotonically. This means they are initially large but get smaller and smaller, ensuring that the series doesn't diverge (i.e., its sum doesn't become infinitely large or jump freely without settling). Understanding convergence is crucial because it affects whether the series has a finite sum and whether subsequent analysis, like error estimates, is valid.

Partial Sum Representation

The partial sum is an important concept when dealing with series. It represents the sum of the first n terms of the series. For the given series, the partial sum $S_n$ of the first n terms is given by $\transformed_sum_{1}^{n}(-1)^{j} a_{j} $. This allows us to approximate the total sum $S$ by adding up a finite number of terms. Think of it as a snapshot of the series up to a certain point. Each partial sum is a step toward fully understanding the behavior of the series as a whole. As you increment n, $S_n$ gets closer to the true value of $S$. The difference between $S_n$ and $S$ helps us understand how well we've approximated the series' total sum.

Remainder Theorem

The remainder theorem, particularly the Alternating Series Remainder Theorem, is a crucial tool in series analysis. It gives us an upper bound for the error made when approximating an infinite series with its partial sum. According to this theorem, for an alternating series that meets certain criteria (where the terms decrease monotonically and approach zero), the absolute value of the remainder is less than or equal to the first omitted term. Specifically, for our series, if we've summed up to the nth term, the remaining error $R_n$, which is the difference between $S$ and $\transformed_sum_{1}^{n}(-1)^{j} a_{j}$, can be bounded by the (n+1)th term. Mathematically, $|R_n| \leq a_{n+1} $. This is tremendously helpful in estimating how close our partial sum is to the actual sum of the series.

Absolute Value

The concept of absolute value plays a vital role in understanding the differences and errors in series. The absolute value of a number is its distance from zero on the number line, ignoring direction. In the context of series, when we talk about the absolute value of the difference between the total series and its partial sum, $|S - S_n|$, we are measuring how far off our partial sum is from the actual sum without considering whether we've overestimated or underestimated. The absolute value ensures that we only consider the magnitude of the error. This makes the estimation process straightforward because we are only focusing on how big the error is, not whether it's positive or negative. For our series, the theorem tells us that this absolute difference $|S - \transformed_sum_{1}^{n}(-1)^{j} a_{j}| $ is less than or equal to $a_{n+1}$, ensuring our approximation is within a known range.

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$Problem 2 Let \(S=\sum_{1}^{\infty}(-1)^{j... [FREE SOLUTION] (3)$

Most popular questions from this chapter

Let $\alpha=\operatorname{Min}\left[\underset{|x| \leq1}{\operatorname{Max}}\left|x^{6}-x^{3}-p_{5}(x)\right|\right]$, where theminimum is taken over all polynomials of degree $\leq 5$. (a) Find $\alpha$. (b) Find the polynomial $p_{5}(x)$ for which the minimum $\alpha$ is attained.(a) Show that the linear minimax approximation to $\sqrt{1+x^{2}}$ on $[0,1]$is $$ q_{1}^{*}(x)=.955+.414 x $$ (b) Using part (a), derive the approximation $$ \sqrt{y^{2}+z^{2}} \doteq .955 z+.414 y \quad 0 \leq y \leq z $$ and determine the error.Let $f(x)=\cos ^{-1}(x)$ for $-1 \leq x \leq 1$ (the principal branch $\left.0\leq f \leq \pi\right)$. Find the polynomial of degree two, $$ p(x)=a_{0}+a_{1} x+a_{2} x^{2} $$ which minimizes $$ \int_{-1}^{1} \frac{[f(x)-p(x)]^{2}}{\sqrt{1-x^{2}}} d x $$Verify that $$ \varphi_{n}(x)=\frac{(-1)^{n}}{n !} e^{x} \frac{d^{n}}{d x^{n}}\left(x^{n}e^{-x}\right) $$ for $n \geq 0$ are orthogonal on the interval $[0, \infty)$ with respect tothe weight function $w(x)=e^{-x} .\left(\right.$ Nòte: $\int_{0}^{\infty}e^{-x} x^{m} d x=m !$ for $\left.m=0,1,2 \ldots\right)$For the Chebyshev expansion (4.7.2), show that if $f(x)$ is even (odd) on$[-1,1]$, then $c_{j}=0$ if $j$ is odd (even).

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$Problem 2 Let \(S=\sum_{1}^{\infty}(-1)^{j... [FREE SOLUTION] (2024)$

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