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Evaluate the following expressions at the given point. Use your calculator and your computer (such as Maple). Then use series expansions to find an approximation to the value of the expression to as many places as you trust. a. \(\frac{1}{\sqrt{1+x^{3}}}-\cos x^{2}\) at \(x=0.015\). b. \(\ln \sqrt{\frac{1+x}{1-x}}-\tan x\) at \(x=0.0015\). c. \(f(x)=\frac{1}{\sqrt{1+2 x^{2}}}-1+x^{2}\) at \(x=5.00 \times 10^{-3}\). d. \(f(R, h)=R-\sqrt{R^{2}+h^{2}}\) for \(R=1.374 \times 10^{3} \mathrm{~km}\) and \(h=1.00 \mathrm{~m}\). e. \(f(x)=1-\frac{1}{\sqrt{1-x}}\) for \(x=2.5 \times 10^{-13}\)

Short Answer

Expert verified
The short answer will be a set of two values for each of the five expressions: one obtained from direct calculation with specified x-value and the other obtained from series expansion. The values will vary based on the specific calculator or computer program used for calculation, and the number of terms taken in series expansion. Series expansion will give more accurate result with more terms taken into account.

Step by step solution

01

Direct Calculation

Use a calculator or computer to directly substitute the given x-values into the expressions and perform the calculations.
02

Series Expansion Approximation

Perform a Taylor or Maclaurin series expansion depending on the point of expansion, usually around zero (Maclaurin) or a given specific point (Taylor), for each expression up to a desired number of terms. Then substitute the given x-values into these approximated series expressions and evaluate.
03

Comparison of Results

Compare the results obtained from direct calculation and series expansion. The difference between these two results gives an estimation of the error committed in series approximation. Repeat the process until the desired precision is reached.
04

Application to Each Problem

Apply Steps 1, 2 and 3 to each of the following expressions: a. \(\frac{1}{\sqrt{1+x^{3}}}-\cos x^{2}\) at \(x=0.015\), b. \(\ln \sqrt{\frac{1+x}{1-x}}-\tan x\) at \(x=0.0015\), c. \(f(x)=\frac{1}{\sqrt{1+2 x^{2}}}-1+x^{2}\) at \(x=5.00 \times 10^{-3}\), d. \(f(R, h)=R-\sqrt{R^{2}+h^{2}}\) for \(R=1.374 \times 10^{3} \mathrm{~km}\) and \(h=1.00 \mathrm{~m}\), e. \(f(x)=1-\frac{1}{\sqrt{1-x}}\) for \(x=2.5 \times 10^{-13}\).

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Key Concepts

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

Taylor and Maclaurin Series
The Taylor series is a powerful tool in calculus, used for approximating functions with an infinite sum of terms calculated from the values of its derivatives at a single point. The more terms we use, the better the approximation becomes. A Taylor series is represented as
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + ... + \frac{f^{(n)}(a)}{n!}(x-a)^n + ... \]
where \(n!\) denotes the factorial of \(n\), and \(f^{(n)}(a)\) is the \(n\)th derivative of \(f\) evaluated at the point \(a\).
A special case of the Taylor series is the Maclaurin series, which is simply a Taylor series centered at zero (\(a=0\)). It's particularly useful when working with functions where evaluating derivatives at zero is straightforward.
The Maclaurin series for the cosine function, for example, is given by
\[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... + (-1)^n\frac{x^{2n}}{(2n)!} + ... \]
In practice, we often truncate the series at a certain number of terms to get an approximation of the function, as was done in the solutions to the textbook exercises. Carrying out the series expansion up to the necessary number of terms can provide a very close approximation to the value of a function for small values of \(x\).
Approximation Methods in Calculus
Approximation methods are essential in calculus as they allow us to represent complex functions and calculations in a more manageable form. One of the most common methods is the use of polynomial approximations, such as those provided by Taylor and Maclaurin series.
When dealing with functions that are difficult to compute directly, a series expansion can offer a reasonable estimate. The process involves expanding the function into a series of terms, where each term is a product of a derivative evaluated at a point and a power of \(x\) or \(x-a\), depending on whether Maclaurin or Taylor series is used.

Direct Calculation vs. Series Approximation

Direct calculation can be challenging or computationally expensive, especially for points close to singularities or functions without simple analytical solutions. Series expansions become helpful in these cases. By selecting a series expansion as an approximation method, we can evaluate complex expressions with various degrees of accuracy, depending on the number of terms used from the series.

Finding Error in Approximations

It's also vital to understand the limitations of these approximations. Calculating the difference between a direct calculation and a series approximation gives us the error margin, which is critical in scientific computations where precision matters. This error typically decreases as we include more terms from the series.
Limits and Continuity
The concepts of limits and continuity are closely related and fundamental to the study of calculus. A limit describes the value that a function approaches as the input approaches a particular point. Limits are the basis for defining continuity, derivatives, and integrals.
A function \(f(x)\) is continuous at a point \(x=a\) if three conditions are met: it is defined at \(a\), the limit of \(f(x)\) as \(x\) approaches \(a\) exists, and the value of the limit equals \(f(a)\). In other words, we can say
\[ \lim_{{x \to a}} f(x) = f(a) \]
In the context of the Taylor and Maclaurin series, continuity plays a vital role in ensuring the success of the series expansion. If a function is not continuous or does not have a derivative at a point, then the Taylor series cannot be formed around that point.
The exercises discussed in this article required the use of limits to determine whether the series would provide good approximations at given points. When using polynomial approximations like Taylor or Maclaurin series, knowing where a function is continuous helps determine where these expansions can be applied successfully—for instance, avoiding points where a function has discontinuities, singularities, or other undefined behaviors.

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Most popular questions from this chapter

Test for pointwise and uniform convergence on the given set. [The Weierstraß M-Test might be helpful.] a. \(f(x)=\sum_{n=1}^{\infty} \frac{\ln n x}{n^{2}}, x \in[1,2]\) b. \(f(x)=\sum_{n=1}^{\infty} \frac{1}{3^{n}} \cos \frac{x}{2^{n}}\) on \(R\).

Recall that the alternating harmonic series converges conditionally. a. From the Taylor series expansion for \(f(x)=\ln (1+x)\), inserting \(x=1\) gives the alternating harmonic series. What is the sum of the alternating harmonic series? b Because the alternating harmonic series does not converge absolutely, a rearrangement of the terms in the series will result in series whose sums vary. One such rearrangement in alternating \(p\) positive terms and \(n\) negative terms leads to the following sum \(^{10}\) : $$ \begin{gathered} \frac{1}{2} \ln \frac{4 p}{n}=\underbrace{\left(1+\frac{1}{3}+\cdots+\frac{1}{2 p-1}\right)}_{p \text { terms }}-\underbrace{\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right)}_{n \text { terms }} \\ +\underbrace{\left(\frac{1}{2 p+1}+\cdots+\frac{1}{4 p-1}\right)}_{p \text { terms }}-\underbrace{\left(\frac{1}{2 n+2}+\cdots+\frac{1}{4 n}\right)}_{n \text { terms }}+\cdots . \end{gathered} $$ Find rearrangements of the alternating harmonic series to give the following sums; that is, determine \(p\) and \(n\) for the given expression and write down the above series explicitly; that is, determine \(p\) and \(n\) leading to the following sums. i. \(\frac{5}{2} \ln 2\). ii. \(\ln 8\). iii. \(0 .\) iv. A sum that is close to \(\pi\).

Consider the sum \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}\) a. Use an appropriate convergence test to show that this series converges. b. Verify that $$ \sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right) $$ c. Find the \(n\)th partial sum of the series \(\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right)\) and use it to determine the sum of the resulting telescoping series.

Use deMoivre's Theorem to write \(\sin ^{3} \theta\) in terms of \(\sin \theta\) and \(\sin 3 \theta .\)

For those sequences that converge, find the limit \(\lim _{n \rightarrow \infty} a_{n}\) a. \(a_{n}=\frac{n^{2}+1}{n^{3}+1} .\) b. \(a_{n}=\frac{3 n+1}{n+2}\). c. \(a_{n}=\left(\frac{3}{n}\right)^{1 / n}\). d. \(a_{n}=\frac{2 n^{2}+4 n^{3}}{n^{3}+5 \sqrt{2+n^{6}}} .\) e. \(a_{n}=n \ln \left(1+\frac{1}{n}\right)\). f. \(a_{n}=n \sin \left(\frac{1}{n}\right)\). g. \(a_{n}=\frac{(2 n+3) !}{(n+1) !} .\)

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