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Chapter 1: Problem 8

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$$

Short Answer

Expert verified
\(\frac{1}{2}\)

Step by step solution

01

- Recall the Maclaurin Series for Cosine

The Maclaurin series for \(\text{cos}(x)\) is \(\text{cos}(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...\).
02

- Use the Approximation for Small x

For small values of \x\, higher-order terms become negligible. So, \(\text{cos}(x) \approx 1 - \frac{x^2}{2}\).
03

- Substitute the Approximation into the Expression

Substitute the approximation into the given limit expression: \(\frac{1 - \text{cos}(x)}{x^2} \approx \frac{1 - \big(1 - \frac{x^2}{2}\big)}{x^2}\).
04

- Simplify the Expression

Simplify the numerator: \(\frac{1 - 1 + \frac{x^2}{2}}{x^2} = \frac{\frac{x^2}{2}}{x^2}\).
05

- Evaluate the Limit

Simplify the fraction and then take the limit as \(x \rightarrow 0 \): \(\frac{\frac{x^2}{2}}{x^2} = \frac{1}{2}\). Thus, \(\text{lim}_{x \rightarrow 0}\frac{1 - \text{cos}(x)}{x^2} = \frac{1}{2}\).

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Key Concepts

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

limit evaluation
The goal of limit evaluation is to find what value a function approaches as the variable within the function approaches a certain point.
In this exercise, we want to determine the limit of \(\text{lim}_{x \rightarrow 0}\frac{1 - \text{cos}(x)}{x^2}\) as \x\ tends to 0.
To evaluate this limit effectively, we can use the Maclaurin series approximation for the cosine function.
This helps simplify the expression, making it easier to handle mathematically.
cosine function
The cosine function, \text{cos}(x)\, is a fundamental trigonometric function with applications in various fields such as physics, engineering, and computer science.
For this exercise, we use the Maclaurin series, which is a special case of the Taylor series.
The Maclaurin series for \text{cos}(x)\ around 0 is given by:
\[ \text{cos}(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \text{...} \]
This series helps us evaluate the cosine function for small values of \x\ more easily.
series approximation
Series approximation involves using a subset of a function's series expansion to estimate the function's value.
For small values of \x\, higher-order terms in the series become negligible.
In our case, the Maclaurin series for the cosine function can be approximated as:
\[ \text{cos}(x) \approx 1 - \frac{x^2}{2} \]
This is because the higher-order terms (\frac{x^4}{4!}, \frac{x^6}{6!}, ...) contribute very little when \x\ is close to 0.
Hence, this approximation makes calculations simpler and more straightforward.
small x approximation
For functions involving small values of \x\ (near zero), we can simplify calculations by ignoring higher-order terms in their series expansions.
For the cosine function, the small \x\ approximation simplifies it to:
\[ \text{cos}(x) \approx 1 - \frac{x^2}{2} \]
By substituting this approximation into our limit expression, we get:
\[ \frac{1 - \text{cos}(x)}{x^2} \approx \frac{1 - (1 - \frac{x^2}{2})}{x^2} \]
This simplifies further to \[ \frac{\frac{x^2}{2}}{x^2} = \frac{1}{2} \]
limit calculation
Finally, we perform the limit calculation based on our simplified expression.
With the small \x\ approximation, the limit expression is now:
\[ \frac{\frac{x^2}{2}}{x^2} = \frac{1}{2} \]
As \x\ approaches 0, the fraction remains constant, so our limit is \frac{1}{2}\.
Hence, we conclude:
\[ \text{lim}_{x \rightarrow 0}\frac{1 - \text{cos}(x)}{x^2} = \frac{1}{2} \]
This result shows how using the Maclaurin series and small \x\ approximations can significantly simplify limit evaluations.

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Most popular questions from this chapter

There are 9 one-digit numbers ( 1 to 9 ), 90 two-digit numbers ( 10 to 99 ). How many three-digit, four-digit, etc., numbers are there? The first 9 terms of the harmonic series \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{9}\) are all greater than \(\frac{1}{10}\); similarly consider the next 90 terms, and so on. Thus prove the divergence of the harmonic series by comparison with the series \(\left[\frac{1}{10}+\frac{1}{10}+\cdots\left(9 \text { terms each }=\frac{1}{10}\right)\right]+\left[90 \text { terms each }=\frac{1}{100}\right]+\cdots\) $$ =\frac{9}{10}+\frac{90}{100}+\cdots=\frac{9}{10}+\frac{9}{10}+\cdots $$ The comparison test is really the basic test from which other tests are derived. It is probably the most useful test of all for the experienced mathematician but it is often hard to think of a satisfactory \(m\) series until you have had a good deal of experience with series. Consequently, you will probably not use it as often as the next three tests.

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\sec x=\frac{1}{\cos x}$$

(a) It is clear that you (or your computer) can't find the sum of an infinite series just by adding up the terms one by one. For example, to get \(\zeta(1.1)=\) \(\sum_{n=1}^{\infty} 1 / n^{1.1}(\text { see Problem } 15.22)\) with error \(<0.005\) takes about \(10^{33}\) terms. To see a simple alternative (for a series of positive decreasing terms) look at Figures 6.1 and 6.2. Show that when you have summed \(N\) terms, the sum \(R_{N}\) of the rest of the series is between \(I_{N}=\int_{N}^{\infty} a_{n} d n\) and \(I_{N+1}=\int_{N+1}^{\infty} a_{n} d n\) (b) Find the integrals in (a) for the \(\zeta(1.1)\) series and verify the claimed number of terms needed for error \(<0.005 .\) Hint: Find \(N\) such that \(I_{N}=0.005 .\) Also find upper and lower bounds for \(\zeta(1.1)\) by computing \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N}^{\infty} n^{-1.1} d n\) and \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N+1}^{\infty} n^{-1.1} d n\) where \(N\) is far less than \(10^{33} .\) Hint: You want the difference between the upper and lower limits to be about \(0.005 ;\) find \(\mathrm{N}\) so that term \(a_{N}=0.005\).

Find the interval of convergence, including end-point tests: $$\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}$$

Test for convergence: $$\sum_{n=2}^{\infty} \frac{2 n^{3}}{n^{4}-2}$$
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