/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Use series to evaluate the limit... [FREE SOLUTION] | 91Ó°ÊÓ

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Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{\theta \rightarrow 0} \frac{\sin \theta-\theta+\left(\theta^{3} / 6\right)}{\theta^{5}} $$

Short Answer

Expert verified
The limit is \( \frac{1}{120} \).

Step by step solution

01

Expand Sine in Taylor Series

The Taylor series expansion of \( \sin \theta \) around \( \theta = 0 \) is:\[ \sin \theta = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \cdots \] We need this expansion to substitute \( \sin \theta \) in the limit expression.
02

Substitute Series into the Limit Expression

Substitute the Taylor series expansion of \( \sin \theta \) from Step 1 into the original limit:\[ \lim _{\theta \rightarrow 0} \frac{(\theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \cdots) - \theta + \frac{\theta^3}{6}}{\theta^5} \] This simplifies to:\[ \lim _{\theta \rightarrow 0} \frac{\frac{\theta^5}{120} - \cdots}{\theta^5} \]
03

Simplify the Expression

The expression from Step 2 is already simplified to:\[ \frac{\theta^5}{120} + \text{higher order terms} \]Terminating at \( \theta^5 \), we ignore higher-order terms as they become negligible for \( \theta \rightarrow 0 \).
04

Evaluate the Limit

Focus on the leading term of the expression:\[ \lim _{\theta \rightarrow 0} \frac{\frac{\theta^5}{120}}{\theta^5} \]This simplifies to:\[ \lim _{\theta \rightarrow 0} \frac{1}{120} = \frac{1}{120} \]
05

Calculate Final Answer

Since all higher-order terms tend to zero, the final value of the limit is simply the coefficient found in Step 4:\[ \boxed{\frac{1}{120}} \]

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Key Concepts

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

Taylor Series
The Taylor series is a mathematical concept used to approximate a function as an infinite sum of terms. It is particularly useful when working with functions that are difficult to evaluate directly. Around a point, typically denoted as \( heta = 0 \), the Taylor series represents a function as a sum of its derivatives at that point, multiplied by powers of the function's variable.

For instance, for the sine function \( ext{sin}(\theta) \), the Taylor series expansion is:
  • \( ext{sin}(\theta) = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \ldots \)
This series is essential in calculus as it provides a framework to handle functions with more ease during integration, differentiation, or evaluation near certain points.
By using Taylor series in calculus, you can approximate complicated functions using polynomials, which are generally simpler to manipulate. This is especially helpful in solving problems that require precision, like calculating limits or deriving solutions in physics and engineering.
Limits
In calculus, the concept of a limit is fundamental. Limits help us understand the behavior of functions as the input approaches a particular value. They are the foundation of defining most calculus operations, such as derivatives and integrals.

The limit evaluates what happens to a function when the input variable gets infinitely close to a certain point, but not necessarily at that point. For example:
  • \( \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \)
This is a classic example where the limit shows that as the angle \( \theta \) becomes extremely small, the ratio of \( \sin \theta \) to \( \theta \) approaches 1.

Understanding limits helps in approximating values, especially when facing indeterminate forms like \(0/0\). Calculus problems often require evaluating such limits to determine the behavior of a function in a given scenario.
Series Expansion
Series expansion is a technique where functions are expressed as the sum of a sequence of simpler functions. This approach is particularly advantageous in calculus for analyzing and approximating functions in a more manageable form.

The series can be finite or infinite, with each term in the sequence bringing the function closer to its desired value. One popular type of series expansion is the Taylor series, often used to approximate functions like sine, cosine, or exponential functions.

For example, the sine function's series expansion is:
  • \( ext{sin}(\theta) = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \cdots \)
In practice, the series allows calculus students to expand a function into a polynomial, making limits easier to tackle. By substituting these expansions into expressions or equations, limits can be evaluated even when direct calculation is challenging.
Sine Function
The sine function, often denoted as sin, is a fundamental component in trigonometry and calculus. When expressed as a function of an angle, it describes the ratio of the length of the opposite side to the hypotenuse in a right triangle.

Mathematically, it oscillates between -1 and 1, making it periodic with a cycle every \(2\pi\). In terms of calculus, the sine function is widely used due to its simple derivative and integral properties.

Sine Function in Taylor Series

The Taylor series expansion of the sine function around \( \theta = 0 \) is immensely useful in calculus, especially for handling small angle approximations. The series is:
  • \( ext{sin}(\theta) = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \frac{\theta^7}{5040} + \ldots \)

Applications in Problem Solving

An accurate approximation like that offered by the Taylor expansion helps in solving limits where evaluating directly seems tough. It provides a method to reduce the problem to polynomial algebra, a simpler calculation method. This example of limit evaluation benefits from the periodic and smooth nature of the sine function and its mathematical characteristics.

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Most popular questions from this chapter

Show that the sum of the first 2\(n\) terms of the series $$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\cdots$$ is the same as the sum of the first \(n\) terms of the series $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+\cdots$$ Do these series converge? What is the sum of the first \(2 n+1\) terms of the first series? If the series converge, what is their sum?

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{y \rightarrow 0} \frac{y-\tan ^{-1} y}{y^{3}} $$

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+y=0, \quad y^{\prime}(0)=0 \text { and } y(0)=1 $$

a. Series for sinh \(^{-1} x\) Find the first four nonzero terms of the Taylor series for $$ \sinh ^{-1} x=\int_{0}^{x} \frac{d t}{\sqrt{1+t^{2}}} $$ b. Use the first three terms of the series in part (a) to estimate \(\sinh ^{-1} 0.25\) . Give an upper bound for the magnitude of the estimation error.

Compound interest, deposits, and withdrawals If you invest an amount of money \(A_{0}\) at a fixed annual interest rate \(r\) compounded \(m\) times per year, and if the constant amount \(b\) is added to the account at the end of each compounding period (or taken from the account if \(b<0 ),\) then the amount you have after \(n+1\) compounding periods is $$ A_{n+1}=\left(1+\frac{r}{m}\right) A_{n}+b $$ a. If \(A_{0}=1000, r=0.02015, m=12,\) and \(b=50\) , calculate and plot the first 100 points \(\left(n, A_{n}\right) .\) How much money is in your account at the end of 5 years? Does \(\left\\{A_{n}\right\\}\) converge? Is \(\left\\{A_{n}\right\\}\) bounded? b. Repeat part (a) with \(A_{0}=5000, r=0.0589, m=12,\) and \(b=-50 .\) c. If you invest 5000 dollars in a certificate of deposit (CD) that pays 4.5\(\%\) annually, compounded quarterly, and you make no further investments in the CD, approximately how many years will it take before you have \(20,000\) dollars? What if the CD earns 6.25\(\% ?\) d. It can be shown that for any \(k \geq 0\) , the sequence defined recursively by Equation \((1)\) satisfies the relation $$ A_{k}=\left(1+\frac{r}{m}\right)^{k}\left(A_{0}+\frac{m b}{r}\right)-\frac{m b}{r} $$ For the values of the constants \(A_{0}, r, m,\) and \(b\) given in part (a), validate this assertion by comparing the values of the first 50 terms of both sequences. Then show by direct substitution that the terms in Equation \((2)\) satisfy the recursion formula in Equation ( 1\()\) .

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