/*! 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 51 Use series to evaluate the limit... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The limit is \( \frac{1}{3} \).

Step by step solution

01

Understand the Function Involved

The function being evaluated is \( \frac{y - \tan^{-1}y}{y^3} \). As \( y \to 0 \), direct substitution yields an indeterminate form \( \frac{0 - 0}{0^3} = \frac{0}{0} \). This suggests using a series expansion for a more insightful evaluation.
02

Apply the Series Expansion of \(\tan^{-1}y\)

The Taylor series for \( \tan^{-1} y \) around \( y = 0 \) is \( \tan^{-1} y = y - \frac{y^3}{3} + \frac{y^5}{5} - \cdots \). For this problem, we'll use the first few terms of this series: \( \tan^{-1} y \approx y - \frac{y^3}{3} \).
03

Substitute Series into Function

Substitute the series expansion into the original function: \(\frac{y - \tan^{-1} y}{y^3} = \frac{y - (y - \frac{y^3}{3})}{y^3} = \frac{y - y + \frac{y^3}{3}}{y^3} = \frac{\frac{y^3}{3}}{y^3}. \)
04

Simplify the Expression

Simplify the expression: \(\frac{\frac{y^3}{3}}{y^3} = \frac{1}{3} \). Here, the \( y^3 \) terms cancel out, leaving a constant value, which is independent of \( y \).
05

Evaluate the Limit

As \( y \) approaches 0, the simplified expression \( \frac{1}{3} \) remains constant. Thus, the limit is \( \frac{1}{3} \) as there are no variables left to consider.

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Key Concepts

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

Taylor Series
When solving limits, especially those resulting in indeterminate forms like \( \frac{0}{0} \), the Taylor series becomes extremely helpful. The Taylor series is a way of representing functions as infinite sums of terms calculated from the values of its derivatives at a single point.
For example, the Taylor series expansion for the inverse tangent function \( \tan^{-1}y \) about \( y = 0 \) is:
  • \( \tan^{-1}y = y - \frac{y^3}{3} + \frac{y^5}{5} - \cdots \)
This series gives us a polynomial approximation of the function near the point it was expanded about.
For calculations around \( y = 0 \), getting the first few terms helps in breaking down complex function expressions into simpler polynomial terms, which can be quite useful in evaluating limits.
Indeterminate Forms
In calculus, limits can sometimes become tricky when they take on the form of undefined mathematical expressions known as indeterminate forms.
One common indeterminate form is the \( \frac{0}{0} \) form. This happens when both the numerator and the denominator of a fraction approach zero as a variable approaches a particular point.
Such indeterminate forms require special techniques to evaluate, such as:
  • Applying L'Hôpital's Rule
  • Using Taylor series expansions
  • Simplifying algebraically by factoring or canceling terms
In our problem, replacing \( y - \tan^{-1}y \) with its series expansion allowed us to simplify the expression, moving away from that indeterminate form.
Trigonometric Functions
Trigonometric functions, such as sine, cosine, and tangent, and their inverses are deeply rooted in calculus problems, especially those involving periodic behavior.
In our exercise involving \( \tan^{-1} y \), which is the inverse tangent function, understanding its behavior is crucial for series expansion.
The inverse tangent has a straightforward series representation:
  • \( \tan^{-1}y = y - \frac{y^3}{3} + \frac{y^5}{5} - \cdots \)
Understanding trigonometric functions and their power series can thus simplify many calculations, providing a path to evaluate the limits involving these functions effectively.
Series Expansion
Series expansion is the method of expressing functions as a sum of their polynomial terms. This technique simplifies complex expressions, making it easier to evaluate limits and functions near specific points.
For the function \( \tan^{-1} y \), using a series expansion helped in approximating the function near \( y = 0 \).
The expression simplifies as follows:
  • \( \tan^{-1} y \approx y - \frac{y^3}{3} \)
By substituting this approximation back into the function \( \frac{y - \tan^{-1} y}{y^3} \), it simplifies the process of calculating the limit.
Series expansion provides a useful tool in calculus for tackling complex limits and functions by turning them into simpler polynomials.

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Most popular questions from this chapter

In Exercises \(33-36\) , use series to estimate the integrals' values with an error of magnitude less than \(10^{-3} .\) (The answer section gives the integrals' values rounded to five decimal places.) $$ \int_{0}^{0.2} \sin x^{2} d x $$

Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{1}{n\left(1+\ln ^{2} n\right)} $$

a. Show that $$ \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant }) $$ converges if and only if \(p>1\) b. What implications does the fact in part (a) have for the convergence of the series $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ? $$ Give reasons for your answer.

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime}-2 y=0, \quad y(0)=1 $$

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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