/*! 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 15 Find the first four terms of the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the first four terms of the Taylor series for the function about the point \(a\). $$1 / x, \quad a=-1$$

Short Answer

Expert verified
The first four terms are: \(-1 - (x+1) - (x+1)^2 - (x+1)^3\).

Step by step solution

01

Understand the Taylor Series formula

The Taylor series of a function \(f(x)\) about a point \(a\) is given by: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots\]This formula will help us find the first four terms of the series for the function \(f(x) = \frac{1}{x}\) around \(a = -1\).
02

Calculate the function value at \(a\)

Calculate \(f(a) = f(-1)\). Since \(f(x) = \frac{1}{x}\), \[f(-1) = \frac{1}{-1} = -1 \]
03

Find the first derivative and evaluate at \(a\)

The first derivative of \(f(x) = \frac{1}{x}\) is \(f'(x) = -\frac{1}{x^2}\). Evaluate it at \(a=-1\):\[f'(-1) = -\frac{1}{(-1)^2} = -1\]
04

Find the second derivative and evaluate at \(a\)

The second derivative of \(f(x)\) is \(f''(x) = \frac{2}{x^3}\). Evaluate it at \(a = -1\):\[f''(-1) = \frac{2}{(-1)^3} = -2\]
05

Find the third derivative and evaluate at \(a\)

The third derivative is \(f'''(x) = -\frac{6}{x^4}\). Evaluate it at \(a = -1\):\[f'''(-1) = -\frac{6}{(-1)^4} = -6\]
06

Construct the first four terms of the Taylor series

Plug the derivatives into the Taylor series formula:\[f(x) = -1 + (-1)(x+1) + \frac{-2}{2!}(x+1)^2 + \frac{-6}{3!}(x+1)^3\]Simplifying each term:1. Constant term: \(-1\)2. First term: \(-1(x+1) = -x-1\)3. Second term: \(-1(x+1)^2\)4. Third term: \(-1(x+1)^3\)Combine these to get:\[-1 - (x+1) - \frac{(x+1)^2}{1} - \frac{(x+1)^3}{1}\]Thus, the first four terms are: \(-1 - (x+1) - (x+1)^2 - (x+1)^3\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Calculus
Calculus is a significant part of modern mathematics, used to explore changes and compute values like areas or volumes. It involves two main components: differentiation and integration.
Differentiation helps us understand how things change by figuring out the slope or rate of change of a curve at a specific point. This leads to finding derivatives, which are essential in many scientific fields.
On the other hand, integration is about adding up parts to find the whole, often associated with areas under curves or three-dimensional volumes.
Calculus allows us to develop mathematical models and solve problems across various sciences, engineering, and economics. It's an invaluable tool that helps transition from a theoretical framework to practical, real-world applications.
Series Expansion
Series expansion refers to expressing a function as a sum of terms derived from its values and derivatives at a single point. This is an extremely useful concept in mathematics, allowing us to approximate complex functions.
One common type of series expansion is the Taylor series, where a function is expanded around a point, usually denoted as 'a'. This series provides a polynomial approximation of the function, which can be as accurate as desired by including more terms.
Series expansions simplify calculations and enable easier manipulation, analysis, and understanding of the behavior of functions near a particular point. They play a crucial role in numerical computations and simulations, where utilizing a finite number of terms can yield highly accurate approximations.
Derivatives
Derivatives are central to calculus and measure how a function changes as its input changes. The derivative of a function at a point reflects the slope of the tangent line to the function's graph at that point.
For the function given in the exercise, \(f(x) = \frac{1}{x}\), the derivative is computed using calculus rules, specifically the power rule and quotient rule. The first derivative, \(f'(x) = -\frac{1}{x^2}\), represents the rate at which \(f(x)\) changes with respect to \(x\).
Higher-order derivatives, like the second (\(f''(x) = \frac{2}{x^3}\)) and third derivatives (\(f'''(x) = -\frac{6}{x^4}\)), provide insights into the curvature and concavity of the function's graph, which are essential in constructing Taylor series.
Taylor Polynomials
Taylor polynomials are finite approximations of Taylor series and are used to represent functions using a limited number of terms, making complex computations more manageable.
In our example, the Taylor series for \(f(x) = \frac{1}{x}\) about \(-1\) involves finding successive derivatives and evaluating them at that point. Each derivative contributes to a term in the polynomial, giving us structured approximations of the original function.
The polynomial \{-1 - (x+1) - (x+1)^2 - (x+1)^3\}, derived in the solution, exemplifies how Taylor polynomials provide simple, yet powerful, approaches to approximate functions around a point of interest. Independently these approximation techniques are widely applied in science and engineering to solve differential equations, optimize systems, and predict behavior using computational models.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Find the Taylor polynomial approximation of degree 4 about \(x=0\) for the function \(f(x)=e^{x^{2}}\) (b) Compare this result to the Taylor polynomial approximation of degree 2 for the function \(f(x)=e^{x}\) about \(x=0 .\) What do you notice? (c) Use your observation in part (b) to write out the Taylor polynomial approximation of degree 20 for the function in part (a). (d) What is the Taylor polynomial approximation of degree 5 for the function \(f(x)=e^{-2 x} ?\)

Decide if the statements are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. If \(f^{(n)}(0) \geq n !\) for all \(n,\) then the Taylor series for \(f\) near \(x=0\) diverges at \(x=0\).

Given that the Taylor series for \(\tan x=x+x^{3} / 3+\) \(21 x^{5} / 120+\cdots,\) then that of \(3 \tan (x / 3)\) is (a) \(3 x+x^{3}+21 x^{5} / 120+\cdots\) (b) \(3 x+x^{3}+21 x^{5} / 40+\cdots\) (c) \(x+x^{3} / 27+7 x^{5} / 3240+\cdots\) (d) \(x+x^{3} / 3+21 x^{5} / 120+\cdots\)

Decide if the statements in Problems are true or false. Give an explanation for your answer. The quadratic approximation to \(f(x)\) for \(x\) near 0 is better than the linear approximation for all values of \(x\)

Given that the radius of convergence of the Taylor series for \(\ln (1-x)\) about \(x=0\) is \(1,\) what is the radius of convergence of the Taylor series about \(x=0\) for the following functions \(?\) (a) \(\ln (4-x)\) (b) \(\ln (4+x)\) (c) \(\ln \left(1+4 x^{2}\right)\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.