/*! 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 25 Find the Taylor series generated... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 25

Find the Taylor series generated by \(f\) at \(x=a\). $$f(x)=x^{4}+x^{2}+1, \quad a=-2$$

Short Answer

Expert verified
The Taylor series is \(T(x) = 21 - 36(x + 2) + 25(x + 2)^2 - 8(x + 2)^3 + (x + 2)^4\).

Step by step solution

01

Identify the Formula for a Taylor Series

The Taylor series of a function \(f(x)\) about a point \(x = a\) is given by \( T(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \). This means we will need the derivatives of \(f(x)\) evaluated at \(x = a\).
02

Calculate the Function Value at \(x = -2\)

Substitute \(x = -2\) into \(f(x) = x^4 + x^2 + 1\) to find \(f(-2)\). Calculate: \((-2)^4 + (-2)^2 + 1 = 16 + 4 + 1 = 21\). So, \(f(-2) = 21\).
03

Find the First Derivative and Evaluate at \(x = -2\)

Differentiate to find \(f'(x) = 4x^3 + 2x\). Substitute \(x = -2\) to find \(f'(-2) = 4(-2)^3 + 2(-2) = -32 - 4 = -36\).
04

Find the Second Derivative and Evaluate at \(x = -2\)

Differentiate again to find \(f''(x) = 12x^2 + 2\). Substitute \(x = -2\) to get \(f''(-2) = 12(-2)^2 + 2 = 48 + 2 = 50\).
05

Find the Third Derivative and Evaluate at \(x = -2\)

Differentiate once more to obtain \(f'''(x) = 24x\). Calculate \(f'''(-2) = 24(-2) = -48\).
06

Find the Fourth Derivative and Evaluate at \(x = -2\)

Differentiating once again results in \(f^{(4)}(x) = 24\). Therefore, \(f^{(4)}(-2) = 24\).
07

Write the Taylor Series Expansion

Using the results from the previous steps, substitute into the Taylor series formula: \[T(x) = 21 - 36(x + 2) + \frac{50}{2!}(x + 2)^2 - \frac{48}{3!}(x + 2)^3 + \frac{24}{4!}(x + 2)^4.\] Simplifying gives: \[T(x) = 21 - 36(x + 2) + 25(x + 2)^2 - 8(x + 2)^3 + 1(x + 2)^4.\]

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 91影视!

Key Concepts

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

Polynomial Functions
Polynomial functions form a crucial part of mathematics, especially in calculus and algebra. A polynomial function is an expression made up of variables, coefficients, and the operations of addition, subtraction, multiplication, and non-negative integer exponents.
  • They are expressed in the form: \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \), where each \( a \) is a coefficient and \( n \) is the degree of the polynomial.

  • The degree of the polynomial is determined by the highest exponent present, which influences the curve's shape and end behavior.

  • For example, the function \( f(x) = x^4 + x^2 + 1 \) is a polynomial of degree 4, known as a quartic polynomial.
Polynomial functions are essential for modeling real-world situations because they can approximate smooth curves, making them useful in several fields such as physics, engineering, and economics. Their versatility and simplicity make them valuable tools in problem-solving and analysis.
Derivatives
Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. If you're familiar with slopes of lines, derivatives take this idea further by dealing with curves.
  • They provide the rate of change or the slope of the tangent line to the curve at a specific point.

  • To find derivatives, you apply the differentiation rules, like the power rule, product rule, or chain rule, depending on the function's complexity.

  • For a polynomial like \( f(x) = x^4 + x^2 + 1 \), differentiating would give \( f'(x) = 4x^3 + 2x \).
Derivatives are crucial in finding the Taylor series of a function. By evaluating derivatives at a particular point \( a \), you can construct the series that provides a polynomial approximation of the function near \( a \). Understanding derivatives also helps in analyzing critical points, determining maxima and minima, and solving optimization problems.
Mathematical Modeling
Mathematical modeling involves simplifying and representing real-world scenarios through mathematical expressions and formulas. It's about crafting a mathematical representation that can describe and predict outcomes in real-life situations.
  • Taylor series act as an effective mathematical model because they approximate complex functions using polynomials, which are much simpler to work with.

  • They can model natural phenomena where functions are too complicated to handle in their original forms.

  • By expanding a function into its Taylor series around a point \( a \), the series allows the function to be expressed as an infinite sum of terms.
In practical applications, the Taylor series can provide highly accurate estimates by considering just a few terms and are used in fields like physics, engineering, and economics to solve differential equations and optimize processes. Understanding how to use Taylor series in mathematical modeling equips you with the tools to tackle complex problems using simplified 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

Use the definition of convergence to prove the given limit. \(\lim _{n \rightarrow \infty} \frac{\sin n}{n}=0\)

Use a geometric series to represent each of the given functions as a power series about \(x=0,\) and find their intervals of convergence. a. \(f(x)=\frac{5}{3-x}\) b. \(\quad g(x)=\frac{3}{x-2}\)

Show that if \(M_{1}\) and \(M_{2}\) are least upper bounds for the sequence \(\left\\{a_{n}\right\\},\) then \(M_{1}=M_{2}\) That is, a sequence cannot have two different least upper bounds.

Linearizations at inflection points Show that if the graph of a twice- differentiable function \(f(x)\) has an inflection point at \(x=a,\) then the linearization of \(f\) at \(x=a\) is also the quadratic approximation of \(f\) at \(x=a\). This explains why tangent lines fit so well at inflection points.

Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step 1: Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(\bar{x}=0\) Step 3: Calculate the ( \(n+1\) )st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=(\cos x)(\sin 2 x), \quad|x| \leq 2$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.