/*! 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 76 Compute the coefficients for the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Compute the coefficients for the Taylor series for the following functions about the given point \(a\), and then use the first four terms of the series to approximate the given number. $$f(x)=\sqrt{x} \text { with } a=36 ; \text { approximate } \sqrt{39}$$.

Short Answer

Expert verified
Answer: The approximation of \(\sqrt{39}\) using the first four terms of the Taylor series is approximately \(6.25087\).

Step by step solution

01

Find the first four derivatives of f(x)

First, we need to find the first four derivatives of the function \(f(x) = \sqrt{x}\). 1st derivative: \(f'(x) = \frac{1}{2\sqrt{x}}\) 2nd derivative: \(f''(x) = -\frac{1}{4x\sqrt{x}}\) 3rd derivative: \(f^{(3)}(x) = \frac{3}{8x^2\sqrt{x}}\) 4th derivative: \(f^{(4)}(x) = -\frac{15}{16x^3\sqrt{x}}\)
02

Evaluate the derivatives at a

Now, we will evaluate the derivatives at \(a = 36\). \(f(36) = \sqrt{36} = 6\) \(f'(36) = \frac{1}{2\sqrt{36}} = \frac{1}{12}\) \(f''(36) = -\frac{1}{4\cdot36\sqrt{36}} = -\frac{1}{288}\) \(f^{(3)}(36) = \frac{3}{8\cdot36^2\sqrt{36}} = \frac{1}{27648}\) \(f^{(4)}(36) = -\frac{15}{16\cdot36^3\sqrt{36}} = -\frac{5}{6,127,872}\)
03

Construct the first four terms of Taylor series

We will now use the Taylor series formula to create the first four terms of the series for the function \(f(x) = \sqrt{x}\) about the point \(a=36\). \(P_3(x) = f(36) + f'(36)(x-36) + \frac{f''(36)}{2!}(x-36)^2 + \frac{f^{(3)}(36)}{3!}(x-36)^3\) \(P_3(x) = 6 + \frac{1}{12}(x-36) - \frac{1}{576}(x-36)^2 + \frac{1}{165,888}(x-36)^3\)
04

Approximate sqrt(39)

Lastly, we will use our Taylor polynomial \(P_3(x)\) to approximate \(\sqrt{39}\) by evaluating the polynomial at \(x=39\). \(P_3(39) = 6 + \frac{1}{12}(39-36) - \frac{1}{576}(39-36)^2 + \frac{1}{165,888}(39-36)^3\) \(P_3(39) = 6 + \frac{1}{4} - \frac{3^2}{576} + \frac{3^3}{165,888}\) \(P_3(39) \approx 6.250868055555556\) Thus, the approximation of \(\sqrt{39}\) using the first four terms of the Taylor series is approximately \(6.25087\).

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.

Derivatives
To understand the process of approximating functions using Taylor series, it's crucial to grasp the concept of derivatives. Derivatives tell us how a function changes as its input changes. They are fundamental tools in calculus used to describe the rate of change or slope of a curve. In the context of the Taylor series, derivatives help us approximate functions by providing coefficients for the polynomial terms.

For the function \( f(x) = \sqrt{x} \), we started by finding the first four derivatives. We calculated the first derivative as \( f'(x) = \frac{1}{2\sqrt{x}} \), the second as \( f''(x) = -\frac{1}{4x\sqrt{x}} \), and so on. Each derivative builds on the previous one, becoming increasingly complex. These derivatives are evaluated at a specific point \( a \), in this case, \( a = 36 \), to facilitate the construction of the Taylor series.

Calculating derivatives is a step-by-step process, and understanding each derivative's role helps simplify using these terms in approximations in real-world scenarios.
Approximation
Approximation involves estimating a complex value with a more straightforward representation for use in calculations or analysis. In mathematics, particularly in calculus, Taylor series are powerful tools for approximation. They allow complex functions to be expressed as a sum of polynomial terms. This simplification is immensely valuable when direct calculation is cumbersome or when using computers to perform tasks quickly.

In our example, the goal was to approximate \( \sqrt{39} \) using the Taylor series centered at \( a = 36 \). By using the derivatives we computed earlier, we formed a Taylor polynomial. The approximation constant and the polynomial's terms (deriving from derivatives) play a vital role, delivering an estimate that becomes more accurate as more terms are included.
Power Series
A power series is a series of the form \( \sum_{n=0}^{\infty} c_n (x - a)^n \), where \( c_n \) represents the coefficients, and \( (x - a) \) indicates the term arrangement based on a center point \( a \). Unlike regular polynomials of fixed degree, power series can be infinite, providing flexibility in expressing functions with great accuracy.

Taylor series are a specific type of power series. They provide an elegant way to expand functions into polynomials by utilizing derivatives as coefficients. For instance, in the Taylor polynomial \( P_3(x) \) for \( \sqrt{x} \) at \( a = 36 \), the constant and linear terms are accompanied by higher powers and their coefficients, forming a composite expression. This characteristic makes power series significantly adaptable for approximating functions around a given point.
Expansion
Expanding a function into its Taylor series means expressing it as a sum of increasing powers of \( (x - a) \). This expansion helps simplify complex functions, making them easier to work with. Each term in the expansion provides more detail about the function's behavior near the center point \( a \).

In building the Taylor series for \( \sqrt{x} \), expansion begins with the zeroth term, \( f(36) = 6 \), then to linear, quadratic, and cubic terms, utilizing derivatives computed earlier. Each successive term refines the function's approximation. The polynomial \( P_3(x) = 6 + \frac{1}{12}(x-36) - \frac{1}{576}(x-36)^2 + \frac{1}{165,888}(x-36)^3 \) includes terms up to the cubic level, showing how these expansions offer a comprehensive approximation.

Expanding functions using Taylor series provides a framework for understanding and estimating behaviors of functions, especially in intermediate math and engineering fields.

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

Probability: sudden-death playoff Teams \(A\) and \(B\) go into sudden-death overtime after playing to a tie. The teams alternate possession of the ball, and the first team to score wins. Assume each team has a \(1 / 6\) chance of scoring when it has the ball, and Team A has the ball first. a. The probability that Team A ultimately wins is \(\sum_{k=0}^{\infty} \frac{1}{6}\left(\frac{5}{6}\right)^{2 k}\) Evaluate this series. b. The expected number of rounds (possessions by either team) required for the overtime to end is \(\frac{1}{6} \sum_{k=1}^{\infty} k\left(\frac{5}{6}\right)^{k-1} .\) Evaluate this series.

Matching functions with polynomials Match functions a-f with Taylor polynomials \(A-F\) (all centered at 0 ). Give reasons for your choices. a. \(\sqrt{1+2 x}\) b. \(\frac{1}{\sqrt{1+2 x}}\) c. \(e^{2 x}\) d. \(\frac{1}{1+2 x}\) e. \(\frac{1}{(1+2 x)^{3}}\) f. \(e^{-2 x}\) A. \(p_{2}(x)=1+2 x+2 x^{2}\) B. \(p_{2}(x)=1-6 x+24 x^{2}\) C. \(p_{2}(x)=1+x-\frac{x^{2}}{2}\) D. \(p_{2}(x)=1-2 x+4 x^{2}\) E. \(p_{2}(x)=1-x+\frac{3}{2} x^{2}\) F. \(p_{2}(x)=1-2 x+2 x^{2}\)

Use the methods of this section to find the power series centered at 0 for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{-3 x}$$

Use the remainder term to find a bound on the error in the following approximations on the given interval. Error bounds are not unique. $$\ln (1+x)=x-\frac{x^{2}}{2} \text {on } [-0.2,0.2]$$

Inverse hyperbolic sine The inverse of hyperbolic sine is defined in several ways; among them are $$ \sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})=\int_{0}^{x} \frac{d t}{\sqrt{1+t^{2}}} $$ Find the first four terms of the Taylor series for \(\sinh ^{-1} x\) using these two definitions (and be sure they agree).

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.