/*! 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 6 Find the third Taylor polynomial... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the third Taylor polynomial at \(x=0\) of each function. $$ f(x)=2-x+3 x^{2}-x^{3} $$

Short Answer

Expert verified
The third Taylor polynomial is: \( 2 - x + 3x^2 - x^3 \).

Step by step solution

01

Determine the Function and Its Derivatives

We begin by identifying the function given: \( f(x) = 2 - x + 3x^2 - x^3 \). In order to find the third Taylor polynomial centered at \( x = 0 \), we need the derivatives of the function up to the third order. Calculate:- 0th derivative: \( f(x) = 2 - x + 3x^2 - x^3 \)- 1st derivative: \( f'(x) = -1 + 6x - 3x^2 \)- 2nd derivative: \( f''(x) = 6 - 6x \)- 3rd derivative: \( f'''(x) = -6 \)
02

Evaluate the Derivatives at x = 0

Next, evaluate each derivative at \( x = 0 \).- \( f(0) = 2 \)- \( f'(0) = -1 \)- \( f''(0) = 6 \)- \( f'''(0) = -6 \)
03

Construct the Taylor Polynomial

The third Taylor polynomial centered at \( x = 0 \) is given by:\[ P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \]Plugging in the values we found:\[ P_3(x) = 2 - x + \frac{6}{2}x^2 + \frac{-6}{6}x^3 \]Simplify the terms to get:\[ P_3(x) = 2 - x + 3x^2 - x^3 \]
04

Verify the Polynomial

Compare the Taylor polynomial \( P_3(x) = 2 - x + 3x^2 - x^3 \) with the original function. Notice that the function \( f(x) = 2 - x + 3x^2 - x^3 \) is already a polynomial of degree 3, making the third Taylor polynomial exactly the same as the function itself.

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Key Concepts

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

Calculus
Calculus is a branch of mathematics focused on the study of change and motion. It is fundamentally divided into two main areas: differential calculus and integral calculus. Differential calculus deals with the concept of a derivative, which measures how a function changes as its input changes.
Integral calculus, on the other hand, deals with the accumulation of quantities and the areas under and between curves. Calculus is essential in various fields such as physics, engineering, and economics because it provides tools for analyzing very dynamic systems. Understanding these concepts can help in identifying patterns and predicting future outcomes in real-world problems.
Polynomial Functions
Polynomial functions are mathematical expressions consisting of variables, coefficients, and non-negative integer exponents. A general form of a polynomial function is expressed as:
\[ f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \] where \( a_n, a_{n-1}, ..., a_1, a_0 \) are coefficients and \( n \) represents the highest power of the variable \( x \).
Polynomials are fundamental in algebra and calculus due to their simplicity and ease of manipulation. They can model a variety of real-world phenomena, such as trends in data and physical processes. In calculus, polynomial functions are often used to approximate more complex functions due to their predictability and ease of differentiation and integration.
Derivatives
Derivatives are a core concept in calculus that describes the rate at which a function changes at any given point. The derivative of a function \( f(x) \) at a point \( x \) is given by the limit:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] - Derivatives provide important information such as speed, acceleration, and can reveal the maximum and minimum values of a function.- In our example, we calculated derivatives up to the third order for the polynomial \( f(x) = 2 - x + 3x^2 - x^3 \).
These derivatives: \( f'(x) \), \( f''(x) \), and \( f'''(x) \), respectively, help in constructing the Taylor polynomial.
Taylor Series
The Taylor series is a way to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For a function \( f(x) \) centered at \( x = a \), the Taylor series is:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + ... \] This series helps in approximating functions that are otherwise difficult to calculate.
In the given exercise, we found the third Taylor polynomial by considering derivatives up to the third order at \( x = 0 \). Interestingly, the polynomial \( f(x) = 2 - x + 3x^2 - x^3 \) is already a degree 3 polynomial, so its third Taylor polynomial is the same as the function itself. This illustrates the usefulness of Taylor polynomials in approximating functions and simplifying complex expressions.

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Most popular questions from this chapter

True or False: The \(n\) th Taylor approximation of a function agrees with the function for at least one value of \(x\).

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Let \(a\) be a given number and suppose that the function \(f\) is \(n+1\) times differentiable between \(a\) and \(x\). Since \(a\) and \(x\) are now fixed, we will use \(t\) for a variable taking values between them. Show that if \(\left|f^{(n+1)}(t)\right| \leq M\) for all \(t\) between \(a\) and \(x\), then $$ \left|R_{n}(x)\right| \leq \frac{M}{(n+1) !}|x-a|^{n+1} $$ This establishes the error estimate on page 664

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For each function: a. Find the fourth Taylor polynomial at \(x=0\). b. Graph the original function and the Taylor polynomial together on the indicated window. $$ \begin{aligned} &f(x)=\cos x\\\ &\text { (for (b), use the window }[-\pi, \pi] \text { by }[-2,2]) \end{aligned} $$

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