/*! 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 119 Find the Taylor polynomials of d... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the Taylor polynomials of degree two approximating the given function centered at the given point. $$ f(x)=\sin (2 x) \text { at } a=\frac{\pi}{2} $$

Short Answer

Expert verified
The Taylor polynomial of degree two is \(-2(x-\frac{\pi}{2}) \).

Step by step solution

01

Understand the Taylor Polynomial Formula

The Taylor polynomial of degree two for a function \( f(x) \) centered at \( a \) is given by: \[ P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 \] We need to calculate the values of \( f(a) \), \( f'(x) \), and \( f''(x) \).
02

Evaluate \( f(a) \)

Compute \( f(a) = f\left(\frac{\pi}{2}\right) \). Since \( f(x) = \sin(2x) \), substitute \( \frac{\pi}{2} \) for \( x \): \[ f\left(\frac{\pi}{2}\right) = \sin\left(2 \times \frac{\pi}{2}\right) = \sin(\pi) = 0 \].
03

Compute the First Derivative and Evaluate

Find the derivative \( f'(x) \). Since \( f(x) = \sin(2x) \), the derivative is: \[ f'(x) = 2\cos(2x) \]. Evaluate \( f'(a) = f'\left(\frac{\pi}{2}\right) \): \[ f'\left(\frac{\pi}{2}\right) = 2\cos\left(2 \times \frac{\pi}{2}\right) = 2\cos(\pi) = 2(-1) = -2 \].
04

Compute the Second Derivative and Evaluate

Find the second derivative \( f''(x) \). The derivative of \( f'(x) = 2\cos(2x) \) is: \[ f''(x) = -4\sin(2x) \]. Evaluate \( f''(a) = f''\left(\frac{\pi}{2}\right) \): \[ f''\left(\frac{\pi}{2}\right) = -4\sin\left(2 \times \frac{\pi}{2}\right) = -4\sin(\pi) = 0 \].
05

Construct the Taylor Polynomial of Degree Two

Substitute \( f(a) \), \( f'(a) \), and \( f''(a) \) into the Taylor polynomial formula: \[ P_2(x) = 0 - 2(x-\frac{\pi}{2}) + \frac{0}{2}(x-\frac{\pi}{2})^2 = -2(x-\frac{\pi}{2}) \]. Therefore, the Taylor polynomial of degree two is \(-2(x-\frac{\pi}{2}) \).

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Key Concepts

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

Second Derivative
The second derivative is a crucial concept in calculus, particularly when dealing with Taylor polynomials. It provides information about the curvature of a function and how it changes over its domain. Let's break this down:- The **first derivative** of a function, denoted as \( f'(x) \), gives the rate of change or the slope of the function at any point \( x \).- The **second derivative**, \( f''(x) \), tells us how the rate of change itself is changing. It's the derivative of the first derivative.
For a Taylor polynomial, the second derivative is used to determine the quadratic (or second-degree) term's coefficient. In our exercise, the function is \( f(x) = \sin(2x) \). First, we found:- The first derivative: \( f'(x) = 2\cos(2x) \).- Then, the second derivative: \( f''(x) = -4\sin(2x) \).
Calculating this second derivative helps in shaping the curve of the polynomial approximation, making it fit the original function more closely around the designated point. In this exercise, evaluating \( f'' \) at \( x = \pi/2 \) resulted in zero, showing no quadratic term contributes to the second-degree Taylor polynomial.
Trigonometric Functions
Trigonometric functions like sine and cosine are fundamental in mathematics, dealing with the relationships between the angles and sides of triangles. They are periodic functions, which means they repeat their values in regular intervals.- **Sine function**: Represented as \( \sin(x) \), it oscillates between -1 and 1.- **Cosine function**: Represented as \( \cos(x) \), it also oscillates between -1 and 1, but starts with a value of 1 at \( x = 0 \).
In this exercise, we focused on \( f(x) = \sin(2x) \). Here, the function's frequency is doubled compared to a regular \( \sin(x) \), leading to a faster oscillation. This alteration affects both the function values and the derivatives used in the Taylor polynomial.
When approximating a trigonometric function with a Taylor series, understanding these periodic properties is essential as they influence how accurate or suitable the approximation will be over different intervals.
Function Approximation
Approximating a function means finding a simpler function that closely mimics the behavior of a complex one around a specific point. Taylor polynomials are powerful tools for such an approximation.- **Taylor Polynomial**: A mathematical expression that estimates a function using its derivatives at a single point. The degree indicates how many derivatives are used, with higher degrees providing better approximations.- **Second Degree Taylor Polynomial**: Utilizes the function value, first derivative, and second derivative at a central point.
In our problem, the function \( f(x) = \sin(2x) \) is approximated at \( x = \pi/2 \). The process involves:- Evaluating the function and its derivatives at the center point.- Substituting these into the formula \( P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 \).- The resulting approximation here is \( -2(x-\frac{\pi}{2}) \), which simplifies greatly due to higher derivatives contributing zero.
Function approximation, especially using Taylor polynomials, is key in fields like physics and engineering, allowing us to simplify calculations for functions that are otherwise complex to work with directly.

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Most popular questions from this chapter

Use the expansion \((1+x)^{1 / 3}=1+\frac{1}{3} x-\frac{1}{9} x^{2}+\frac{5}{81} x^{3}-\frac{10}{243} x^{4}+\cdots\) to write the first five terms (not necessarily a quartic polynomial) of each expression. $$ (1+4 x)^{1 / 3} ; a=0 $$

The exercise make use of the functions \(S_{5}(x)=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}\) and \(C_{4}(x)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}\) on \([-\pi, \pi]\). (Taylor approximations and root finding.) Recall that Newton's method \(x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}\) approximates solutions of \(f(x)=0\) near the input \(x_{0}\). a. If \(f\) and \(g\) are inverse functions, explain why a solution of \(g(x)=a\) is the value \(f(a)\) of \(f\). b. Let \(p_{N}(x)\) be the \(N\) th degree Maclaurin polynomial of \(e^{x}\). Use Newton's method to approximate solutions of \(p_{N}(x)-2=0\) for \(N=4,5,6\) c. Explain why the approximate roots of \(p_{N}(x)-2=0\) are approximate values of \(\ln (2)\)

Compute the Taylor series of each function around \(x=1\). $$ f(x)=(x-2)^{2} $$

The exercise make use of the functions \(S_{5}(x)=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}\) and \(C_{4}(x)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}\) on \([-\pi, \pi]\). Plot \(\left|2 S_{5}(x) C_{4}(x)-\sin (2 x)\right|\) on \([-\pi, \pi]\).

Find the Maclaurin series of \(F(x)=\int_{0}^{x} f(t) d t\) by integrating the Maclaurin series of \(f\) term by term. If \(f\) is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero. $$ F(x)=\int_{0}^{x} \frac{\ln (1+t)}{t} d t ; \quad f(t)=\sum_{n=0}^{\infty}(-1)^{n} \frac{t^{n}}{n+1} $$

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