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Chapter 10: Problem 42

The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a .\) In Exercises \(41-46,\) find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0\) $$f(x)=e^{\sin x}$$

Short Answer

Expert verified
Linearization is \(1 + x\) and quadratic approximation is \(1 + x + \frac{x^2}{2}\).

Step by step solution

01

Understand and Extract Information

We need to find the Taylor polynomials of order 1 and 2 for the function \( f(x) = e^{\sin x} \) at \( x = 0 \). These are also known as the linearization and the quadratic approximation.
02

Identify Function Values and Derivatives

Calculate \( f(0) \), \( f'(0) \), and \( f''(0) \).- First, \( f(0) = e^{\sin 0} = e^0 = 1 \).- Next, find the first derivative: \( f'(x) = e^{\sin x} \cdot \cos x \). So, \( f'(0) = e^0 \cdot \cos 0 = 1 \cdot 1 = 1 \).- Now calculate the second derivative: \( f''(x) = -e^{\sin x} \cdot \sin x + e^{\sin x} \cdot \cos^2 x \). At \( x = 0 \), \( f''(0) = -e^0 \cdot 0 + e^0 \cdot 1 = 1 \cdot 1 = 1 \).
03

Construct the Linearization (Order 1 Polynomial)

The Taylor polynomial of order 1, or the linearization, is given by \( P_1(x) = f(0) + f'(0)x \) which is from \( f(x) = e^{\sin x} \).- Substituting the values, \( P_1(x) = 1 + 1 \cdot x = 1 + x \).
04

Construct the Quadratic Approximation (Order 2 Polynomial)

The Taylor polynomial of order 2 is \( P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 \).- Using \( f(0) = 1 \), \( f'(0) = 1 \), and \( f''(0) = 1 \):- \( P_2(x) = 1 + 1x + \frac{1}{2}x^2 = 1 + x + \frac{x^2}{2} \).

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Key Concepts

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

Quadratic Approximation
Imagine trying to predict what a function looks like near a specific point. That's where quadratic approximation comes into play. It uses a parabolic approach based on Taylor's theorem to approximate a function near a particular point using a polynomial of degree 2. This approximation is known as a Taylor polynomial of order 2.

For a given function \( f(x) \), the quadratic approximation at \( x = a \) is formulated as:
  • \( P_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 \)
Let's break this down:
  • The term \( f(a) \) gives the function's value at the point \( a \).
  • The term \( f'(a)(x-a) \) provides the linear slope (like a tangent line).
  • Lastly, \( \frac{f''(a)}{2}(x-a)^2 \) accounts for the curvature, making it capture the typical parabolic shape.
Quadratic approximations are handy since they account for the changes in the slope as the graph curves, offering a better fit around the approximation point compared to linear approximations.
Linearization
Linearization is akin to zooming in on a tiny, straight-line section of a function's graph. This technique uses a linear polynomial, essentially a tangent line, to approximate a function near a specific point.

The formula for linearization (or Taylor polynomial of order 1) at \( x = a \) for a function \( f(x) \) is:
  • \( P_1(x) = f(a) + f'(a)(x-a) \)
Here's how it works:
  • \( f(a) \) marks the function's value at the point of interest.
  • \( f'(a)(x-a) \) gives the function's rate of change at \( a \), akin to the slope of a line at that point.
Linearization is beneficial because straight lines are much simpler to work with than curves. It gives a rough, but quick, approximation that's particularly useful for initial estimates or when the point of approximation is near the point of interest.
Derivatives Calculation
Derivatives are the backbone of both linearization and quadratic approximation. They're like the levers that adjust the slope and curvature of your approximate functions.

How do derivatives come into play?
  • The first derivative, \( f'(x) \), indicates the slope or rate of change of the function at any point \( x \).
  • The second derivative, \( f''(x) \), provides insight into how the rate of change itself changes, giving us the curvature information.
In practical terms, to approximate a function \( f(x) \) around \( x = a \) using derivatives, you'd need to:
  • Calculate \( f(a) \), which gives the function's value at the point.
  • Find \( f'(a) \) for the slope of the tangent, informing the linear component.
  • Determine \( f''(a) \), essential for constructing the quadratic approximation as it informs about the curvature.
Each derivative builds upon the previous one, making your approximation more precise as you increase their order.

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Most popular questions from this chapter

Use the identity \(\sin ^{2} x=(1-\cos 2 x) / 2\) to obtain the Maclaurin series for \(\sin ^{2} x\). Then differentiate this series to obtain the Maclaurin series for \(2 \sin x \cos x .\) Check that this is the series for \(\sin 2 x\).

Show that the error \(\left(L-s_{n}\right)\) obtained by replacing a convergent geometric series with one of its partial sums \(s_{n}\) is \(a r^{n} /(1-r)\)

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. \(a_{n}=\int_{1}^{n} \frac{1}{x^{p}} d x, \quad p>1\)

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) \(a_{n}=\frac{\sin n}{n}\)

Derive the series $$ \begin{array}{l} \tan ^{-1} x=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x>1 \\ \tan ^{-1} x=-\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x<-1 \end{array} $$ by integrating the series $$ \frac{1}{1+t^{2}}=\frac{1}{t^{2}} \cdot \frac{1}{1+\left(1 / t^{2}\right)}=\frac{1}{t^{2}}-\frac{1}{t^{4}}+\frac{1}{t^{6}}-\frac{1}{t^{8}}+\cdots $$ in the first case from \(x\) to \(\infty\) and in the second case from \(-\infty\) to \(x\).
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