Chapter 9: Problem 21
Find the \(n\) th Taylor polynomial of \(f\) for the given values of \(n\). $$ f(x)=\ln (\cos x) ; n=2 $$
Short Answer
Expert verified
The second Taylor polynomial is \(-\frac{x^2}{2}\).
Step by step solution
01
Understand the Taylor Series Formula
The Taylor series for a function \( f(x) \) about \( a \) is given by:\[T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n\]Here, we are expanding around \( a=0 \), so the formula becomes:\[T_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \ldots + \frac{f^{(n)}(0)}{n!}x^n\]We need to find the first three derivatives of \( f(x) = \ln(\cos x) \).
02
Calculate the First Derivative
Start by finding the first derivative of \( f(x) = \ln(\cos x) \):\[f'(x) = \frac{d}{dx}[\ln(\cos x)] = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x\]Thus, \( f'(0) = \tan(0) = 0 \).
03
Calculate the Second Derivative
Now find the second derivative:\[f''(x) = \frac{d}{dx}[-\tan x] = -\sec^2 x\]Evaluate at \( x = 0 \):\[f''(0) = -\sec^2(0) = -1 \]
04
Formulate the Second Taylor Polynomial
The second Taylor polynomial centered at zero with derivatives computed is:\[T_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2\]With \( f(0) = \ln(1) = 0 \), \( f'(0) = 0 \), and \( f''(0) = -1 \), the polynomial is:\[T_2(x) = 0 + 0 \cdot x + \frac{-1}{2!}x^2 = -\frac{x^2}{2}\]
05
State Final Taylor Polynomial
Thus, the second Taylor polynomial for \( f(x) = \ln(\cos x) \) is:\[T_2(x) = -\frac{x^2}{2}\]
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.
Taylor Series
The Taylor series is a powerful concept in calculus that provides a way to represent functions as infinite sums of terms. These terms are calculated based on the function's derivatives at a single point, typically denoted by \( a \). When the series is truncated to a finite number of terms, it forms a Taylor polynomial, which approximates the function around that point.
To create a Taylor series for a function \( f(x) \), you need to find the function’s value and derivatives at the chosen expansion point. The formula for a Taylor series is:
In many practical scenarios, we expand the function about zero, making \( a = 0 \). This simplifies our calculations significantly and yields a Maclaurin series, which is specifically a Taylor series centered at zero. For example, if we are to expand \( f(x) = \ln(\cos x) \) around zero, our task involves finding up to the nth derivatives at 0 and substituting them into the generalized formula. This yields a polynomial approximation of the actual function.
To create a Taylor series for a function \( f(x) \), you need to find the function’s value and derivatives at the chosen expansion point. The formula for a Taylor series is:
- \( T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n \)
In many practical scenarios, we expand the function about zero, making \( a = 0 \). This simplifies our calculations significantly and yields a Maclaurin series, which is specifically a Taylor series centered at zero. For example, if we are to expand \( f(x) = \ln(\cos x) \) around zero, our task involves finding up to the nth derivatives at 0 and substituting them into the generalized formula. This yields a polynomial approximation of the actual function.
Derivative Calculation
Derivative calculations play a crucial role in constructing a Taylor series. Each derivative provides the coefficients for the corresponding term in the Taylor polynomial. In our example with \( f(x) = \ln(\cos x) \), we start by determining the first few derivatives.
Let's break down the process:
Let's break down the process:
- **First Derivative:** Applying the chain rule provides \( f'(x) = \frac{d}{dx}[\ln(\cos x)] = -\tan x \). Evaluating this at \( x = 0 \) gives \( f'(0) = 0 \).
- **Second Derivative:** Continuing with the derivatives, we calculate \( f''(x) = \frac{d}{dx}[-\tan x] = -\sec^2 x \), and at \( x = 0 \), \( f''(0) = -1 \).
Polynomial Expansion
Polynomial expansion via the Taylor series is a way to approximate complex functions using simpler polynomial expressions. Once we have the necessary derivatives, as shown with our example \( f(x) = \ln(\cos x) \), we can use them to construct the polynomial.
In the case of a Taylor polynomial of degree 2, we use the first two derivatives and the original function value:
This expanded form provides a quadratic approximation of \( \ln(\cos x) \) around \( x = 0 \), offering insight into the function's behavior near zero. Polynomial expansion simplifies calculations by allowing us to use a polynomial function that is much easier to evaluate than the logarithmic cosine function.
In the case of a Taylor polynomial of degree 2, we use the first two derivatives and the original function value:
- The Taylor polynomial formula at \( a = 0 \) is \( T_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 \).
This expanded form provides a quadratic approximation of \( \ln(\cos x) \) around \( x = 0 \), offering insight into the function's behavior near zero. Polynomial expansion simplifies calculations by allowing us to use a polynomial function that is much easier to evaluate than the logarithmic cosine function.