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Chapter 11: Problem 18

Find the Taylor polynomials \(p_{1}, \ldots, p_{5}\) centered at \(a=0\) for \(f(x)=e^{-x}\).

Short Answer

Expert verified
Answer: The Taylor polynomials of degree 1 to 5 for the function \(f(x) = e^{-x}\) centered at \(a = 0\) are: 1. \(p_1(x) = 1 - x\) 2. \(p_2(x) = 1 - x + \frac{1}{2}x^2\) 3. \(p_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\) 4. \(p_4(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{24}x^4\) 5. \(p_5(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{24}x^4 - \frac{1}{120}x^5\)

Step by step solution

01

Find the first five derivatives of f(x) at a=0

We start by finding the first five derivatives of \(f(x) = e^{-x}\) and their values at \(a = 0\). Using the chain rule, we obtain: 1. \(f'(x) = -e^{-x}\) 2. \(f''(x) = e^{-x}\) 3. \(f^{(3)}(x) = -e^{-x}\) 4. \(f^{(4)}(x) = e^{-x}\) 5. \(f^{(5)}(x) = -e^{-x}\) Now, we must evaluate these derivatives at \(a = 0\): 1. \(f'(0) = -e^{-0} = -1\) 2. \(f''(0) = e^{-0} = 1\) 3. \(f^{(3)}(0) = -e^{-0} = -1\) 4. \(f^{(4)}(0) = e^{-0} = 1\) 5. \(f^{(5)}(0) = -e^{-0} = -1\)
02

Construct the Taylor polynomials

Now that we have the derivative values, we can construct the Taylor polynomials of degree 1 to 5 using the general formula for Taylor polynomials centered at \(a = 0\): \(p_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}x^k\) For each degree, we use the corresponding derivative values we found in step 1 as the coefficients for the terms in the polynomial. 1. \(p_1(x) = f(0) + f'(0)x = e^{-0} - 1x = 1 - x\) 2. \(p_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 = 1 - x + \frac{1}{2}x^2\) 3. \(p_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\) 4. \(p_4(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{24}x^4\) 5. \(p_5(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{24}x^4 - \frac{1}{120}x^5\) Now we have found the Taylor polynomials \(p_1\), \(p_2\), \(p_3\), \(p_4\), and \(p_5\) centered at \(a = 0\) for the function \(f(x) = e^{-x}\), which are: 1. \(p_1(x) = 1 - x\) 2. \(p_2(x) = 1 - x + \frac{1}{2}x^2\) 3. \(p_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\) 4. \(p_4(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{24}x^4\) 5. \(p_5(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{24}x^4 - \frac{1}{120}x^5\)

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Key Concepts

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

Chain Rule
The chain rule is an essential tool in calculus for finding the derivative of a composite function. It is used when one function is nested inside another function, and it allows us to differentiate the complex function in a step-by-step fashion. To apply the chain rule, you take the derivative of the outer function and multiply it by the derivative of the inner function.

For example, in our exercise, the function is an exponential function, and we're interested in its derivatives. When finding derivatives of the function \(f(x) = e^{-x}\), we treat \(e^{-x}\) as the outer function and \(-x\) as the inner function. So, the derivative of \(e^{-x}\) with respect to \(x\) would involve the exponential function's derivative multiplied by the derivative of \(-x\), hence \(f'(x) = -e^{-x}\).
Derivatives of Exponential Functions
When dealing with exponential functions, such as \(f(x) = e^x\), their derivatives have a unique property; the derivative of \(e^x\) with respect to \(x\) is itself \(e^x\). This is a critical concept because it simplifies many calculus problems.

In our problem, we deal with \(f(x) = e^{-x}\), a variation of the exponential function. The negative sign in the exponent changes the behavior of the derivatives as we differentiate multiple times. However, due to the nature of exponential functions, the derivative forms a repetitive pattern, alternately positive and negative, as seen in the exercise.
Taylor Polynomial Formula
Taylor polynomials offer a way to approximate more complex functions with polynomials, which are simpler to use and understand. The Taylor polynomial formula for a function \(f\) centered at a point \(a\) is given by:

\[p_n(x) = \biggl\(f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \biggr\)... + \frac{f^{(n)}(a)}{n!}(x - a)^n\]
This formula means that we take the function's value and its derivatives at the point \(a\) and construct a polynomial that matches the function's value and the values of its derivatives up to a certain degree at \(a\). The factorial term in the denominator serves to scale the influence of higher-order derivatives. In practice, the degree of the Taylor polynomial reflects the level of approximation accuracy. A higher degree polynomial will more closely approximate the function around the point \(a\).
Maclaurin Series
A Maclaurin series is a specific case of the Taylor series centered at zero, \(a = 0\). It is used to represent a function with an infinite sum of terms calculated from the derivatives of the function at zero. In essence, the series provides a powerful approach to approximating functions near the origin.

The general form of a Maclaurin series for a function \(f(x)\) is:

\[f(x) = \biggl\(f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \biggr\)...\]
In the exercise given, by finding the value of the function and its first five derivatives at zero, and then applying these values to the Maclaurin series, we construct the polynomials \(p_1(x)\) to \(p_5(x)\). This showcases how the Maclaurin series serves as an indispensable tool for understanding the behavior of functions near the point \(a=0\) and how we can approximate functions using polynomials.

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Most popular questions from this chapter

Find the remainder in the Taylor series centered at the point a for the following functions. Then show that \(\lim R_{n}(x)=0\) for all \(x\) in the interval of convergence. $$f(x)=\cos x, a=\frac{\pi}{2}$$

Bessel functions Bessel functions arise in the study of wave propagation in circular geometries (for example, waves on a circular drum head). They are conveniently defined as power series. One of an infinite family of Bessel functions is $$ J_{0}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2^{2 k}(k !)^{2}} x^{2 k} $$ a. Write out the first four terms of \(J_{0}\) b. Find the radius and interval of convergence of the power series for \(J_{0}\) c. Differentiate \(J_{0}\) twice and show (by keeping terms through \(x^{6}\) ) that \(J_{0}\) satisfies the equation \(x^{2} y^{\prime \prime}(x)+x y^{\prime}(x)+x^{2} y(x)=0\)

Show that the first five nonzero coefficients of the Taylor series (binomial series) for \(f(x)=\sqrt{1+4 x}\) centered at 0 are integers. (In fact, all the coefficients are integers.)

Suppose you want to approximate \(\sqrt{72}\) using four terms of a Taylor series. Compare the accuracy of the approximations obtained using the Taylor series for \(\sqrt{x}\) centered at 64 and \(81 .\)

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. $$f(x)=\frac{1}{\sqrt{x}} \text { with } a=4 ; \text { approximate } \frac{1}{\sqrt{3}}$$.
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