Problem 16 - Calculus Early Transcendentals

Question

a. Find the nth-order Taylor polynomials of the given function centered at \(0,\) for \(n=0,1,\) and 2 b. Graph the Taylor polynomials and the function. $$f(x)=e^{-x}$$

Step-by-Step Solution

Answer

#Answer# For the function \(f(x) = e^{-x}\) centered at \(0\), the nth-order Taylor polynomials are: - \(P_0(x) = 1\) - \(P_1(x) = 1 - x\) - \(P_2(x) = 1 - x + \frac{x^2}{2}\)

Step 1: Find the nth derivative of the function

We need to find the derivatives of the function \(e^{-x}\) with respect to x, up to the second order: - \(f(x) = e^{-x}\) - \(f'(x) = -e^{-x}\) - \(f''(x) = e^{-x}\)

Step 2: Write down the Taylor polynomial formula

The general formula for the nth-order Taylor polynomial centered at \(a\) is given by: $$P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$$ In our case, the center is at \(a=0\). So, the Taylor polynomial will be: $$P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}x^k$$

Step 3: Calculate the Taylor polynomial coefficients

Now, let's find the values of the derivatives at \(x = 0\): - \(f(0) = 1\) - \(f'(0) = -1\) - \(f''(0) = 1\) For \(n=0, 1, 2\), the Taylor polynomials are: - \(P_0(x) = 1\) - \(P_1(x) = 1 - x\) - \(P_2(x) = 1 - x + \frac{x^2}{2}\)

Step 4: Graph the Taylor polynomials and the original function

Using graphing software, such as Desmos or any other online tool, graph the original function \(f(x) = e^{-x}\) and the Taylor polynomials \(P_0(x) = 1\), \(P_1(x) = 1-x\), and \(P_2(x) = 1-x+\frac{x^2}{2}\). You will observe that as the degree of the Taylor polynomial increases, it provides a better approximation to the original function within a range around the center value.