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Chapter 9: Problem 51

Find the remainder \(R_{n}\) for the nth-order Taylor polynomial centered at a for the given functions. Express the result for a general value of \(n\). $$f(x)=e^{-x}, a=0$$

Short Answer

Expert verified
The remainder of the nth-order Taylor polynomial for the function \(f(x) = e^{-x}\) centered at \(a=0\) is given by: $$R_n(x) = \frac{(-1)^{n+1}}{(n+1)!}x^{n+1}$$

Step by step solution

01

Find the first few derivatives of \(f(x)\)

To find the Taylor series, let's first find the first few derivatives of \(f(x)\): $$f(x) = e^{-x}$$ $$f'(x) = -e^{-x}$$ $$f''(x) = e^{-x}$$ $$f'''(x) = -e^{-x}$$ As we can observe, the derivatives of \(f(x)\) alternate between positive and negative. This pattern will continue for higher-order derivatives.
02

Evaluate the derivatives at \(a=0\)

Now let's evaluate the derivatives we found at \(a=0\): $$f(0) = e^{-0} = 1$$ $$f'(0) = -e^{-0} = -1$$ $$f''(0) = e^{-0} = 1$$ $$f'''(0) = -e^{-0} = -1$$
03

Write the general term for the Taylor series

Now that we have the values of the derivatives at \(a=0\), we can write the general term for the Taylor series: $$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$ The general term of this series is given by: $$T_n(x) = \frac{(-1)^n}{n!}x^n$$
04

Write the general term for the remainder

Now we can identify the general term for the remainder, \(R_n\), which is: $$R_n(x) = \frac{f^{(n+1)}(0)}{(n+1)!}x^{n+1} = \frac{(-1)^{n+1}}{(n+1)!}x^{n+1}$$ The remainder \(R_n(x)\) for the nth-order Taylor polynomial centered at \(a=0\) for the function \(f(x)=e^{-x}\) is: $$R_n(x) = \frac{(-1)^{n+1}}{(n+1)!}x^{n+1}$$

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