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Chapter 11: Problem 1

Find Taylor's formula for the given function \(f\) at \(a=0 .\) Find both the Taylor polynomial \(P_{n}(x)\) of the indicated degree \(n\) and the remainder term \(R_{n}(x)\). \(f(x)=e^{-x}, \quad n=5\)

Short Answer

Expert verified
Taylor polynomial is \(1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120}\) and remainder is \(\frac{e^{-c}}{720}x^6\).

Step by step solution

01

Understand Taylor's Formula

Taylor's formula allows us to express a function as a sum of its derivatives at a specific point. For a function \(f(x)\), the nth degree Taylor polynomial at \(a\) is given by \(P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k\). The remainder term \(R_n(x)\) represents the difference between the function and the polynomial and is given by the remainder theorem.
02

Calculate Derivatives of the Function

We need to compute the derivatives of \(f(x) = e^{-x}\) at \(x = 0\). The derivatives are: - \(f(x) = e^{-x}\) - \(f'(x) = -e^{-x}\) - \(f''(x) = e^{-x}\) - \(f'''(x) = -e^{-x}\) - \(f^{(4)}(x) = e^{-x}\) - \(f^{(5)}(x) = -e^{-x}\).
03

Evaluate Derivatives at the Point a = 0

Substitute \(x = 0\) into the derivatives calculated in Step 2: - \(f(0) = e^{0} = 1\) - \(f'(0) = -1\) - \(f''(0) = 1\) - \(f'''(0) = -1\) - \(f^{(4)}(0) = 1\) - \(f^{(5)}(0) = -1\).
04

Write the Taylor Polynomial of Degree 5

Using the formula, the Taylor polynomial \(P_5(x)\) at \(a = 0\) is expressed as: \[P_5(x) = 1 - \frac{1}{1!}x + \frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \frac{1}{4!}x^4 - \frac{1}{5!}x^5\] Simplifying, we get: \[P_5(x) = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120}\].
05

Determine the Remainder Term R_n(x)

The remainder term for the Taylor polynomial is given by Lagrange's remainder formula: \[R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\] for some \(c\) between \(a\) and \(x\). For \(n = 5\), \(R_5(x) = \frac{f^{(6)}(c)}{6!}x^6\). Since \(f^{(6)}(x) = f''(x)= e^{-x}\), this becomes: \[R_5(x) = \frac{e^{-c}}{720}x^6\].
06

Conclusion

The Taylor polynomial of degree 5 for \(f(x) = e^{-x}\) at \(a = 0\) is \(P_5(x) = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120}\). The remainder term is \(R_5(x) = \frac{e^{-c}}{720}x^6\) where \(c\) is some value between \(0\) and \(x\).

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Key Concepts

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

Derivative
In the process of constructing a Taylor series, understanding derivatives is crucial. A derivative represents the rate of change of a function. Essentially, it's how much the function's value changes when its input changes. When dealing with Taylor series, we often compute several derivatives of the function at a particular point. For this particular exercise, the function given is \(f(x) = e^{-x}\).

Derivatives are structured hierarchically: first derivative \(f'(x)\), second derivative \(f''(x)\), and so on.
  • The first derivative, \(f'(x) = -e^{-x}\), tells us about the slope of the function.
  • The second derivative, \(f''(x) = e^{-x}\), tells us how the slope itself is changing.
  • This alternation continues, with each derivative highlighting a different aspect of the function's behavior.
For Taylor's formula, each of these derivatives is evaluated at a particular point, like \(x = 0\) in this exercise, providing the coefficients for the polynomial.
Remainder Term
The remainder term is a key component of the Taylor series, important for understanding how well the Taylor polynomial approximates the function. When we approximate a function with a Taylor polynomial, we're essentially cutting off the infinite series after a certain number of terms. What's left out is the remainder.

For this exercise, to determine the remainder \(R_n(x)\), we use Lagrange's remainder formula:
  • \(R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\)
  • This indicates the error between our Taylor polynomial and the actual function.
  • Our polynomial is of degree 5, so we analyze \(f^{(6)}(x) = e^{-x}\), leading us to \(R_5(x) = \frac{e^{-c}}{720}x^6\).
Understanding and estimating this remainder term helps gauge the accuracy of our function's approximation.
Taylor Polynomial
A Taylor polynomial is essentially an approximation of a function using a sum of its derivatives evaluated at a specific point. For this exercise, we're tasked with finding the Taylor polynomial of degree 5 for \(f(x) = e^{-x}\) at \(a = 0\). This involves constructing the polynomial using derivatives we calculated earlier.

The Taylor polynomial \(P_5(x)\) is given by:
  • \(P_5(x) = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120}\)
  • Each term in the polynomial is derived from the corresponding derivative: \(1\) from \(f(0)\), \(-x\) from \(f'(0)\), and so on.
  • The degree of the polynomial \(n\) indicates how many terms we include, influencing the accuracy of the approximation.
With this polynomial, we get a practical and manageable way to express the function around \(x = 0\), capturing the essential behavior of \(e^{-x}\) with just a handful of terms.

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