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Chapter 11: Problem 24

Determine the \(n\) th Taylor polynomial of \(f(x)=1 / x\) at \(x=1\).

Short Answer

Expert verified
The \(n\)th Taylor polynomial of \(f(x) = \frac{1}{x}\) at \(x=1\) is \(P_n(x) = \sum_{k=0}^{n} (-1)^{k} (x-1)^k\).

Step by step solution

01

Recall the Definition of the Taylor Series

The Taylor series of a function centered at a point \(a\) is given by: f(x) = f(x) f(a)+f^{ a)+ \frac{f'(a)}{1!}(x-a)+\frac{f'(a)}{2!} (x-a^2)+\ldots\frac{ f^{n}( F^ =x-)= ! (x-a)^{ f}}.( x^n )
02

Find the Derivatives of \(\displaystyle f(x)=\frac{1}{x}\)

To find the Taylor polynomial, calculate the necessary derivatives of \(f(x) = \frac{1}{x}\).First derivative: \(f'(x) = \frac{-1}{x^2}\)Second derivative: \(f''(x) = \frac{2}{x^3}\)Third derivative: \(f'''(x) = \frac{-6}{x^4}\)Notice the pattern in the derivatives.
03

Evaluate the Derivatives at \(\displaystyle x= 1\)

Evaluate each derivative at \(x = 1\):\(f(1) = 1$$f'(1) = -1$$f''(1) = 2$$f'''(1) = -6\)Use these values in the Taylor series formula.
04

Construct the Taylor Polynomial

Using the results from the previous steps, the general form of the \(n\)th Taylor polynomial at \(x=1\) is:\(P_n(x) = 1 + (-1)(x-1) + \frac{2}{2!}(x-1)^2 + \frac{-6}{3!}(x-1)^3 + \ldots + \frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n\)Simplify each term:
05

Simplify the Polynomial

Simplify the polynomial terms of \(P_n(x)\):\(P_n(x) = 1 - (x-1) + (x-1)^2 - (x-1)^3 + \ldots + (-1)^{n}(x-1)^n\)This gives the explicit form of the Taylor polynomial.

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Key Concepts

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

Taylor series
The Taylor series is a fascinating way to represent a function as an infinite sum of terms. Each term is derived from the function's derivatives at a single point. The Taylor series centered at a point \(a\) for a function \(f(x)\) can be written as:
\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \frac{f^{(n)}(a)}{n!}(x-a)^n + \text{...}\]

This series involves:
  • \(f(a)\) - the function's value at \(a\)
  • \(f'(a)\) - the first derivative value at \(a\)
  • Higher derivatives like \(f''(a)\), \(f'''(a)\), and so forth
By using these values, the Taylor series approximates the function around the point \(a\) more accurately as more terms are added.
derivatives computation
Critical to forming a Taylor polynomial are the derivatives of the function. Derivatives describe how a function changes at different points. For the function \(f(x) = \frac{1}{x}\), we compute the first few derivatives as follows:
  • The first derivative: \(f'(x) = \frac{-1}{x^2}\)
  • The second derivative: \(f''(x) = \frac{2}{x^3}\)
  • The third derivative: \(f'''(x) = \frac{-6}{x^4}\)

Understanding these derivatives helps predict how the function behaves and is crucial for constructing the Taylor series.
polynomial simplification
Once you've computed the derivatives and evaluated them at the desired point, you can insert these values into the Taylor series formula. Let's take our example function at \(x=1\):

The Taylor series for \(f(x) = \frac{1}{x}\) at \(x=1\) becomes:
\[P_n(x) = 1 + (-1)(x-1) + \frac{2}{2!}(x-1)^2 + \frac{-6}{3!}(x-1)^3 + \text{...}\text{up to}\ (x-1)^n\]
Next, simplify each term:
  • \(\frac{2}{2!} = 1\)
  • \(\frac{-6}{3!} = -1\)
This simplification leads to the polynomial:
\[P_n(x) = 1 - (x-1) + (x-1)^2 - (x-1)^3 + \text{...} + (-1)^n(x-1)^n\]
higher-order derivatives
Higher-order derivatives are the repeated derivatives of a function. They shine in Taylor series, capturing intricate details about the function's behavior. For the function \(f(x) = \frac{1}{x}\), the derivatives follow a noticeable pattern:
  • The first derivative: \(f'(x) = \frac{-1}{x^2}\)
  • The second derivative: \(f''(x) = \frac{2}{x^3}\)
  • The third derivative: \(f'''(x) = \frac{-6}{x^4}\)
By exploring these higher-order derivatives, we can predict how the function changes more precisely, leading to a more accurate Taylor polynomial around the point of interest.

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Most popular questions from this chapter

Use the integral test to determine whether the infinite series is convergent or divergent. (You may assume that the hypotheses of the integral test are satisfied.) $$\sum_{k=1}^{\infty} \frac{1}{(3 k)^{2}}$$

Draw the graph of \(f(x)=x^{4}-2 x^{2},[-2,2]\) by \([-2,2] .\) The function has zeros at \(x=-\sqrt{2}, x=0\), and \(x=\sqrt{2} .\) By looking at the graph, guess which zero will be approached when you apply the Newton-Raphson algorithm to each of the following initial approximations: (a) \(x_{0}=1.1\) (b) \(x_{0}=.95\) (c) \(x_{0}=.9\) Then, test your guesses by actually carrying out the computations.

Use the integral test to determine whether the infinite series is convergent or divergent. (You may assume that the hypotheses of the integral test are satisfied.) $$\sum_{k=2}^{\infty} \frac{1}{k \sqrt{\ln k}}$$

Graph the function \(\mathbf{Y}_{1}=e^{x}\) and its fourth Taylor polynomial in the window \([0,3]\) by \([-2,20] .\) Find a number \(b\) such that graphs of the two functions appear identical on the screen for \(x\) between 0 and \(b .\) Calculate the difference between the function and its Taylor polynomial at \(x=b\) and at \(x=3\).

Rely on the fact that $$ \lim _{n \rightarrow \infty} \frac{|x|^{n+1}}{(n+1) !}=0 $$ The proof of this fact is omitted. Let \(R_{n}(x)\) be the \(n\) th remainder of \(f(x)=\cos x\) at \(x=0\). (See Section 11.1.) Show that, for any fixed value of \(x,\left|R_{n}(x)\right| \leq|x|^{n+1} /(n+1) !\), and hence, conclude that \(\left|R_{n}(x)\right| \rightarrow 0\) as \(n \rightarrow \infty .\) This shows that the Taylor series for \(\cos x\) converges to \(\cos x\) for every value of \(x\).
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