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Find a formula for an arbitrary Taylor polynomial of \(f\). $$ f(x)=\frac{1}{1+2 x} $$

Short Answer

Expert verified
The Taylor polynomial is \( P_n(x) = \sum_{k=0}^{n} (-2x)^k \).

Step by step solution

01

Recognize the Function Form

The function given is \( f(x) = \frac{1}{1+2x} \). This can be recognized as a geometric series of the form \( \sum_{n=0}^{\infty} ar^n \) with \( a = 1 \) and \( r = -2x \).
02

Identify the Pattern of Derivatives

To find the Taylor series expansion, discover the derivatives of \( f(x) \). The first derivative \( f'(x) = \frac{-2}{(1+2x)^2} \). The second derivative \( f''(x) = \frac{8}{(1+2x)^3} \). Continuing this pattern helps determine a general formula for the \( n \)-th derivative.
03

General Formula for the n-th Derivative

Observe the pattern, which suggests that the \( n \)-th derivative is \( f^{(n)}(x) = \frac{(-1)^n n! (2)^n}{(1+2x)^{n+1}} \). This pattern arises since each differentiation introduces a factor of \( -2 \).
04

Construct the Taylor Polynomial

The Taylor polynomial is given by \( P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} (x-a)^k \). For \( a=0 \), \( f^{(k)}(0) = (-1)^k (2)^k k! \), so \( P_n(x) = \sum_{k=0}^{n} (-1)^k 2^k x^k \).
05

Taylor Polynomial for Arbitrary Degree

Thus, the formula for the Taylor polynomial is \( P_n(x) = \sum_{k=0}^{n} (-2x)^k \). This polynomial is constructed by summing the powers of the series expansion, terminating at the chosen degree \( n \).

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Key Concepts

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

Geometric Series
A geometric series is a sum of terms where each term after the first is multiplied by a constant. This constant is called the common ratio, denoted as \( r \). A basic example is the series \( a + ar + ar^2 + ar^3 + \ldots \) which can be written in an expanded form as \( \sum_{n=0}^{\infty} ar^n \).

For the function \( f(x) = \frac{1}{1+2x} \), it can be rearranged to represent a geometric series with \( a = 1 \) and \( r = -2x \).
  • The terms of the series will follow with powers of \( -2x \).
  • This approach allows us to express the function as an infinite series that is easier to analyze and differentiate.
This transformation leverages the formula for the sum of an infinite geometric series when \(|r| < 1\), given by \( \frac{1}{1-r} \).

By recognizing this form, you can efficiently develop related Taylor series expansions.
Derivatives
Derivatives are fundamental in calculus, representing the rate of change of a function. When working with Taylor polynomials, derivatives allow us to approximate functions using polynomials, focusing on the behavior near a specific point.

For our function \( f(x) = \frac{1}{1+2x} \), calculating derivatives involves recognizing a pattern:
  • First derivative: \( f'(x) = \frac{-2}{(1+2x)^2} \)
  • Second derivative: \( f''(x) = \frac{8}{(1+2x)^3} \)
  • General pattern observed: \( f^{(n)}(x) = \frac{(-1)^n n! (2)^n}{(1+2x)^{n+1}} \)
Each derivative introduces a factor of \( -2 \), resulting in the pattern observed across further derivatives.

Recognizing this consistent pattern is crucial for constructing the Taylor polynomial, as it simplifies the computation of higher-order derivatives.
Taylor Series Expansion
The Taylor series expansion is a method for representing a function as an infinite sum of terms calculated from its derivatives at a single point, often around \( x = 0 \).

The general formula for a Taylor series expanded around \( a \) is:
  • \( P(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x-a)^k \)
For our exercise, using \( a = 0 \), we derived the Taylor polynomial:
  • \( P_n(x) = \sum_{k=0}^{n} (-1)^k (2)^k x^k \)
This equation approximates \( f(x) \) by summing derivatives weighted by the factorial term \( \frac{1}{k!} \), ensuring accuracy.

Taylor series are incredibly powerful because they enable approximations of complex functions with polynomials, which are easier to handle computationally and analytically.
Pattern Recognition
Pattern recognition is a key skill in mathematics and calculus, used to identify consistent trends or rules within a sequence of operations or series.

In this exercise, a clear pattern emerges from the derivatives of the function \( f(x) = \frac{1}{1+2x} \). Each derivative introduces factors leading to a sequence:
  • First derivative contributes \( -2 \)
  • Each subsequent derivative follows with \( (-1)^n n! (2)^n \)
Recognizing this allows us to form a general formula for all derivatives.

This ability to spot patterns makes the process of finding Taylor polynomials more efficient, simplifying the calculation of each term of the series. By observing these patterns, one can quickly predict the behavior of the function's derivatives, aiding in constructing accurate series expansions.

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Most popular questions from this chapter

The Sierpinski carpet is constructed as follows: Begin with a square region \(R\) with sides of length \(1 .\) Divide \(R\) into 9 subsquares of equal area, and remove the interior but not the boundary of the middle square (Figure 9.12(a)). For each of the remaining subsquares, perform the same operation, leaving a region consisting of 64 smaller squares (Figure \(9.12(\mathrm{~b}))\). Let \(R_{n}\) be the region that remains after performing the same operation \(n\) times (Figure \(9.12(\mathrm{c}))\). The Sierpinski carpet \(S\) consists of all the points in \(R\) that are not removed by any of the operations; in other words, \(S\) consists of all points that are in \(R_{n}\) for every \(n \geq 1\). a. Find the area \(A_{n}\) of \(R_{n}\) for any given \(n \geq 1\).

Find the Taylor series of the given function about \(a\). Use the series already obtained in the text or in previous exercises. $$ f(x)=x \ln \left(1+x^{2}\right) ; a=0 $$

The Swiss mathematician Leonhard Euler used ideas expressed in the Geometric Series Theorem to deduce that if \(r>0\), then $$ \sum_{n=0}^{\infty}\left(\frac{1}{r}\right)^{n}=\frac{1}{1-1 / r}=\frac{-r}{1-r} \text { and } \sum_{n=1}^{\infty} r^{n}=\frac{r}{1-r} $$ Then he concluded that $$ \begin{gathered} \left(\cdots+\frac{1}{r^{4}}+\frac{1}{r^{3}}+\frac{1}{r^{2}}+\frac{1}{r}+1\right)+\left(r+r^{2}+\cdots\right) \\\ =\sum_{n=0}^{\infty}\left(\frac{1}{r}\right)^{n}+\sum_{n=1}^{\infty} r^{n}=\frac{-r}{1-r}+\frac{r}{1-r}=0 \end{gathered} $$ Since all terms in the series on the left side of the equation are positive, this is absurd. Why is this argument invalid?

Approximate the value of the integral with an error less than the given error, first using the Integration Theorem to express the integral as an infinite series and then approximating the infinite series by an appropriate partial sum. $$ \int_{0}^{1 / 2} \frac{x^{2}}{1+x} d x ; 10^{-3} $$

The probability of getting a 7 or 11 on a roll of two dice is \(\frac{8}{36}=\frac{2}{9}\) and the probability of getting some other sum is \(1-\frac{2}{9}=\frac{7}{9}\). This implies that if the dice are rolled repeatedly, then for any integer \(n \geq 1\), the probability of rolling a 7 or 11 for the first time on the \(n\) th roll is \(\left(\frac{7}{9}\right)^{n-1} \frac{2}{9}\). The expected number of rolls required to roll a 7 or 11 the first time is therefore $$ \sum_{n=1}^{\infty} n\left(\frac{7}{9}\right)^{n-1} \frac{2}{9} $$ Find the sum \(N\) of the series.

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