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Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a\). $$f(x)=\ln x, \quad a=1$$

Short Answer

Expert verified
The Taylor polynomials are: \(P_0(x) = 0\), \(P_1(x) = x-1\), \(P_2(x) = x-1 - \frac{1}{2}(x-1)^2\), \(P_3(x) = x-1 - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3\).

Step by step solution

01

Understand the Function and Point for Expansion

The given function is \(f(x) = \ln x\), and we are asked to find Taylor polynomials of different orders centered at \(a = 1\). This means we have to expand the function around \(x = 1\).
02

Calculate the Function Value at a

Evaluate \(f(a) = \ln(1) = 0\). This value will be used in all orders of the Taylor polynomial.
03

Derivative Calculations

Find the derivatives of \(f(x) = \ln x\).\[\begin{align*}f'(x) &= \frac{1}{x}\rightarrow f'(1) = 1,\f''(x) &= -\frac{1}{x^2}\rightarrow f''(1) = -1,\f'''(x) &= \frac{2}{x^3}\rightarrow f'''(1) = 2.\end{align*}\]
04

Zero Order Taylor Polynomial

The zero order Taylor polynomial is just the function value at \(a = 1\): \[P_0(x) = f(1) = 0.\]
05

First Order Taylor Polynomial

The first order Taylor polynomial includes the first derivative term: \[P_1(x) = f(1) + f'(1)(x-1) = 0 + 1 \cdot (x-1) = x-1.\]
06

Second Order Taylor Polynomial

The second order Taylor polynomial includes up to the second derivative term: \[P_2(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 = 0 + (x-1) - \frac{1}{2}(x-1)^2.\]
07

Third Order Taylor Polynomial

The third order Taylor polynomial includes up to the third derivative term: \[P_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3.\]

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Key Concepts

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

Derivative Calculations
In mathematics, derivatives play an essential role in understanding how functions change. Derivative calculations are crucial for finding Taylor polynomials, as they provide information about the curvature and steepness of the function at a specific point. To find a Taylor series, we need to calculate several derivatives of the function in question.
For example, consider the function \( f(x) = \ln x \). To expand this into a Taylor polynomial around \( x = 1 \), we first find the derivatives at this point:
  • The first derivative \( f'(x) = \frac{1}{x} \), evaluated at \( x = 1 \), gives \( f'(1) = 1 \).
  • The second derivative \( f''(x) = -\frac{1}{x^2} \), evaluated at \( x = 1 \), results in \( f''(1) = -1 \).
  • The third derivative \( f'''(x) = \frac{2}{x^3} \), at \( x = 1 \), gives \( f'''(1) = 2 \).
Each of these derivatives will influence the terms of the Taylor polynomial, with each successive derivative adding more detail to the approximation.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a vital function in calculus and often appears in diverse areas of science. The natural logarithm of a number is the power to which the base \( e \) (approximately 2.718) must be raised to obtain that number. In simple terms, \( \ln(x) \) is the inverse of the exponential function \( e^x \).
A unique property is that \( \ln(1) = 0 \) because any number to the power of zero is one. This property is particularly useful in calculating Taylor polynomials centered at \( x = 1 \). In this context, the logarithmic function exhibits behavior suitable for approximation using polynomial terms.
The steepness and rate of change captured through derivatives help in crafting an accurate polynomial that mirrors the logarithmic curve closely, especially around the expansion point. The utility of \( \ln(x) \) in simplifying complex multiplicative relationships into additive ones is a reason for its prevalence in calculus and beyond.
Polynomial Approximation
Polynomial approximation is a technique where we use polynomials to represent more complex functions. It's particularly helpful because polynomials are easier to work with regarding computation and analysis. The Taylor polynomial is a common form of polynomial approximation.
When we approximate \( f(x) = \ln x \) around \( x = 1 \), we're creating a series of polynomials that get more accurate as we add more terms. Here's how it works:
  • Zero Order: This starts with \( P_0(x) = f(1) = 0 \).
  • First Order: Adding the first derivative term gives \( P_1(x) = (x-1) \).
  • Second Order: Adding the second derivative term, \( -\frac{1}{2}(x-1)^2 \), we get \( P_2(x) = (x-1) - \frac{1}{2}(x-1)^2 \).
  • Third Order: Including the third derivative term, \( \frac{1}{3}(x-1)^3 \), the polynomial becomes \( P_3(x) = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 \).
Each successive term refines the accuracy of the approximation, capturing more of the function's nuances. This method can approximate many types of functions, not just logarithms, proving highly valuable across mathematical and engineering fields.

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Most popular questions from this chapter

If \(\cos x\) is replaced by \(1-\left(x^{2} / 2\right)\) and \(|x|<0.5,\) what estimate can be made of the error? Does \(1-\left(x^{2} / 2\right)\) tend to be too large, or too small? Give reasons for your answer.

a. Show that if two power series \(\Sigma_{n=0}^{\infty} a_{n} x^{n}\) and \(\Sigma_{n=0}^{\infty} b_{n} x^{n}\) are convergent and equal for all values of \(x\) in an open interval \((-c, c),\) then \(a_{n}=b_{n}\) for every \(n\) \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n} .\) Differentiate term by term to show that \(a_{n}\) and \(b_{n}\) both equal \(f^{(n)}(0) /(n !)\) b. Show that if \(\Sigma_{n=0}^{\infty} a_{n} x^{n}=0\) for all \(x\) in an open interval \((-c, c),\) then \(a_{n}=0\) for every \(n\)

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To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

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