/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 154 Compute the Taylor series of eac... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 154

Compute the Taylor series of each function around \(x=1\). $$ f(x)=\ln x $$

Short Answer

Expert verified
The Taylor series for \( \ln x \) around \( x=1 \) is: \( (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots \)

Step by step solution

01

Identify Initial Terms of the Series

The Taylor series for a function \( f(x) \) about \( x = a \) is given by the formula: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]First, we need to identify the value of \( f(1) \) for \( f(x) = \ln x \). So, \( f(1) = \ln 1 = 0 \). Thus, the constant term of the series is 0.
02

Calculate the First Derivative

The first derivative of \( f(x) = \ln x \) is given by:\[ f'(x) = \frac{1}{x} \]Evaluating this at \( x = 1 \), we get:\[ f'(1) = \frac{1}{1} = 1 \]This gives us the first term after the constant, which is \( 1 \cdot (x-1) \).
03

Calculate the Second Derivative

The second derivative is found by differentiating \( f'(x) \):\[ f''(x) = -\frac{1}{x^2} \]Evaluating at \( x=1 \), we find:\[ f''(1) = -\frac{1}{1^2} = -1 \]This gives us the second term in the series: \(-\frac{1}{2!}(x-1)^2\).
04

Calculate the Third Derivative

The third derivative is:\[ f'''(x) = \frac{2}{x^3} \]Evaluating at \( x=1 \), we get:\[ f'''(1) = \frac{2}{1^3} = 2 \]This contributes the term \( \frac{2}{3!}(x-1)^3 \) to the series.
05

Generalize the Pattern

Observing the pattern for higher derivatives, each derivative can be expressed as:\[ f^{(n)}(x) = (-1)^{n-1} \frac{(n-1)!}{x^n} \]Evaluating each at \( x=1 \), we get:\[ f^{(n)}(1) = (-1)^{n-1} (n-1)! \]Contributing the term \( \frac{(x-1)^n}{n} \) to the series due to factorial simplifications.
06

Write the Taylor Series

Combining all the computed terms, the Taylor series for \( \ln x \) about \( x=1 \) is given by:\[ \ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots \]This series continues with alternating signs and increasing powers of \( (x-1) \).

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Key Concepts

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

ln x
The function \( \ln x \) is the natural logarithm of \( x \). It is defined only for positive values of \( x \). This function grows very slowly compared to polynomial functions, which means it can be really useful to express it as a series for approximation around a specific point, such as \( x = 1 \). When dealing with logarithmic functions, remember:
  • \( \ln 1 = 0 \)
  • \( \ln e = 1 \)
Knowing these basic values helps when evaluating the Taylor series at specific points.
derivatives
Derivatives play a crucial role in constructing Taylor series. They give us the rate at which functions change, allowing us to capture how the function behaves close to a certain point. For \( \ln x \), the first few derivatives are:
  • \( f'(x) = \frac{1}{x} \)
  • \( f''(x) = -\frac{1}{x^2} \)
  • \( f'''(x) = \frac{2}{x^3} \)
  • \( f^{(n)}(x) = (-1)^{n-1} \frac{(n-1)!}{x^n} \)
To evaluate a Taylor series at \( x = 1 \), calculate these derivatives at \( x = 1 \). This gives coefficients for the series.
function expansion
Function expansion involves rewriting a function into an infinite sum of terms. Each term is based on the function's derivatives and powers of \( (x - a) \). For \( \ln x \) expanded about \( x=1 \), function expansion uses derivatives evaluated at \( x=1 \), leading to the following terms:
  • \( f(1) = 0 \)
  • \( f'(1) = 1 \cdot (x-1) \)
  • \( f''(1) = -\frac{1}{2!}(x-1)^2 \)
  • \( f'''(1) = \frac{2}{3!}(x-1)^3 \)
These terms continue indefinitely, forming the series.
Maclaurin series
The Maclaurin series is a special case of the Taylor series where it is expanded around \( x = 0 \). While our original problem deals with a Taylor series expanded around \( x=1 \), the Maclaurin series is similarly useful because it simplifies function representation when \( x \) is near zero. For \( \ln x \), the series cannot directly be calculated at \( x=0 \) due to its undefined nature. Instead, we adapt the Taylor series approach for non-zero values. The methodology remains similar:
  • Calculate the function value at a specific point
  • Use successive derivatives to find coefficients
  • Use the expansion principle of adding powers of \( (x - a) \)
This understanding aids in simplifying functions around points other than zero effectively.

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Most popular questions from this chapter

Find the Taylor series of the given function centered at the indicated point. $$ x^{4} \text { at } a=-1 $$

Suppose that \(\sum_{n=0}^{\infty} a_{n} x^{n}\) converges to a function \(y\) such that \(y^{\prime \prime} y^{\prime}+y=0\) where \(y(0)=1\) and \(y^{\prime}(0)=0 .\) Find a formula that relates \(a_{n+2}, a_{n+1},\) and \(a_{n}\) and compute \(a_{0}, \ldots, a_{5}\).

Find the smallest value of \(n\) such that the remainder estimate \(\left|R_{n}\right| \leq \frac{M}{(n+1) !}(x-a)^{n+1}\), where \(M\) is the maximum value of \(\left|f^{(n+1)}(z)\right|\) on the interval between \(a\) and the indicated point, yields \(\left|R_{n}\right| \leq \frac{1}{1000}\) on the indicated interval. $$ f(x)=e^{-2 x} \text { on }[-1,1], a=0 $$

Find the Maclaurin series of \(F(x)=\int_{0}^{x} f(t) d t\) by integrating the Maclaurin series of \(f\) term by term. If \(f\) is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero. $$ F(x)=\tanh ^{-1} x ; \quad f(t)=\frac{1}{1-t^{2}}=\sum_{n=0}^{\infty} t^{2 n} $$

Compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of \(f\). $$ f(x)=e^{x} \cos x $$
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