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Chapter 10: Problem 27

Using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$\begin{aligned} &1\\\ &1-\ln (1+t) \end{aligned}$$

Short Answer

Expert verified
The first four nonzero terms are: \( 1, -t, \frac{t^2}{2}, -\frac{t^3}{3} \).

Step by step solution

01

Recognize the Taylor Series of the Function

We know that the Taylor series expansion of \( \ln(1 + t) \) around zero is given by: \[ \ln(1 + t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \cdots \] This series alternates signs and increases in power.
02

Develop the Expression for \( 1 - \ln(1 + t) \)

We need to find the series for \( 1 - \ln(1 + t) \). Substitute the Taylor series of \( \ln(1+t) \) from Step 1: \[ 1 - \ln(1 + t) = 1 - \left( t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \cdots \right) \] Distribute the negative sign: \[ 1 - t + \frac{t^2}{2} - \frac{t^3}{3} + \frac{t^4}{4} - \cdots \]
03

Collect the First Four Nonzero Terms

From the series derived, identify and list the first four non-zero terms: \( 1, -t, \frac{t^2}{2}, -\frac{t^3}{3} \). These are the first four terms in the series of \( 1 - \ln(1 + t) \).

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Key Concepts

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

Series Expansion
Understanding series expansion is fundamental in calculus. It helps us break down complex functions into simpler forms. A series expansion enables you to express a function as a sum of terms from a sequence.
These terms are typically polynomial-like, making calculations easier or approximations more intuitive.In Taylor series specifically, you expand a function about a point, often zero. This means:
  • The function can be expressed as an infinite sum of terms.
  • Each term involves derivatives of the function evaluated at the point of expansion.
  • This allows us to approximate a function near this point.
The Taylor series formula generally is: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]This involves powers and factorials, applying them iteratively results in more accurate approximations.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithm with the base \( e \), where \( e \) is approximately 2.718. It's widely used in mathematics, particularly calculus. This logarithm is natural due to its fundamental properties in continuous compounding and growth processes.In the context of series expansions, we use the natural logarithm expansion at zero:\[\ln(1 + t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \cdots\]This expansion represents the natural logarithm as an infinite series that:
  • Alternates in sign.
  • Includes a sequence of terms that increase in power.
This representation is essential for approximating \( \ln(1 + t) \) when \( t \) is close to zero, where the series converges quickly.
Polynomial Approximation
Polynomial approximation is a technique to simplify functions using polynomials. It allows more manageable calculations and aids understanding of function behavior. By using polynomial functions:
  • We can approximate complex functions closely in specific intervals.
  • This provides insight into the function鈥檚 nature using simple, finite expressions.
In our specific example鈥攖he function \( 1 - \ln(1 + t) \)鈥攑olynomial approximation comes from subtracting the natural logarithm series from 1. This results in:\[1 - t + \frac{t^2}{2} - \frac{t^3}{3} + \cdots\]This series is a polynomial with terms that correspond to increasing powers of \( t \). It's crucial for finding the simplest model of a function to understand its local behavior near the point of expansion. Such approximations are invaluable for calculations involving small intervals.

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Most popular questions from this chapter

Use a Taylor polynomial with the derivatives given to make the best possible estimate of the value. \(f(1.1),\) given that \(f(1)=3, f^{\prime}(1)=2, f^{\prime \prime}(1)=-4\)

The pulse train of width \(c\) is the periodic function \(f\) of period \(2 \pi\) given by $$f(x)=\left\\{\begin{array}{ll} 0 & -\pi \leq x<-c / 2 \\ 1 & -c / 2 \leq x< c / 2 \\ 0 & c / 2 \leq x<\pi \end{array}\right.$$ Suppose that \(f\) is the pulse train of width 0.4. (a) What fraction of the energy of \(f\) is contained in the constant term of its Fourier series? In the constant term and the first harmonic together? (b) Find a formula for the energy of the \(k^{\text {th }}\) harmonic of \(f .\) Use it to sketch the energy spectrum of \(f\). (c) What fraction of the energy of \(f\) is contained in the constant term and the first five harmonics of \(f ?\) (The constant term and the first thirteen harmonics are needed to capture \(90 \%\) of the energy of \(f .\) ) (d) Graph \(f\) and its fifth Fourier approximation on the interval \([-3 \pi, 3 \pi]\).

True or false? If \(f\) is an even function, then the Fourier series for \(f\) on \([-\pi, \pi]\) has only cosines. Explain your answer.

Decide if the statements in Problems \(65-71\) are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. If the Taylor series for \(f(x)\) around \(x=0\) has a finite number of terms and an infinite radius of convergence, then \(f(x)\) is a polynomial.

A hydrogen atom consists of an electron, of mass \(m,\) or biting a proton, of mass \(M,\) where \(m\) is much smalle than \(M .\) The reduced mass, \(\mu,\) of the hydrogen atom defined by $$\mu=\frac{m M}{m+M}$$ (a) Show that \(\mu \approx m.\) (b) To get a more accurate approximation for \(\mu,\) express \(\mu\) as \(m\) times a series in \(m / M.\) (c) The approximation \(\mu \approx m\) is obtained by disregarding all but the constant term in the series. The first-order correction is obtained by including the linear term but no higher terms. If \(m \approx M / 1836\) by what percentage does including the linear term change the estimate \(\mu \approx m ?\)
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