/*! 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 18 where the primes denote successi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 18

where the primes denote successive derivatives of \(f(x)\). (Depending on the function this series may converge for any increment \(\delta\) or only for values of \(\delta\) less than some nonzero "radius of convergence.") This theorem is enormously useful, especially for small values of \(\delta\), when the first one or two terms of the series are often an excellent approximation. \(^{11}\) (a) Find the Taylor series for \(\ln (1+\delta)\). (b) Do the same for cos \(\delta\). (c) Likewise sin \(\delta\). (d) And \(e^{\delta}\).

Short Answer

Expert verified
(a) \( \ln(1+\delta) = \delta - \frac{\delta^2}{2} + \frac{\delta^3}{3} - \ldots \); (b) \( \cos \delta = 1 - \frac{\delta^2}{2} + \frac{\delta^4}{4!} - \ldots \); (c) \( \sin \delta = \delta - \frac{\delta^3}{3!} + \frac{\delta^5}{5!} - \ldots \); (d) \( e^{\delta} = 1 + \delta + \frac{\delta^2}{2!} + \ldots \)."

Step by step solution

01

Taylor Series Formula

The Taylor series for a function \( f(x) \) about \( x = a \) is given by \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \). In these problems, we will consider \( a = 0 \) and express each function in terms of \( \delta \).
02

Taylor Series for \( \ln(1 + \delta) \)

The function \( f(x) = \ln(1 + x) \). Compute its derivatives: \( f'(x) = \frac{1}{1 + x} \), \( f''(x) = -\frac{1}{(1 + x)^2} \), \( f'''(x) = \frac{2}{(1 + x)^3} \). At \( x = 0 \), these are \( f(0) = 0 \), \( f'(0) = 1 \), \( f''(0) = -1 \), \( f'''(0) = 2 \), and so on. The Taylor series becomes \( \ln(1 + \delta) = \delta - \frac{\delta^2}{2} + \frac{\delta^3}{3} - \ldots \).
03

Taylor Series for \( \cos \delta \)

The function \( f(x) = \cos(x) \). Compute its derivatives: \( f'(x) = -\sin(x) \), \( f''(x) = -\cos(x) \), \( f'''(x) = \sin(x) \). At \( x = 0 \), these are \( f(0) = 1 \), \( f''(0) = -1 \), \( f''''(0) = 1 \), etc. The Taylor series becomes \( \cos(\delta) = 1 - \frac{\delta^2}{2} + \frac{\delta^4}{4!} - \ldots \).
04

Taylor Series for \( \sin \delta \)

The function \( f(x) = \sin(x) \). Compute its derivatives: \( f'(x) = \cos(x) \), \( f''(x) = -\sin(x) \), \( f'''(x) = -\cos(x) \). At \( x = 0 \), these are \( f(0) = 0 \), \( f'(0) = 1 \), \( f'''(0) = 0 \), \( f'''(0) = -1 \). The Taylor series becomes \( \sin(\delta) = \delta - \frac{\delta^3}{3!} + \frac{\delta^5}{5!} - \ldots \).
05

Taylor Series for \( e^{\delta} \)

The function \( f(x) = e^x \). All derivatives \( f'(x), f''(x), f'''(x) \), etc., are \( e^x \). At \( x = 0 \), \( f(0) = f'(0) = f''(0) = 1 \). The Taylor series becomes \( e^{\delta} = 1 + \delta + \frac{\delta^2}{2!} + \frac{\delta^3}{3!} + \ldots \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Differentiation and the Taylor Series
Differentiation is a fundamental concept in calculus, used to find the rate at which a function is changing. In the context of a Taylor series, differentiation helps us derive the various terms of the series. A Taylor series approximates a function as an infinite sum of terms calculated from the values of its derivatives at a single point. This is especially useful because it allows us to express a complex function in simpler polynomial form.

To find a Taylor series for a function, you:
  • Compute derivatives of the function at a particular point.
  • Use those derivatives to form terms like \( f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \).

This series helps in calculating approximate values, simplifying complex expressions, and analyzing behavior near the expansion point.
Decoding the Radius of Convergence
The radius of convergence tells us the range of values for which a Taylor series converges to the actual value of the function. For any given function, the Taylor series may converge anywhere within this radius.

To understand this concept:
  • Think of the radius of convergence as a limit beyond which the series may not provide a reliable approximation.
  • It can be calculated using the ratio test, which involves the limit of the absolute values of the ratios of successive terms.

If a Taylor series has an infinite radius of convergence, it means the series converges for any value of the variable. Otherwise, it converges only within a specific interval centered at the expansion point.
Making Approximations with Taylor Series
Taylor series are powerful tools for approximating functions, especially when dealing with small values of the variable near the expansion point. These approximations make complex functions more manageable and easier to compute by focusing on just the first few terms of the series.

Approximation steps:
  • Use the first term or first few terms of the Taylor series to get an estimated value.
  • Check the accuracy of the approximation by considering the magnitude of the discarded higher-order terms.

For functions like \(\ln(1+\delta)\), \(\cos(\delta)\), and \(\sin(\delta)\), the first few terms can provide highly accurate results if \(\delta\) is small, making them extremely practical for real-world calculations and analyses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For any complex number \(z=x+i y,\) the real and imaginary parts are defined as the real numbers \(\operatorname{Re}(z)=x\) and \(\operatorname{Im}(z)=y .\) The modulus or absolute value is \(|z|=\sqrt{x^{2}+y^{2}}\) and the phase or angle is the value of \(\theta\) when \(z\) is expressed as \(z=r e^{i \theta} .\) The complex conjugate is \(z^{*}=x-i y\) (This last is the notation used by most physicists; most mathematicians use \(\bar{z}\).) For each of the following complex numbers, find the real and imaginary parts, the modulus and phase, and the complex conjugate, and sketch \(z\) and \(z^{*}\) in the complex plane: (a) \(z=1+i \quad\) (b) \(z=1-i \sqrt{3}\) (d) \(z=5 e^{i \omega t}\) (c) \(z=\sqrt{2} e^{-i \pi / 4}\) In part (d), \(\omega\) is a constant and \(t\) is the time.

Problem 2.7 is about a class of one-dimensional problems that can always be reduced to doing an integral. Here is another. Show that if the net force on a one-dimensional particle depends only on position, \(F=F(x),\) then Newton's second law can be solved to find \(v\) as a function of \(x\) given by $$v^{2}=v_{\mathrm{o}}^{2}+\frac{2}{m} \int_{x_{0}}^{x} F\left(x^{\prime}\right) d x^{\prime}$$ [Hint: Use the chain rule to prove the following handy relation, which we could call the " \(v\) dv/d \(x\) rule": If you regard \(v\) as a function of \(x,\) then $$\dot{v}=v \frac{d v}{d x}=\frac{1}{2} \frac{d v^{2}}{d x}$$ Use this to rewrite Newton's second law in the separated form \(m d\left(v^{2}\right)=2 F(x) d x\) and then integrate from \(x_{\mathrm{o}}\) to \(x .\) ] Comment on your result for the case that \(F(x)\) is actually a constant. (You may recognise your solution as a statement about kinetic energy and work, both of which we shall be discussing in Chapter 4.)

A charged particle of mass \(m\) and positive charge \(q\) moves in uniform electric and magnetic fields, \(\mathbf{E}\) and \(\mathbf{B}\), both pointing in the \(z\) direction. The net force on the particle is \(\mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B})\). Write down the equation of motion for the particle and resolve it into its three components. Solve the equations and describe the particle's motion.

Use the series definition (2.72) of \(e^{z}\) to prove that \(^{12} d e^{z} / d z=e^{z}\).

(a) Using Euler's relation (2.76), prove that any complex number \(z=x+\) iy can be written in the form \(z=r e^{i \theta},\) where \(r\) and \(\theta\) are real. Describe the significance of \(r\) and \(\theta\) with reference to the complex plane. (b) Write \(z=3+4 i\) in the form \(z=r e^{i \theta} .\) (c) Write \(z=2 e^{-i \pi / 3}\) in the form \(x+i y.\)
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electromagnetism

Read Explanation

Mechanics and Materials

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.