/*! 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 5 a) Investigate whether the funct... [FREE SOLUTION] | 91Ó°ÊÓ

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a) Investigate whether the function \(f(z)=\frac{1}{1+z^{2}}\) is continuous at all points of the plane \(\mathbb{C}\) b) Expand the function \(\frac{1}{1+z^{2}}\) in a power series around \(z_{0}=0\) and find its radius of convergence. c) Solve parts a) and b) for the function \(\frac{1}{1+\lambda^{2} z^{2}}\), where \(\lambda \in \mathbb{R}\) is a parameter. Can you make a conjecture as to how the radius of convergence is determined by the relative location of certain points in the plane \(\mathbb{C}\) ? Could this relation have been understood on the basis of the real line alone, that is, by expanding the function \(\frac{1}{1}\), where \(\lambda \in \mathbb{R}\) and \(x \in \mathbb{R} ?\)

Short Answer

Expert verified
If not, where is it discontinuous? Additionally, what is the radius of convergence for the power series expansion of the given function and the function \(g(z) = \frac{1}{1+\lambda^2 z^2}\)? Answer: The function \(f(z) = \frac{1}{1+z^2}\) is continuous at all points of the complex plane except \(z=\pm i\), where the denominator is zero. The radius of convergence for the power series expansion of the given function is 1. For the function \(g(z) = \frac{1}{1+\lambda^2 z^2}\), the radius of convergence is \(\frac{1}{|\lambda|}\).

Step by step solution

01

Part a: Investigation of Continuity

A function is continuous at a point if its limit exists and equals the function value at that point. Given the function \(f(z) = \frac{1}{1+z^2}\), as a rational function of a complex variable, it is continuous at any point \(z\) where its denominator is non-zero. Let's find the values of z where the denominator is zero: \(1+z^2=0\) \(z^2=-1\) Here, \(z = \pm i\). Since the denominator is zero only at these two points, the function is continuous at all other points in the complex plane.
02

Part b: Expanding the Function and Finding the Radius of Convergence

To find the power series expansion, we make use of the geometric series formula: \(\frac{1}{1-z} = 1 + z + z^2 + \cdots = \sum_{n=0}^\infty z^n\) when \(|z| < 1\). Let's now substitute \(z^2\) with \(-z^2\): \(f(z) = \frac{1}{1 + (-z^2)} = \sum_{n=0}^\infty (-1)^n z^{2n}\) This is the power series expansion of the given function around \(z_0=0\). The power series converges inside the circle |z|< 1 because every complex number except \(z = \pm i\) results in convergence. The radius of convergence, R, is given by: \(R = \frac{1}{\limsup_{n\to\infty} \sqrt[2n]{|a_{2n}|}}\) Here, \(a_{2n} = (-1)^n\), so: \(R = \frac{1}{\limsup_{n\to\infty} \sqrt[2n]{|-1|^n}} = 1\) The radius of convergence is 1.
03

Part c: Expanding the Function and Finding the Radius of Convergence for \(\frac{1}{1+\lambda^2 z^2}\)

Given \(g(z) = \frac{1}{1+\lambda^2 z^2}\), let's follow a similar process to part b: \(g(z) = \frac{1}{1-\left(-\lambda^2 z^2\right)} = \sum_{n=0}^\infty (-\lambda^2 z^2)^n = \sum_{n=0}^\infty (-1)^n \lambda^{2n} z^{2n}\) The power series converges inside the circle \(|\lambda^2 z^2|< 1\). The radius of convergence, R, is given by: \(R = \frac{1}{\limsup_{n\to\infty} \sqrt[2n]{|(-1)^n \lambda^{2n}|}} = \frac{1}{|\lambda|}\) The radius of convergence is \(\frac{1}{|\lambda|}\). We can make a conjecture that the radius of convergence is determined by the distance of the closest singular points in the complex plane, and this relationship holds in the real line case also.

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Key Concepts

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

Continuity in Complex Functions
In complex analysis, continuity is a fundamental property of functions. A function is continuous at a point if the limit of the function as it approaches that point is equal to the function's value at that point. This concept helps understand how the function behaves locally.
For the function \(f(z) = \frac{1}{1+z^2}\), it is a rational function, meaning it is expressed as the ratio of two polynomials. Rational functions in complex analysis are continuous everywhere except where the denominator is zero.
In this specific problem, we are asked to check for continuity in the entire complex plane \(\mathbb{C}\). To do this, we identify where the denominator, \(1+z^2\), becomes zero, hence finding the potential points of discontinuity. Solving \(1+z^2=0\), we get \(z = \pm i\).
Therefore, \(f(z)\) is continuous everywhere on \(\mathbb{C}\), except at the points where \(z = i\) and \(z = -i\). These points are known as singularities, and specifically, they are poles of order 1 for this function. Away from these singularities, the function behaves in a "smooth" manner, without any jumps or breaks.
Power Series Expansion
Power series play a significant role in complex analysis as they allow complex functions to be represented in terms of simpler polynomial functions. A power series is an infinite sum of terms in the form \(a_n z^n\), where \(a_n\) are the coefficients, and \(z\) is the complex variable.
To find the power series expansion of \(\frac{1}{1+z^2}\) around \(z_0 = 0\), utilize the geometric series formula, \(\frac{1}{1-z} = \sum_{n=0}^\infty z^n\), which converges when \(|z|<1\). By substituting \(z^2\) with \(-z^2\), the series becomes \(\frac{1}{1-(-z^2)} = \sum_{n=0}^\infty (-1)^n z^{2n}\).
Each term \((-1)^n z^{2n}\) simplifies the function representation and displays how the function behaves in the vicinity of \(z_0 = 0\). This power series representation sheds light on the nature of the complex function by allowing calculations involving derivatives and integrals to be more straightforward.
Radius of Convergence
The radius of convergence is an essential concept involved when discussing power series. It determines the region within which the series representation of a function reliably converges to the function itself.
The convergence of a power series \(\sum_{n=0}^\infty a_n z^n\) is limited to the region \(|z| < R\), where \(R\) is the radius of convergence. This is calculated using the Cauchy-Hadamard theorem, given by \( R = \frac{1}{\limsup_{n\to\infty} \sqrt[n]{|a_n|}} \).
For the function \(\frac{1}{1+z^2}\), the series \(\sum_{n=0}^\infty (-1)^n z^{2n}\) indicates that the radius of convergence \(R = 1\). This is because the closest singularity from the origin, the point of expansion, is at \(z = \pm i\), which are at a distance 1 from the origin on the complex plane.
Therefore, the power series is valid only inside the disk \(|z| < 1\). As for the modified function \(\frac{1}{1+\lambda^2 z^2}\), the radius of convergence is determined by the parameter \(\lambda\), specifically \(R = \frac{1}{|\lambda|}\).
This underscores the idea that the singular points and their distances from the center of expansion play a critical role in defining the limits of convergence, which cannot be concluded only from a real line perspective but must consider the full complex plane.

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Most popular questions from this chapter

Motion under the action of a Hooke \(^{34}\) central force (the plane oscillator). To develop Eq. (5.156) for a linear oscillator in Sect. 5.6.6 and in Problem 5 let us consider the equation \(m \ddot{\mathbf{r}}(t)=-k \mathbf{r}(t)\) satisfied by the radius-vector \(\mathbf{r}(t)\) of a point of mass \(m\) moving in space under the attraction of a centripetal force proportional to the distance \(|\mathbf{r}(t)|\) from the center with constant of proportionality (modulus) \(k>0\). Such a force arises if the point is joined to the center by a Hooke elastic connection, for example, a spring with constant \(k\). a) By differentiating the vector product \(\mathbf{r}(t) \times \dot{\mathbf{r}}(t)\), show that the motion takes place in the plane passing through the center and containing the initial position vector \(\mathbf{r}_{0}=\mathbf{r}\left(t_{0}\right)\) and the initial velocity vector \(\dot{\mathbf{r}}_{0}=\dot{\mathbf{r}}\left(t_{0}\right)\) (a plane oscillator). If the vectors \(\mathbf{r}_{0}=\mathbf{r}\left(t_{0}\right)\) and \(\dot{\mathbf{r}}_{0}=\dot{\mathbf{r}}\left(t_{0}\right)\) are collinear, the motion takes place along the line containing the center and the vector \(\mathbf{r}_{0}\) (the linear oscillator considered in Sect. 5.6.6). b) Verify that the orbit of a plane oscillator is an ellipse and that the motion is periodic. Find the period of revolution. c) Show that the quantity \(E=m \dot{\mathbf{r}}^{2}(t)+k \mathbf{r}^{2}(t)\) is conserved (constant in time). d) Show that the initial data \(\mathbf{r}_{0}=\mathbf{r}\left(t_{0}\right)\) and \(\dot{\mathbf{r}}_{0}=\dot{\mathbf{r}}\left(t_{0}\right)\) completely determine the subsequent motion of the point.

10\. Show that if a function \(f\) has derivatives up to order \(n+1\) inclusive at a point \(x_{0}\) and \(f^{(n+1)}\left(x_{0}\right) \neq 0\), then in the Lagrange form of the remainder in Taylor's formula, $$ r_{n}\left(x_{0} ; x\right)=\frac{1}{n !} f^{(n)}\left(x_{0}+\theta\left(x-x_{0}\right)\right)\left(x-x_{0}\right)^{n} $$ where \(0<\theta<1\) and the quantity \(\theta=\theta(x)\) tends to \(\frac{1}{4}\) as \(x \rightarrow x_{0}\).

Let \(f \in C^{(\infty)}(\mathbb{R})\). Show that for \(x \neq 0\) $$ \frac{1}{x^{n+1}} f^{(n)}\left(\frac{1}{x}\right)=(-1)^{n} \frac{\mathrm{d}^{n}}{\mathrm{~d} x^{\mathrm{n}}}\left(x^{n-1} f\left(\frac{1}{x}\right)\right) $$

Write a Taylor polynomial of \(\mathrm{e}^{x}\) at zero that makes it possible to compute the values of \(\mathrm{e}^{x}\) on the closed interval \(-1 \leq x \leq 2\) within \(10^{-3}\).

A body that can be regarded as a point mass is sliding down a smooth hill under the influence of gravity. The hill is the graph of a differentiable function \(y=f(x)\). a) Find the horizontal and vertical components of the acceleration vector that the body has at the point \(\left(x_{0}, y_{0}\right)\). b) For the case \(f(x)=x^{2}\) when the body slides from a great height, find the point of the parabola \(y=x^{2}\) at which the horizontal component of the acceleration is maximal.

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