Chapter 15: Problem 1
Find the Taylor or Maclaurin series of the given function with the given point as center and determine the radius of convergence. $$\epsilon^{-2 z}, 0$$
Short Answer
Expert verified
The Maclaurin series of \(e^{-2z}\) is \(\sum_{n=0}^{\infty} \frac{(-2)^n}{n!} z^n\) with radius of convergence \(\infty\).
Step by step solution
01
Recognize the Function Type
The given function is \[e^{-2z}\].We need to determine its Maclaurin series, which means we expand around the point 0.
02
Write the General Formula for a Maclaurin Series
The formula for a Maclaurin series of a function \(f(z)\) is:\[f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n\]where \(f^{(n)}(0)\) is the \(n\)-th derivative of \(f\) evaluated at 0.
03
Calculate the Derivatives of the Function
For the function \(e^{-2z}\), the first few derivatives are:\[f(z) = e^{-2z}\]\[f'(z) = -2 e^{-2z}\]\[f''(z) = 4 e^{-2z}\]\[f^{(n)}(z) = (-2)^n e^{-2z}\].Evaluating these at \(z=0\): \[f(0) = 1\]\[f'(0) = -2\]\[f''(0) = 4\]\[f^{(n)}(0) = (-2)^n\].
04
Construct the Series with the Derivatives
Plug these derivatives back into the Maclaurin series formula:\[e^{-2z} = \sum_{n=0}^{\infty} \frac{(-2)^n}{n!} z^n\].
05
Determine the Radius of Convergence
The radius of convergence for a power series \(\sum a_n z^n\) is found using:\[\frac{1}{R} = \limsup_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\].For this series, \(a_n = \frac{(-2)^n}{n!}\).Thus:\[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-2)^{n+1}/(n+1)!}{(-2)^n/n!} \right| = \left| \frac{-2}{n+1} \right|\rightarrow 0 \text{ as } n \rightarrow \infty\].Therefore, the radius of convergence \(R\) is \(\infty\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor series
The Taylor series is a method of expressing a function as an infinite sum of terms calculated from the values of its derivatives at a certain single point. Imagine you have a smooth curve, and you want to approximate it using a series of terms that involve powers of the variable. That's what a Taylor series does! It provides a way to approximate a complex function using polynomial terms.The Taylor series centered around a point, say 'a', for a function \(f(z)\) is written as:
- \(f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (z-a)^n\)
- \(f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n\)
radius of convergence
In the realm of power series, convergence is key. Not all infinite series converge to a finite sum. The radius of convergence is a way to determine how far from the center of the series we can move before the series stops converging.For a power series \( \sum a_n z^n \), where \(a_n\) are constants, the radius of convergence \(R\) is found using:
- \(\frac{1}{R} = \limsup_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)
power series
A power series is like a magic formula that mathematicians use to rewrite functions as infinite series. Unlike typical polynomial equations, power series go on forever. They are written in the form \( \sum_{n=0}^{\infty} a_n z^n \), where each term uses a constant \(a_n\) multiplied by \(z\) raised to a power \(n\).The beauty of power series is in their adaptability:
- They can represent functions that are tricky to deal with using other methods.
- For example, exponential, logarithmic, and trigonometric functions can be expanded into power series.
- The Maclaurin series is just a special type of power series centered at zero.