Problem 12
Question
In Problems 1-12, expand the given function in a Maclaurin series. Give the radius of convergence of each series. $$ f(z)=\cos ^{2} z $$
Step-by-Step Solution
Verified Answer
The Maclaurin series of \( \cos^2 z \) is \( \sum \left( \frac{(-1)^n}{2} \frac{2^{2n} z^{2n}}{(2n)!} \right) + \frac{1}{2} \) with radius of convergence infinity.
1Step 1: Definition of Maclaurin Series
A Maclaurin series is a special case of a Taylor series centered at 0. It is given by the formula:\[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \]where \( f^{(n)}(0) \) is the \( n \)-th derivative of the function evaluated at 0.
2Step 2: Applying the Double Angle Identity
We know from trigonometric identities that \( \cos^2 z = \frac{1 + \cos(2z)}{2} \). So, first, use this identity:\[ f(z) = \frac{1 + \cos(2z)}{2} \]
3Step 3: Maclaurin Series for Cosine Function
Recall that the Maclaurin series for \( \cos(z) \) is:\[ \cos(z) = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!} \]Hence, the series for \( \cos(2z) \) would be:\[ \cos(2z) = \sum_{n=0}^{\infty} (-1)^n \frac{(2z)^{2n}}{(2n)!} \]
4Step 4: Substitute and Simplify
Substitute the series of \( \cos(2z) \) into the equation from Step 2:\[ \cos^2 z = \frac{1}{2} + \frac{1}{2} \sum_{n=0}^{\infty} (-1)^n \frac{(2z)^{2n}}{(2n)!} \]This simplifies to:\[ \cos^2 z = \sum_{n=0}^{\infty} \left( \frac{(-1)^n}{2} \frac{2^{2n} z^{2n}}{(2n)!} \right) + \frac{1}{2} \]
5Step 5: Calculate Specific Derivatives if Needed
Since the trigonometric identity approach is used here, we don't need to calculate specific derivatives for the series expansion; instead, we have directly used known series. But remember, you could differentiate \( \cos^2 z \) to use the typical Maclaurin series process.
6Step 6: Determine Radius of Convergence
The radius of convergence for \( \cos(z) \) and hence \( \cos(2z) \) is infinity since it can be expanded around all complex numbers. Therefore, the radius of convergence of \( \cos^2(z) \) is also infinity.
Key Concepts
Trigonometric IdentitiesTaylor SeriesRadius of ConvergenceSeries Expansion
Trigonometric Identities
Trigonometric identities are essential tools in simplifying and transforming complex trigonometric expressions. They allow us to rewrite expressions in different forms, making calculations easier and revealing useful properties of the functions. In this exercise, the identity used is the double angle identity for cosine:
- \( \cos^2 z = \frac{1 + \cos(2z)}{2} \)
Taylor Series
A Taylor series is a representation of a function as an infinite sum of terms. These terms are calculated from the values of the function's derivatives at a single point. The Maclaurin series is a special case of the Taylor series, centering the expansion at zero. For any function \( f(z) \), the Maclaurin series is given by:\[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \]The Taylor series is a powerful tool because it allows us to approximate functions with polynomials. This is particularly useful for analytic functions, like polynomials, exponentials, and trigonometric functions, which can be expanded to approximate their values over some interval.
Radius of Convergence
The radius of convergence indicates the interval within which a series converges to the function it represents. For a series centered at a point, it defines how far the series can be trusted to approximate the function. For example, the series for \( \cos(z) \), and thereby \( \cos(2z) \), converges for all values of \( z \), as it can be expanded around any complex number. Hence, its radius of convergence is infinity.Generally, if we are given a power series:\[ \sum_{n=0}^{\infty} a_n z^n \]The radius of convergence \( R \) can be found using the formula:
- \( \frac{1}{R} = \limsup_{n \to \infty} \sqrt[n]{|a_n|} \)
Series Expansion
Series expansion is a method of expressing a function as a sum of terms from a sequence, typically involving powers of a variable. This approach is used for functions that can be difficult to handle in their original form. In this exercise, we took the function \( \cos^2 z \) and expressed it using trigonometric identities, then expanded it into a Maclaurin series.By substituting the known series for \( \cos(2z) \):\[ \cos(2z) = \sum_{n=0}^{\infty} (-1)^n \frac{(2z)^{2n}}{(2n)!} \]we were able to write \( \cos^2 z \) as:\[ \cos^2 z = \sum_{n=0}^{\infty} \left( \frac{(-1)^n 2^{2n}}{2 \cdot (2n)!} z^{2n} \right) + \frac{1}{2} \]Series expansions help in approximating complex functions with simpler polynomial forms, which are easier to evaluate and analyze mathematically.
Other exercises in this chapter
Problem 12
In Problems \(9-12\), the indicated number is a zero of the given function. Use a Maclaurin or Taylor series to determine the order of the zero. $$ f(z)=1-\pi i
View solution Problem 12
In Problems 7-12, expand \(f(z)=\frac{1}{z(z-3)}\) in a Laurent series valid for the indicated annular domain. $$ 1
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In Problems 11 and 12, show that the given sequence \(\left\\{z_{n}\right\\}\) converges to a complex number \(L\) by computing \(\lim _{n \rightarrow \infty} \
View solution Problem 13
Expand the given function in a Taylor series centered at the indicated point. Give the radius of convergence of each series. \(f(z)=1 / z, z_{0}=1\)
View solution