Problem 8
Question
Adding the series for \(e^{z}\) and \(e^{-z}\) gives \(\cosh z=\frac{1}{2}\left(e^{z}+e^{-z}\right)=\sum_{k=0}^{\infty} \frac{z^{2 k}}{(2 k) !}\) where \(R=\infty\)
Step-by-Step Solution
Verified Answer
The series describes \( \cosh z \), with even powers of \( z \).
1Step 1: Understanding the Exponential Series
The functions, \( e^z \) and \( e^{-z} \), can be expressed as their corresponding series expansions: \( e^z = \sum_{k=0}^{\infty} \frac{z^k}{k!} \) and \( e^{-z} = \sum_{k=0}^{\infty} \frac{(-z)^k}{k!} \). Let's analyze these expressions step by step.
2Step 2: Deriving the Expression for \( e^z + e^{-z} \)
We add the series for \( e^z \) and \( e^{-z} \):\[e^z + e^{-z} = \sum_{k=0}^{\infty} \frac{z^k}{k!} + \sum_{k=0}^{\infty} \frac{(-z)^k}{k!}.\]
3Step 3: Analyzing the Resulting Terms in the Summation
Separate the terms in the summation by their parity (i.e., even or odd powers of \( z \)), because they affect the cancellation or addition in terms:\[= \left( \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!} \right) + \left( \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!} \right) + \left( \sum_{n=0}^{\infty} \frac{(-z)^{2n}}{(2n)!} \right) + \left( \sum_{n=0}^{\infty} \frac{(-z)^{2n+1}}{(2n+1)!} \right).\]
4Step 4: Cancellation of Odd Terms
The odd terms in the series cancel each other out due to the alternating signs in \( e^z \) and \( e^{-z} \):\[e^z + e^{-z} = 2 \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!}. \]
5Step 5: Identifying the Expression for \( \cosh z \)
The function \( \cosh z \) is defined as:\[\cosh z = \frac{1}{2} (e^z + e^{-z}) = \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!}\] noting that the \( \frac{1}{2} \) factor cancels with the multiplied 2.
Key Concepts
Exponential SeriesCosh FunctionSeries ExpansionEven and Odd Functions
Exponential Series
The exponential series is a way to represent the exponential function using an infinite sum of terms. This series expands the mathematical expression of the exponential function, which is frequently encountered in mathematics due to its applications in many fields such as physics, engineering, and statistics. The exponential function, often denoted as \(e^z\), can be expanded into a series:
The series expansion is particularly useful for approximating the value of the exponential function and helps simplify complex calculations involving exponentials.
- \(e^z = \sum_{k=0}^{\infty} \frac{z^k}{k!}\)
- \(e^{-z} = \sum_{k=0}^{\infty} \frac{(-z)^k}{k!}\)
The series expansion is particularly useful for approximating the value of the exponential function and helps simplify complex calculations involving exponentials.
Cosh Function
The hyperbolic cosine function, known as the \(\cosh\) function, is an important hyperbolic function that mirrors many properties of the regular cosine function but differs in its mathematical construction. It is defined for all real numbers \(z\), and its relation to the exponential function is as follows:
This behavior can be observed in its graph, which is symmetric about the y-axis, reinforcing its status as an even function. Applications of \(\cosh z\) are pervasive across many areas of science and engineering, such as the description of hyperbolic curves and in analyses involving wave motion.
- \(\cosh z = \frac{1}{2}(e^z + e^{-z})\)
This behavior can be observed in its graph, which is symmetric about the y-axis, reinforcing its status as an even function. Applications of \(\cosh z\) are pervasive across many areas of science and engineering, such as the description of hyperbolic curves and in analyses involving wave motion.
Series Expansion
Series expansion refers to the method of expressing more complex functions as an infinite sum of simpler terms, often powers of a variable. This is extremely useful in calculus and other advanced branches of mathematics because it allows one to analyze functions in a more digestible form.The expression of \(e^z + e^{-z}\) as a series allows us to identify the hyperbolic cosine function, \(\cosh z\), in a more explicit form:
The technique of series expansion is also foundational in developing approximations for other functions, particularly in contexts where a graphical depiction or numerical calculation is needed.
- \(e^z + e^{-z} = 2 \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!}\)
The technique of series expansion is also foundational in developing approximations for other functions, particularly in contexts where a graphical depiction or numerical calculation is needed.
Even and Odd Functions
Understanding even and odd functions is crucial in mathematics, especially when dealing with symmetry in function graphs. An even function, by definition, satisfies \(f(-x) = f(x)\) for all \(x\). In contrast, an odd function meets the condition \(f(-x) = -f(x)\).
The hyperbolic cosine function is a classic example of an even function because \(\cosh(-z) = \cosh(z)\). This symmetry induces specific consequences in its series expansion:
The hyperbolic cosine function is a classic example of an even function because \(\cosh(-z) = \cosh(z)\). This symmetry induces specific consequences in its series expansion:
- For even functions like \(\cosh z\), only the terms involving even powers of \(z\) remain after simplification.
- The formula \(\cosh z = \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!}\) reflects this evenness, showcasing the exclusion of any odd-powered terms that would otherwise cancel out.
Other exercises in this chapter
Problem 7
Write \(f(z)=e^{z}\left(e^{z}-1\right)\) to see that \(2 n \pi i, n=0,\pm 1,\pm 2, \ldots\) are zeros of \(f .\) Now \(f^{\prime}(z)=2 e^{2 z}-e^{z}\) and \(f^{
View solution Problem 7
Subtracting the series for \(e^{z}\) and \(e^{-z}\) gives \(\sinh z=\frac{1}{2}\left(e^{z}-e^{-z}\right)=\sum_{k=0}^{\infty} \frac{z^{2 k+1}}{(2 k+1) !}\) where
View solution Problem 8
The zeros of \(f\) are the zeros of \(\sin z,\) that is, \(n \pi, n=0,\pm 1,\pm 2, \ldots .\) From \(f^{\prime}(z)=2 \sin z \cos z\) we see \(f^{\prime}(n \pi)=
View solution Problem 8
$$f(z)=\frac{1}{z^{2}} \cdot \frac{1}{1-\frac{3}{z}}=\frac{1}{z^{2}}\left[1+\frac{3}{z}+\frac{3^{2}}{z^{2}}+\frac{3^{3}}{z^{3}}+\dots\right]=\frac{1}{z^{2}}+\fr
View solution