Problem 12
Question
The indicated number is a zero of the given function. Use a Maclaurin or Taylor series to determine the order of the zero. $$ f(z)=z-\sin z ; z=0 $$
Step-by-Step Solution
Verified Answer
The zero of \( f(z) \) at \( z=0 \) is of order 3.
1Step 1: Determine Maclaurin Series for f(z)
The Maclaurin series for a function is given by expanding the function around zero. The function given is \( f(z) = z - \sin z \). We know the Maclaurin series for \( \sin z \) is \( \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \ldots \). Therefore, the Maclaurin series for \( f(z) \) is \( f(z) = z - (z - \frac{z^3}{3!} + \frac{z^5}{5!} - \ldots) \).
2Step 2: Simplify the Series Expression
Simplifying the series, we find that the \( z \) terms cancel out: \( f(z) = z - z + \frac{z^3}{3!} - \frac{z^5}{5!} + \ldots \) which simplifies to \( f(z) = \frac{z^3}{3!} - \frac{z^5}{5!} + \ldots \).
3Step 3: Identify the Order of the Zero
The order of a zero is determined by the first non-zero term in the series expansion of the function. From the series \( f(z) = \frac{z^3}{3!} - \frac{z^5}{5!} + \ldots \), the first non-zero term is \( \frac{z^3}{3!} \). This indicates that \( z = 0 \) is a zero of order 3, as the first non-zero derivative at \( z=0 \) is of degree 3.
Key Concepts
Maclaurin SeriesTaylor SeriesOrder of Zero
Maclaurin Series
The Maclaurin series is a special case of the Taylor series, used for approximating functions around the point zero. It's derived from the Taylor series by setting the center of the expansion at 0, thereby simplifying the process. To construct a Maclaurin series, you repeatedly take derivatives of the function and evaluate them at zero.
For instance, the Maclaurin series of a function \( f(z) \) is represented as:
In providing an approximation for functions like \( \sin z \), the series becomes rather simple:
For instance, the Maclaurin series of a function \( f(z) \) is represented as:
- \( f(0) + \frac{f'(0)}{1!}z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \ldots \)
In providing an approximation for functions like \( \sin z \), the series becomes rather simple:
- \( \sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \ldots \)
Taylor Series
The Taylor series is a powerful tool in complex analysis, giving us a way to represent functions as an infinite sum of terms computed from the values of their derivatives at a single point. While this series can be centered at any point \( a \), when \( a = 0 \) it specifically becomes the Maclaurin series.
The general formula for a Taylor series expansion of a function \( f \) around the point \( a \) is:
This series is particularly helpful in determining behavior near certain points. For example, the exercise demonstrates how using a Taylor or Maclaurin series reveals the order of a zero by highlighting the first non-zero term in its expansion.
The general formula for a Taylor series expansion of a function \( f \) around the point \( a \) is:
- \( f(a) + \frac{f'(a)}{1!}(z-a) + \frac{f''(a)}{2!}(z-a)^2 + \ldots \)
This series is particularly helpful in determining behavior near certain points. For example, the exercise demonstrates how using a Taylor or Maclaurin series reveals the order of a zero by highlighting the first non-zero term in its expansion.
Order of Zero
In mathematics, the term 'order of zero' refers to the smallest power of \( z \) that appears in a function’s expansion, which is not zero. This provides insight into how many times a function 'touches' or 'crosses' the zero axes at a particular point.
To determine the order of zero for a function like \( f(z) = z - \sin z \), you must analyze its series expansion around the point \( z = 0 \). As seen in the previous steps, if the series expansion begins with \( \frac{z^3}{3!} \), it means the function has a zero of order 3 at that point.
The significance of the order of zero can be seen in:
To determine the order of zero for a function like \( f(z) = z - \sin z \), you must analyze its series expansion around the point \( z = 0 \). As seen in the previous steps, if the series expansion begins with \( \frac{z^3}{3!} \), it means the function has a zero of order 3 at that point.
The significance of the order of zero can be seen in:
- Identifying critical points where graph behavior changes.
- Solving equations where such zeros could dramatically alter outcomes.
- Determining multiplicity in solutions, which can have implications in stability analysis in engineering and physics.
Other exercises in this chapter
Problem 11
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}
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Use the theory of residues to compute the inverse Laplace transform \(\mathscr{L}^{-1}\\{F(s)\\}\) for the given function \(F(s)\). $$ \frac{s}{\left(s^{2}+1\ri
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Expand \(f(z)=\frac{1}{z(z-3)}\) in a Laurent series valid for the indicated annular domain. $$ 1
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