Problem 32 - Advanced Engineering Mathematics

Question

(a) \(R=1,\) which is the distance from the origin to \(z=-1\) (b) Using Taylor's Theorem [or integrating the series for \(1 /(1+z)]\) we obtain for \(R=1\) $$\operatorname{Ln}(1+z)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} z^{k}$$ (c) By replacing \(z\) in part $$\operatorname{Ln}(1+z)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} z^{k}$$ (c) By replacing \(z\) in part (b) by \(-z\) we obtain for \(R=1\) $$\operatorname{Ln}(1-z)=-\sum_{k=0}^{\infty} \frac{z^{k}}{k}$$ (d) One way of obtaining the Maclaurin series for \(\operatorname{Ln}\left(\frac{1+z}{1-z}\right)\) is to use Taylor's Theorem. Alternatively, let us write $$\operatorname{Ln}\left(\frac{1+z}{1-z}\right)=\operatorname{Ln}(1+z)-L(1-z)$$ and subtract the series in parts (b) and (c). This gives for the common circle of convergence \(|z|=1\) $$\operatorname{Ln}\left(\frac{1+z}{1-z}\right)=2 z+\frac{2}{3} z^{3}+\frac{2}{5} z^{5}+\frac{2}{7} z^{7}+\cdots=2 \sum_{k=0}^{\infty} \frac{1}{(2 k+1)} z^{2 k+1}$$ But recall that in general \(\operatorname{Ln}\left(z_{1} / z_{2}\right) \neq \operatorname{Ln} z_{1}-\operatorname{Ln} z_{2}\) since \(\operatorname{Ln} z_{1}\) and \(\operatorname{Ln} z_{2}\) could differ by a constant multiple of \(i .\) That is, \(\operatorname{Ln} z_{1}-\operatorname{Ln} z_{2}=C i\) for some \(C .\) So $$\operatorname{Ln}\left(\frac{1+z}{1-z}\right)=\operatorname{Ln}(1+z)-\operatorname{Ln}(1-z)-C i$$ When \(z=0\) we obtain \(\operatorname{Ln} 1=\operatorname{Ln} 1-\operatorname{Ln} 1-C i .\) since \(\operatorname{Ln} 1=0\) we get \(C=0\)

Step-by-Step Solution

Answer

The Maclaurin series for \(\operatorname{Ln}\left(\frac{1+z}{1-z}\right)\) is \(2z + \frac{2}{3}z^3 + \frac{2}{5}z^5 + \cdots\).

Step 1: Identify the Series Expansions

We are given the series expansions for \(\operatorname{Ln}(1+z)\) and its transformation \(\operatorname{Ln}(1-z)\). These series are derived using Taylor's Theorem: \[\operatorname{Ln}(1+z) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} z^{k}\] \[\operatorname{Ln}(1-z) = -\sum_{k=0}^{\infty} \frac{z^{k}}{k} \] where the substitution from part (b) by \(-z\) results in the series in part (c).

Step 2: Define Series for \(\operatorname{Ln}\left(\frac{1+z}{1-z}\right)\)

To find \(\operatorname{Ln}\left(\frac{1+z}{1-z}\right)\), we subtract the series expansions from part (b) and part (c). This results in the expression: \[\operatorname{Ln}(1+z) - \operatorname{Ln}(1-z) = 2 \sum_{k=0}^{\infty} \frac{1}{(2k+1)} z^{2k+1} \] which is analogously simplified to: \[2z + \frac{2}{3}z^3 + \frac{2}{5}z^5 + \cdots\]

Step 3: Consider Potential Constant Term

While subtracting logarithmic series, we must consider potential constants. We identify that the subtraction \(\operatorname{Ln}(1+z) - \operatorname{Ln}(1-z)\) may include a constant term \(Ci\) because logarithmic terms differ by multiples of \(i\). In this context: \[\operatorname{Ln}\left(\frac{1+z}{1-z}\right) = \operatorname{Ln}(1+z) - \operatorname{Ln}(1-z) - Ci\]

Step 4: Determine Constant C

At \(z=0\), we evaluate the equation to ensure no constants remain: \[\operatorname{Ln}1 = \operatorname{Ln}1 - \operatorname{Ln}1 - Ci\] gives us \(0 = 0 - 0 - Ci\), which implies \(C=0\). Therefore, the result follows directly from the series.

Key Concepts

Series Expansions

Series expansions are mathematical tools used for approximating complex functions by simpler ones—typically infinite sums. In this exercise, we primarily deal with Taylor series, which are a type of series expansion. Taylor's Theorem allows us to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. This makes it easier to understand and work with functions that are otherwise difficult to compute.

For logarithmic functions like \(\operatorname{Ln}(1+z)\), a series expansion helps in representing the logarithm as a sum that is easier to manipulate and evaluate, particularly for values of \(|z| < 1\).

• Starting with Taylor's Theorem, a function is expanded around a point (usually 0 for Maclaurin series).
• The series summands involve the derivatives of the function.
• For \(\operatorname{Ln}(1+z)\), the expansion is: \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} z^{k}\).

Series expansions are not only useful for simplifying calculations but also provide insights into the behavior of functions within certain radii of convergence.

Maclaurin Series

Maclaurin series are a special case of Taylor series, expanded around zero—a common choice for simplicity and practicality. When expanding a function using a Maclaurin series, it aids in capturing the behavior of the function near the origin.

For \(\operatorname{Ln}(1+z)\), which has its importance in computing log functions, a Maclaurin series offers a simpler and more computable expression:

• The expansion starts at zero (\(z=0\)) and sums infinitely many terms.
• For logarithmic functions, we follow the derivatives of \(\operatorname{Ln}(1+z)\) as a series: \(-z+\frac{z^2}{2}-\frac{z^3}{3}+\ldots\).
• This expansion is valid within a radius of convergence, typically \(|z| < 1\).

Maclaurin series make complex functions easier to compute and approximate over a small range, especially in contexts like calculus and numerical analysis.

Logarithmic Functions

Logarithmic functions are an essential part of mathematics, often appearing in problems involving exponential growth or decay, such as finance, biology, and physics.

A logarithmic function, specifically the natural logarithm \(\operatorname{Ln}\), has properties that make it unique among other functions:

• It is the inverse of the exponential function, meaning \(\operatorname{Ln}(e^x) = x\).
• It exhibits a unique curvature; increasing slowly and never reaches an asymptote.
• Logs can convert multiplication into addition, a handy property for simplifying expressions.

When expanded using series such as Taylor or Maclaurin, logarithmic functions provide deeper insights into their behaviors and values over specific ranges. Understanding these expansions broadens our ability to handle real-world problems involving logarithms.