Question

Find the Taylor series expansion about the origin of the function \(f(z)\) defined by $$ f(z)=\sum_{r=1}^{\infty}(-1)^{r+1} \sin \left(\frac{p z}{r}\right) $$ where \(p\) is a constant. Hence verify that \(f(z)\) is a convergent series for all \(z\).

Short answer

The Taylor series of \(f(z)\) about the origin is \( f(z) = \left( \sum_{r=1}^{\infty}(-1)^{r+1} \frac{p}{r} \right) z - \left( \sum_{r=1}^{\infinity}(-1)^{r+1} \frac{p^3}{r^3 3!} \right) z^3 + \cdots \).The series is convergent for all z.

Step-by-step solution

- Understand the Function

The function given is \( f(z)=\sum_{r=1}^{\infty}(-1)^{r+1} \sin \left(\frac{p z}{r}\right) \) and will be expanded using the Taylor series about the origin (z=0).

- Taylor Series Expansion of Sin Function

Recall the Taylor series expansion for \( \sin(x) \) around the origin, which is \[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]

- Apply the Series to Function

Substitute \( x = \frac{p z}{r} \) into the Taylor series: \[ \sin \left(\frac{p z}{r}\right) = \frac{p z}{r} - \frac{\left(\frac{p z}{r}\right)^3}{3!} + \frac{\left(\frac{p z}{r}\right)^5}{5!} - \cdots \]

- Substitute Back into Function

Expand \( f(z) \) by substituting the series expansion back: \[ f(z) = \sum_{r=1}^{\infty}(-1)^{r+1} \left( \frac{p z}{r} - \frac{1}{3!} \left(\frac{p z}{r}\right)^3 + \frac{1}{5!} \left(\frac{p z}{r}\right)^5 - \cdots \right) \]

- Rearrange the Series

Combine like powers of \( z \): \[ f(z) = \left( \sum_{r=1}^{\infty}(-1)^{r+1} \frac{p}{r} \right) z - \left( \sum_{r=1}^{\infty}(-1)^{r+1} \frac{p^3}{r^3 3!} \right) z^3 + \cdots \]

- Check Convergence

To verify convergence, observe that the terms \( \sum_{r=1}^{\infty}(-1)^{r+1} \frac{1}{r^k} \) converge for all positive integers \( k \). This follows from the alternating series test and the fact that \( \frac{1}{r^k} \to 0 \) as \( r \to \infty \).

Key concepts

Convergent Series

A convergent series is a series whose terms approach a specific value as the number of terms goes to infinity. In simpler terms, if you keep adding terms of the series, you get closer and closer to a certain number.

For instance, consider the series \(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...\). Each term is half of the previous term. As you add more terms, the sum gets closer to 1, even though it may never actually reach 1. This series is called convergent because it sums up to a finite number.

In our exercise, the function \( f(z) = \sum_{r=1}^{\infty}(-1)^{r+1} \left(\frac{p z}{r} \right)\) can be considered a convergent series for all values of \(z\). We verify this with the alternating series test.

This property is crucial, as it ensures that the function has a meaningful Taylor series expansion. This implies that we can trust the expansion to give a good approximation of the function for all values of \(z\).

Alternating Series Test

The alternating series test is a handy tool for checking if an infinite series converges. Here's how it works:

• The series must have terms that alternate in sign.
• The absolute value of the terms must decrease monotonically (each term is smaller than the one before).
• The limit of the terms must be zero as the number of terms goes to infinity.

Take our series \(\sum_{r=1}^{\infty}(-1)^{r+1}\frac{1}{r^k}\) for example. It alternates in sign because of the \((-1)^{r+1}\) part. The terms \(\frac{1}{r^k}\) decrease monotonically and approach zero as \( r \) goes to infinity for any positive integer \( k \) .

So, using the alternating series test, we can conclude that the series \( \sum_{r=1}^{\infty}(-1)^{r+1}\frac{1}{r^k} \) converges, meaning it sums up to a finite value. This confirms the convergence of each term in the Taylor series expansion of \( f(z) \).

Sin Function Expansion

The Taylor series expansion of the sine function around the origin is a commonly used series in calculus. The Taylor series for \( \sin(x) \) is given by:
\[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]
This series involves all the odd-powered terms of \(x\) and alternates between positive and negative. The factorials in the denominators grow very quickly, which ensures that higher-order terms contribute less to the sum, making the series converge faster.

In our problem, we substitute \( x = \frac{p z}{r} \) into the sine Taylor series:
\[ \sin \left( \frac{p z}{r} \right) = \frac{p z}{r} - \frac{1}{3!} \left( \frac{p z}{r} \right)^3 + \frac{1}{5!} \left( \frac{p z}{r} \right)^5 - \cdots \]
By substituting this back into our function \( f(z) \), we get:
\[ f(z) = \sum_{r=1}^{\infty}(-1)^{r+1} \left( \frac{p z}{r} - \frac{1}{3!} \left( \frac{p z}{r} \right)^3 + \frac{1}{5!} \left( \ \frac{p z}{r} \right)^5 - \cdots \right) \]
This expansion helps us express the original complex series in a more manageable form, making it easier to analyze, especially regarding its convergence.