Question

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges. $$\sum_{0}^{\infty}(\sin x)^{n}(-1)^{n} 2^{n}$$

Short answer

The series converges when \(-\frac{1}{2} < \sin x < \frac{1}{2}\).

Step-by-step solution

Define the series and identify the variable transformation

The series given is \(\sum_{n=0}^{\infty} (\sin x)^{n}(-1)^{n} 2^{n}\). Notice it includes a term in the form of \(\sin x\) raised to the power of \(n\). The goal is to transform this series into a power series by making an appropriate change of variable. Let's set \(y = -2\sin x\).

Rewrite the series using the new variable

Substitute \(y = -2\sin x\) into the original series. The series becomes \(\sum_{n=0}^{\infty} (-1)^{n} 2^{n} (\sin x)^{n} = \sum_{n=0}^{\infty} (\sin x)^{n}(-2)^{n} = \sum_{n=0}^{\infty} y^{n} \).

Identify the convergence criteria for the power series

A power series \(\sum_{n=0}^{\infty} y^{n}\) converges when \(|y| < 1 \). Therefore, the series \(\sum_{n=0}^{\infty}y^{n}\) converges if \(-1 < y < 1\).

Transform back to the original variable

Substitute back \(y = -2\sin x\). So the series converges when \(-1 < -2\sin x < 1\). This simplifies to \(-\frac{1}{2} < \sin x < \frac{1}{2}\).

Determine the final interval of convergence

The series will converge for \(-\frac{1}{2} < \sin x < \frac{1}{2}\). This represents the interval where \(\sin x\) takes values that preserve the convergence of the power series.

Key concepts

Power Series Mastery

Power series are extremely powerful in mathematics because they allow us to represent complex functions as sums of simpler polynomial terms. A power series is generally written as \(\sum_{n=0}^{\infty} a_n (x-c)^n\), where \(a_n\) are coefficients, \(x\) is the variable, and \(c\) is the center of the series. The series converges within a certain radius around \(c\), called the radius of convergence.
Power series are used extensively in calculus for functions like exponential, trigonometric, and logarithmic functions. Knowing how to transform a given series into a power series can make it easier to analyze and solve. This enables you to understand their convergence and properties better.
Key points include:

• Identifying the coefficients \(a_n\).
• Determining the center \(c\).
• Finding the interval of convergence.

These properties simplify complex mathematical operations significantly.

Variable Substitution for Series Transformation

Variable substitution is a critical tool to transform non-power series into power series. By correctly choosing a new variable, we can simplify complex expressions. A common technique is to find a substitution that converts the terms of the original series into a more familiar power series form.
In the given problem, the series \(\sum_{0}^{\infty}(\sin x)^{n}(-1)^{n} 2^{n}\) was transformed using the substitution \( y = -2 \sin(x)\).

This simplifies the series to \(\sum_{n=0}^{\infty} y^{n}\), which is a standard geometric series. The choice of substitution is critical because it directly impacts the convergence criteria and simplifies the problem substantially.

• Select an appropriate variable (here \(y\)).
• Rewrite the original series.

This allows us to focus on convergence in a simpler form.

Decoding Convergence Criteria

Understanding convergence criteria is essential for determining where a series converges. For a power series \(\sum_{n=0}^{\infty} y^{n}\), it converges when \(|y| < 1\).
This fundamental criterion applies to geometric series and more complex power series. For the transformed series in the exercise, we determined the series converged if \(|y| < 1\).

By substituting back the original variable, we found the range of \(|\sin x|\) which ensures convergence. Here, solving \text{-}1 < y < 1\ for \( y = -2 \sin(x)\) leads to the interval \(-\frac{1}{2} < \sin x < \frac{1}{2}\).
This simple rule extends to more complex series, taking

• The absolute value of the new variable.
• Ensuring it lies between -1 and 1.

Understanding these criteria allows us to find where our original series converges.

Unraveling Mathematical Series

Mathematical series are sums of terms that form an important part of calculus and higher mathematics. They can represent functions, solve equations, and model real-world situations. There are various kinds of series, such as arithmetic, geometric, and power series. In our exercise, we worked with a geometric series that became a power series.
These series can often be tough to tackle directly. However, transforming them into a common power series form simplifies their analysis. Key concepts include:

• Understanding the definitions of different series.
• Using substitution to simplify series.
• Applying convergence criteria to determine the valid range.

Mathematicians have developed these tools to break down complex series into manageable parts. Hence, transforming non-power series to power series helps simplify and solve many mathematical problems.