Question

Suppose there's an integer \(M\) such that \(b_{m} \neq 0\) for \(m \geq M,\) and $$ \lim _{m \rightarrow \infty}\left|\frac{b_{m+1}}{b_{m}}\right|=L $$ where \(0 \leq L \leq \infty .\) Show that the radius of convergence of $$ \sum_{m=0}^{\infty} b_{m}\left(x-x_{0}\right)^{2 m} $$ is \(R=1 / \sqrt{L},\) which is interpreted to mean that \(R=0\) if \(L=\infty\) or \(R=\infty\) if \(L=0 .\) HINT: Apply Theorem 7.1.3 to the series \(\sum_{m=0}^{\infty} b_{m} z^{m}\) and then let \(z=\left(x-x_{0}\right)^{2}\).

Short answer

Answer: The radius of convergence (R) for the power series is given by \(R = \frac{1}{\sqrt{L}}\).

Step-by-step solution

The Ratio Test (Theorem 7.1.3)

According to Theorem 7.1.3, we need to find the limit of \(|\frac{b_{m+1}z^{m+1}}{b_mz^m}|\) as \(m \to \infty\). We are given that the limit of \(|\frac{b_{m+1}}{b_m}| = L\). So, we have: $$ \lim_{m \to \infty} \left|\frac{b_{m+1}z^{m+1}}{b_mz^m}\right| = \lim_{m \to \infty} \left|\frac{b_{m+1}}{b_m}\right| \cdot \left|z\right| = L |z| $$

The Radius of Convergence (R)

By the definition of the Radius of Convergence, the series converges if \(L|z| < 1\) and diverges if \(L|z| > 1\). Therefore, the radius of convergence is given by: $$ R = \frac{1}{L} $$

Substituting \(z\) with \((x - x_0)^2\)

Now, let's substitute the given expression of \(z = (x - x_0)^2\) into our radius of convergence formula: $$ R = \frac{1}{L} = \frac{1}{\sqrt{L}\sqrt{(x -x_0)^2}} $$

Final Radius of Convergence

Simplifying the expression, we get the final radius of convergence: $$ R = \frac{1}{\sqrt{L}} $$ which is interpreted to mean that \(R = 0\) if \(L = \infty\) or \(R = \infty\) if \(L = 0\).

Key concepts

Ratio Test

The Ratio Test is a handy tool for determining the convergence or divergence of a series. Here’s how it works: given a series \(\sum a_m\), calculate the limit of the absolute value of the ratio of consecutive terms. Specifically, you find:\[\lim_{m \to \infty} \left| \frac{a_{m+1}}{a_m} \right|\]

• If this limit is less than 1, the series converges.
• If the limit is greater than 1, the series diverges.
• If the limit equals 1, the test is inconclusive.

For power series, this test helps in finding the radius of convergence by providing insights on when the series behaves nicely.

Power Series

A power series is a series of the form \(\sum_{m=0}^\infty c_m(x-x_0)^m\). This structure is incredibly flexible and powerful in mathematics, especially when dealing with functions.

• Parameter \( x_0 \): Known as the center of the series.
• Coefficients \( c_m \): Determine the role each term plays in the series.

A key property of power series is their radius of convergence, \(R\). This tells us over which values of \(x\) the series converges. It’s crucial in analyzing the behavior and applicability of the series for solving problems.

Limits

In calculus, limits help us understand how a function behaves as it approaches a certain point. When evaluating the convergence of a series, limits are indispensable.

• Limits in Series: In the context of the ratio test, you calculate the limit of the ratio of consecutive terms.
• Efficiency: This limit gives crucial information about whether the series converges or diverges.

Think of limits as the microscope that allows you to zoom into how series terms relate to each other, helping define convergence under specific conditions.

Differential Equations

Differential equations involve functions and their derivatives and are used to model a plethora of real-world phenomena. Although not directly part of the series discussed in the exercise, understanding power series' convergence helps solve differential equations.

• Application: Power series are used to find solutions to differential equations, especially when simple algebraic methods are inefficient.
• Series Solutions: By knowing the radius of convergence, you can ensure solutions are valid within specific intervals.

The use of power series expands the toolbox for mathematicians and engineers dealing with complex differential equations.