Question

Using the information given in (7), write a Maclaurin series for the given function \(f(t)\). Determine the radius of convergence of the series. $$ f(t)=\frac{1}{1-t^{2}} $$

Short answer

Question: Determine the Maclaurin series representation and the radius of convergence for the function \(f(t) = \frac{1}{1-t^2}\). Answer: The Maclaurin series representation for the function \(f(t)\) is given by \(f(t) = 1 + 2t^2 + \cdots\). The radius of convergence for the series is \(R = \infty\).

Step-by-step solution

Compute the derivatives of the function

To find the Maclaurin series representation, we first need to compute the derivatives of the given function \(f(t)\). We have: $$ f(t)=\frac{1}{1-t^{2}} $$ Calculate the first few derivatives and evaluate them at \(t=0\): $$ f'(t) = \frac{d}{dt} \left(\frac{1}{1-t^2}\right) = \frac{2t}{(1-t^2)^2} \Rightarrow f'(0) = 0 $$ $$ f''(t) = \frac{d^2}{dt^2} \left(\frac{1}{1-t^2}\right) = \frac{2(1+t^2)}{(1-t^2)^3} \Rightarrow f''(0) = 2 $$ Continue this process to calculate higher derivatives.

Write the Maclaurin series representation

Use the derivatives evaluated at \(t=0\) to write the Maclaurin series representation of the function: $$ f(t) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} t^n $$ Since we have calculated the first few derivatives and evaluated them at \(t=0\) in step 1, we can write the series representation as: $$ f(t) = \frac{1}{1} - \frac{0}{1!}t + \frac{2}{2!}t^2 - \cdots = 1 + 2t^2 + \cdots $$

Determine the radius of convergence

Now, we will use the ratio test to determine the radius of convergence \(R\) of the series. Consider the absolute value of the ratio of consecutive terms: $$ \lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{\frac{2}{(n+1)!}t^{n+1}}{\frac{2}{n!}t^n} = \lim_{n\to\infty} \frac{t}{n+1} $$ For the series to converge, the limit should be less than 1: $$ \lim_{n\to\infty} \frac{t}{n+1} < 1 $$ As \(n \to \infty\), we see that for any finite \(t\), the limit will be less than 1 and the series converges for all \(t \in \mathbb{R}\), so the radius of convergence is \(R = \infty\).

Key concepts

Radius of Convergence

When dealing with power series, one crucial aspect to understand is the radius of convergence. It tells us the range of values for which the series converges, meaning where it represents the function accurately.

The radius of convergence is calculated through methods such as the ratio test. In the context of the Maclaurin series for the function \( f(t) = 1/(1 - t^2) \), it can seem daunting to find the limit of a fraction involving factorials and powers. A simplification is to realize the higher the power of \(n\), the less impact the variable \(t\) has on the result. This leads to the realization that for any finite value of \(t\), the limit of the ratio test approaches zero, indicating the series converges for all real values of \(t\). Thus, we can determine the radius of convergence to be infinite, which means the Maclaurin series for \(f(t)\) converges for all real values of \(t\).

Understanding the implications of a radius of convergence is not only essential for theoretical mathematics but also for practical applications such as engineering and physics where series approximations are often used.

Derivative Computation

To find the Maclaurin series of a function, we start with derivative computation. This process involves finding the derivatives of the function at \(t=0\) up to the desired degree. These derivatives are what we then use to build the power series.

For the function \(f(t)\), we calculated derivatives such as \(f'(t)\) and \(f''(t)\) and evaluated them at \(t=0\). In exercises like this, the pattern in the derivatives can often be spotted, which simplifies the process. However, computing these derivatives can show a level of complexity when dealing with higher-powered terms and factorials. To ensure that students can follow along and don't get lost in the process, instructors should emphasize the pattern recognition in the derivatives and how these patterns contribute to the formation of the power series representation. By understanding the relationship between a function and its derivatives, students will have a more robust framework for tackling calculus problems.

Power Series Representation

The power series representation of a function is like a treasure map, guiding us to the function's value for a given input within the radius of convergence. By expanding functions into an infinite sum of terms, power series allow us to approximate functions with polynomials.

In this case, the Maclaurin series is a type of power series centered at \(t=0\). It is effectively the ‘DNA’ of the function \(f(t)\), enabling us to recreate it through its derivatives at the origin. Representing \(f(t) = 1/(1 - t^2)\) in its series form, we have an infinite set of terms: \(1 + 2t^2 + 2t^4 + \)… and so on.

It's important to recognize that each term in this series consists of derivatives evaluated at \(t=0\) divided by factorial terms, then multiplied by the variable to the power of the term's degree. Presenting this concept in steps and giving clear examples help students understand how the series builds up, term by term, and demonstrates the role of derivatives in its construction — ultimately demystifying the concept and building confidence in their mathematical understanding.