Question

Suppose you know the Maclaurin series for \(f\) and it converges for \(|x| < 1 .\) How do you find the Maclaurin series for \(f\left(x^{2}\right)\) and where does it converge?

Short answer

Answer: The Maclaurin series for the function \(f(x^2)\) is: \(f(x^2) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{2n}\) The domain of convergence for the Maclaurin series is \(0 \leq x < 1\).

Step-by-step solution

Recall the Maclaurin series definition

A Maclaurin series is a Taylor series expansion of a function \(f(x)\) around the point \(x = 0\). It is given by the formula: \begin{equation} f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{n} \end{equation} where \(f^{(n)}(0)\) denotes the \(n\)-th derivative of \(f\) evaluated at \(0\). The first step is to start from the given Maclaurin series of the function \(f(x)\) and then perform a substitution for \(x^2\) to find the Maclaurin series for the function \(f(x^2)\).

Substitute \(x^2\) into the Maclaurin series

To find the Maclaurin series for \(f(x^2)\), we will substitute \(x^2\) into the Maclaurin series of \(f(x)\). That is, \begin{equation} f(x^2) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x^2)^{n} \end{equation}

Simplify the new series

Now, simplify the series by combining the exponents of \(x\): \begin{equation} f(x^2) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{2n} \end{equation} This is the Maclaurin series for the function \(f(x^2)\).

Determine the convergence of the new series

Since we know that the original series converges for \(|x| < 1\), we can determine the convergence of the new series by analyzing the domain for \(x^2\). In this case, we have: \begin{equation} |x^2| < 1 \end{equation} Since \(x^2\) is always non-negative, we can re-write this as: \begin{equation} 0 \leq x^2 < 1 \end{equation} Taking square roots of both sides, we have: \begin{equation} 0 \leq x < 1 \end{equation} This means that the Maclaurin series for \(f(x^2)\) converges for \(0 \leq x < 1\). In conclusion, the Maclaurin series for the function \(f(x^2)\) is: \begin{equation} f(x^2) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{2n} \end{equation} and it converges for \(0 \leq x < 1\).