Problem 5
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?
Step-by-Step Solution
Verified 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 \(0 \leq x < 1\).
1Step 1: 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 \(f(x^2)\).
2Step 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}
3Step 3: 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 \(f(x^2)\).
4Step 4: 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 \(0 \leq x < 1\).
Other exercises in this chapter
Problem 4
In general, how many terms do the Taylor polynomials \(p_{2}\) and \(p_{3}\) have in common?
View solution Problem 5
If \(f(x)=\sum_{k=0}^{\infty} c_{k} x^{k}\) and the series converges for \(|x|
View solution Problem 5
Do the interval and radius of convergence of a power series change when the series is differentiated or integrated? Explain.
View solution Problem 5
How is the remainder in a Taylor polynomial defined?
View solution