Suppose you know the Maclaurin series for f and that it converges to f(x) for |x|<1. How do you find the Maclaurin series for f(x²) and where does it converge?

Verified step by step guidance

1. Recall that the Maclaurin series for a function \(f(x)\) is given by \(f(x) = \sum_{n=0}^{\infty} a_n x^n\), where \(a_n = \frac{f^{(n)}(0)}{n!}\) and the series converges for \(|x| < 1\).
2. To find the Maclaurin series for \(f(x^2)\), substitute \(x^2\) in place of \(x\) in the original series. This gives \(f(x^2) = \sum_{n=0}^{\infty} a_n (x^2)^n\).
3. Simplify the powers: \((x^2)^n = x^{2n}\), so the series becomes \(f(x^2) = \sum_{n=0}^{\infty} a_n x^{2n}\).
4. Analyze the radius of convergence for the new series. Since the original series converges for \(|x| < 1\), the substitution \(x^2\) means the new series converges when \(|x^2| < 1\), which simplifies to \(|x| < 1\).
5. Therefore, the Maclaurin series for \(f(x^2)\) is \(\sum_{n=0}^{\infty} a_n x^{2n}\), and it converges for \(|x| < 1\).

Key Concepts

Maclaurin Series

A Maclaurin series is a Taylor series expansion of a function about zero, expressed as an infinite sum of terms involving derivatives of the function at zero. It represents the function as f(x) = Σ (f⁽ⁿ⁾(0)/n!) xⁿ, valid within its radius of convergence.

Substitution in Power Series

To find the Maclaurin series for f(x²), substitute x² in place of x in the original series. This changes each term xⁿ to (x²)ⁿ = x^(2n), effectively altering the powers and coefficients but preserving the series structure.

Radius of Convergence

The radius of convergence defines the interval where the power series converges to the function. For f(x²), the convergence depends on |x²| < 1, which implies |x| < 1, so the radius of convergence remains the same as the original series.