Shifting power series If the power series f(x)=∑ cₖ xᵏ has an interval of convergence of |x|<R, what is the interval of convergence of the power series for f(x−a), where a ≠ 0 is a real number?

Verified step by step guidance

1. Recall that the original power series is given by \(f(x) = \sum_{k=0}^{\infty} c_k x^k\) and it converges for \(|x| < R\) where \(R\) is the radius of convergence.
2. When we consider the shifted function \(f(x - a)\), we rewrite the series as \(f(x - a) = \sum_{k=0}^{\infty} c_k (x - a)^k\).
3. The key observation is that the power series is now centered at \(x = a\) instead of \(x = 0\), because the variable inside the series is \((x - a)\).
4. The radius of convergence remains the same, \(R\), but the interval shifts accordingly. So the new interval of convergence is all \(x\) such that \(|x - a| < R\).
5. Therefore, the interval of convergence for \(f(x - a)\) is the open interval \((a - R, a + R)\), which is the original interval shifted by \(a\).

Key Concepts

Power Series and Interval of Convergence

A power series is an infinite sum of the form ∑ cₖ (x - x₀)ᵏ, centered at x₀, where cₖ are coefficients. The interval of convergence is the set of x-values for which the series converges. For a series centered at 0, the interval is |x| < R, where R is the radius of convergence.

Effect of Shifting the Center of a Power Series

Shifting the argument of a power series from f(x) to f(x - a) changes the center of the series from 0 to a. This means the new power series is centered at x = a, and the interval of convergence shifts accordingly, typically becoming |x - a| < R, preserving the radius of convergence.

Radius of Convergence and Its Invariance Under Translation

The radius of convergence R depends on the coefficients cₖ and remains unchanged when the series is shifted horizontally. Translation of the center does not affect R, only the center point changes, so the interval of convergence moves but retains the same radius.