Question

Exponential function In Section 11.3, we show that the power series for the exponential function centered at 0 is

eˣ = ∑ₖ₌₀∞ (xᵏ)/k!, for −∞ < x < ∞

Use the methods of this section to find the power series centered at 0 for the following functions. Give the interval of convergence for the resulting series.

f(x) = x²eˣ

Accepted Answer

Recall the power series expansion for the exponential function centered at 0:  
$$e^{x} = \sum_{k=0}^{\infty} \frac{x^{k}}{k!} To $$find the power series for $$f(x) = x^{2} e^{x}$$, multiply the entire series for $$e^{x} by$$ $$x^{2}$$:  
$$f(x) = x^{2} \sum_{k=0}^{\infty} \frac{x^{k}}{k!} = \sum_{k=0}^{\infty} \frac{x^{k+2}}{k!}$$ Rewrite the series to express it in a standard power series form $$\sum_{n=0}^{\infty} a_{n} x^{n} by $$changing the index of summation. Let $$n = k + 2$$, so when $$k=0$$, $$n=2. $$Thus,  
$$f(x) = \sum_{n=2}^{\infty} \frac{x^{n}}{(n-2)!}$$ Identify the coefficients $$a_{n} of $$the power series:  
For $$n \geq 2$$, $$a_{n} = \frac{1}{(n-2)!}$$, and for $$n \u003c 2$$, $$a_{n} = 0$$ since the series starts at $$n=2. $$Determine the interval of convergence. Since the original series for $$e^{x}$$ converges for all real $$x (i.e$$., $$(-\infty, \infty)$$), multiplying by $$x^{2}$$ does not change the radius or interval of convergence. Therefore, the interval of convergence for $$f(x) is $$also $$(-\infty, \infty).$$

Verified video answer for a similar problem:

Key Concepts

Power Series Representation

Manipulation of Power Series

Interval of Convergence