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Determine the interval of convergence of ∑ xn / n! and show that it satisfies y′ = x + y

by RoRi

Consider the function f(x) defined by the power series

    \[ f(x) = \sum_{n=2}^{\infty} \frac{x^n}{n!}. \]

Determine the interval of convergence for f(x) and show that it satisfies the differential equation

    \[ y' = x + y. \]

(We might notice that this is almost the power series expansion for the exponential function e^x and deduce the interval of convergence and the differential equation from properties of the exponential that we already know. We can do it from scratch just as easily though.)

First, we apply the ratio test

    \begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{n+1}}{(n+1)!} \right) \left( \frac{n!}{x^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x}{n+1} \\[9pt] &= 0. \end{align*}

Hence, the series converges for all x. Next, we take a derivative

    \begin{align*} y &= \sum_{n=2}^{\infty} \frac{x^n}{n!} \\[9pt] y' &= \sum_{n=2}^{\infty} \frac{nx^{n-1}}{n!} \\[9pt] &= \sum_{n=2}^{\infty} \frac{x^{n-1}}{(n-1)!} \\[9pt] &= \sum_{n=1}^{\infty} \frac{x^n}{n!}. \end{align*}

Then we have

    \begin{align*} x + y &= x + \sum_{n=2}^{\infty} \frac{x^n}{n!} \\[9pt] &= \sum_{n=1}^{\infty} \frac{x^n}{n!} \\[9pt] &= y'. \end{align*}

Now, to compute the sum we can solve the given differential equation

    \[ y' = x + y \quad \implies \quad y' - y = x. \]

This is a first order linear differential equation of the form y' + P(x)y = Q(x) with P(x) = -1 and Q(x) = x. We also know that f(0) = 0; therefore, this equation has a unique solution satisfying the given initial condition which is given by

    \[ y = be^{-A(x)} + e^{-A(x)} \int_a^x Q(t) e^{A(t)} \, dt. \]

Where a = b = 0 and

    \[ A(x) = \int_0^x P(t) \, dt = - \int_0^x dt = -x. \]

Therefore, we have

    \begin{align*} f(x) &= e^{x} \left( \int_0^x t e^{-t} \, dt \\[9pt] &= e^x \left( -te^{-t} \Bigr \rvert_0^x + \int_0^x e^{-t} \, dt \right) \\[9pt] &= e^x \left( -xe^{-x} - e^{-x} + 1 \right) \\[9pt] &= e^x - x - 1. \end{align*}

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