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Compute the coefficients of a given power series

Consider the power series

    \[ f(x) = \sum_{n=0}^{\infty} a_n x^n \]

with coefficients determined by the identity

    \[ e^{-2x} = \sum_{n=0}^{\infty} (2a_n + (n+1)a_{n+1})x^n. \]

Compute the coefficients a_1, a_2, a_3 and determine the sum of the series.


We know the power-series expansion for e^{-2x} is given by

    \[ e^{-2x} = 1 + (-2x) + \frac{(-2x)^2}{2!} + \frac{(-2x)^3}{3!} + \cdots. \]

Starting with a_0= 1 and the given identity we can compute the coefficients a_1, a_2, a_3 by equating the coefficients of like powers of x,

    \begin{align*}  2a_0 + a_1 &= 1 & \implies && a_1 &= -1 \\[9pt]  2a_1 + 2a_2 &= -2 & \implies && a_2 &= 0 \\[9pt]  2a_2 + 3a_3 &= 2 & \implies && a_3 &= \frac{2}{3}. \end{align*}

Then from the identity for the coefficients (and noting that the series converges absolutely for all real x so we may split the sum into separate sums without worry),

    \begin{align*}  e^{-2x} &= \sum_{n=0}^{\infty} (2a_n + (n+1)a_{n+1})x^n \\[9pt]  &= 2 \sum_{n=0}^{\infty} a_n x^n + \sum_{n=0}^{\infty} (n+1)a_{n+1} x^n \\[9pt]  &= 2 f(x) + f'(x). \end{align*}

This is a first order linear differential equation of the form y' + P(x) y = Q(x). Furthermore, the initial condition a_0 = 1 implies that y = 1 when x = 0. Therefore, the solution is

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

where

    \[ a = 0, \quad b= 1, \quad A(x) = \int_0^x P(t) \,dt = 2x. \]

So, we have

    \begin{align*}   f(x) &= e^{-2x} +  e^{-2x} \int_0^x e^{-2t} e^{2t} \, dt \\[9pt]  &= e^{-2x} + e^{-2x} \int_0^x \, dt \\[9pt]  &= (x+1)e^{-2x}. \end{align*}

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