Finding a recursive relation from a differential equation.

Question

We consider the following differential equation: $$(1-x)\psi'(x) - 1 = 0$$ We then define: $\psi(x) = \sum_{n = 0}^{\infty}a_nx^n$.(power series)

Plugging it into the formula we get to this step: $$\sum_{n=0}^{\infty}a_nn(x^{n-1}-x^{n}) - 1= 0$$

How can we then identify the recursive relation between the summands thus determine $a_{n+1}$ ?

Answer 1

Assuming $\psi(x) = a_0 + a_1 x+a_2 x^2+\ldots$, then it follows \begin{align} (1-x)\psi'(x) =&\ (1-x)(a_1 + 2a_2x+\ldots + na_n x^{n-1}+\ldots)\\ =&\ a_1+ (2a_2-a_1)x + (3a_3-2a_2)x^2+\ldots + ((n+1)a_{n+1}-na_n)x^n+\ldots \end{align} So, we have \begin{align} (1-x)\psi'(x)-1 = (a_1-1) + (2a_2-a_1)x + \ldots + ((n+1)a_{n+1}-na_n)x^n+\ldots = 0 \end{align} We know that in order for a power series to be identicially zero, then it must be that all of its coefficients are zeros. Hence we have \begin{align} a_1-1 = 0, 2a_2-a_1=0, \ldots, (n+1)a_{n+1}-n a_n = 0. \end{align} So the recursion formula is \begin{align} a_{n+1} = \frac{n}{n+1}a_n \ \ \ \text{ with } \ \ \ a_1=1. \end{align}

Solving the recursion, we see that \begin{align} a_n = \frac{n-1}{n}a_{n-1} = \frac{n-1}{n}\frac{n-2}{n-1}a_{n-2}= \frac{n-1}{n}\frac{n-2}{n-1}\cdots\frac{1}{2}a_1 = \frac{1}{n}a_1 = \frac{1}{n}. \end{align} Thus, the solution is given by \begin{align} \psi(x) = a_0+\sum^\infty_{n=1} \frac{x^n}{n} = a_0 - \log(1-x) \end{align} for some constant $a_0$.

Answer 2

Note that there is no contribution from the $n=0$ term, and shift the second terms index by $1$ \begin{eqnarray*} a_1 -1 + \sum_{n=2}^{\infty} (n a_n x^{n-1} - (n-1)a_{n-1} x^{n-1}) =0. \\ \end{eqnarray*} You can now read off & solve the recurrence relation \begin{eqnarray*} \underbrace{a_1 -1}_{a_1=1} + \sum_{n=2}^{\infty} (\underbrace{n a_n - (n-1)a_{n-1} }_{na_n=(n-1)a_{n-1}=1} )x^{n-1} =0. \\ \end{eqnarray*}

Answer 3

Hint:$$ \begin{align} \sum_{n \ge 0} a_nn(x^{n-1}-x^{n}) &= \sum_{n \ge 0} a_n nx^{n-1}- \sum_{n \ge 0} a_n n x^{n} \\[5px] &= \sum_{n \ge \color{red}{1}} a_n nx^{n-1}- \sum_{n \ge 0} a_n n x^{n} \\[5px] &= \sum_{n \ge \color{red}{0}} a_{n\color{red}{+1}} (n\color{red}{+1})x^{\color{red}{n}}- \sum_{n \ge 0} a_n n x^{n} \\ &= \sum_{n \ge 0} \big((n+1)a_{n+1}-na_n\big)x^n \end{align} $$