# What is the solution if $a=1$

We have the equation $$x_n = a^nx_0+b\,\left(\dfrac{1-a^n}{1-a}\right).$$

I had to find the solution if $a=1$ and $a=-1$

• For $a=1$ we must divide by $0$ which is of course impossible, but I wonder; does that render $a=1$ 'solutionless' or does that term just disappear, leaving $x_n = x_0$ for $a=1$?

• Am I correct to say that for $a=-1$ we have the solution $x_n = - x_0 + B$ for uneven $n$ and $x_0$ for even $n$?

• Hint: $a^n - 1 = (a-1)(a^{n-1}+a^{n-2}+\ldots + a + 1)$, but you have to be a bit careful with this because technically $a=1$ is not in the domain. Jun 15, 2013 at 14:10
• @JohnMartin So that means $x_n = x_0$ for $a=1$? Jun 15, 2013 at 14:15
• Not quite. Given what I wrote you have $\frac{1-a^n}{1-a} = a^{n-1}+a^{n-2}+\ldots+a +1$, so you will be evaluating $a^nx_0 + b(a^{n-1}+\ldots+a+1)$ at $a=1$. Jun 15, 2013 at 14:19
• But you do have to be careful, because in order to make this statement precise, one has to take a limit; do you know about limits and discontinuities of functions? Jun 15, 2013 at 14:21
• @JohnMartin A bit Jun 15, 2013 at 14:22

For $a=1$, as you have typed the problem, they will all be undefined.
For $a=-1$, and even $n$, we get $$x_0+b\frac0{1-a}=x_0$$ So you are correct their. For odd $n$, we get $$-x_0+b\frac22=-x_0+b$$ So you are correct again.