If p(z) has a double root $s_1$, show that $s_1^n$ and $ns_1^n$ are both solutions of $\sum_{j=0}^r \alpha_j U^{(n+j)}=0$

Question

If p(z) has a double root $s_1$, show that $s_1^n$ and $ns_1^n$ are both solutions of $\sum_{j=0}^r \alpha_j U^{(n+j)}=0$

I was able to prove the no repeated root case. I have a hint to use that $\rho'(S_1)=0$ but I can't figure out the significance of this. Thank you. My thought is that if we have a repeated root, the polynomial will still vanish at that root due to the (S-s_1) term of the poly $\rho(S)=(S-s_1)(S-ns_1)(S-s_2)...(S-s_r)$. I am not seeing how the hint would be used here though.

Non repeated root case. First, let us consider a linear, rth order homogeneous difference equation $$\sum_{j=0}^r \alpha_j U^{(n+j)}=0$$ where we have the $\alpha_j$ given. We wish to look for a solution of the form $$U^{(n)}=S^n$$ This would be a solution if we have for $n=0,1,2...$ \begin{align*} \sum_{j=0}^r \alpha_j S^{n+j}&=0\\ \Rightarrow S^n \sum_{j=0}^r \alpha_jS^j=0 \end{align*}

We are looking for a nontrivial solution if and only if $$\rho(S)=\sum_{j=0}^r \alpha_jS^j=0$$ which is a polynomial of degree $r$ in $S$. We see that if $S_i$ is a root of $\rho(S)$, then $U^{(n)}=S_p^n$ solves $\sum_{j=0}^r \alpha_j U^{(n+j)}=0$.\ We have shown that any linear combination of $S^n_{i_1},S^n_{i_2}...$ are solutions. Therefore, if $\rho(S)$ has $r$ distinct roots $S_1,S_2,...,S_r$, then we have a solution of the form $$U^{(n)}=c_1S_1^n+C_2S_2^n+...+c_rS_r^n$$ solves $\sum_{j=0}^r \alpha_j U^{(n+j)}=0$ for any numbers $c_1,c_2,...,c_r$.

Answer 1

If $$\rho(s)=\sum_{j=0}^r\alpha_js^j=(s-s_j)^2q(s)$$ Since $s_1$ is a double zero of $\rho(s)$. Then $$\begin{align}\rho^{\prime}(s)&=\sum_{j=0}^r\alpha_jjs^{j-1}=\sum_{j=1}^r\alpha_jjs^{j-1}=\sum_{j=0}^{r-1}\alpha_{j+1}(j+1)s^j\\ &=2(s-s_1)q(s)+(s-s_1)^2q^{\prime}(s)=(s-s_1)\left[2q(s)+(s-s_1)q^{\prime}(s)\right]\end{align}$$ So as promised, $\rho^{\prime}(s_1)=0$. Then we have of course if $U^{(n)}=s_1^n$, $$\sum_{j=0}^r\alpha_jU^{(n+j)}=\sum_{j=0}^r\alpha_js_1^{n+j}=s_1^n\sum_{j=0}^r\alpha_js_1^j=s_1^n\rho(s_1)=0$$ And now if $U^{(n)}=ns_1^n$ then $$\begin{align}\sum_{j=0}^r\alpha_jU^{(n+j)}&=\sum_{j=0}^r\alpha_j(n+j)s_1^{n+j}=ns_1^n\sum_{j=0}^r\alpha_js_1^j+\sum_{j=0}^r\alpha_jjs_1^{n+j}\\ &=ns_1^n\rho(s_1)+\sum_{j=1}^r\alpha_jjs_1^{n+j}=\sum_{j=0}^{r-1}\alpha_{j+1}(j+1)s_1^{n+j+1}\\ &=s_1^{n+1}\sum_{j=0}^{r-1}\alpha_{j+1}(j+1)s_1^j=s_1^{n+1}\rho^{\prime}(s_1)=0\end{align}$$