Why a root $\alpha_i$ with multiplicity $m_i$ will satisfy up to $(m_i-1)$-th derivatives of polynomial of degree $n$?

I'm reading a proof of linear recurrence relation with constant coefficients of order $$k$$, which gives the formula of $$a_n$$ when some roots of its characteristic equation has multiplicity $$\ge2$$ . In one step it says:

\begin{align*} &\textrm{Since }\alpha_i\textrm{ satisfy}\\ &C_n\alpha^n+C_{n-1}\alpha^{n-1}+\dots+C_{n-k}\alpha^{n-k}=0,\,\,\,\,\,\,\,\textrm{(A)}\\ &\textrm{and }\alpha_i\textrm{ has multiplicity }m_i,\textrm{it satisfies up to }(m_i-1)\textrm{-th derivative of (A).} \end{align*}

So what's the reason the last sentence makes sense?

The theorem:

\begin{align*} &\textrm{Let }k\in\mathbb Z^+, C_n,C_{n-1},\cdots,C_{n-k}\in\mathbb R,\textrm{and }C_n,C_{n-k}\not=0;\\ &C_na_n+C_{n-1}a_{n-1}+\dots+C_{n-k}a_{n-k}=f(n)\\ &\textrm{Assume }\alpha_1,\alpha_2,\dots,\alpha_t\textrm{ are its characteristic roots, and }\alpha_i\textrm{ has multiplicity }m_i,\\ & 1\le i\le t\textrm{ and let }u_i(n)=(d_{i_0}+d_{i_1}n+\dots+d_{i_{m_i-1}}n^{m_i-1})\alpha_i^n,\textrm{where }d_k\textrm{ are any constant, then}\\ &a_n=u_1(n)+u_2(n)+\dots+u_t(n). \end{align*}

• This looks like a problem you have collected from / inspired by some source. According to recent discussions in Meta, we are looking forward to including sources for all applicable questions. Can you provide the source by editing the question?Refer-math.meta.stackexchange.com/questions/29290/… – tatan Oct 23 '18 at 6:19
• @tatan: Sorry I have tried my best to translate the problem from my language in English, and (if you have time) could you point out which part is not clear? Because I think the part lead to my confusion is about calculus thing(or maybe I'm wrong) that why a root with multiplicity $m$ will satisfy $0,1,\dots,m-1$-th derivative of the original polynomial equation. – Postal Model Oct 23 '18 at 6:22

Let r be a root of p(x) in R[x] with multiplicity k.
Then p(x) = (x - r)$$^k$$ q(x) for some q(x) in R[x].
By induction show r is a root of the j-th derivative
for all j's upto k - 1.

If $$\alpha_i$$ is a root of

$$p(x) = 0 \tag 1$$

of multiplicity $$m_i \ge 2$$, then

$$p(x) = (x - \alpha_i)^{m_i} q(x), \tag 2$$

where $$q(x)$$ is some polynomial with

$$\deg q = \deg p - m_i; \tag 3$$

we see that

$$p'(x) = m_i(x - \alpha_i)^{m_i - 1}q(x) + (x - \alpha_i)^{m_i} q'(x), \tag 4$$

whence

$$p'(\alpha_i) = m_i(\alpha_i - \alpha_i)^{m_i - 1}q(x) + (\alpha_i - \alpha_i)^{m_i} q'(x) = 0, \tag 5$$

and $$\alpha_i$$ is a root of $$p'(x)$$; we observe that (4) may be written

$$p'(x) = (x - \alpha_i)^{m_i - 1} (m_iq(x) + (x - \alpha_i)q'(x)), \tag 6$$

which is in the same form as (2) but with $$m_i$$ replaced by $$m_i - 1$$; the similarity 'twixt (2) and (6) suggests a simple induction: taking $$m_i = 2$$ as the base case, we assume that $$m_i = k$$, and that for any $$p(x)$$ such that

$$p(x) = (x - \alpha_i)^k q(x) \tag 7$$

$$\alpha_i$$ is a root of both $$p(x)$$ and it's first $$k - 1$$ derivatives:

$$p^{(j)}(\alpha_i) = 0, \; 0 \le j \le k - 1; \tag 8$$

then if

$$p_1(x) = (x - \alpha_i)^{k + 1} q_1(x), \tag 9$$

as in (6) we have

$$p_1'(x) = (x - \alpha_i)^k ((k + 1)q_1(x) + (x - \alpha_i)q_1'(x)); \tag{10}$$

clearly $$\alpha_i$$ is a zero of $$p_1(x)$$, and applying our inductive hypothesis to (10) we affirm that $$\alpha_i$$ is a root of both $$p_1'(x)$$ and its first $$k - 1$$ derivatives, that is, $$\alpha$$ satisfies $$p_1(x)$$ and the first $$k$$ derivatives thereof, the precise affirmation we seek. Thus we have inductively demonstrated that for all $$m_i$$,

$$p(x) = (x - \alpha_i)^{m_i}q(x) \Longrightarrow p^{(j)}(\alpha_i) = 0, \; 0 \le j \le m_i - 1. \tag{11}$$