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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*}

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    $\begingroup$ 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/… $\endgroup$ – tatan Oct 23 '18 at 6:19
  • $\begingroup$ @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. $\endgroup$ – Postal Model Oct 23 '18 at 6:22
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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.

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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 $m_i$,

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

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