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Question:

Assume that $m$ is a positive integer, define the sequence $$\{u_{k}\},\{v_{k}\},u_{0}=v_{0}=u_{1}=v_{1}=1$$ and for any real number $a_{i},i=\{1,2,\cdots,m-1\}$, $$\begin{cases} u_{k+1}=u_{k}+a_{k}u_{k-1}\\ v_{k+1}=v_{k}+a_{m-k}v_{k-1} \end{cases}$$

show that $$u_{m}=v_{m}$$

My try: if

(1):$m=2$,since $$u_{2}=u_{1}+a_{1}u_{0}=1+a_{1}$$ and $$v_{2}=v_{1}+a_{1}v_{0}=1+a_{1}$$ so $$u_{2}=v_{2}=1+a_{1}$$

(2):if $m=3$, then $$u_{2}=1+a_{1},\Longrightarrow u_{3}=u_{2}+a_{2}u_{1}=1+a_{1}+a_{2}$$ and $$v_{2}=v_{1}+a_{2}v_{0}=1+a_{2},\Longrightarrow v_{3}=v_{2}+a_{1}v_{1}=1+a_{2}+a_{1}$$ so $$u_{3}=v_{3}=1+a_{1}+a_{2}$$

(3):if $m=4$,so $$u_{4}=u_{3}+a_{3}u_{2}=1+a_{1}+a_{2}+a_{3}(1+a_{1})=1+a_{1}+a_{2}+a_{3}+a_{3}a_{1}$$ $$v_{2}=v_{1}+a_{3}v_{0}=1+a_{3}\Longrightarrow v_{3}=v_{2}+a_{2}v_{1}=1+a_{3}+a_{2}$$ $$\Longrightarrow v_{4}=v_{3}+a_{1}v_{2}=1+a_{3}+a_{2}+a_{1}+a_{1}a_{3}$$ also $$u_{4}=v_{4}$$

this problem fact show $u_{n}$ is symmetry. then How prove it in general can you help ?Thank you

math110
  • 93,304

1 Answers1

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Lemma 1 : $ u_k = 1 + \sum a_i + \sum a_i a_{i-2} + \sum a_i a_{i-2} a_{i-4} + \ldots $

This is obvious by induction on $u_k$. Note that $u_k$ are independent of $m$. $_\square$

Note: 1 is the "sum of the empty product", IE $1 = \sum {\text{( product of 0 terms )} } $. This is the algebraic explanation behind why the formula works, though you can verify it by expanding too.

Lemma 2 : Fix $m$. The sequence of $v_k$ obtained from $a_1, a_2, \ldots a_{m-1}$ is equal to the sequence of $u^* _k$ obtained from setting $a^*_1 = a_{m-1}, a^* _2 = a_{m-2}, \ldots , a^*_{m-1} = a_1$.

This is obvious from the recurrence definition. Apply your favorite symmetry argument. $_\square$

Lemma 3: Fix $m$. $u_m = v_m$.

Use Lemma 2 to calculate $v_m$ via Lemma 1. Compare it to $u_m$ via Lemma 1. $_\square$

Calvin Lin
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