# 例子(3.42)的验证

[按：下文是邮件笔记的内容，标题是原有的。有点应景哦 ^_^]

“窈窕淑女，君子好逑。”

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(3.15)       α(0) = β(0) = 2-1/2.

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It follows that any choice of 2π-periodic function α and β satisfying (3.14),(3.15) and ∑|an|<,∑|bn|<, leads, via (3.13),(3.9) and (3.7), to two filter operators H and G satisfying (3.1)-(3.4). These filter operators can then be used for a decomposition and reconstruction algorithm "a la Mallat", without reference to multiresolution analysis.

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(3.14)      |α(ξ)|2+|β(ξ)|= 1.

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Example 3.2. The next simplest example is

α(ξ)=2-1/2[ν(ν-1)+(ν+1)e]/(ν2+1),

β(ξ)=2-1/2[(1-ν)+ν(ν+1)e]/(ν2+1),

where ν is an arbitrary real number.

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α(ξ)=2-1/2[(ν+1)e]/(ν+1),

β(ξ)=2-1/2[(ν+1)e]/(ν+1),

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One can show (see [27]) that, up to an arbitrary phase factor ±eiKξ，K∈Z, all the polynomials B satisfying (4.11) are necessarily of the form (4.12).

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...For N∈lN, N>1 fixed, this determines Q unambiguously, up to a phase factor eiKξ，K∈Z. For the sake of definiteness we fix this phase factor so that Q contains only positive frequencies, starting from zero,...

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(4.13)          PN(y)=∑j=0,N-1C jN-1+jyj.

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① ν=1/√3，代入例子3.2(见上文高亮处)，得：

α(ξ)=[1-√3+(√3+3)e]/(4√2),

β(ξ)=[3-√3+(1+√3)e]/(4√2),

eei3ξ的项分到一个组，提取出e((3-√3)+(1+√3)e)ei2ξ。好像行不通。为了凑出(1+e)，尝试方括号内全部打开1-√3+3e-√3e+√3ei2ξ+3ei2ξ+ei3ξ+√3ei3ξ。分组如下——

1-√3-√3e3e+3ei2ξ√3ei2ξ+√3ei3ξei3ξ

(1+e)(1 - e  + ei2ξ) -√3(1+e) + 3(1+e)e + √3(1+e)ei2ξ =  (1+e)[1 - √3 + 2e  + (1+ √3)ei2ξ

1 - √3 + 2e  + (1+ √3)ei2ξ

= [(1+ √3)2ei2ξ + 2 (1+ √3)e +1 -3]/(1+ √3)

= {[(1+ √3)e+ 1]2 -(√3)2 }/(1+ √3)

= [(1+ √3)e+ 1 + √3][(1+ √3)e+ 1 - √3]/(1+ √3)

= (1+e)[(1+ √3)e+ 1 - √3]

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For the second example (3.42), we find (see (4.4)) m0(ξ)=[1/2 (1+e)]2 ·1/2[(1∓ √3)e], corresponding to N=2 and |Q(e)|2=2 – cosξ = 1 + 2sin2 ξ/2; hence P(y) = 1 + 2y = P2(y). (原著笔误由灰色字体标出)

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=1/4 [(1+ √3)e+ 1 - √3][(1+ √3)e-iξ+ 1 - √3]

=1/4 [(1+ √3)2+ (1 - √3)2 -2e -2 e-iξ]

=2 - 1/2(e + e-iξ)

=2-cosξ

=1 + 2sin2 ξ/2

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= y2(3-2y) + (1-y)2(1 + 2y)

= 3y2 - 2y3 + (1 - 2y + y2)(1 + 2y)

= 3y2 - 2y3 + (1 - 4y2 + y2 + 2y3)

=1

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② ν= -1/√3，情况应该是类似的。值得考虑的是，Q(e)是否只差个相位因子eiKξ，其中K是某个整数。待后续探究。

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m0之谜与特异形态 2024/6/9

https://wap.sciencenet.cn/blog-315774-1443522.html

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