英语翻译then it is clear that the basis vector ϕ−1,k

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英语翻译
then it is clear that the basis vector ϕ−1,k is the kth column of the matrix [H(−1)]t.Obviously,
these basis functions are again orthonormal:[H(−1)][H(−1)]t=I4.Thus,we can easily ﬁnd
the projection x(−1)ofx onto the subspace V−1.It is given by x(−1)= [H(−1)]tX(−1) with
X(−1)=H(−1)X(0).Thus
1
X(−1)=√
2
[x1+ x2,x3+ x4,x5+ x6,x7+ x8]t.
This projection gives a low resolution picture of the original signal because in the original
picture we could distinguish between function values that were 1 unit apart in the horizontal
direction,while in the new picture we have a resolution of 2 units in the horizontal direction.
If the original signal is plotted on a screen that can not distinguish between two adjacent
pixels in the horizontal direction,then Figure 1.3 is the best we can get as a picture of the
signal.It is a low resolution approximation of the original.

刚刚高考完
看到它们呢就头疼
呼``
等等看能帮你找来不
找来了：
然后,这是明确表示,根据向量φ - 1 , k是k故障栏的矩阵[H （ -1 ） ]t.很明显,
这些职能的基础上再次正交： [H （ -1 ） ] [H （ -1 ） ]t= i4 .因此,我们可以很容易地找到
投影× （ -1 ） ofx上子V-1 .这是由于用X （ -1 ） = [H （ -1 ） ]tX（ -1 ）
× （ -1 ） =[H （ -1 ） × （ 0 ） .因而
1
× （ -1 ） = √
2
[ x1 + x2 , x3 + x4 , x5 + x6 , x7 + x8 ]t.
这一预测提供了低解析度的照片的原始信号,因为在原来的
图片我们可以区分函数值分别为1个单元,除了在横向
方向,而在新的图片我们有一个解决2个单位在水平方向.
如果原始信号是策划对一个屏幕,不能区分两个相邻的
像素在水平方向,那么图1.3是最好的,我们可以得到一个图片的
信号.这是一个分辨率低的逼近原.

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