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Foundations of Data Science

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11.5 Conditions on the Dilation Equation<br />

We would like a basis for a vector space <strong>of</strong> functions where each basis vector has<br />

finite support and the basis vectors are orthogonal. This is achieved by a wavelet system<br />

consisting <strong>of</strong> a shifted version <strong>of</strong> a scale function that satisfies a dilation equation along<br />

with a set <strong>of</strong> wavelets <strong>of</strong> various scales and shifts. For the scale function to have a nonzero<br />

integral, Lemma 11.1 requires that the coefficients <strong>of</strong> the dilation equation sum to two.<br />

Although the scale function φ(x) for the Haar system has the property that φ(x) and<br />

φ(x − k), k > 0, are orthogonal, this is not true for the scale function for the dilation<br />

equation φ(x) = 1φ(2x) + φ(2x − 1) + 1 φ(2x − 2). The conditions that integer shifts <strong>of</strong> the<br />

2 2<br />

scale function be orthogonal and that the scale function has finite support puts additional<br />

conditions on the coefficients <strong>of</strong> the dilation equation. These conditions are developed in<br />

the next two lemmas.<br />

Lemma 11.2 Let<br />

∑d−1<br />

φ(x) = c k φ(2x − k).<br />

k=0<br />

If φ(x) and φ(x − k) are orthogonal for k ≠ 0 and φ(x) has been normalized so that<br />

∫ ∞<br />

−∞ φ(x)φ(x − k)dx = δ(k), then ∑ d−1<br />

i=0 c ic i−2k = 2δ(k).<br />

Pro<strong>of</strong>: Assume φ(x) has been normalized so that ∫ ∞<br />

φ(x)φ(x − k)dx = δ(k). Then<br />

−∞<br />

Since<br />

∫ ∞<br />

x=−∞<br />

∫ ∞<br />

φ(x)φ(x − k)dx =<br />

∫ ∞<br />

x=−∞ i=0<br />

∑d−1<br />

∑d−1<br />

∫ ∞<br />

= c i c j<br />

i=0<br />

j=0<br />

x=−∞<br />

φ(2x − i)φ(2x − 2k − j)dx = 1 2<br />

∑d−1<br />

∑d−1<br />

c i φ(2x − i) c j φ(2x − 2k − j)dx<br />

= 1 2<br />

x=−∞<br />

∫ ∞<br />

x=−∞<br />

∫ ∞<br />

x=−∞<br />

j=0<br />

= 1 δ(2k + j − i),<br />

2<br />

φ(2x − i)φ(2x − 2k − j)dx<br />

φ(y − i)φ(y − 2k − j)dy<br />

φ(y)φ(y + i − 2k − j)dy<br />

∫ ∞<br />

x=−∞<br />

−∞<br />

i=0<br />

j=0<br />

∑d−1<br />

∑d−1<br />

1<br />

φ(x)φ(x − k)dx = c i c j<br />

2 δ(2k + j − i) = 1 ∑d−1<br />

c i c i−2k . Since φ(x) was nor-<br />

2<br />

malized so that<br />

∫ ∞<br />

∑d−1<br />

φ(x)φ(x − k)dx = δ(k), it follows that c i c i−2k = 2δ(k).<br />

i=0<br />

i=0<br />

361

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