Differential Equations Solutions 24

# Differential Equations Solutions 24 - 34 Chapter 6...

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Unformatted text preview: 34 Chapter 6. Solutions: Case Study: Image Deblurring: I Can See Clearly Now 2 UT g − ΣVT f −αVT f = = g − Σf −αf = g 0 2 2 Σ αI − 2 2 f . 2 CHALLENGE 6.3. Let’s write the answer for a slightly more general case: K of dimension m × n with m ≥ n. g 0 − Σ αI 2 f 2 = g − Σf 2 2 + α2 f n 2 2 m ˆ (ˆi − σi fi )2 + g = i=1 n gi + α2 ˆ2 i=n+1 ˆ fi2 . i=1 Setting the derivative with respect to fi to zero we obtain ˆ ˆ −2σi (ˆi − σi fi ) + 2α2 fi = 0, g so ˆ fi = ˆ σi gi . + α2 2 σi CHALLENGE 6.4. From Challenges 2 and 3 above, with α = 0, the solution is ˆ σi gi gi ˆ ˆ . fi = 2 = σi σi ˆ Note that gi = uT g. Now, since f = Vf , we have ˆ i ˆ ˆ f = v1 f1 + . . . + vn f n and the formula follows. CHALLENGE 6.5. See the posted program. Some comments: ...
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