Lecture 2

2. Lecture 2

Last time we showed that if P (D) is elliptic of order m and u ∈ C −∞ (Ω), for Ω ⊂ R n open, satisfies P (D)u ∈ H s loc (Ω) then u ∈ H m+s loc (R n) and for any φ, ψ ∈ C ∞ c (Ω) with φ = 1 in a neighbourhood of supp(φ), (2.1) ψu s+m ≤ CψP (D)u s + C φu s+m−1 for any M ∈ R, with C depending only on ψ, φ, M and P (D) and C only on P (D) (so neither depends on u). We proceed to try to do the same ...

Lecture 2

Solution One quick way to get the answer (but not the proof) is to test the question for very small numbers. Suppose n = k = 2 and the bin contains one red and one blue ball. Then the possible outcomes of the first experiment are RR, RB, BR, and BB – all with equal probabilities. Hence the entropy of the first experiment is 2 bits. The possible outcomes of the second experiment are RB and BR, a...

Lecture 2

Last time we have seen that the proof of Fermat’s Last Theorem for the exponent 4 provides us with two elliptic curves (y = x + x and y = x − 4x) in the guise of the quartic equations (X + Y 4 = Z and X − 4Y 4 = Z). Moreover, in the actual proof we used connections between these quartic curves and the ‘underlying conic’, namely the quadratic curves x+ y = z and x− 4y = z, which actually both re...

Lecture 2:

In this lecture I will put further flesh on the notion of a mechanism and the idea that adducing a mechanism can provide an explanation by examining one of the few cases in cognitive neuroscience in which the basic structure of the relevant mechanism is known, the visual processing system. Since my interest is not just in mechanisms as finished explanatory products, but in the process of develo...

Lecture 2