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Derivative of an L1 norm of transform of a vector.

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Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite 1 I have to take derivative of the l-1 norm. L1 is the function R in the following expression: $$ R(psi Fx) $$ where x is a vector, F is the inverse Fourier transform, and $psi$ is a wavelet transform. If I define a variable C such that $$C = psi F$$ then my l1 norm is defined as: $$||Cx||^1_1$$ I know that taking a derivative of an l1 norm is not possible. The l1 norm is defined as: $$sumnolimits|x_i^1_1$$ To take a derivative of the l1 term, I addd a small positive number, call it $epsilon$. Therefore, $$sumnolimits|x_1^1_1 = sumnolimitssqrtx^*_ix_i + epsilon$$ My question is, what is the derivative of the l1 norm Cx and what would be the elements of the matrix C? linear-algebra functional-analysis wavelets share | cite | improve this question edited Feb 20 '15 at 2:45 Mathemagician1234 13.6k 2 40 54 asked Feb 20 '15 at 2:29 user212257 11 2 Why do you need...

What enables hash function to produce uniform distribution given any distribution of input

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Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite I always take the uniformity of hash output as a given and didn't think much of it. Now I am kind of curious, how does good hash function like sha guarantees output uniformity. Intuitively, given 1:1 input cardinality to output cardinality, the same amount distribution that has high entropy (uniform) in hash output must be equal to the amount of distribution that has low entropy in hash output. So that means approximately 50% of the possible input distribution will experience less uniformity after hashing. This cannot be right, can someone point out where my logic is wrong? And guide me on the how good hash function output uniformity for any random distribution? Edit: I realize my question is a bit ambiguous, so I think I should try to clarify. For simplification assuming sha1 is a function that map Integer 2^64 -> Integer 2^64 My question is that given a stream of n random number where n << 2^62, th...