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PDF Editor FAQ
How can I prove that the likelihood function of exponential family is convex?
You can’t, because it’s false. What is true is that the log likelihood is concave, and that’s not too difficult too show. There’s a very brief sketch at http://www.stats.ox.ac.uk/~steffen/teaching/bs2siMT04/si6bw.pdf, but you’ll have to do some work to fill in the details.
Does it make sense to talk about the pi-th root of a number?
To illustrate what Paul Olaru said.The "πth roots"—the 1/π powers—of 2 + i would be the exponentials of[math]{1\over \pi}( \log|2+i| [/math][math][/math][math]+ i\cdot\tan^{-1}({1\over2}) [/math][math][/math][math]+ 2\pi k \cdot i [/math][math][/math][math])[/math][math][/math][math] [/math][math][/math][math] [/math][math][/math][math] [/math][math][/math][math] = {1\over \pi}\log\sqrt{5} [/math][math][/math][math]+ i\cdot\tan^{-1}({1\over2})) [/math][math][/math][math]+ 2k\cdot i[/math]where k ranges over the integers.In the first diagram below I have shown 2+i together with its πth roots corresponding to k=0, 1, 2, 3. You can see the counterclockwise advance by (radian) angle 2 from one to the next. As k ranges through the integers, these points never coincide with another, and all together they densely fill the circle of radius [math](\sqrt{5})^{1/\pi}[/math]. The second diagram shows the image of 2+i under the multivalued log function in the complex plane, and the scaling of those points by 1/π; the latter when exponentiated yield the points going around the circle. Staying in the original complex plane you can visualize the principal root (shown in blue) as the scaling (shrink) of the point 2+i by the factor 1/π and a rotation back to 1/π of the angle arctan(1/2) along which 2+i sits.
How do I solve the Fibonacci Returns problem from IOPC 2014?
Editorials are supposed to be only that intuitive. It gives you the direction to proceed, and you fill in the details yourself. I will try to explain the solution a little anycase.First of all, modulo any integer M, the Fibonacci sequence is periodic. This is not difficult to understand. Consider the consecutive terms of the sequence modulo M. (F_i mod M,F_i+1 mod M) can take only M² different values. Thus eventually, some pair has to repeat. Since (F_i mod M,F_i+1 mod M) uniquely defines (F_i+1 mod M,F_i+2 mod M) and vice versa, F_i is periodic modulo M.Let P be the period of F modulo M. Then, F_n mod M = F_(n mod P) mod M. This is what you use to calculate the answer. Finding F_n mod M in O(log(n)) using matrix exponentiation is well known.This leaves us with two further steps needed to find the solution:Given M, how does one find P?Given N and R, how does one find N choose R modulo P?The first question is answered by the pdf linked to in the editorial. It essentially uses the prime factorisation of M. The second question is answered in the editorial itself, again involving prime factorisation and Chinese Remainder Theorem.If something is unclear, comment on the answer and I will try to explain further.
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