Jaime
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 Oct 10 awarded Yearling Apr 2 awarded Nice Answer Jan 14 awarded Editor Jan 14 revised Vectors and Loci of points added 1 characters in body Jan 14 answered Solving $C_1=4y^2+(y')^2+8y$ Jan 14 answered Vectors and Loci of points Jan 13 answered What is the probability that the 2002 mean salary of a random sample of 50 baseball players was within $20,000 of the population mean, μ . Dec 31 awarded Yearling Dec 28 comment geometric series with probability The sum of the digits of your student number is 11, and its largest digit is 1, so you should actually solve it for k=9. Dec 19 comment Minimizing a functional definite integral Look up calculus of variations (en.wikipedia.org/wiki/Calculus_of_variations), to see how these type of problems are normally dealt with. In your case, as Vibert points out, either$f(g)=0$or$f(g) = -\inf$, depending on how you define minimisation, is the solution to your problem. Dec 15 answered Obtaining the$\frac{1}{2\pi}$factor in the Fourier transform Dec 11 comment What is the necessary condition for a matrix to have eigenvalue 1? There must be a vector that is unchanged by multiplication with the matrix, I don't really think there is much more to it... Dec 10 answered Probability that distance between$X$and$Y$is$>L/3$Dec 10 answered How to mathematically express square to not “lose” sign of a vector Dec 8 answered Improving Newton's iteration where the derivative is near zero? Dec 7 comment Improving Newton's iteration where the derivative is near zero? You could approximate your function at$x$by a parabola, using$f(x)$,$f'(x)$and$f''(x)\$, instead of a line using just the first two... Dec 7 answered Intersection of a line segment and a paraboloid in 3D Dec 7 comment Counting strings with given numbers of occurrences of 0 and 1, and containing a given substring Figuring out the sequences with two, but not more, consecutive 1s is the difficult thing... Dec 7 comment Counting strings with given numbers of occurrences of 0 and 1, and containing a given substring Ok, so it's not the same, but if instead of 3 consecutive 1s you are after only 2 consecutive 1s, the formula for "in how many ways can you arrange m 1s in n positions without having two consecutive 1s" is Binomial[n-m, m]. So you will have at least two consecutive ones in Binomial[n,m] - Binomial[n-m,m]. Figuring out how many have two, but not three consecutive 1's is beyond me right now... Dec 7 comment Counting strings with given numbers of occurrences of 0 and 1, and containing a given substring I thought about that one, but then you are counting 1 (111) ... and (111) 1 ... as different, which they are not. Your formula doesn't produce the right result for his first example