# MhAcKNI

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An aspiring programmer and a budding researcher :-)

 Apr26 comment Bilinearity: what does it mean?+ for the link to currying Mar8 comment Is it possible to find a 2D distribution function such that the higher order moments always exist?Yes. Sorry. So i should have phrased it 'moments of any order, but non-zero. I ve edited the question. Mar8 comment Is it possible to find a 2D distribution function such that the higher order moments always exist?thanks for your answer. But every distribution whose support is bounded has moments of every order might not be true. As an example, consider a distribution which is symmetric. In this case, some of the 3rd order moments become zero. Is it not? Feb12 comment Is it possible to find a 2D distribution function such that the higher order moments always exist?And can we ensure those moments are always non-zero even if the support is possitive i.e., for the supports [a,b] and [c,d]? Aug20 comment How many unit hexagonal tiles can be placed inside a larger hexagon of sides $a,b,c,a,b,c$?And thanks a lot for answering the philosophical questions. To Read: Martin Gardner's math books is on my to do list :-) What an awesome community by the way! Aug20 comment How many unit hexagonal tiles can be placed inside a larger hexagon of sides $a,b,c,a,b,c$?Brilliant illustration. I loved it. I get it now. The number of tiles on the visible faces is (ab+bc+ca). Then the number of tiles that have been counted multiple times in the previous product (a+b+c) + 1 (the tile at the intersection of the red, green and blue lines). Voilà. Aug19 comment How many unit hexagonal tiles can be placed inside a larger hexagon of sides $a,b,c,a,b,c$?oops. Sorry Sir.My fault.I have removed that comment saying your answer is wrong. Now, I have upvoted your answer. Although before accepting your answer, i would like to clarify the logic behind. So, $a+(a+1)+...$ is the number of unit hexagons in 1 trapezoid. right? This is an arithmetic progression. How did you find out the last term (a+b-1)? Second, I dont understand how you arrived at $(c−b−1)(a+b−1)$? One more thing is about your assumption. The formula works even when your assumption is violated. a = 985;b=2;c=2 Aug19 comment How many unit hexagonal tiles can be placed inside a larger hexagon of sides $a,b,c,a,b,c$?hmm.thanks for your quick response. But Sorry the answer is wrong. For a=7,b=8,c=13, the correct answer is 224. Your expression gives 84. Aug19 comment How many unit hexagonal tiles can be placed inside a larger hexagon of sides $a,b,c,a,b,c$?@copper.hat: I dont get it. How does a>1 remove tiles? Jul23 comment Multi binomial theorem applicationyes. I did mean that $v<=alpha$. term. thanks. I got it. But still not sure of how to proceed with the expansion. Jul23 comment Multi binomial theorem applicationThats what i m not sure. Under the assumptions that i have given, we can approximate each of the terms above as (a1/c1)x+(b1/c1)y+1. So should i expand it using multi multinomial theorem? Apr19 comment Computing the moments of a triangleright! but do you know if i can find a copy of that file somewhere or if you have one, i ll be glad to take it. Apr19 comment Computing the moments of a triangleThe second link is not working.. Oct28 comment Odd order moments of a symmetrical distributionthat was superb! Oct14 comment Symbolic computation of the derivative of dot product of 2 vectors@Bill Cook: Thank you. I tried using what you ve sent me. The expressions seem to be complicated though. Well. fair enough. But i was wondering if i could retain a and b and its derivatives a' and b' (w.r.t time)as vectors itself in the final expression. Jan24 comment How to multiply vector and matrix expressions involving transposes@Rahul.Thanks again. I should have googled for "matrix transpose properties". Jan24 comment How to multiply vector and matrix expressions involving transposes@Rahul.Thanks. can you give me some links having all these basic identities involving vectors and matrices? Like a'b+ba'=? and so on. Jan20 comment Solving matrix equations of the form $XA = XB$Title corrected..