<t>Check Roger B. Nelsen book “An Introduction to Copulas” for a mathematical presentation of Archimedean family of copulas (Clayton is a member of this family). In finance it can be used for building joint distributions from marginal distributions. A typical example is the single mixture-factor mod...

I notice that in GSL (GNU Scientific Library in C) there are a lot of functions declarations that pass by value const arguments.What is the advantage to use a declaration as Func ( const double x );instead of Func ( double x );?

I agree with Triptaka,In LaTexMay be the author meant shorter over longer. In that case there is a finite solution

<t>I agree with Ouyang. The only number is 21200. However I use a different line of reasoning. Since a+b+c+d+e=5, e can be only 1 or 0. 1 cannot be since one of the a, b, c, or d must by 4 and the rest 0 (that is 3 zeros)So e = 0Same, d can be only 1 or 0 1 cannot be since we have at least one 1’s a...

mat,For me = 1.76084 not 2!!!A proof for this problem was posted on July 15. Dose anybody know a shorter one?

alexanderC,I have a proble with the line 3. Do I read well a^(1/b)=a^(-b)? Then for a=b=2 we will have sqrt(2)=1/4!!!. Cool, but wrong.

Light first rope at both ends simultaneously and in the same time the second rope at one end.When the first rope had burn completely light the second rope at the other end. or use a watch

alexandreC,I guess the confusion comes from the fact that (a^b^c)^d = a^(b^c*d) and not a^b^c^d. (Hint: a^b^c^d = a^(b^(c^d)) )So if you take the power 1/x on both sides at best you get 2^(1/x) = 2^(1/x) which is OK but proves nothing.

<t>EStealth,You miss the point. So again, if the sequence {f(n)} is convergent with limit b then x=b^(1/b). That was proved since the very first replay.Then the sequence {f(n)} is convergent if and only if b is in (0,e] (that implies x is in (0,e^(1/e)]. This is what is proved.I cannot help you more...

EStealth,For arbitrary finite precision arithmetic try Numerical Recipes and use your home PC.

<t>My conjecture wasThe eq x^x^x^…=b has a solution if and only if b is in (0,e] and that solution is x=b^(1/b).The proof is in four steps. First we will show that the conjecture is true in (0,1], second that for x=e^(1/e) the limit x^x^x^… exist and is equalt to e, third that the conjecture is true...

<t>Estealth,I guess we got lost on notations. Here is how I understand them. 1. When I see a notation like x^y^z, I understand x^(y^z)2. It is not common to use the notation (x^y)^z since this is equal to x^(y*z) which is a lot more simple. Thus when you posted the problem x^x^x^…=2, I understood x^...

Estealth,Let’s define the series y_k = x^(x^(x^..) k times. We know how to compute y_kThen, we define x^x^x^… to be its limit (ie lim_(k->infty) y_k ). If (when) the limit exists, it’s a valid question. So, the problem is well defined, as well as the fun

FV,No offence there, the classification was done below by zerdna.

Anthony. I guess you are right. I was suspecting that. Thanks a lot