In relation with a probabilistic problem for concrete numbers I came upon the generalization of an idea and there’s a problem:

Let’s suppose we are throwing by special cubes with sides numbered serially by 1,2,…,m and the probability of toss of each number is the same according to the classical definition of probability). Let denote the sum of all numbers on the sides of all cubes shortly after the last toss. Which sum has the highest probability.

Let be random variables that represents the value of the toss on cubes in this order and . Then should be the value of the most probable sum on the sides of cubes (or in case that isn’t integer let’s take the closest two integers which E X is situated between). At last it’s clear that , then:

But I would like to solve it more classically and exactly, so first I need to solve this:

Let be some natural numbers such that , then the esencial question is how many (orderly) somes does there exist such that each x_i is a natural number from 1 to maximally and:

.
Could anybody give me any analytic formula (depending on ) for number of all convenient some (or any hint).

Obviously, the number of all convenient somes is also the number of all adic combinations with reprise of elements such that each element can appear minimally once and maximally times.

I’ve tried to use the principle of inclusion and exclusion, but it doesn’t give me the right result.

Thank you….

1. Let S be the set . Is S dense in ?
2. Solve the functional equation for (May be R->R?)
3. Solve
4. Solve the recurrance relations with .
5. Does there exist a limit such that but ľHôpital’s rule cannot be used (repeatly)?
   (If the rule yields a non-existsing limit or can be directly simplified like , it is considered to be valid).
6. In , it is known that the hyperarea of each facet of a polytope is fixed. Is it true that a concyclic (?) polytop yields the maximum content?
7. Given . Find the probability that .

Q7… It isn’t 2/π, is it? -.-

Last edited by Kenny TM~ on Thu Sep 30, 2004 5:03 pm; edited 3 times in total

Suppose Q = field of rationals, R = ring, f, g: Q -> R are ring homomorphism that maps 1 to the unity (multiplicative identity) of R. Prove f = g. Comment: It is easy to show that if n is an integer, f(n) = g(n). The trickest part is to show f(x) = g(x) in general.

Notice: R may not be a division ring/integral domain.

Surprisingly, I managed to prove the following result:
For any rational , there exists an irrational such that is rational.

Suppose . consider the equation . We aim at proving that there is no rational satisfies this equation.

Assume in contrary that such an exists.
(1) : This gives , i.e., , violating our assumption.

(2) : Write for some positive coprime integers and .

Since is rational, is also rational. Therefore for some positive coprime integers and with .
Moreover, implies . So .
Substituting these formulae into the original equation, we get , or .
Since is irreducible and is also irreducible, we have and .
For , i.e., , since , we have or .
For , we have ; when , we have , which has or as its integer solutions.
(i) . Since , such does not exist. There is no solution.
(ii) . Put into the equation, and get , which does not have integer solution.
(iii) , . We have .
Try : , which is impossible. Try : is, again, impossible. So there is no solution either.
In conclusion, there is no rational satisfies the equation.

(3) : Write , where and are positive coprime integers. Again, for the same reason.

Plug in all these substitution to the equation: , or .
Again, by irreducibility, we have and .
(i) . Then and .
Note that . So . So , which is impossible.
(ii) . Then . Hence , which is impossible.
In conclusion, there is no rational number satisfies the original equation.

Now suppose . Consider . If there exists a rational satisfying this equation, then satisfies , contradicts to the above conclusion.

Now we prove that, for any rational , is rational for some irrational .

For , since , there exists with . It is proved that cannot be rational. So must be irrational.
For , we have . So there exists with . Again, cannot be rational by previous arguments. So must be irrational.
For , it is trivial that , for which is irrational but is rational.

This ‘proof’ is too clumsy and perhaps there are lots of errors. Please help me check if the arguments are valid.

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世上沒有完美的人完美的事，而我們的責任就是要令自己的表演達至最精彩最完美。所以魔術一直都沒有停頓下來，與時代一起進步去追求無止境的完美。

Let us perfrom such an “experiment”. There is an urn containing some balls such that each ball is labelled by a natural number and every natural number is used in labelling exactly one ball. Now we draw one ball from the urn. The sample space S={1,2,3…} By Probability Axiom, we have P(S)=1 Then it seems to me that the probability of drawing out some balls must be different from that of drawing out some other balls. For if the probability is the same for all balls, it can neither be positive nor zero; if it is positive, we can use another Probability Axiom to argue that P(S) is unbounded; if it is zero, we can argue that P(S)=0.