Hain

Howdy folks, I have an interesting math problem that eludes 6 math geeks and 2 college professors, although we can't assume that the profs are working on this problem.
Now s does exist... and supposedly it is 1/2. I wont discuss any of my work---other than it is wrong. I understand why it is wrong, finally. I would love some ideas to be spit-balled.


SOLUTION:

I finally solved this beast to my satisfaction. First, one needs to see that the limit has ties to the Poisson Distribution. One can prove that the asymmetric Poisson Distribution becomes symmetric as the expectant value, lambda or here denoted as n, goes to infinity. This means that the probability above n and below n are equal to one another. Next, one can use Stirling's Approximation to show that the probability of n occurrences occurring as n goes to infinity is also zero. Finally, it is simple algebra to show that the sum of P"below n", P"of n", and P"above n" equals 1, where P"below n" is equal to P"above n" and P"of n" is zero, then 2 * P"below n" = 1, and P"below n" is 1/2. Therefor, the limit is equal to P"below n" plus P"of n", and is equal to 1/2.

Dilbert1234567

we'll I'm not quite sure it could be anything other then 0

e^-n is the same as 1/(e^n) as n goes to infinity, it goes to zero.

using the ratio test, the summation converges absolutely, so it's finite (i think, this is as long as i can treat n as a constant) so since it's a finite number, multiplied by 0, it should be zero... if the summation does not converge, then the summation equals infinity. with the first e^-n going to zero, i don't see any possibility of it going to 1/2.

or at least that's what my 2 semesters of calculus say (well actually 5, 2 that i passed)

I'll forward it to my math professor, he likes challenges.

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Hain

Using similar logic [n is constant], I came up with s=1. The summation approaches exp(n) with every increase in n. The rate that the summation approaches exp(n) is the problem. Since that n controls the length of the sum and is inside the sum, we cannot treat n as constant.

All my work ends up with either s=0 or s=1, and it all was based on integer values of n being constant.

Dilbert1234567

one more thought of mine, if the series diverges, then it could be possible to reach a number.

in the case limit as n goes to infinity of 2x/x, we use l'Hôpital's rule, we take the limit and it goes to 2. This could be similar to your's, since the bottom (e^n) goes to infinity, and the top (the summation) could go to infinity (divergent). it is my experience that factorials go to infinity the fastest, as apposed to exponential, but this is when the number is not infinity...

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Hain

The summation is an incomplete Maclaurin Series for exp(n) (see equation 31 on linked page). Since it is incomplete form of exp(n) is must be always less than exp(n).
The fact there there is a (n!)^-1 * n^n term in the sum makes l'Hôpital's Rule difficult to apply.

EDIT: The simplified proof I have above means only that any value of n less than infinity will be between 0 and 1. However, the limit can be those. The range of values for the limit: 0≤s≤1 .

Dilbert1234567

Maclaurin are next in my class, i can't be of much more help than this... yet...

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Hain

This post has been delorted, much worse than being deleted.

Dilbert1234567

my proff agreed its 1/2, but did not have the time to explain it to me. if he gets back to me I'll let you know.

i wish i could help you more.

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Hain

The only "proof" that the originator of the problem has is a statistical analysis with a high confidence that the limit can be 1/2! ... Ridiculous ...

Dilbert1234567

i seem to recall that any convergent sum can actually converge to any number depending on how you organize the parts...

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Hain

Could you elaborate on your statement. Do you mean: any sum has a means of organization such that its sum can be found, or; the sum can be whatever I want it to be? The only example I can possibly think you are referencing is a sum that diverges due to oscillation:
This can be arranges two ways:
Neither of those organizations is correct since the sum of that series is indeterminate due to the oscillation.

Dilbert1234567

the sum can be what ever it wants to be, although i don't know if i heard my professor correctly. I'll send him an email and see what he says.

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Hain

I don't like this solution. The problem is reminiscent of the Poisson Distribution. In short, this determines the probability that k counts (called "arrivals") occurring if you expect λ counts occurring. The Poisson Distribution's form is:
As shows it can be manipulated to looks like terms of this limit.

Some properties of the Poisson distribution is that it is asymptotic to zero as you reach infinity, which makes sense since the odds that you will have infinitely many things occurring when you expect a finite number is zero. The Poisson distribution has a peak probability at λ, shown here [click image to enlarge]:

I've shown before that this problem's solution lies between 0 and 1, which is confirmed as probability is used. The proper way to compute the probability of a range of occurrences, is to add the individual probabilities together. For instance, if you average 10 people arriving per day in a travel office, and you want to know the probability of 9, 10, or 11 people arriving, then you: Looking at this, if we find the probability of all possible occurrences, from none to infinitely many occurrences, then the probability will be 1.

My limit goes from zero to n occurrences, with the expected value being n. Since we are going from 0 to n, the point of the peak, we are only analyzing the left range of the total values. Since n is going to infinity, and the end is already at infinity, we are analyzing only the left half of all expected values. If we are only analyzing half the values, then the probability is 1/2.

Dilbert1234567

*whoooosh* that flew right over my head, koodoo's to you though!

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Hain

Not even my stats professor liked that one. The only thing he agrees with is that the domain of the answer is from zero to one, including the ends. While he is intrigued with that kind of visual answer, he thinks that it can be shown that with n going to infinity the distribution becomes "symmetric" proving that solution.

A calculus based solution is still in the works.

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Hain

I just cranked a solution out to this problem...

I proved the goddamn distribution becomes symmetric as the expectancy goes to infinity...

... Re ... goddamn ... duckulous ...

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albania

That’s a quirky problem; offhand, if the limit, L, exists then log of the limit exists and in particular the log of the limit is the limit of the log of the expression, which in this case yields log L = -log (e^n) + log(Sum_k_n (n^k/k!)); which gives the condition that L is between 0 and 1; in fact it seems to me that it says L= 1, since that’s wrong I must have made a bad turn somewhere, perhaps these logs cancel in some non trivial way to get L=.5, idk.

Weird, glad you solved it though. Maybe I should send this one to my math advisor he’s an analyst, I bet he does this stuff with his eyes closed.

Hain

albania, I'd gladly welcome your profs input on this. If you didn't notice, I am an obsessive SOB.

This is my proof that demonstrates the symmetry of the Poisson distribution as the expectancy goes to infinity.

Unfortunately, this only proves that the sum from zero to n-1 is 1/2, as the initial assumption for the symmetry starts from the fact that Pi(n-1,n) and Pi(n,n) are equal. As they are equal, I have a symmetry line at k=n+1/2. I then have to prove that Pi(n,n) goes to zero as n goes to infinity, allowing that the original sum from zero to n is still 1/2. I could just not be so picky about it, and change the symmetry line to exist at k=n, but I know it is not there.

This is why I still don't like this proof, as having the answer before hand allowed me to begin this line of inquiry. I would much rather see an actual proof, that does not require me to know the answer previously.

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Hain

My obsessive nature is finally satisfied with this problem. After proving that Poisson Distribution becomes symmetric as the expected value, lambda, approached infinity, something never sat right with me. It came to me that I never proved the probability of lambda occurrences occurring as lambda goes to infinity was zero. The graphs above show that as lambda increases, the probability decreases. I assumed it goes to zero. I can think of a couple hand-waves that state it can't be non-zero, but nothing analytically grounded. I started attacking that limit only to be stumped by different ways to simplify the limit without gaining any ground. I found the Stirling Approximation, which approximates large factorials, solved it easily.

It was quite serendipitous how this happened, too. I was talking with my intro to astronomy/astrophysics professor, and he had what seemed to be a work book of math problems. It was actually a table that helped calculation something or other and he was investigating how good the approximations of the chart were when using large numbers. He simultaneously had an actual math book out that explained the Stirling Approximation to help him with the calculations. Curious about this Stirling Approximation, he explained it was used to approximated large factorials. At the time, this limit hadn't crossed my mind. When checking the Approximation on Wikipedia, I immediately recognized the terms to be part of one of my simplifications.

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