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I would like to see if $ b_l:=4^{-l} \sum_{j=0}^l \frac{\binom{2 l}{2 j} \binom{n}{j}^2}{\binom{2 n}{2 j}}\text{.} $ is decreasing when $l$ is large enough say around $10^6$. I dont need any theoretical derivations though I wrote the following code part


b=zeros(1,10^7-10^6);  for l=10^6:10^7-1      for j=0:l          b(l-999999) = b(l-999999)             + (nchoosek(2*l,2*j)*nchoosek(10^7,j)^2)             / (nchoosek(2*10^7,2*j));      end  end 

EDIT:

This question is a simplified version of the original one. I intended to carry the matter here since what I found might imply that the function is no more decreasing for larger $l$. Please see the discussion over there.

Inequality involving sums of fractions of products of binomial coefficients

I couldnt make use of Stirlings approximation. Here is the changed code part:


b=zeros(1,10^7-10^6);

for l=10^6:10^7-1

for j=0:l      b(l-999999) = b(l-999999)  + ((sqrt(2*pi*2*l)*(2*l/exp(1))^(2*l))/((sqrt(2*pi*2*j)*(2*j/exp(1))^(2*j)*((sqrt(2*pi*2*l)*(2*l/exp(1))^(2*l))-(sqrt(2*pi*2*j)*(2*j/exp(1))^(2*j)))))*...                                   ((sqrt(2*pi*10^7)*(10^7/exp(1))^(10^7))/((sqrt(2*pi*j)*(j/exp(1))^(j)*((sqrt(2*pi*10^7)*(10^7/exp(1))^(10^7))-(sqrt(2*pi*j)*(j/exp(1))^(j))))))/...                                  (sqrt(2*pi*2*10^7)*(2*10^7/exp(1))^(2*10^7))/((sqrt(2*pi*2*j)*(2*j/exp(1))^(2*j)*((sqrt(2*pi*2*10^7)*(2*10^7/exp(1))^(2*10^7))-(sqrt(2*pi*2*j)*(2*j/exp(1))^(2*j))))));  end 

end

which cannot provide me any result due to the accuracy of nchoosek, i.e., big numbers are creating problems. Do you have any idea how I can deal with this problem? I only want to know if the function is decreasing or not.

Any help will be appreciated.

Thanks in advance.

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    @Sasha I will edit. Please have a look at the conversations over there. I have some findings on the question and now from there I have another claim. Finally what i do is constructive. I am not interested in answer. I am interested in helping the owner of the original question.2012-08-27

2 Answers 2

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You can rewrite nchoosek to return the log of nchoosek using Stirling's approximation. It is plenty accurate for your needs.

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    sorry my mistake. I looked only stirling's appx. Didn't understand that you explicitly mentioned about log-exp.2012-08-27
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First hack is to compute using long integers instead of regular integers, doubling the precision.

Also, depending on the value of $n$, an approach could be to transform the fraction. Denote $L=2l, N=2n, J=2j$. Assuming $n \geq l$,

$ \binom{L}{J}/\binom{N}{J} = \frac{L! (N-J)!}{N! (L-J)!} = \frac{L! (N-J)! (N-L)!}{N! (L-J)! (N-L)!} = \frac{(N-J)!}{(L-J)! (N-L)!} / \binom{N}{L} = \binom{N-J}{L-J} / \binom{N}{L} $

and if $l > n$, similar trick would work. Perhaps these would come out better.

Third alternative is to try to cancel the fractions before multiplying, or switch to real numbers and do calculations in floating-point arithmetic, which allows much larger numbers.

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    I don't agree. The code parts above are working. If you can provide me 100 results for $l$ I can believe.2012-08-27