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April 22nd, 2012, 02:39 AM   #1
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Which one is bigger?

[attachment=0:3o87a6ix]which one is bigger.jpg[/attachment:3o87a6ix]

Sorry for bothering!

Albert
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 April 22nd, 2012, 06:43 AM #2 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 1 Re: Which one is bigger? Interesting. But imho, I think the difference between A and B is huge. I'll show you my solution, but only after I finish watching one of my favorite tv series: Touch - by Kiefer Sutherland.
 April 22nd, 2012, 08:43 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,425 Thanks: 640 Re: Which one is bigger? It sure is huge: 315 ...... 125 ; 80 digits! "divisions" start off this way: 1/2, 3/8, 5/16, 35/128 .... ...back to the hockey game...
 April 22nd, 2012, 09:50 AM #4 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 1 Re: Which one is bigger? We're given $A=\frac{1\times 3\times 5\times 7\times.......\times99}{2\times 4\times 6\times 8\times.....\times 100}$ $A=\frac{\prod_{n=1}^{50} (2n-1)}{\prod_{n=1}^{50} (2n)}$ But we know that the product of positive odd/even integers can be rewritten as $A=\frac{[\frac{(2k)!}{k!.2^k}]}{[k!.2^k]}$ where k=50 $\therefore A=\frac{(2k)!}{(k!)^2.(2^k)^2}$ $A=\frac{[\frac{(2k)!}{k!k!}]}{[(2^k)^2]}$ $A=\frac{\binom{2k}{k}}{2^{2k}}$ I then take the natural log of both sides to obtain $\ln A=\ln [{\frac{\binom{2k}{k}}{2^{2k}}}]$ $\ln A=\ln \binom{2k}{k}- \ln 2^{2k}$ $\ln A=\ln \binom{2k}{k}- 2k\ln 2$ Now I substitute k=50 into the above equation to get $\ln A=\ln \binom{100}{50}- 100\ln 2$ $\ln A=-2.530876404$ $A=e^{-2.530876404}$ $A=0.079589237$ Last, I compare the value of A to the value of B (which is 0.1). Obviously, B is larger than A. Now that I see the remark given by Denis, I believe we solve this problem in different way.
April 24th, 2012, 02:43 AM   #5
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Re: Which one is bigger?

[attachment=0:194zisrt]B is bigger.jpg[/attachment:194zisrt]
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 April 24th, 2012, 02:49 AM #6 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Which one is bigger? Is it possible to prove (or disprove) that A < 1/5 ???
 April 24th, 2012, 03:49 AM #7 Senior Member   Joined: Dec 2011 Posts: 277 Thanks: 1 Re: Which one is bigger? Well done, Albert! Look at my solution again I understand that there is no reason to find the exact value for A. I even think on how to approximate $\ln \binom{2k}{k}$ by using Stirling approximation...
April 24th, 2012, 03:53 AM   #8
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Re: Which one is bigger?

Quote:
 mathbalarka said: Is it possible to prove ( or disprove) that A < 1/5 ???
Of course:
since A<B=1/10,
A must be less than 1/5.

 April 24th, 2012, 04:17 AM #9 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Which one is bigger? Thanks, albert. Now for another solution to this problem. I'm logging out now, and will tell you the method a few hours later.

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