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The previous question.

My goal is to find a function of the difference (error) between F and G generally.

F = $\Pi_{0}^{n-1}\frac{365-b}{365}$

G = $ \frac {364}{365}^{\frac {n^2-n}{2}}$

Now F can be rewritten as $\prod _{b=0}^{n-1}1-{\frac {1}{365}}\,b$

To find the error for eat value of b. F-G

F-G is $\prod _{b=0}^{n-1}1-{\frac {1}{365}}\,b- \left( {\frac {364}{365}} \right) ^{1/2\,n \left( n-1 \right) } $

So the value of F will be positive as long as $\frac {b}{365} \leq 1$ or $ b \leq N$ (just saying this so if I replace 365 with N in both F and G)

I understand that if 365 is replace with N (a very large number) the value will be quite small.

Is this the correct method for finding the error?

Also, I used Maple to take the derivative of F-G and got $ {\frac {\partial }{\partial n}}\prod _{b=0}^{n-1}1-{\frac {1}{365}}\,b -1/2\, \left( {\frac {364}{365}} \right) ^{1/2\,n \left( n-1 \right) } \mbox {D} \left( n \right) \left( n-1 \right) \ln \left( {\frac {364 }{365}} \right) $

What does the D(n)(n-1) mean?

ps. If you find fault with my question, please explain.

share|improve this question
    
"I took the derivative" - the derivative of what? I assume you took the derivative of $G$ with respect to $b$. The answer should be $-(364/365)^{(b^2-b)/2}((2b-1)/2)\log(364/365)$. I don't know what the $D$ means. –  Gerry Myerson Jan 26 '12 at 23:01
    
You have $F$ a function of $n$ and $G$ a function of $b$. In order to compare them, they should be functions of the same variable. –  Byron Schmuland Jan 26 '12 at 23:17
    
@GerryMyerson Thanks, didn't realize that I didn't type G. As for the D(b)(b-1)...I can't find what it means in the Maple help, nor online, so I thought somebody here would understand the Maple output and could explain it. I am lost to what it means, as well. –  yiyi Jan 26 '12 at 23:18
    
@ByronSchmuland didn't realize that till now. thanks, I'll fix that. –  yiyi Jan 26 '12 at 23:19
    
You want to remove $1-$ from $G$ or to insert it into $F$, or to look at $F+G$ instead of $F-G$. –  Henry Jan 26 '12 at 23:29

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