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 Subject: Large numbers (follow up for MathTalk) Category: Science > Math Asked by: mrsneaky-ga List Price: \$30.00 Posted: 05 Jan 2004 23:50 PST Expires: 04 Feb 2004 23:50 PST Question ID: 293570
 ```Thank you VERY much for your previous answer! I would like to know the value of n to make 481176!^n => 481177! I'm VERY impressed with your last answer. Thanks!``` Request for Question Clarification by mathtalk-ga on 06 Jan 2004 07:00 PST ```Thanks, mrsneaky-ga, but I want to make sure I'm not oversimplifying what you are asking here... Clearly n = 1 does not work, as: 481176! < 481177! but n = 2 can be shown to satisfy: (481176!)^n > 481177! Is that the result you are looking for? I'd be happy to post the demonstration as an Answer, but as a survivor of many in-class math exams, I'm always wary of the case where I've interpreted the Question in too easy a fashion (or too difficult!). regards, mathtalk-ga``` Clarification of Question by mrsneaky-ga on 06 Jan 2004 12:37 PST ```Well that is the answer that I wanted... 2 satisfies the equation. Yes it was a simple question, but I was making up for a GREAT answer before. THANKS``` Clarification of Question by mrsneaky-ga on 06 Jan 2004 13:51 PST `So I wanted to add a little 481176! squared is bigger than 481177!?`
 ```Hi, mrsneaky-ga: Your kind words are greatly appreciated, but I feel like I've been rewarded twice because I enjoyed the earlier problem as well! To polish off this last question, we first need to show: 481176! > 700 * 700 = 490000 > 481177 Then multiplying both sides by 481176! gives us the result: (481176!)^2 > 481176! * 481177 = 481177! Here's a little more information about my earlier answer. When we need to get a lower bound on the sum of logarithms, I claimed that due to the concave downness of ln(x) we had: INTEGRAL ln(x) [OVER x = N - 1/2 to N + 1/2] < ln(N) We can think of this an application of the Midpoint Rule for numerical integration, ie. the area under the curve is approximately equal to the area under the rectangle that passes through the curve over the midpoint of the interval. But to get the rigorous inequality above, we need to think of a trapezoid instead of a rectangle. Instead of drawing the horizontal top of the rectangle through the "midpoint", put a tangent line at the same point. Now the top lines "slopes" but it doesn't change the area, since it "gains" the same amount of area on the right of the midpoint as it "loses" on the left. Furthermore since the logarithm curve is concave down, it curves below that tangent line to both sides. Hence: area under curve < area under trapezoid = area under rectangle Finally there's one other topic that I'll provide as a link, and that Stirling's Approximation to the factorial. Actually Stirling considered a variety of formulas for approximating factorials, but a couple of the most popular are given here: [Stirling's Approximation] http://hyperphysics.phy-astr.gsu.edu/hbase/math/stirling.html [Stirling's Approximation (with error terms)] http://ece.colorado.edu/~bart/book/stirling.htm So this is yet another way that in practical terms we might go about solving your original question. FOR FUTURE RESEARCH: The factorial function x! can be identified with a translation of the gamma function, Gamma(x+1), defined continuously for all x > 0 (and more). We know: (2!)^ < 3! and (3!)^2 > 4! For what real value x > 1 is (x!)^2 = (x+1)! [NB: I'm excluding x = 0 as too easy!] [Gamma Function] http://mathworld.wolfram.com/GammaFunction.html regards, mathtalk-ga```
 mrsneaky-ga rated this answer: and gave an additional tip of: \$25.00 `AS USUAL, GREAT. THANKS!!!!`