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Q: Large numbers (follow up for MathTalk) ( Answered 5 out of 5 stars,   0 Comments )
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.


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


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!?
Subject: Re: Large numbers (follow up for MathTalk)
Answered By: mathtalk-ga on 06 Jan 2004 14:46 PST
Rated:5 out of 5 stars
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]

[Stirling's Approximation (with error terms)]

So this is yet another way that in practical terms we might go about
solving your original question.


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]

regards, mathtalk-ga
mrsneaky-ga rated this answer:5 out of 5 stars and gave an additional tip of: $25.00

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