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 |