Nick Lewis New poster Post Number: 1	Posted on Thursday, 06 January, 2005 - 03:37 pm: Which is bigger, 100^300 or 300!? Please help i am truly stuck.
David Loeffler Veteran poster Post Number: 1179	Posted on Thursday, 06 January, 2005 - 04:03 pm: This is quite a tough question. Well, 300!/100300 is We'd like to show that this is > 1, so it's sufficient to prove its logarithm is > 0. So take the logarithm of the product: Now, draw yourself a graph and convince yourself that's greater than Now 3 > e, so this is positive; so the original product was greater than 1. (Similar, but rather more elaborate, manipulations give an approximate formula for n! for large n: which is known as Stirling's formula.) David
Roy Madar Regular poster Post Number: 45	Posted on Thursday, 06 January, 2005 - 08:56 pm: Also, a kind of cheating way using excel 100300 = 10600 300! cant be done, its too big but 170! is about 7.25*10306 and (300!)/(171!) is about 4.2*10307 so by multiplying them together you can see that the product will be to the order 10613 , so will be bigger than 10600
Graeme McRae Veteran poster Post Number: 505	Posted on Thursday, 06 January, 2005 - 09:52 pm: Roy, I think you meant to say (300!)/(170!) is about 4.2*10307 , which is true. Other than that, I agree with everything you said, and you came to the correct conclusion. If you are using Excel anyway, you can get approximations of very large factorials using the GAMMALN function. It gives the natural logarithm of the Gamma function. GAMMALN(300+1)/LN(10) = 614.485803, so 300! is approximately 10614.485803 , which is about 3*10614 , considerably more than 100300 =10600 . --Graeme
Roy Madar Regular poster Post Number: 46	Posted on Thursday, 06 January, 2005 - 10:31 pm: (300!)/(170!) is also too big for excel, (300!)/(171!) isnt, so i know what you mean, i just omitted the fact that what ever the product is you'll have to multiply it by 171. just done a bit of reading up on the gamma function (didnt know what was), seems a bit strange, cant quite understand how they define z! when z is a complex number, and if theyre saying n! is valid even when n isnt an integer. am i miss understanding something on this page? http://mathworld.wolfram.com/GammaFunction.html Roy
Graeme McRae Veteran poster Post Number: 506	Posted on Thursday, 06 January, 2005 - 11:01 pm: Gamma(x+1) = x*Gamma(x), which is a property of the Gamma function for all x. Gamma(1) = 1 These two properties of Gamma can be combined very easily to show that Gamma(n+1) = n! It is tempting, then, to extend the meaning of factorial (!) to real or even complex numbers. The Mathworld article you cite calls the Gamma function an "extension" of factorial to complex and real arguments. However the symbol, !, is generally not used with arguments other than nonnegative integers. --Graeme
Arun Iyer Veteran poster Post Number: 735	Posted on Friday, 07 January, 2005 - 10:38 am: Alternately, (for this question in particular) we can calculate the number of digits in each and have a simpler answer. arun
Panna Lal Patodia Regular poster Post Number: 36	Posted on Friday, 07 January, 2005 - 10:46 am: If computer is allowed, the answer is very simple, when you ask gp-pari: 300! > 100^300, it returns 1 which means true. However, it is not easy to find without using computer. Can somebody give method to find it without using computer?
Arun Iyer Veteran poster Post Number: 738	Posted on Friday, 07 January, 2005 - 10:52 am: Panna, Look at the solution given by David or look at my post(which follows similar veins). arun P.S -> Number of digits in a number can be found using logs.
Nick Lewis New poster Post Number: 2	Posted on Friday, 07 January, 2005 - 11:54 am: Thanks all, Nick
Terry Barker Frequent poster Post Number: 83	Posted on Friday, 07 January, 2005 - 02:07 pm: For what it's worth, here's something I spotted. I (very briefly) looked at the two functions 100n and n!, and compared these. It was "obvious" that the ratio (100n )/n! was getting larger as n increased, and so I was fully expecting the answer to be that 100300 is larger than 300! When I came back to see how everyone was getting on, I was quite surprised. Then, after playing around a bit more (well, ok, bashing a couple of formulae into Excel), it now appears that the ratio peaks at n=100 (or n=101) and then drops again - obviously in time for n=300. Moral: Never let a guess get in the way of calculation!
David Loeffler Veteran poster Post Number: 1188	Posted on Friday, 07 January, 2005 - 02:22 pm: You might find it interesting to note that $\sum _{i=0}^{\infty }{100}^n/n!$ converges (to $e^{100}$) so the terms had better become small eventually.
Roy Madar Regular poster Post Number: 47	Posted on Friday, 07 January, 2005 - 02:47 pm: Just out of interest how do you know the above converges to $e^{100}$? Is that a general formula, i.e., does ${\displaystyle \sum _{i=1}^{\infty }{k}^n/n!}$ diverge to $e^k$? Roy
James Veteran poster Post Number: 690	Posted on Friday, 07 January, 2005 - 03:22 pm: Yes, that is the Maclaurin expansion (actually the sum starts from n=0, not 1), and it converges to ek , not diverges.
David Loeffler Veteran poster Post Number: 1189	Posted on Friday, 07 January, 2005 - 03:44 pm: As for why the maximum is at 100 (ish): this is not unconnected with the Poisson distribution in probability. For a Poission random variable X with mean 100, the probability that X = n is e-100 100n /n!. Now (essentially because of the Central Limit Theorem) if the mean is large (in practice this means greater than about 30) the shape of the Poisson distribution closely resembles the normal distribution: symmetric and bell-shaped. This implies that its mean and its mode are close together: that is, the value occurring with greatest probability should be 100, as you observed in this case.