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Topic: WHICH IS LARGER? (Read 1487 times) |
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pcbouhid
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WHICH IS LARGER?
« on: Nov 25th, 2005, 11:33am » |
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Which is larger, 100^300 or 300! (factorial of 300)?
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Michael Dagg
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Re: WHICH IS LARGER?
« Reply #1 on: Nov 25th, 2005, 11:41am » |
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The latter by many orders of 10.
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JohanC
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Re: WHICH IS LARGER?
« Reply #2 on: Nov 25th, 2005, 12:16pm » |
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You could make use of Stirling's approximation for ln(n!) .
Then you get:
ln(300!) is approx. 300*ln(300)-300=1411.134...
which is larger than
ln(100^300)=300*ln(100)=1381.551...
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Eigenray
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Re: WHICH IS LARGER?
« Reply #3 on: Nov 25th, 2005, 1:24pm » |
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The most elementary way to make that precise is to consider
log(300!) = sum(log n, n=1..300)
as an over-approximation to
int(log x, x=1..300) = 300 log 300 - 300 + 1
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JocK
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Re: WHICH IS LARGER?
« Reply #4 on: Nov 26th, 2005, 1:29am » |
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on Nov 25th, 2005, 11:33am, pcbouhid wrote:| Which is larger, 100^300 or 300! (factorial of 300)? |
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What about 111^300 vs 300! ...?
(or: 367882^999999 vs 999999! for that matter...?)
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| « Last Edit: Nov 26th, 2005, 1:38am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Deedlit
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Re: WHICH IS LARGER?
« Reply #5 on: Nov 26th, 2005, 3:12pm » |
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To approximate factorials more closely, you can use a more complicated version of Stirling's formula:
log (n!) = n log n - n + log n / 2 + log (2 pi) / 2 + [sum]k=1m (B2k / (2k (2k - 1) n2k-1 )) + x (B2m+2 / ((2m + 2)(2k + 1) n2m+1 )),
where x is between 0 and 1, and Bk is the kth Bernoulli number. You can cut off the series at any point for the desired level of accuracy. For example, with m = 2:
log (n!) = n log n - n + log n / 2 + log (2 pi) / 2 + 1/(12n) - 1/(360 n3 + x / (1260 n5), with 0 < x < 1.
Applying the above, we get that log (999999!) lies between
12,815,504.56914761165997697178501711315368718154145293529
and
12,815,504.56914761165997697178501711315368797519621485195
and so 999999! lies between
367881.9525828613742098353957252121140716300871695073894999999
and
367881.9525828613742098353957252121140719220587249695342999999
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JocK
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Re: WHICH IS LARGER?
« Reply #6 on: Nov 26th, 2005, 11:45pm » |
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Hmmm.... should have asked:
"What is larger: 367881.9525828613742098353957252121140718999999 or 999999! ...?"
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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srn437
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Re: WHICH IS LARGER?
« Reply #7 on: Sep 1st, 2007, 3:34pm » |
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Quit answering with questions. 300! is higher, but by one digit!
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mikedagr8
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Re: WHICH IS LARGER?
« Reply #8 on: Sep 1st, 2007, 4:26pm » |
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Quote:| Quit answering with questions. 300! is higher, but by one digit! |
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By one digit, do you mean that 4>3 or 15>4? Because one digit makes an awful lot of difference.
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cool_joh
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We first show that the product of n consecutive natural numbers is greater than the nth power of the square root of the product of the first and last of these numbers. Let the n integers be a, a+1, ..., a+n-1. Then the kth number from the beginning will be a+k-1, and the kth number from the end will be a+n-k. Their product is: (a+k-1)(a+n-k) = a^2+an-a+(k-1)(n-k) ≥ a^2+an-a = a(a+n-1)
where the equality is obtained only for k=1 or k=n. That is the product of two positive integers equidistant respectively from each end of the sequence (for odd n, the two integers are taken to be the common middle one) always exceeds the product of the two extreme (first and last) integers. But then we have, the product of all the numbers,
a(a+1)...(a+n-1) ≥ [a(a+n-1)]^(n/2)
where the equality holds only if n=1 or n=1.
We shall show now that 300!>100^300.
We have 1*2*3*...*25>25^(25/2)=5^25
26*27*...*50>(26*50)^(25/2)>35^25
51*...*100>(51*100)^(25)>70^50
101*...*200>100^50 * 200^50 = 100^200 * 2^50
201*...*300>200^50 * 300^50 = 10^200 * 2^50 * 3^50
If we multiply together all the left members of these inequalities and compare the results with the product of all the right members, we obtain
300!>5^25 * 35^25 * 70^50 * 10^400 * 2^100 * 3^50 = 5^50*7^25*5^50*14^50*10^400*2^100*3^50 = 10^500^21^25*42^25*14^25 >10^500*20^25*40^25*14^25
=10^550*2^25&4^25&14^25 = 10^550*112^25 = 10^600*1,12^25 > 10^600=100^300
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ThudnBlunder
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Re: WHICH IS LARGER?
« Reply #10 on: Sep 1st, 2007, 6:18pm » |
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cool_joh, not a bad answer for someone who doesn't know the difference between 'dave' and 'deaf'. 
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| « Last Edit: Sep 1st, 2007, 7:05pm by ThudnBlunder » |
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srn437
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Re: WHICH IS LARGER?
« Reply #11 on: Sep 1st, 2007, 7:18pm » |
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Maybe he meant daft.
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cool_joh
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on Sep 1st, 2007, 6:18pm, ThudanBlunder wrote:| cool_joh, not a bad answer for someone who doesn't know the difference between 'dave' and 'deaf'. |
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I have solved this a few months ago. But in Indonesian (I'm really poor in English) 
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srn437
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Re: WHICH IS LARGER?
« Reply #13 on: Sep 25th, 2007, 5:03pm » |
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Calculaters don't help in these questions since it makes such high numbers become something e+something(whatever that represents).
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ThudnBlunder
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Re: WHICH IS LARGER?
« Reply #14 on: Sep 25th, 2007, 5:46pm » |
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on Sep 25th, 2007, 5:03pm, srn347 wrote:| Calculaters don't help in these questions since it makes such high numbers become something e+something(whatever that represents). |
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Such questions are not meant to be solved with calculators.
e+some integer means 10 to the power of 'some integer'.
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srn437
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Re: WHICH IS LARGER?
« Reply #15 on: Sep 25th, 2007, 7:49pm » |
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So it's scientific notation. Why don't the calculaters just show the exponent in the form of an exponent?
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mikedagr8
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Re: WHICH IS LARGER?
« Reply #16 on: Sep 25th, 2007, 8:06pm » |
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on Sep 25th, 2007, 7:49pm, srn347 wrote:| So it's scientific notation. Why don't the calculaters just show the exponent in the form of an exponent? |
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Well, if you plan to display every single digit, go for it. The calculator isn't stupid enough to waste it's time, so it just simplifies it, making it easier for all.
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srn437
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Re: WHICH IS LARGER?
« Reply #17 on: Sep 25th, 2007, 8:09pm » |
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I meant why not represent n10x as n10x instead of n+ex.
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Sameer
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Re: WHICH IS LARGER?
« Reply #18 on: Sep 25th, 2007, 8:48pm » |
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on Sep 25th, 2007, 8:09pm, srn347 wrote:| I meant why not represent n10x as n10x instead of n+ex. |
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The E in calculator is not same as the constant e. But you should know that already, right?
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srn437
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Re: WHICH IS LARGER?
« Reply #19 on: Sep 25th, 2007, 8:55pm » |
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I know. It isn't the same as 14 in base 15 or higher either. But why use it instead of exponents?
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ima1trkpny
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Re: WHICH IS LARGER?
« Reply #20 on: Sep 25th, 2007, 10:34pm » |
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on Sep 25th, 2007, 8:55pm, srn347 wrote:| I know. It isn't the same as 14 in base 15 or higher either. But why use it instead of exponents? |
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Because if you know E means multiplied by 10 to the power of some integer, the format of say 6E-7 is less work than showing it as 6(10-7). It is just a simplification.
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| « Last Edit: Sep 25th, 2007, 10:37pm by ima1trkpny » |
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Obob
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Re: WHICH IS LARGER?
« Reply #21 on: Sep 27th, 2007, 9:38pm » |
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Calculators use E instead of exponents because simple calculators with digital displays can't possibly express exponents (i.e. superscripts), and because the E notation takes less room. When space is at a premium (as it is on an 8 to 10 digit calculator display) it is important to make the best use possible of the space given.
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