
#1
Feb2713, 11:18 AM

P: 106

Hello,
I want to verify which is bigger between: y=2013! and z=1007[itex]^{2013}[/itex] What I've tried was taking log on both equations and subtracting one from the other. I represented y=log(2013)+log(2012)+...log(1) by its equivalent Taylor series of y=[itex]\sum xx^{2}/2 + x^{3}/6 . . . [/itex] But I am getting road blocks. 



#2
Feb2713, 11:23 AM

Mentor
P: 10,801

You can write both numbers as products and look for some clever way to group the factors.
Alternatively, use the Stirling formula to approximate 2013!. 



#3
Feb2713, 12:56 PM

P: 106

I know I can use Stirling's Formula but this a sort of a proof and according to what was written there is a way to group them.




#4
Feb2713, 01:10 PM

Sci Advisor
HW Helper
PF Gold
P: 2,907

Which is bigger ?
Try applying this inequality to appropriate pairs of factors:
$$ab \leq \left(\frac{a + b}{2}\right)^2$$ Under what conditions on ##a## and ##b## does equality hold? 



#6
Feb2713, 08:12 PM

P: 4,570

If you can use a computer, then just program a simple loop and use a high enough precision data type to make a comparison.




#7
Feb2713, 10:32 PM

P: 234

2013! == 0 mod(2014), 1007^2013 == 1007 mod(2014) or is this not right because 2014 is a composite??
The above proves n! is less than 1007^2013, so please ignore 



#8
Feb2713, 11:50 PM

P: 106

Actually I got this question from our school olympiad and was one of the questions I couldn't solve in the allocated time.
Of course I had an idea that the 2013 was chosen to mean this year and the 1007 was there because of this relationship: a!<((a+1)/2)^2 I think there is some number theory proof for that ? 



#9
Feb2813, 12:16 AM

Sci Advisor
HW Helper
PF Gold
P: 2,907

OK, since it's not homework I'll elaborate on my hint. Write ##2013 = (1)(2)\cdots(2013)##. Let ##a_1## and ##b_1## be the first and last factor, that is, ##a_1 = 1## and ##b_1 = 2013##. Then ##(a_1 + b_1)/2 = 2014/2 = 1007##. We may apply the inequality
$$a_1b_1 \leq \left(\frac{a_1 + b_1}{2}\right)^2$$ to conclude that ##a_1 b_1 \leq (1007)^2##. Now repeat this process with the next pair of factors, ##a_2 = 2## and ##b_2 = 2012##. Once again, ##(a_2+b_2)/2 = 1007## and we again have $$a_2b_2 \leq \left(\frac{a_2 + b_2}{2}\right)^2 = (1007)^2$$ In general, let ##a_n = n## and ##b_n = 2014n##, for ##n = 1,2,\ldots,1006##. For each ##n##, we have $$a_nb_n \leq (1007)^2$$ Thus $$\prod_{n=1}^{1006} a_n b_n \leq (1007)^{2(1006)} = (1007)^{2012}$$ I'll let you finish the proof. 



#10
Feb2813, 09:58 AM

P: 106

Thanks . . . OMG it looks a lot like the Gauss proof for sum of AP series !! That was neat !! I feel so foolish now !!
BTW if you can be a bit more generous by providing me with some advices. Actually I had no business participating in the competition because it was for mathematics students and I am in the engineering stream where as you know maths is more lenient. But after feeling complacent with Calc 1,2,3 with vector calc . . Linear Algebra . . Differential Eq . . Complex Analysis . . . I thought why not see where I stand. Ok . . . I am planning for another assault next year and I hope you can give me an advice, being an engineer, what is there to be done to make the transition from a guy used to modeling systems to one scattering a problem. Will reading a book on Number Theory help or one in real analysis and if so do you have anything in mind ?? I hope you take my request seriously and thank you. 



#11
Feb2813, 12:04 PM

Sci Advisor
HW Helper
PF Gold
P: 2,907

So if your goal is to participate in competitions, I think your time would be best spent practicing the types of problems that show up in competitions. There are many books containing such problems along with their solutions. I'm sure if you ask in the "Academic Guidance / Science Textbook Discussion" section, people will be able to make recommendations. 



#12
Feb2813, 02:35 PM

P: 106

Indeed when I saw the technique you applied, I almost hid myself under the bed out of shame. To say I went all the way to use Gamma/Beta functions is an understatement !!




#13
Mar113, 04:37 AM

P: 25

This can be decided by induction. First, note that 2013=2*10071. Let s=1007 then the terms of the problem may be written as s^{2s1} and
(2s1)! Test the values s=1 and s=2. Use induction on s. 



#14
Mar113, 08:59 AM

P: 234

I also came up with noting that 2013! has mid term 1007. Each number preceding and following are (10071) and (1007+1). The next set are (10072) and 1007+2).
multiplying each set is 1007^2  1^2 which is less than 1007^2, the following set is 1007^2  2^2 which is less than 1007^2. Getting the same result as all the above. 



#15
Mar113, 09:43 AM

Mentor
P: 10,801

It is easy to show that ##(2s+1)! < (2s+1)(2s)*s^{2s1}##, but then you have to mess around with the base and the exponent to include the prefactors. It is probably possible in some way, but it looks quite complicated. 



#16
Mar313, 12:56 AM

P: 315

medwatt,
As you can see, 1007^2013 is very much greater than 2013! . So by using the technique of raising the factors of each number to the factors of the powers, very large results can be calculated with the limited range of a hand calculator. Ratch 



#17
Mar313, 04:14 AM

P: 159

I won't reveal any actual numbers but I can tell you that
2013! has 5779 digits whereas 1007^{2013} has 6046 digits. 



#18
Mar313, 08:47 AM

P: 315

uperkurk,
Isn't that what I calculated in the post just before yours, where I calculated the actual numbers? Ratch 


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