Which is Bigger: y=2013! or z=1007^{2013}?

[medwatt]

Hello,
I want to verify which is bigger between:
y=2013! and z=1007$^{2013}$

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=$\sum x-x^{2}/2 + x^{3}/6 . . .$

But I am getting road blocks.

 

[mfb]

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!.

 

[medwatt]

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.

 

[jbunniii]

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?

 

[coolul007]

1007 seems to be cleverly chosen to be the mid term of 2013

 

[chiro]

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

 

[coolul007]

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

 

[medwatt]

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 ?

 

[jbunniii]

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 $b_n = 2014-n$, 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.

 

[medwatt]

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.

 

[jbunniii]

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.

I could be wrong, but it is my impression that most competition problems don't rely on a very deep knowledge of any particular field of mathematics, such as number theory, but rather on your ability to apply basic facts such as standard inequalities in clever ways. Of course, you have to first have the "aha" moment when you see what to do, which requires luck (probably the case for me with this problem) and/or a lot of practice doing these sorts of problems.

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.

 

[medwatt]

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 !

 

[mathsman1963]

This can be decided by induction. First, note that 2013=2*1007-1. Let s=1007 then the terms of the problem may be written as s^2s-1 and
(2s-1)!

Test the values s=1 and s=2. Use induction on s.

 

[coolul007]

I also came up with noting that 2013! has mid term 1007. Each number preceding and following are (1007-1) and (1007+1). The next set are (1007-2) 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.

 

[mfb]

This can be decided by induction. First, note that 2013=2*1007-1. Let s=1007 then the terms of the problem may be written as s^2s-1 and
(2s-1)!

Test the values s=1 and s=2. Use induction on s.

How do you compare $(2s+1)! < (s+1)^{2s+1}$ with the help of $(2s-1)! < s^{2s-1}$?
It is easy to show that $(2s+1)! < (2s+1)(2s)*s^{2s-1}$, 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.

 

[Ratch]

medwatt,

Which is bigger ?

--------------------------------------------------------------------------------
Hello,
I want to verify which is bigger between:
y=2013! and z=10072013

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=x−x2/2+x3/6...

But I am getting road blocks.

You are getting roadblocks because your hand calculator cannot handle those hellaciously high powers that Stirling's formula and "z" require. Therefore, you have to break down each of those numbers into bitesized chunks. Here is the way to do it. First, notice that 2013 can be factored into 33 x 61. Start with Stirling's formula and find (2013/e)^33 = 4.955947203E94 . Then take 4.955947203^61*1E94^61 and get 2.52770819010E5776 . Next complete Stirlings formula by multiplying by sqrt(2*pi*2013) to get 2.84275053010E5778. Now that is one big number. Do the same for "y". 1007^33 = 1.25884578010E99 . And finally 1.258845780^61*10E99^61 = 1.25407707910E6045 .

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

 

[uperkurk]

I won't reveal any actual numbers but I can tell you that

2013! has 5779 digits whereas 1007²⁰¹³ has 6046 digits.

 

[Ratch]

uperkurk,

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

Ratch

 

[uperkurk]

uperkurk,

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

Ratch

Yes sorry I didn't read your post. Besides I just wanted to say which was bigger without actually revealing to the OP how to find the answer, leaving the working out to him.