wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> hard >> Prove this!
        
(Message started by: pcbouhid on Nov 25^th, 2005, 10:55am)

Title: Prove this!
Post by pcbouhid on Nov 25^th, 2005, 10:55am

Prove that if m > n, where m and n are rational numbers:

(1 + 1/m)^m > (1 + 1/n)^n

Title: Re: Prove this!
Post by Michael_Dagg on Nov 25^th, 2005, 11:21am

When m=2 > n=1, (3/2)^1 > (2)^2 is not true.

Now it holds.

Title: Re: Prove this!
Post by pcbouhid on Nov 25^th, 2005, 11:36am

Sorry, I forgot.........my glasses. The problem is fixed.

Title: Re: Prove this!
Post by JohanC on Nov 25^th, 2005, 12:02pm

It is clear that neither m nor n may be equal to 0.
But I also get the impression that negative numbers aren't allowed?

Title: Re: Prove this!
Post by Eigenray on Nov 25^th, 2005, 2:00pm

Just consider two savings accounts with the same interest rate, but one compounded more often than the other.

Title: Re: Prove this!
Post by Icarus on Nov 26^th, 2005, 7:42am

If m=1, (1+1/m)^m = 2
If n=-2, (1+1/n)ⁿ = 4

So the condition has to be for m > n > 0. Also, the function f(x) = (1+1/x)^x is continuous for x>0, so there is no reason for the restriction to rational values only.

Therefore the result desired is that the function f is strictly increasing. This in turn is equivalent to saying that g(x) = log f(x) is strictly increasing.

g'(x) = log(1+1/x) - 1/(1+x). Note that g'(x) --> 0 as x --> oo.

Now g''(x) = -1/x(1+x)², which is < 0 when x>0. Therefore g'(x) is strictly decreasing, and so g'(x) > 0 for all x > 0.

And so g(x), and f(x) are strictly increasing.

Title: Re: Prove this!
Post by Eigenray on Nov 26^th, 2005, 1:47pm

Here's another way (that works when m/n is rational): if y is an integer, consider
(1+z/y)^y = 1 + z + (y-1)/2 z² + ...
= [sum]_k=0^y (^yC_k)z^k/y^k
The k-th term is
z^k/k! *1(1-1/y)(1-2/y)...(1-(k-1)/y),
which is obviously increasing as a function of y. Since the number of terms also increases, the sum is strictly increasing, and therefore
(1+z/y)^y/z
is also strictly increasing as a function of y, with z fixed. But since we can write m=x/z, n=y/z over a common denominator, with x>y, the result follows.