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Topic: Prove this! (Read 424 times) |
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pcbouhid
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Prove this!
« on: Nov 25th, 2005, 10:55am » |
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Prove that if m > n, where m and n are rational numbers:
(1 + 1/m)^m > (1 + 1/n)^n
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| « Last Edit: Nov 25th, 2005, 11:36am by pcbouhid » |
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Michael Dagg
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Re: Prove this!
« Reply #1 on: Nov 25th, 2005, 11:21am » |
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When m=2 > n=1, (3/2)^1 > (2)^2 is not true.
Now it holds.
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| « Last Edit: Nov 25th, 2005, 12:43pm by Michael Dagg » |
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Michael Dagg
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pcbouhid
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Re: Prove this!
« Reply #2 on: Nov 25th, 2005, 11:36am » |
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Sorry, I forgot.........my glasses. The problem is fixed.
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JohanC
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Re: Prove this!
« Reply #3 on: Nov 25th, 2005, 12:02pm » |
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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?
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Eigenray
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Re: Prove this!
« Reply #4 on: Nov 25th, 2005, 2:00pm » |
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Just consider two savings accounts with the same interest rate, but one compounded more often than the other.
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Icarus
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Re: Prove this!
« Reply #5 on: Nov 26th, 2005, 7:42am » |
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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.
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Eigenray
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Re: Prove this!
« Reply #6 on: Nov 26th, 2005, 1:47pm » |
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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=0y (yCk)zk/yk
The k-th term is
zk/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.
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