RE: Math Problem

From: Lee Corbin (lcorbin@tsoft.com)
Date: Wed Feb 12 2003 - 00:31:03 MST

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    Zahir writes

    > sounds like proof by deduction.

    I thought that you meant "induction", and indeed I
    had been thinking of integers. A proof similar to
    Hal's but not so general does indeed work for
    integers. But maybe you did mean "deduction" because

    Hal writes
    >
    > > I am very sure that for n >= 5, the nth root of n+1
    > > is less than (2n)/(n+1). But how can it be shown in
    > > a nice way?
    >
    > As n approaches infinity, the nth root of n+1 approaches 1, while
    > (2n)/(n+1) approaches 2. So obviously this will be true for some
    > sufficiently large n. You can establish it manually for 5. The
    > remaining piece would be to show that (n+1)**(1/n) is a decreasing
    > function for numbers of this size, which I think you could show by
    > taking the derivative.
    >
    > My calculus is a little rusty so I will take the natural log of this
    > expression instead, which is ln(n+1)/n, and I will show that it is
    > decreasing (I remember how to take the derivative of fractions but not
    > of exponentials). The derivative of this is (n/(n+1) - ln(n+1))/n^2,
    > which I claim is negative. For it to be negative ln(n+1) > n/n+1,
    > which will be true if ln(n+1) > 1 since n/n+1 < 1. ln(n+1) > 1 if n+1 > e,
    > hence n > 2. So at least for values n > 2, the nth root of n+1 is a
    > decreasing function and the whole thing follows.

    Thanks for your impeccable logic.

    Lee



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