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I've got a function whose general form is

$f(x) = \frac{1}{x^\alpha}$

where $x > 0$ and $\alpha > 0$. I would like to show that if $0< \alpha < 1$ $f(x)$ has slow decay and if $\alpha > 1$ the $f(x)$ has rapid decay. (I've already verified these properties of $f(x)$ using a graphing application.)

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1  
Well, what definitions of "rapid" and "slow decay" are you using? –  Mariano Suárez-Alvarez Oct 27 '10 at 17:28
    
Slow/fast relative to what? –  J. M. Oct 27 '10 at 17:29
    
I'm trying to understand why $\sum_{x = 0}^\infty f(x) = \infty$ if $\alpha > 1$ and why $\sum_{x = 0}^\infty f(x) < \infty$ if $0 < \alpha < 1$. (If it matters, this is in relation to long range dependence -- en.wikipedia.org/wiki/Long_range_dependence.) –  Olumide Oct 27 '10 at 17:43
    
If it is the convergence/divergence of the series you are looking for then you should read about the "integral test". –  AD. Oct 27 '10 at 18:18
    
Thanks. A search for the integral test led me to Paul Dawkins' notes (tutorial.math.lamar.edu/Classes/CalcII/IntegralTest.aspx) which I am currently studying. But I'd like to ask if (in the absence of growth) divergence implies slow decay in some sense. –  Olumide Oct 27 '10 at 22:15

3 Answers 3

If you start at 1 (your sum is not defined at x=0), you can bound the sums as integrals: $ \int_{x=2}^{\infty}x^{-\alpha} dx \lt \sum_{x = 1}^\infty x^{-\alpha} \lt \int_{x=1}^{\infty}x^{-\alpha} dx$. One will solve it for $\alpha \le 1$ and one for $\alpha \gt 1$

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Another approach is to use Cauchy's condensation test which states that a series $\sum a_n$, where the terms $a_n$ are positive and decreasing, is convergent if and only if $\sum 2^m a_{2^m}$ is convergent.

In this example the line $\mathbf{Re}(s)=1$ is the edge of the region of ansolute convergence of the usual formula $\sum_{n=1}^\infty 1/n^s$ for the Riemann zeta function.

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According to the Wikipedia article, the terms $a_{n}$ should be non-increasing (which, of course, is not necessarily 'decreasing'). –  Robert Smith Oct 28 '10 at 0:58

I'm probably talking nonsense but

$\int_1^\infty \frac{1}{x^\alpha} = \left. \frac{x^{1 - \alpha}}{1 - \alpha} \right|_1^\infty$

The integral diverges if $1 - \alpha > 0$ i.e. if $\alpha < 1$, but converges if $1 - \alpha < 0$ i.e. $\alpha > 1$.

My difficulty now is in showing that the integral is a bound on the sum, which from what I've studied can only be done if the terms of the sum are taken as follows alt text

instead of

alt text

which is more natural and intuitive.

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You are not talking nonsense at all. That is the reason for the change in the lower limit in my post. The first integral will overestimate the sum and can be used to prove convergence. The second integral will underestimate the sum and can be used to prove divergence. In fact, when you do an integral like this, it really doesn't matter where you start, as the fact of convergence only depends upon what happens at infinity. But to use the bounding theorem, it is nice to be strictly greater or less. –  Ross Millikan Oct 27 '10 at 23:25

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