Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Growth order of function.

  1. Apr 1, 2010 #1
    let be the integral equation

    [tex] h(x)= \int_{0}^{\infty} \frac{dy}{y}K(y/x)f(y) [/tex]

    here the kernel is always a positive , then if [tex] h(x)=O(x^{a}) [/tex] and the integral


    [tex] \int_{0}^{\infty} \frac{dy}{y}K(y)y^{a} [/tex] exists and is a positive real number then also [tex] f(x)= O(x^{a}) [/tex]
     
  2. jcsd
  3. Apr 1, 2010 #2
    I'll restrict to positive x as the negative case can be tackled similarly.
    With a change of variable y/x =z , we get
    [tex] \frac{h(x)}{x^{a}} = \int_{0}^{\inftz} \frac{dz}{z}K(z)z^{a}F(xz)[/tex]

    where F(t) =f(t) /t^a. As [tex] h(x)=O(x^{a}) [/tex] , the right hand side is bounded,O(1).
    I don't see why F =O(1) . F could tend to infinity much slower than
    [tex] K(z)z^{a -1}[/tex] & the first equation could still hold.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Growth order of function.
  1. Growth of functions (Replies: 5)

  2. Order of element (Replies: 1)

  3. Growth rate (Replies: 0)

Loading...