Function to satisfy this integral for all parameters?

Click For Summary

Discussion Overview

The discussion revolves around the possibility of finding a function f(x) that satisfies a specific integral condition involving arbitrary parameters a and b. The integral in question is defined as \(\int_{-\infty}^\infty e^{-x^2}f(ax+b)\mathrm{d}x=0\), with additional conditions on the behavior of f at infinity.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant wonders if a function f(x) can be found that meets the integral condition for arbitrary parameters a and b, suggesting that f should oscillate and satisfy f(±∞) = 0.
  • Another participant argues that while f(x) = 0 satisfies the integral, it is likely not the intended solution. They explain that changing variables reveals the integral's dependence on the value of f at b, implying that f must be zero everywhere for the integral to be zero for all a and b.
  • A third participant inquires whether estimates for a and b are known, suggesting that this could influence the search for a suitable function.
  • Another participant proposes redefining f(ax + b) as g(x) and suggests that if g is an antisymmetrical function, it could satisfy the integral condition, although they later acknowledge that their initial example does not meet the requirement of f(±∞) = 0.
  • They also suggest that a function symmetrical about x = 0, multiplied by x, could potentially satisfy the conditions.

Areas of Agreement / Disagreement

Participants express differing views on the existence of a suitable function f(x) that meets the integral condition. There is no consensus on a definitive solution, and the discussion includes various hypotheses and approaches.

Contextual Notes

Participants note the importance of the behavior of f at infinity and the implications of the parameters a and b on the integral's outcome. There are unresolved questions regarding the continuity and oscillatory nature of potential functions.

Gerenuk
Messages
1,027
Reaction score
5
I wonder:

Is it possible to find a function f(x) which satisfies this integral for *arbitrary* parameters a,b (but maybe of known domain)?

[tex]\int_{-\infty}^\infty e^{-x^2}f(ax+b)\mathrm{d}x=0[/tex]

I want [tex]f(\pm\infty)=0[/tex] and I can deduce
[tex]\int_{-\infty}^\infty x^n e^{-x^2}f^{(m)}(ax+b)\mathrm{d}x=0[/tex]

It would have to be oscillary I guess?!
 
Last edited:
Physics news on Phys.org
Yes, and no. Obviously, f(x)=0 satisfies the criterion, but this is probably not what you were looking for. If you do a change of variable in the expression, you'll see that the integrand is exp(-((x-b)/a)^2)*f(x), i.e., a Gaussian times some function. However, if you let a and b be arbitrary, you're allowing the Gaussian's position and width to vary as well, and by making a arbitrarily large, you can make the Gaussian as thin as you want. The net effect is that the integral would "pick out" the value of f(b), so unless f is 0 everywhere, the integral cannot be 0 everywhere.

This, of course, assumes that f is well-behaved. Any function which did satisfy the criterion would be nowhere continuous, as it would have to oscillate infinitely at every point.
 
And I an estimate for b and a is known? Say it's in a range around some a0 and b0.
 
Since coming here I have found questions I could answer, but from the other answers it looked like I hadn't understood the question. :redface: With that reservation:

Firstly, why f(ax + b) ? Call it g(x). If you can get that you can translate into a function of (ax + b) afterwards.

Now [tex]e^{-x^2}[/tex] is symmetrical about x = 0. So if g is any continuous and finite function of x that is antisymmetrical i.e. g(x) = - g(-x) , and defined from - to + infinity, it will satisfy your requirement won't it? Simplest example g(x) = x.

If this shows I have not understood the question (guess no-one would ask one that easy) someone please tell me as other times I was left suspended.:confused::smile:
 
Last edited:
Ps

:redface: sorry, g(x) = x doesn't satisfy your [tex]f(\pm\infty)=0[/tex] condition, but you can easily find one that does, e.g. any function symmetrical about x = 0 that does, multiplied by x.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K