# Math Help - Can 2 functions be equal for a continuous range?

1. ## Can 2 functions be equal for a continuous range?

I've been wondering this for a while. Given two C-infinity functions f(x) and g(x) such that f(x) != g(x), is it possible for f(x) = g(x) for some continuous range (a, b)?

Basically, I first wanted to know if it was possible for any functions, then realized that there are cheap ways out like absolute value functions and defining the function to be different for different ranges. That's why I'm asking specifically about C-infinity functions (all derivatives, and derivatives of derivatives, and so-on are continuous).

2. Well, what about two functions that have equal asymptotes as they approach infinity?

3. Originally Posted by 1005
Well, what about two functions that have equal asymptotes as they approach infinity?
No; for example f(x)=1/x and g(x)=1/x^2.

4. there are smooth functions where it is identically 1 in [-1, 1] and 0 in (- \infinite, -2] and [ 2, +\infinite)

5. Originally Posted by xxp9
there are smooth functions where it is identically 1 in [-1, 1] and 0 in (- \infinite, -2] and [ 2, +\infinite)
Which functions are those?

6. Originally Posted by JvJ
I've been wondering this for a while. Given two C-infinity functions f(x) and g(x) such that f(x) != g(x), is it possible for f(x) = g(x) for some continuous range (a, b)?

Basically, I first wanted to know if it was possible for any functions, then realized that there are cheap ways out like absolute value functions and defining the function to be different for different ranges. That's why I'm asking specifically about C-infinity functions (all derivatives, and derivatives of derivatives, and so-on are continuous).
$f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}$

7. Originally Posted by Jose27
$f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}$
This actually looks promising. It's still a piecewise function, but it looks like all derivatives approach 0 from the right, and it will never be discontinuous.

8. Originally Posted by JvJ
This actually looks promising. It's still a piecewise function, but it looks like all derivatives approach 0 from the right, and it will never be discontinuous.
It is $C^{\infty}$ and this can be proven by noting that $\lim_{x\rightarrow \infty} p(x)e^{-x}=0$ where $p(x)$ is any polynomial.

9. Originally Posted by Jose27
$f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}$
It is famous Cauchy function... $f^{(n)}(0)=0$ for every $n\in \mathbb{N}$

10. Which, by the way, shows that the Taylor's series of a function does NOT necessarily converge to that function! The Taylor's series, about x= 0, for this function. is just "0" and converges to 0 for all x.

Of course, this function is not "analytic" in any neighborhood of 0.