# Can 2 functions be equal for a continuous range?

• July 9th 2010, 11:39 AM
JvJ
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).
• July 9th 2010, 01:05 PM
1005
Well, what about two functions that have equal asymptotes as they approach infinity?
• July 9th 2010, 01:12 PM
Mathelogician
Quote:

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.
• July 9th 2010, 05:44 PM
xxp9
there are smooth functions where it is identically 1 in [-1, 1] and 0 in (- \infinite, -2] and [ 2, +\infinite)
• July 13th 2010, 07:32 AM
JvJ
Quote:

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?
• July 13th 2010, 07:42 AM
Jose27
Quote:

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).

$f(x)= \{ \begin{array}{cc} 0 \ \mbox{if} \ x\leq0 \\ e^{-\frac{1}{x^2}} \ \mbox{if} \ x>0 \end{array}$
• July 13th 2010, 09:34 AM
JvJ
Quote:

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.
• July 13th 2010, 09:39 AM
Jose27
Quote:

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.
• July 13th 2010, 09:51 AM
Also sprach Zarathustra
Quote:

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 $n\in \mathbb{N}$
• July 13th 2010, 01:03 PM
HallsofIvy
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.