# Well behaved functions

• Sep 18th 2007, 08:21 PM
myoplex11
Well behaved functions
which of the following are not well behaved funtions:
a)ax^2 + bx+c
b)e^x for all x
c)y= |x|
d)g(x) = cosx + sinx for -pie=< x =< pie and 0 elsewhere
• Sep 18th 2007, 08:28 PM
ThePerfectHacker
Quote:

Originally Posted by myoplex11
which of the following are not well behaved funtions:
a)ax^2 + bx+c
b)e^x for all x
c)y= |x|
d)g(x) = cosx + sinx for -pie=< x =< pie and 0 elsewhere

"Well-behaved" is a term mathematicians use because they are lazy to state all the conditions a function must meet. Hence this term is not well-defined. You need to say what it means over here.
• Sep 19th 2007, 04:56 AM
topsquark
Quote:

Originally Posted by myoplex11
which of the following are not well behaved funtions:
a)ax^2 + bx+c
b)e^x for all x
c)y= |x|
d)g(x) = cosx + sinx for -pie=< x =< pie and 0 elsewhere

(sigh) (Fubar)
Would you eat $\pi$? No. So don't spell it "pie," spell it correctly: "pi."

Okay, this might be what you need: $y = |x|$ is not "smooth." That is to say all the derivatives of y are not continuous. (In fact no derivative of this function is continuous.) In this sense $y = |x|$ is not well behaved.

-Dan
• Sep 19th 2007, 01:10 PM
CaptainBlack
Quote:

Originally Posted by myoplex11
which of the following are not well behaved funtions:
a)ax^2 + bx+c
b)e^x for all x
c)y= |x|
d)g(x) = cosx + sinx for -pie=< x =< pie and 0 elsewhere

Quote:

Originally Posted by topsquark
(sigh) (Fubar)
Would you eat $\pi$? No. So don't spell it "pie," spell it correctly: "pi."

Okay, this might be what you need: $y = |x|$ is not "smooth." That is to say all the derivatives of y are not continuous. (In fact no derivative of this function is continuous.) In this sense $y = |x|$ is not well behaved.

-Dan

d) is not even continuous at +/- pi

RonL