Why is a function not differentiable at a point?

[DeadWolfe]

I'm only in high school, and I was wondering: Why are some functions not differentiable at certain points?

 

[mathman]

"Why" is a funny question. They just are. For example x.sin(1/x) does not have a derivative at x=0, even though it is continuous. There are more ccomplicated examples, like continuous functions with derivatives nowhere.

 

[DeadWolfe]

Originally posted by mathman
"Why" is a funny question. They just are. For example x.sin(1/x) does not have a derivative at x=0, even though it is continuous. There are more ccomplicated examples, like continuous functions with derivatives nowhere.

I guess I can understand how a derivitave f'(x) could have undefined values like any other function... But the thing is, the derivitave, is (as I was thought in my math 31 class) the slope of a tangent line.

My question: How can there be no tangent line? Or, how can it have no slope?

In the example you gave, the original function was undefined, so that doesn't help me that much.

 

[cookiemonster]

Consider the function y = |x|.

For x < 0, the slope is -1. For x > 0, the slope is 1. So what's the slope at 0?

Well, a slope of -1 would be tangent. But 1 would also be tangent. And so would 0! Right along with 1/2, -1/2, 1/4, .3225509809234, -.6548941623189435, etc. So which is it?

cookiemonster

 

[DeadWolfe]

That helps. Thanks.

 

[Chen]

What are continuous functions and how is y = |x| one?

 

[cookiemonster]

A continuous function is one whose graph you can draw without having to lift your pencil.

More precisely, it is a function with domain D such that

$$\lim_{x \to x_0} f(x) = f(x_0)\qquad \forall x \in D$$

cookiemonster

 

[Chen]

Originally posted by mathman
"Why" is a funny question. They just are. For example x.sin(1/x) does not have a derivative at x=0, even though it is continuous. There are more ccomplicated examples, like continuous functions with derivatives nowhere.

So how can $$f(x) = x\sin (x^{-1})$$ be a continuous function? I must lift my pencil because it is not defined at $$x = 0$$, is it?

 

[cookiemonster]

Actually, it is. It goes to 0 as x = 0.

The sin term is bounded by -1 and 1, but the x term goes to 0. 0 times -1 or 1 or anything inbetween is still 0.

cookiemonster

 

[matt grime]

It is continuous with the definition of it as 0 at 0.

any if x_i is any sequence tending to zero then x_i*sin(1/x_i) tends to zero too (sandwich principle).

but taking pen off paper is not the definition of discontiuous:

(x^2 -2)^-1 is continuous when the domain is the rational numbers.

 

[Chen]

Ok, thanks. For some reason in school when I asked about this the teacher said that continuous functions are functions whose derivative is defined for every x.

 

[cookiemonster]

Why should we have to impose the value of 0 at x = 0 on xsin(1/x) in order to make it continuous? The limit still exists, and that's what determines the continuity, right?

cookiemonster

 

[matt grime]

The function as given is not defined at 0. I can define it to be anything I care about. You can either state the function is continuous which implies the value must be zero there, or you can define the value is zero and then conclude it is continuous. Personal preference as to which comes first in your opinion.

 

[cookiemonster]

I'm not following you here.

"The function as given is not defined at 0. I can define it to be anything I care about."

and

"You can either state the function is continuous which implies the value must be zero there, or you can define the value is zero and then conclude it is continuous."

mean different things to me. The first sounds like I can define it to be whatever I want, like 5 for example, whereas the second seems to say it must be 0.

cookiemonster

 

[matt grime]

xsin(1/x) as given is not defined at zero. We can assign any value to it at the origin to make a complete function on the real line. The first time this came up it was stated that the function is continuous. That implicitly states that we chose the value zero there. Any other choice would lead to a discontinuous function. That's all.

 

[Chen]

The way I see it, at x = 0 the function either has a value of zero, or it has an asymptote. But if that's the case then as x approaches zero the function needs to tend to infinity, right? But that can't be right, because as cookiem said the values of sine are limited to be between 1 and -1, so $$-x \leq f(x) \leq x$$ for every x, right? So if x is very small how can the function tend to infinity?

 

[cookiemonster]

Hm... I must be dense because I'm still not following you.

Is

$$f(x) = \left\{ \begin{array}{11} 1 & x \neq 2 \\ 5 & x = 2 \end{array} \right.$$

continuous? Is it a fair analogy to our previous choice of function?

cookiemonster

 

[matt grime]

Two posts in 1.

the reason xsin(1/x) does not come with a value at zero is cos of the 1/x inside the sin.

the function you give cookiemonster is most definitely not a continuous function (assuming you're not doing something bizarre with the topology).

 

[Chen]

Originally posted by matt grime
the reason xsin(1/x) does not come with a value at zero is cos of the 1/x inside the sin.

Doesn't there have to be an asymptote wherever a function is not defined? I may well be wrong here because I never really studied advanced math...

 

[cookiemonster]

I should read what I write more often. Geez, I'd already explained the whole thing to myself in one neat little equation.

Sorry for the trouble, matt.

cookiemonster

 

[matt grime]

Originally posted by Chen
Doesn't there have to be an asymptote wherever a function is not defined? I may well be wrong here because I never really studied advanced math...

All that is required is for there not to be a definition of the function at that point.

sin(1/x) is not trivially defined at zero unless we declare its value to be something specific; no assignment of a value makes it continuous at zero. The real valued function sqrt(x) is not defined for x negative, there are no sinuglarities in sight.

 

[HallsofIvy]

Originally posted by cookiemonster
Why should we have to impose the value of 0 at x = 0 on xsin(1/x) in order to make it continuous? The limit still exists, and that's what determines the continuity, right?

cookiemonster

No, that's not"what determines continuity". To be continuous at a point, a, a function must have 3 properties:
1) lim(x->a)f(x) exists
2) f(a) exists
3) lim(x->a)f(x)= f(a).

For example, if I define f(x)= x for x not equal to 1 and f(1)= 5,
the limit, as x approaches 1, exists (it is equal to 1) and the function is defined at x=1 (and has value 5) but the function is not continuous at x=1 because the two values are not the same.

Similarly, if I define f(x)= (x²-4)/(x-2), it is easy to see that the limit, as x approaches 2, exists and is equal to 4. This f is not continuous at 2 because f(2) does not exists (0/0 is undetermined).

Both of these are example of a "removable discontinuity" because we can make a continuous function by redefining f at a single point. (In the first example, define f(1)= 1, in which case the function is exactly f(x)= x. In the second, define f(2)= 4 so the function is exactly f(x)= x+ 2.

 

[cookiemonster]

Yeah, I kinda realized that when I read my own definition. Thanks for the effort, though.

You have to admit, it's pretty impressive that somebody can throw out the definition of something and then change it in their own mind for no apparent reason not five minutes later.

cookiemonster

 

[HallsofIvy]

Originally posted by cookiemonster
Yeah, I kinda realized that when I read my own definition. Thanks for the effort, though.

You have to admit, it's pretty impressive that somebody can throw out the definition of something and then change it in their own mind for no apparent reason not five minutes later.

cookiemonster

Actually, "impressive" is not the word that springs to my mind.

 

[cookiemonster]

Unusual? Strange? Annoying? Peculiar? Interesting? Or do I not want to know?

cookiemonster

 

[exekil]

"the function you give cookiemonster is most definitely not a continuous function (assuming you're not doing something bizarre with the topology)."

I know I'm going to be so confused at the answer for this, but in any case, what is topology and how could that be used to make that equation continuous?

 

[matt grime]

There is an alternative definition (the 'proper one') that states: a function between topological spaces is continuous iff the inverse image of the open set is open.

As a map between R and R with the usual topology that map given is not continuous. If we were considering it as a map from R with the discrete topology to the two point set with the discrete topology (or R with the discrete topology) then it is continuous. This was just an observation so that no one can pull out the 'ha you've not considered this...' argument, and is just a cya, cos mathematicians are like that., just as we like to consider every possibility, just in case.

 

[exekil]

Originally posted by matt grime
There is an alternative definition (the 'proper one') that states: a function between topological spaces is continuous iff the inverse image of the open set is open.

As a map between R and R with the usual topology that map given is not continuous. If we were considering it as a map from R with the discrete topology to the two point set with the discrete topology (or R with the discrete topology) then it is continuous. This was just an observation so that no one can pull out the 'ha you've not considered this...' argument, and is just a cya, cos mathematicians are like that., just as we like to consider every possibility, just in case.

increadibly enough I understood some of that, it sounds very interesting, what kind of math does topology fall into? Did I just completely forgat about all of topology or do you lear that after calc.?

 

[lethe]

Originally posted by exekil
increadibly enough I understood some of that, it sounds very interesting, what kind of math does topology fall into? Did I just completely forgat about all of topology or do you lear that after calc.?

i usually consider topology an equal sibling to the other basic kinds of pure math. it's just that topology is only about 80 to 100 years old, whereas algebra is, what, 700? 1000? years old, geometry is 2000 or 3000 years old. even analysis is at least 400 years old.

so other branches of math are classical, and people learn them in school. topology is newer, so i don't think it has had enough time to filter down into lower grades. it will though, at least, i believe so.

and in higher education, topology is ubiquitous in math and physics.

 

[HallsofIvy]

I've always thought of topology as being geometry in its most general aspect: The two really main "umbrellas" in mathematics are algebra and geometry or, in other words, "discrete" and "continuous".
Analysis uses both.