is it true that....

[Ken Quirici]

Suppose a function f(x) is continuous in [a,b] and differentiable in
(a,b).
Is it true that f'(x) is continuous in (a,b)? I would suppose yes,
since the limit that defines f'(x) at a point x0 approaches it from
both
above and below, and is equal from both directions.

Regards,

Ken

[Mike Terry]

"Ken Quirici" <kqui...@yahoo.com> wrote in message
news:281efd1a-8187-4084...@db5g2000vbb.googlegroups.com...

No - you can make f oscillate rapidly near a point c in (a,b) so that f'(x)
does not converge as x-->c, and yet f is enveloped in such a way to ensure
f'(c) exists. It's a question of arranging for f to oscillate quicker than
the envelope is flattening the function, if you see what I'm trying to
say...

Mike.

> Regards,
>
> Ken

[Gus Gassmann]

Standard examples for this sort of thing are functions f_n(x) = x^n
sin(1/x). f_2 is differentiable at 0 but not continuously
differentiable.

[David C. Ullrich]

On Tue, 7 Feb 2012 14:44:05 -0800 (PST), Ken Quirici
<kqui...@yahoo.com> wrote:

>Suppose a function f(x) is continuous in [a,b] and differentiable in
>(a,b).
>Is it true that f'(x) is continuous in (a,b)?

No. Say f(x) = x^2 sin(1/x) for x <> 0, f(0) = 0.
Then f is differentiable on all of R, but the derivative
f' is not continuous at the origin.

>I would suppose yes,
>since the limit that defines f'(x) at a point x0 approaches it from
>both
>above and below, and is equal from both directions.

I don't see what that has to do with it. Ok, you just said
you "suppose"... Anyway, whatever argument you have
in mind here is evidently wrong.

>Regards,
>
>Ken

[Frederick Williams]

As you've described it, f_n is not defined at 0.

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

[Ken Quirici]

On Feb 8, 9:29 am, David C. Ullrich <ullr...@math.okstate.edu> wrote:
> On Tue, 7 Feb 2012 14:44:05 -0800 (PST), Ken Quirici
>

I thought a function was a continuous at a point if it approached the
same (finite) value from above and below.

Regards,

Ken

[Ken Quirici]

On Feb 7, 7:43 pm, "Mike Terry"
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "Ken Quirici" <kquir...@yahoo.com> wrote in message

your f must be continuous at c tho, right, since differentiability =>
continuity ? I thought increasingly rapid oscillation towards a point
c
was a way of creating or describing a singularity. I must think about
your interesting picture which I bet can have many variations with
differing (no pun intended) differentiability and continuity outcomes.

But at least I think I see what my mistake was, which I made again in
my reply just now to David Ullrich - that I was confusing the
convergence
of f(x)-f(c)/(x-c) to f'(c) with the convergence of f(x) to f(c).

[Gus Gassmann]

On Feb 8, 2:13 pm, Frederick Williams <freddywilli...@btinternet.com>
wrote:

> Gus Gassmann wrote:
>
> > On Feb 7, 6:44 pm, Ken Quirici <kquir...@yahoo.com> wrote:
> > > Suppose a function f(x) is continuous in [a,b] and differentiable in
> > > (a,b).
> > > Is it true that f'(x) is continuous in (a,b)? I would suppose yes,
> > > since the limit  that defines f'(x) at a point x0 approaches it from
> > > both
> > > above and below, and is equal from both directions.
>
> > > Regards,
>
> > > Ken
>
> > Standard examples for this sort of thing are functions f_n(x) = x^n
> > sin(1/x). f_2 is differentiable at 0 but not continuously
> > differentiable.
>
> As you've described it, f_n is not defined at 0.

Quite so. I hang my head in shame. Of course I needed to add that
f_n(0) = 0. So:

Standard examples for this sort of thing are functions f_n of the form

f_n(x) = x^n sin(1/x) for x =/= 0
= 0 for x = 0.

[Gus Gassmann]

On Feb 9, 9:00 pm, Ken Quirici <kquir...@yahoo.com> wrote:

> I thought a function was a continuous at a point if it approached the
> same (finite) value from above and below.

There might be a kernel of truth in that, but it is definitely too
sloppy to be useful.

First, a function is not necessarily a function of a single variable;
"above" and "below" may not make sense in that case. Second, a
function cannot be continuous at a point unless it is defined at that
point and the function value agrees with the limit.

Third, a function can be _dis_continuous for several reasons: the
function value at the point considered is not equal to the limit, or
the function values approach different limits if you approach the
point considered in different ways, or any of the limits do not exist.
The derivative of the function David Ullrich gave you fails to be
continuous at x=0 for this last reason.

[Mike Terry]

"Ken Quirici" <kqui...@yahoo.com> wrote in message

news:32336935-650a-47ae...@m5g2000yqk.googlegroups.com...

Right. Others have supplied definitions of precise functions doing what I
was describing, but in all cases the motivation for them is as I described.
It definitely helps to draw it all out. E.g. taking c=0 :

1) draw an enveloping function above and below the x-axis.
e.g. first try y = |x|, along with it's reflection below the x-axis
(this won't work, but is instructive)
2) then draw in the oscillating function f between the positive and
negative branches of the enveloping function. The function needs
to oscillate quicker as it approaches 0...

With the suggested enveloping function, observe that f is forced to be
continuous at 0 because it is between the two branches of the enveloping
function that meet at (0,0). However this example fails to work because f
will not be differentiable at x=0. (Check you understand why...)

We can mend this with a better enveloping function...

So now try y = x^2 (and it's reflection below the x-axis) as the enveloping
function. This is better, because regardless of how you draw f, the
envelope forces f to be both continuous and differentiable at x=0. (with
f'(0) = 0. Again make sure you see why...)

However, now there's a danger that f' will turn out to be continuous at 0 if
you've drawn the oscillations so that it's not oscillating quickly enough
(compared to the envelope approaching zero)!

So as you suspect, the general approach gives many possible behaviours
depending on the enveloping function and the oscillating rate. To get a
function that oscillates quicker as it approaches zero, sin(x^-n) is a
standard approach, for a suitable positive n. [Note this is undefined at
x=0, so when defining f you will have to specify the value at zero
separately]. Just multiply this by your choice of envelope function and
play with the alternatives to get the behaviour you're after... :)

Mike.

[Virgil]

In article
<b4eef268-6a87-43a0...@p21g2000yqm.googlegroups.com>,

Not quite! It must also have that finite value at that point.

There is some question as to whether f(x) = x^2 sin(1/x) is properly
defined/continuous at x = 0 (due to the 1/x appearing in it) without
additionally explicitly requiring that f(0) = 0.
--

[David C. Ullrich]

More or less. Yes, f is continuous at a if f approaches f(a) from
above and below.

So what? I don't see what your point is - nothing I said
contradicts what you say here (or rather nothing I said
contradicts the slightly corrected version of what you said.)

>
>Regards,
>
>Ken

[Tonico]

On Feb 10, 5:55 am, Virgil <vir...@ligriv.com> wrote:
> In article
> <b4eef268-6a87-43a0-8d7d-4b4a128a9...@p21g2000yqm.googlegroups.com>,

I don't think there's much question about the defined thing: the
function f(x) = x^2 sin(1/x) is not, and cannot be, defined at x = 0
since substituing this value into the function renders division by
zero, which cannot be done within the usual frame of rules and laws
(I won't even get into that discussion here...).

Now, as f(x) --> 0 when x --> O we can define A NEW FUNCTION F(x)
in such a way that F(x):= f(x), if x is not 0, and F(x) = 0 for x =
0 , and then F(x) is properly, an correctly defined AND continuous on
the whole real line R.

As far as I see it, x^2 sin(1/x) or sin x/x are not "automatically"
defined in some definite way at zero just because they can be
continuous there (i.e., since their limit when x --> 0 exists in a
finite way), though some authors seem to imply that whenever a
removable discontinuity is found for a function we must/should/will/
have to actually remove it and "redefine" the function at that point
in such a way that it is continuous there.

Tonio

[Ken Quirici]

On Feb 9, 9:49 pm, "Mike Terry"

Ah! I was about to reply w/ further confusion when I noticed that the
enveloping
function(s) (above and below) meet at (0,0). Now I can see the
continuity
(the limits from above and below approach the same value - no matter
how close
you want to be to y=0, you can find an x0 such that everything closer
than that to 0 is
closer than your chosen y 'closeness'. Really poor language but I
think I get the
point. But now I have to work on the failure of differentiability.

[David C. Ullrich]

There is no failure of differentiability. The function is
differentiable everywhere; the derivative is not continuous.

[Ken Quirici]

On Feb 11, 11:10 am, David C. Ullrich <ullr...@math.okstate.edu>
wrote:

right. I meant that the function was continuous, not the derivative.
The
existence of the derivative of a function at a point => the function
is