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| Stephen J. Herschkorn |
 Posted: Mon Dec 26, 2005 2:38 pm |
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Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
--
Stephen J. Herschkorn sjherschko@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan |
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| Rouben Rostamian |
| Posted: Mon Dec 26, 2005 2:38 pm |
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In article <u1m0r1tj1h80gcts8nu7m421on57rs8r3k@4ax.com>,
quasi <quasi@null.set> wrote:
[quote:24ef9f2a52]On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
I don't see it.
Can you show more of the details?
[/quote:24ef9f2a52]
No need for the mean value theorem. Restrict the domain to [0,1)
and note that since f'(x) goes to zero as x goes to zero, we have:
For any e>0 there exists b > 0 such that:
|f'(x)| < e whenever 0 < x < b.
Then for any 0 < a < b:
|f(b) - f(a)| = |int f'(x) from a to b|
< int |f'(x)| from a to b
< (b-a) e.
Now let a->0, then divide by b, then let b->0 to conclude f'(0)=0.
Repeat the same argument on (-1,0].
--
Rouben Rostamian |
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| Rouben Rostamian |
| Posted: Mon Dec 26, 2005 2:38 pm |
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In article <hou0r19sd8r0m97le2muar9ci9tqafaojm@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
[quote:c2abd2e576]On Mon, 26 Dec 2005 21:28:07 +0000 (UTC), rouben@pc18.math.umbc.edu
(Rouben Rostamian) wrote:
Then for any 0 < a < b:
|f(b) - f(a)| = |int f'(x) from a to b|
Hold on a minute there - why do you think that
f(b) - f(a) = int f'(x) from a to b ? That's
not true for just any differentiable function;
the FTC has _hypotheses_.
[/quote:c2abd2e576]
Mea culpa. I wasn't thinking. Thanks for straightening me out.
--
Rouben Rostamian |
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| The World Wide Wade |
| Posted: Mon Dec 26, 2005 3:02 pm |
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In article
<1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
[quote:5561ab4779]Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
[/quote:5561ab4779]
Yes, by the mean value theorem. |
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| quasi |
| Posted: Mon Dec 26, 2005 3:54 pm |
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On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
<waderameyxiii@comcast.remove13.net> wrote:
[quote:4d7d6e53b0]In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
[/quote:4d7d6e53b0]
I don't see it.
Can you show more of the details?
quasi |
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| Ron Sperber |
| Posted: Mon Dec 26, 2005 4:42 pm |
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quasi wrote:
[quote:5bf8cbf3f9]On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
I don't see it.
Can you show more of the details?
quasi
I think the argument is something like this. We consider[/quote:5bf8cbf3f9]
lim (f(h)-f(0))/h
h->0
We wish to show this limit is 0. Let e>0. Since f'(x)->0 as x->0, we can
find a d>0 so that if |x|<d, |f'(x)|<e. Suppose 0<|h|<d. If h>0 then we
have f continuous on [0,h], differentiable on (0,h) so by the MVT there
is a c in (0,h) with f'(c)=(f(h)-f(0))/h. Since c is in (0,h)
0<|c|<|h|<d, so |f'(c)|<e. So for 0<h<d, |(f(h)-f(0))/h|<e. When h<0 one
has a virtually identical argument. Thus the limit is 0 and f'(0)=0.
-Ron |
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| Rob Johnson |
| Posted: Mon Dec 26, 2005 5:04 pm |
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In article <1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
[quote:2e4bdf90cc]Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
[/quote:2e4bdf90cc]
This is pretty close to the question at
<http://groups.google.com/groups?threadm=1134589103.378861.41010@z14g2000cwz.googlegroups.com>
The types of discontinuities of a derivative are limited, and if
lim f'(x) = A
x->a
and f is continuous at a, then f'(a) = A.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying |
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| David C. Ullrich |
| Posted: Mon Dec 26, 2005 6:24 pm |
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On Mon, 26 Dec 2005 21:28:07 +0000 (UTC), rouben@pc18.math.umbc.edu
(Rouben Rostamian) wrote:
[quote:5f5fb8e5ee]In article <u1m0r1tj1h80gcts8nu7m421on57rs8r3k@4ax.com>,
quasi <quasi@null.set> wrote:
On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
I don't see it.
Can you show more of the details?
No need for the mean value theorem. Restrict the domain to [0,1)
and note that since f'(x) goes to zero as x goes to zero, we have:
For any e>0 there exists b > 0 such that:
|f'(x)| < e whenever 0 < x < b.
Then for any 0 < a < b:
|f(b) - f(a)| = |int f'(x) from a to b|
[/quote:5f5fb8e5ee]
Hold on a minute there - why do you think that
f(b) - f(a) = int f'(x) from a to b ? That's
not true for just any differentiable function;
the FTC has _hypotheses_.
On the other hand the MVT _is_ true assuming
only that f is differentiable. This is why you
see it used in a lot of proofs; the, um, need
for MVT arises because it applies under weaker
hypotheses than things like FTC.
[quote:5f5fb8e5ee]int |f'(x)| from a to b
(b-a) e.
Now let a->0, then divide by b, then let b->0 to conclude f'(0)=0.
Repeat the same argument on (-1,0].
[/quote:5f5fb8e5ee]
************************
David C. Ullrich |
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| The World Wide Wade |
| Posted: Mon Dec 26, 2005 6:27 pm |
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In article <dopn97$bg7$1@pc18.math.umbc.edu>,
rouben@pc18.math.umbc.edu (Rouben Rostamian) wrote:
[quote:3b4ac0d621]In article <u1m0r1tj1h80gcts8nu7m421on57rs8r3k@4ax.com>,
quasi <quasi@null.set> wrote:
On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
I don't see it.
Can you show more of the details?
No need for the mean value theorem. Restrict the domain to [0,1)
and note that since f'(x) goes to zero as x goes to zero, we have:
For any e>0 there exists b > 0 such that:
|f'(x)| < e whenever 0 < x < b.
Then for any 0 < a < b:
|f(b) - f(a)| = |int f'(x) from a to b|
int |f'(x)| from a to b
(b-a) e.
Now let a->0, then divide by b, then let b->0 to conclude f'(0)=0.
Repeat the same argument on (-1,0].
[/quote:3b4ac0d621]
Unfortunately your MVT-free proof suffers in two ways: 1. It's
more complicated. 2. It's wrong. |
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| The World Wide Wade |
| Posted: Mon Dec 26, 2005 6:29 pm |
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Guest
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In article <u1m0r1tj1h80gcts8nu7m421on57rs8r3k@4ax.com>,
quasi <quasi@null.set> wrote:
[quote:612c36c425]On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
I don't see it.
Can you show more of the details?
[/quote:612c36c425]
[f(x) - f(0)]/(x-0) = f'(c_x) by the MVT. As x -> 0, c_x -> 0,
which by hypothesis implies f'(c_x) -> 0. |
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| quasi |
| Posted: Mon Dec 26, 2005 6:42 pm |
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On Mon, 26 Dec 2005 15:29:32 -0800, The World Wide Wade
<waderameyxiii@comcast.remove13.net> wrote:
[quote:7cfb09a493]In article <u1m0r1tj1h80gcts8nu7m421on57rs8r3k@4ax.com>,
quasi <quasi@null.set> wrote:
On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
I don't see it.
Can you show more of the details?
[f(x) - f(0)]/(x-0) = f'(c_x) by the MVT. As x -> 0, c_x -> 0,
which by hypothesis implies f'(c_x) -> 0.
[/quote:7cfb09a493]
But you want to prove f'(0) exists.
You already know that lim f'(x) --> 0 as x --> 0. That's part of the
hypothesis.
Thus, establishing that f'(c_x) --> 0 seems unproductive.
quasi |
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| The World Wide Wade |
| Posted: Mon Dec 26, 2005 7:01 pm |
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In article <sjv0r1halu60n0apifo8p0u7cc0alof88n@4ax.com>,
quasi <quasi@null.set> wrote:
[quote:13e2a2dc66]On Mon, 26 Dec 2005 15:29:32 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article <u1m0r1tj1h80gcts8nu7m421on57rs8r3k@4ax.com>,
quasi <quasi@null.set> wrote:
On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
I don't see it.
Can you show more of the details?
[f(x) - f(0)]/(x-0) = f'(c_x) by the MVT. As x -> 0, c_x -> 0,
which by hypothesis implies f'(c_x) -> 0.
But you want to prove f'(0) exists.
You already know that lim f'(x) --> 0 as x --> 0. That's part of the
hypothesis.
Thus, establishing that f'(c_x) --> 0 seems unproductive.
[/quote:13e2a2dc66]
I see we're having a little too much eggnog during the holidays.
[f(x)-f(0)]/(x-0) = f'(c_x), right? So if f'(c_x) -> 0, then ... |
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| quasi |
| Posted: Mon Dec 26, 2005 7:17 pm |
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On Mon, 26 Dec 2005 16:01:23 -0800, The World Wide Wade
<waderameyxiii@comcast.remove13.net> wrote:
[quote:c430544ecf]In article <sjv0r1halu60n0apifo8p0u7cc0alof88n@4ax.com>,
quasi <quasi@null.set> wrote:
On Mon, 26 Dec 2005 15:29:32 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article <u1m0r1tj1h80gcts8nu7m421on57rs8r3k@4ax.com>,
quasi <quasi@null.set> wrote:
On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
I don't see it.
Can you show more of the details?
[f(x) - f(0)]/(x-0) = f'(c_x) by the MVT. As x -> 0, c_x -> 0,
which by hypothesis implies f'(c_x) -> 0.
But you want to prove f'(0) exists.
You already know that lim f'(x) --> 0 as x --> 0. That's part of the
hypothesis.
Thus, establishing that f'(c_x) --> 0 seems unproductive.
I see we're having a little too much eggnog during the holidays.
[f(x)-f(0)]/(x-0) = f'(c_x), right? So if f'(c_x) -> 0, then ...
[/quote:c430544ecf]
then lim f'(x) --> 0 as x --> 0.
What am I missing?
quasi |
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| quasi |
| Posted: Mon Dec 26, 2005 8:48 pm |
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On Mon, 26 Dec 2005 19:17:04 -0500, quasi <quasi@null.set> wrote:
[quote:e5585b3ffd]On Mon, 26 Dec 2005 16:01:23 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article <sjv0r1halu60n0apifo8p0u7cc0alof88n@4ax.com>,
quasi <quasi@null.set> wrote:
On Mon, 26 Dec 2005 15:29:32 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article <u1m0r1tj1h80gcts8nu7m421on57rs8r3k@4ax.com>,
quasi <quasi@null.set> wrote:
On Mon, 26 Dec 2005 12:02:55 -0800, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
1135625922.654371.258210@g43g2000cwa.googlegroups.com>,
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote:
Let f be a continuous, real-valued function on the open interval
(-1, 1). Suppose f is differentiable on (-1, 0) U (0, 1) and that
f'(x) approaches 0 as x approaches 0. Is f necessarily
differentiable at 0?
Yes, by the mean value theorem.
I don't see it.
Can you show more of the details?
[f(x) - f(0)]/(x-0) = f'(c_x) by the MVT. As x -> 0, c_x -> 0,
which by hypothesis implies f'(c_x) -> 0.
But you want to prove f'(0) exists.
You already know that lim f'(x) --> 0 as x --> 0. That's part of the
hypothesis.
Thus, establishing that f'(c_x) --> 0 seems unproductive.
I see we're having a little too much eggnog during the holidays.
[f(x)-f(0)]/(x-0) = f'(c_x), right? So if f'(c_x) -> 0, then ...
then lim f'(x) --> 0 as x --> 0.
What am I missing?
[/quote:e5585b3ffd]
I see it now -- sorry I was so dense on this -- a blind spot, I guess.
It's not the eggnog, though -- perhaps it's lack of sleep.
Ok, so to finish the exercise (you've led me right up to the punch
line):
[f(x)-f(0)]/(x-0) = f'(c_x) for some c_x in (0,x).
But since c_x --> 0, it follows, by hypothesis, that f'(c_x) -> 0
Hence [f(x)-f(0)]/(x-0) --> 0 as x --> 0, so f'(0)=0.
It is a clean, simple argument -- I should have seen it right away.
In any case, thanks for leading the horse to water.
quasi |
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| Dave L. Renfro |
| Posted: Mon Dec 26, 2005 9:05 pm |
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The World Wide Wade wrote:
[quote:42afc085fd]I see we're having a little too much eggnog
during the holidays. [f(x)-f(0)]/(x-0) = f'(c_x),
right? So if f'(c_x) -> 0, then ...
[/quote:42afc085fd]
quasi wrote:
[quote:42afc085fd]then lim f'(x) --> 0 as x --> 0.
What am I missing?
[/quote:42afc085fd]
Well, actually you don't know that f'(x) --> 0 as x --> 0
simply from f'(c_x) --> 0 (the stronger limit, where x --> 0,
might not exist, while the weaker limit, where x approaches
zero in such a way that x is always equal to one of the c_x's,
might exist), although of course we do know that f'(x) --> 0
because this was in the original poster's hypothesis. [I'm not
suggesting that you didn't recognize this as in the original
hypothesis, by the way. Indeed, I assume this was part of
your confusion, namely that the argument seems to only prove
something we already knew. My point is that the argument
doesn't even do this as far as I can tell.]
I believe what's going on is that in the equation
[f(x)-f(0)] / (x-0) = f'(c_x),
if we take the limit as x --> 0 of both sides, then the
left side becomes the definition of f'(0) and the right
side becomes 0.
Dave L. Renfro |
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