Does this imply that lim x --> oo f'(x) = 0?

[steine...@gmail.com]

Suppose f:[0, oo) --> R is increasing, differentiable and has a finite limit as x --> oo. Then, must we have lim x --> oo f'(x) = 0? I guess not, but couldn't find a counter example.

Thank you

[Robin Chapman]

Think of a function that's flat, then rapidly jerks upward then
flat again then rapidly jerks up and does this infinitely
often ....

[dull...@sprynet.com]

In particular, jerks upward very fast, but on a very short
interval ("short" interval relative to how fast is fast).
Then flat on a very long interval. Then jerks upward
even faster, on an even shorter interval...

Showing that f' need not even be bounded.

[Bart Goddard]

steine...@gmail.com wrote in
news:287a9b10-3790-4437...@googlegroups.com:

Something like f(x) = sum_1^infinity arctan(2^n ( x-2^n) )/2^n should

work. f'(x) is a sum of terms like 1/(1 + (2^n x -2^(2n))^2.
f'(2^n)=1 plus some small positive terms. But f'(2^n+2^(n-1)) should
be pretty close to zero.

[James Waldby]

On Thu, 23 May 2013 18:55:08 +0000, Bart Goddard wrote:
> steine...@gmail.com wrote ...

>
>> Suppose f:[0, oo) --> R is increasing, differentiable and has a finite
>> limit as x --> oo. Then, must we have lim x --> oo f'(x) = 0? I guess
>> not, but couldn't find a counter example.

> Something like f(x) = sum_1^infinity arctan(2^n ( x-2^n) )/2^n should
>
> work. f'(x) is a sum of terms like 1/(1 + (2^n x -2^(2n))^2.
> f'(2^n)=1 plus some small positive terms. But f'(2^n+2^(n-1)) should
> be pretty close to zero.

As another example, f(x) = ln(x)*sin(x*x)/x goes to zero and
d(f(x))/dx goes to +/-oo as x goes to +oo, since it is dominated
by ln(x)*cos(x*x).

<http://www.wolframalpha.com/input/?i=d%28ln%28x%29*sin%28x*x%29%2Fx%29%2Fdx>
shows a graph for small x of d(f(x))/dx = d/dx((log(x) sin(x x))/x)
= (2 x^2 log(x) cos(x^2)-(log(x)-1) sin(x^2))/x^2.

--
jiw

[Bart Goddard]

James Waldby <n...@valid.invalid> wrote in
news:knm0hh$ugq$3...@dont-email.me:

The OP wanted an _increasing_ function.

B.

[steine...@gmail.com]

Thank you all.

The difficult thing is giving an example with f increasing. If we don't require f to be increasing, another simple example is f(x) = (sinx^2))/x if x <> 0, f(0) = 0

[William Elliot]

On Thu, 23 May 2013, steine...@gmail.com wrote:

> Suppose f:[0, oo) --> R is increasing, differentiable and has a finit

> limit as x --> oo. Then, must we have lim x --> oo f'(x) = 0?

Most likely. What would happen if lim(x->oo) f'(x) /= 0?
What would happen if the limit was infinite or didn't exist.

[bacle...@gmail.com]

On Thursday, May 23, 2013 11:11:12 AM UTC-4, steine...@gmail.com wrote:
> Suppose f:[0, oo) --> R is increasing, differentiable and has a finite limit as x --> oo. Then, must we have lim x --> oo f'(x) = 0? I guess not, but couldn't find a counter example.
>
>
>
> Thank you

I think so; use the MVThm repeatedly. Starting in [0,1]:

f(1)-f(0)=f'(c1)*1 , for c in (0,1)

f(2)-f(1)=f'(c2)*1 ; c in (1,2)

...........

f(n+1)-f(n)=f'(cn)*1

_______________________

Now, if f approaches a finite limit at oo , then , as n-->oo f(n+1)-f(n) =f'(cn) -->0 .

[Graham Cooper]

On May 24, 7:09 am, James Waldby <n...@valid.invalid> wrote:
> On Thu, 23 May 2013 18:55:08 +0000, Bart Goddard wrote:

> > steinerar...@gmail.com wrote ...

>
> >> Suppose f:[0, oo) --> R is increasing, differentiable and has a finite
> >> limit as x --> oo. Then, must we have lim x --> oo f'(x) = 0?  I guess
> >> not, but couldn't  find a counter example.
> > Something like f(x) = sum_1^infinity  arctan(2^n ( x-2^n) )/2^n should
>
> > work.  f'(x) is a sum of terms like 1/(1 + (2^n x -2^(2n))^2.
> > f'(2^n)=1 plus some small positive terms.  But f'(2^n+2^(n-1)) should
> > be pretty close to zero.
>
> As another example, f(x) = ln(x)*sin(x*x)/x goes to zero and
> d(f(x))/dx goes to +/-oo as x goes to +oo, since it is dominated
> by ln(x)*cos(x*x).
>

> <http://www.wolframalpha.com/input/?i=d%28ln%28x%29*sin%28x*x%29%2Fx%2...>

> shows a graph for small x of d(f(x))/dx = d/dx((log(x) sin(x x))/x)
> = (2 x^2 log(x) cos(x^2)-(log(x)-1) sin(x^2))/x^2.
>
> --
> jiw

Like a fractal coastline, if you zoom in at high x

the oscillation continues at all scales!

I have to plot some sine waves for the Lonely Runner Conjecture..

http://blockprolog.com/runner.php

Click [RENDER]


Herc

[William Elliot]

On Thu, 23 May 2013, bacle...@gmail.com wrote:
> On Thursday, May 23, 2013 11:11:12 AM UTC-4, steine...@gmail.com wrote:
> > Suppose f:[0, oo) --> R is increasing, differentiable and has a finite limit as x --> oo. Then, must we have lim x --> oo f'(x) = 0? I guess not, but couldn't find a counter example.
>

> I think so; use the MVThm repeatedly. Starting in [0,1]:
>
> f(1)-f(0)=f'(c1)*1 , for c in (0,1)
>
> f(2)-f(1)=f'(c2)*1 ; c in (1,2)
> ...........
>
> f(n+1)-f(n)=f'(cn)*1

> Now, if f approaches a finite limit at oo , then , as n-->oo f(n+1)-f(n)
> =f'(cn) -->0 .

This can't be right because there's counter examles when f isn't monotone
and nowhere do you use monotonicity.

The problem is proving c_n -> 0 while all that you have is c_n in [0,1].
Even with that, the continuity of f' is needed to complete the proof.

[William Elliot]

No, it doesn't.

For x in [0,1], let f(x) = 0.
For x in [n, n+1/2], let f(x) = 1 - 1/n.

For x in [n+1/2, n+1],
. . let f(x) = 1 - 1/n + (x - n + 1/2)(1/n - 1/(n+1)).

To assure f' exists, round the corners at n, n + 1/2, for all n in N.

To make f strictly increasing, slope slightly upwards the horizontal
portion. gtest

[bacle...@gmail.com]

On Friday, May 24, 2013 3:28:09 AM UTC-4, William Elliot wrote:

> On Thu, 23 May 2013, bacail.com wrote:
>
> > On Thursday, May 23, 2013 11:11:12 AM UTC-4, steine...@gmail.com wrote:
>
> > > Suppose f:[0, oo) --> R is increasing, differentiable and has a finite limit as x --> oo. Then, must we have lim x --> oo f'(x) = 0? I guess not, but couldn't find a counter example.
>
> >
>
> > I think so; use the MVThm repeatedly. Starting in [0,1]:
>
> >
>
> > f(1)-f(0)=f'(c1)*1 , for c in (0,1)
>
> >
>
> > f(2)-f(1)=f'(c2)*1 ; c in (1,2)
>
> > ...........
>
> >
>
> > f(n+1)-f(n)=f'(cn)*1
>
>
>
> > Now, if f approaches a finite limit at oo , then , as n-->oo f(n+1)-f(n)
>
> > =f'(cn) -->0 .
>
>
>
> This can't be right because there's counter examles when f isn't monotone

> and nowhere do you use monotonicity.

How about this: with the same lay out as before: f(n+1)-f(n)=f'(cn).

Then, f(n)-f(0) telescopes to f'(c1)+f'(c2)+....+f'(cn) ; cn in (n,n+1).

Now, f(n)-f(0) is bounded, and increasing, then f'(cn)>0 , so that the

partial sums f'(cj) are a monotone

(since f'(cn)>0 ) bounded sequence, which then converges ( to the LUB of the

sum) , so that the tail --and so the individual terms -- goes to zero.

[Graham Cooper]

On May 24, 5:42 pm, William Elliot <ma...@panix.com> wrote:

OK, so you approach the limit in smaller and smaller bumps!

Another self-similar fractal coastline but increasing!

Any plot?

...


The un-differentiated fractal curve from above!

http://www.wolframalpha.com/input/?i=ln%28x%29*sin%28x*x%29%2Fx


Herc

[steine...@gmail.com]

If inf{f'(x) | x > = 0} > 0, then f(x) --> oo even if f' doesn't have a limit at oo. But if this infimum is 0, then f can have a finite limit at oo. Since f is increasing, the infimum is always >= 0. This can be seen by MVT.

Artur

[William Elliot]

On Fri, 24 May 2013, bacle...@gmail.com wrote:
> On Friday, May 24, 2013 3:28:09 AM UTC-4, William Elliot wrote:

> > > > Suppose f:[0, oo) --> R is increasing, differentiable and has a
> > > > finite limit as x --> oo. Then, must we have lim x --> oo f'(x) =
> > > > 0? I guess not, but couldn't find a counter example.

> How about this: with the same lay out as before: f(n+1)-f(n)=f'(cn).

Give it up, counter examples have been presented.

[Graham Cooper]

On May 25, 12:50 pm, William Elliot <ma...@panix.com> wrote:

I think this one works..

-1/(5+sin(x))/x/x


http://www.wolframalpha.com/input/?i=-1%2F%285%2Bsin%28x%29%29%2Fx%2Fx




Herc
--
www.BLoCKPROLOG.com

[Graham Cooper]

Close, but no cigar?


Herc

[steine...@gmail.com]

No, because f' takes on positive and negative values.

[bacle...@gmail.com]

On Friday, May 24, 2013 8:28:07 PM UTC-7, Graham Cooper wrote:
> On May 25, 12:50 pm, William Elliot <ma...@panix.com> wrote:
>
> > On Fri, 24 May 2013, baclesb...@gmail.com wrote:
>
> > > On Friday, May 24, 2013 3:28:09 AM UTC-4, William Elliot wrote:
>
> > > > > > Suppose f:[0, oo) --> R is increasing, differentiable and has a
>
> > > > > > finite limit as x --> oo. Then, must we have lim x --> oo f'(x) =
>
> > > > > > 0?  I guess not, but couldn't find a counter example.
>
> > >  How about this: with the same lay out as before: f(n+1)-f(n)=f'(cn).
>
> >
>
> > Give it up, counter examples have been presented.

Maybe you should check your counterexamples more carefully.

[bacle...@gmail.com]

On Saturday, May 25, 2013 10:25:30 PM UTC-7, bacle...@gmail.com wrote:
> On Friday, May 24, 2013 8:28:07 PM UTC-7, Graham Cooper wrote:
>
> > On May 25, 12:50 pm, William Elliot <ma...@panix.com> wrote:
>
> >
>
> > > On Fri, 24 May 2013, baclesb...@gmail.com wrote:
>
> >
>
> > > > On Friday, May 24, 2013 3:28:09 AM UTC-4, William Elliot wrote:
>
> >
>
> > > > > > > Suppose f:[0, oo) --> R is increasing, differentiable and has a
>
> >
>
> > > > > > > finite limit as x --> oo. Then, must we have lim x --> oo f'(x) =
>
> >
>
> > > > > > > 0?  I guess not, but couldn't find a counter example.
>
> >
>
> > > >  How about this: with the same lay out as before: f(n+1)-f(n)=f'(cn).
>
> >
>
> > >
>
> >
>
> > > Give it up, counter examples have been presented.
>
>
>
> Maybe you should check your counterexamples more carefully.

Morover, you told me you did not see where monotonicity was used, and

I showed you where, or that I produce an argument where monotonicity

was used, and this is what I did. So don't ask me for something and then

complain when I answer your question.