Differentiability

[William Elliot]

A problem from
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst&task=list

Suppose f(x)= e^(-1/x^2) for x not equal to 0, and f(0)=0.

Without using l'hopital's rule, prove f is differentiable at 0 and
that f'(0)=0.

[Robin Chapman]

This is a very standard example.

A tougher problem would be to prove this *using* The Hospital's rule :-)

[David C. Ullrich]

On Wed, 20 Feb 2013 20:13:24 -0800, William Elliot <ma...@panix.com>
wrote:

You really _should_ try proving this using l'Hopital and see what
happens...

Hint: One way or another, show that

(*) e^x < x (x > 0).

For example, using the power series, or using the fact that

e^x - 1 = int_0^x e^t dt

or whatever.

Now what does (*) imply about e^(-x) for x > 0?

[William Elliot]

On Thu, 21 Feb 2013, David C. Ullrich wrote:
> > Suppose f(x)= e^(-1/x^2) for x not equal to 0, and f(0)=0.
> >
> > Without using l'hopital's rule, prove f is differentiable at 0 and
> > that f'(0)=0.
>
> You really _should_ try proving this using l'Hopital and see what
> happens...
>

f'(0) = lim(x->0) (exp -1/x^2)/x
= lim(x->0) 2(exp -1/x^2)/x^3
= lim(x->0) 4(exp -1/x^2)/x^3 * 1/2x^2

> Hint: One way or another, show that
>
> (*) e^x < x (x > 0).
>

Can't be done. e^x = 1 + x + x^2 / 2 + ... > x for x > 0.

> For example, using the power series, or using the fact that
> e^x - 1 = int_0^x e^t dt
> or whatever.
>

Oh?

> Now what does (*) imply about e^(-x) for x > 0?

Since Dexp is positive, exp is increasing
and exp -x decreasing. So for x > 0
exp -x < exp 0 = 1.

[David C. Ullrich]

On Fri, 22 Feb 2013 00:34:12 -0800, William Elliot <ma...@panix.com>
wrote:

>On Thu, 21 Feb 2013, David C. Ullrich wrote:
>> > Suppose f(x)= e^(-1/x^2) for x not equal to 0, and f(0)=0.
>> >
>> > Without using l'hopital's rule, prove f is differentiable at 0 and
>> > that f'(0)=0.
>>
>> You really _should_ try proving this using l'Hopital and see what
>> happens...
>>
>f'(0) = lim(x->0) (exp -1/x^2)/x
> = lim(x->0) 2(exp -1/x^2)/x^3
> = lim(x->0) 4(exp -1/x^2)/x^3 * 1/2x^2
>
>> Hint: One way or another, show that
>>
>> (*) e^x < x (x > 0).
>>
>Can't be done. e^x = 1 + x + x^2 / 2 + ... > x for x > 0.

What I wrote is so obviously false that you might consider
the possibility that it was a typo or something, and think
about what the correct version is. Of course I meant

(**) e^x > x (x > 0).

>> For example, using the power series, or using the fact that
>> e^x - 1 = int_0^x e^t dt
>> or whatever.
>>
>Oh?
>
>> Now what does (*) imply about e^(-x) for x > 0?
>
>Since Dexp is positive, exp is increasing
>and exp -x decreasing. So for x > 0
>exp -x < exp 0 = 1.

Which is no help with your question.
What does (**) say about e^(-x) for x > 0?

[William Elliot]

On Sat, 23 Feb 2013, David C. Ullrich wrote:

> >> > Suppose f(x)= e^(-1/x^2) for x not equal to 0, and f(0)=0.

> >> > prove f is differentiable at 0 and that f'(0)=0.

> >> Hint: One way or another, show that

> (**) e^x > x (x > 0).

> >> For example, using the power series, or using the fact that
> >> e^x - 1 = int_0^x e^t dt

> What does (**) say about e^(-x) for x > 0?
>

1/x < exp -x

f'(0+) = lim(x->0+) (exp -1/x^2)/x = 0
Proof
0 <= lim(x->0+) (exp -1/x^2)/x
= lim(x->0+) x(exp -1/x^2)/x^2
<= lim(x->0+) x(exp -1/x^2)^2 = 0

f'(0-) = lim(x->0-) (exp -1/x^2)/x = -lim(x->0+) (exp -1/x^2)/x = 0
f'(0) = f'(0+) = f'(0-) = 0

[David C. Ullrich]

On Sat, 23 Feb 2013 18:53:34 -0800, William Elliot <ma...@panix.com>
wrote:

>On Sat, 23 Feb 2013, David C. Ullrich wrote:
>
>> >> > Suppose f(x)= e^(-1/x^2) for x not equal to 0, and f(0)=0.
>> >> > prove f is differentiable at 0 and that f'(0)=0.
>
>> >> Hint: One way or another, show that
>
>> (**) e^x > x (x > 0).
>
>> >> For example, using the power series, or using the fact that
>> >> e^x - 1 = int_0^x e^t dt
>
>> What does (**) say about e^(-x) for x > 0?
>>

> [1] 1/x < exp -x

At first I assumed this was just a typo. But no, [1] is
exactly what you use below, so it must be what you
meant.

Inequality [1] is obvious nonsense. Consider small x > 0
and you see [1] implies ifinity < 0.

What actually follows from (**) is

[2] 1/x > exp(-x) (x > 0).

Because 0 < a < b implies 1/a > 1/b. Like 2 < 3
so 1/2 > 1/3.

>f'(0+) = lim(x->0+) (exp -1/x^2)/x = 0
>Proof
>0 <= lim(x->0+) (exp -1/x^2)/x
> = lim(x->0+) x(exp -1/x^2)/x^2
> <= lim(x->0+) x(exp -1/x^2)^2 = 0

Since [1] is wrong and you use [1] here I assumed at
first there must be an error. But no, this chain of
inequalities is correct (or rather would be correct if [1]
were correct).

The actual proof, using [2]: For any x <> 0, positive
or negative, [2] shows that

exp(-1/x^2) < x^2.

So

|exp(-1/x^2)/x| < x^2/|x| = |x|,

so the limit is 0.

Now for a minute I was puzzled how both [2] and
its opposite [1] could lead to a proof that f'(0) = 0.
The answer: You applied [1] for small x, while the
correct proof applies [2] for large x.

[smn]

for y>0 let g(y)=ye^(-y) ;its derivative has a 0 at y=1 ;is increasing for y<1 and decreasing for y>1 (eg the 3nd derivative of g is positive for all y>0.
Thus for y>1 , g(y)<g(1) so g(y)/y^1/2 = y^1/2 e^(-y)--> 0 as y---> +oo
Put 1/x^2 for y and let x-->0 (x>0) to get your result (the case x<0 follows since what we proved means that |f(x)/x|--> 0 as x-->0 (or |x|-->0). smn

[smn]

On Thursday, February 28, 2013 8:33:58 PM UTC-8, smn wrote:
> On Wednesday, February 20, 2013 8:13:24 PM UTC-8, William Elliot wrote:
>
> > A problem from
>
> >
>
> > http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst&task=list
>
> >
>
> >
>
> >
>
> > Suppose f(x)= e^(-1/x^2) for x not equal to 0, and f(0)=0.
>
> >
>
> >
>
> >
>
> > Without using l'hopital's rule, prove f is differentiable at 0 and
>
> >
>
> > that f'(0)=0.
>
>
>

> for y>0 let g(y)=ye^(-y) ;its derivative has a 0 at y=1 ;is increasing for y<1 and decreasing for y>1 (eg the 2nd derivative of g is positive for all y>0.