# Thread: prove that this function differentiable endles times..

1. ## prove that this function differentiable endles times..

prove that this function differentiable endles times on x=0 ??

i tried to prove for the first derivative
but i dont get a final limit here
$
f(x)=e^\frac{-1}{x^2} \\
$

$
f'(0)=\lim_{x->0^+}\frac{e^\frac{-1}{x^2}-0}{x-0} \\
$

$
f'(0)=\lim_{x->0^-}\frac{e^\frac{-1}{x^2}-0}{x-0}
$

2. Originally Posted by transgalactic
prove that this function differentiable endles times on x=0 ??

i tried to prove for the first derivative
but i dont get a final limit here
$
f(x)=e^\frac{-1}{x^2} \\
$

$
f'(0)=\lim_{x->0^+}\frac{e^\frac{-1}{x^2}-0}{x-0} \\
$
note that this is the same as $\lim_{x \to 0^+} \frac {\frac 1x}{e^{1/x^2}}$

now apply L'Hopital's

$
f'(0)=\lim_{x->0^-}\frac{e^\frac{-1}{x^2}-0}{x-0}
$
coming from the left, the function is zero. you have the limit of (0 - 0)/x = 0

3. $
\lim_{x \to 0^+} \frac {\frac 1x}{e^{1/x^2}}
$

after the lhopital law i still dont have a finite limit
$
\lim_{x \to 0^+} \frac {-\frac{1}{x^2}}{-2\frac{1}{x^3}e^{1/x^2}}
$

??

4. Originally Posted by transgalactic
$
\lim_{x \to 0^+} \frac {\frac 1x}{e^{1/x^2}}
$

after the lhopital law i still dont have a finite limit
$
\lim_{x \to 0^+} \frac {-\frac{1}{x^2}}{-2\frac{1}{x^3}e^{1/x^2}}
$

??
well, we could simplify to get $\frac x{2e^{1/x^2}}$ and use the fact exponentials grow faster than polynomials to conclude the limit goes to zero.

if you do not like that, there is always the option of expressing $e^{-1/x^2}$ as a power series from the begining, or in the new expression we got from L'Hopital's. the same conclusion follows

5. ok we prove the this function is differentiable for the first derivative

ho to proove that its differentiable on every other derivative too

how to find the n'th derivative??

should i use macloren series for that?

6. Originally Posted by transgalactic
ok we prove the this function is differentiable for the first derivative

how to prove that its differentiable on every other derivative too

how to find the n'th derivative??

should i use macloren series for that?
yes, that is fine. once we can express the function as a power series around zero and it is differentiable at zero, we know it is infinitely differentiable, and we can find it's derivative by differentiating the power series term by term. do you have to actually come up with a formula for the nth derivative? if so, it is still beneficial to start with the power series

7. I think there is something wrong in the question !
Every function is differentiable endles times !!

f(x)=3x^2
f(x)=6x
f(x)=6
f(x)=0
..=0
..=0 ------> Because 0 is a costant !! so its DIFF is 0

8. Originally Posted by TWiX
I think there is something wrong in the question !
Every function is differentiable endles times !!

f(x)=3x^2
f(x)=6x
f(x)=6
f(x)=0
..=0
..=0 ------> Because 0 is a costant !! so its DIFF is 0
not necessarily if we consider a certain point as we are doing here. recall, for instance, |x| is NOT differentiable at zero... not even once. and in general, a function is not differentiable where it is not continuous

9. Yes am talking about the conts. function

10. the makloren formula is
$
f(x)=\sum_{n}^{k}\frac{f^{(k)}}{k!}x^k+o(x^n)
$

how to take the n'th derivative from here??

where to put the formula in and get the n'th derivative??

this function
is for approximating so in order for me to get to the n'th members approximation
first i need to do manually one by one n times derivative
so its not helping me
??

11. Originally Posted by TWiX
Yes am talking about the conts. function

yes, i realize that. but continuity is not enough. the example i gave, |x|, is continuous everywhere

Originally Posted by transgalactic
the makloren formula is
$
f(x)=\sum_{n}^{k}\frac{f^{(k)}}{k!}x^k+o(x^n)
$

how to take the n'th derivative from here??

where to put the formula in and get the n'th derivative??

this function
is for approximating so in order for me to get to the n'th members approximation
first i need to do manually one by one n times derivative
so its not helping me
??
no approximations, use the actual power series. once you have it, differentiate it a few times to see a pattern

12. i cant see any pattern here
??
$
f(x)=e^\frac{-1}{x^2} \\
$

$
f'(x)=\frac x{2e^{1/x^2}}\\
$

$
f''(x)=\frac{2e^{1/x^2}+4xe^{1/x^2}\frac{1}{x^3}}{4e^{2/x^2}}=\frac{1+2\frac{1}{x^2}}{2e^{1/x^2}}
$

13. Originally Posted by transgalactic
i cant see any pattern here
??
$
f(x)=e^\frac{-1}{x^2} \\
$

$
f'(x)=\frac x{2e^{1/x^2}}\\
$

$
f''(x)=\frac{2e^{1/x^2}+4xe^{1/x^2}\frac{1}{x^3}}{4e^{2/x^2}}=\frac{1+2\frac{1}{x^2}}{2e^{1/x^2}}
$
as i said, it's probably easier to use the power series.

if you want to continue doing it the way you are doing it, going only to the second derivative is hardly ever enough to find a pattern

14. so you say substitute e^x with its approximation taylor series??

15. Originally Posted by transgalactic
so you say substitute e^x with its approximation taylor series??
not approximation, the actual Taylor series that converges to $e^{-1/x^2}$

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