How many times can we differentiate this function?

[transgalactic]

how many times can we differentiate this function??

how many times can we differentiate this splitted function on point x=0
??
$f(x) = \{ x^{2n} \sin (\frac{1}{x}) ,x \ne 0$
$\{ 0,x = 0}} \\$

 

[tiny-tim]

Hint: for what values of n can we differentiate it once?

 

[transgalactic]

we can differentiate every function enless times
for example
f(x)=x^3
f'(x)=x^2
f''(x)=x^1
f'''(x)=1
f''''(x)=0
f'''''(x)=0
etc..

 

[tiny-tim]

really?

how many times can we differentiate this function at x=0?

$f(x)\ =\ x^2\sin (\frac{1}{x}) ,x \ne 0$

{ 0,x = 0​

 

[transgalactic]

what now?
$$f(x)\; = \;x^2 \sin ({1 \over x}),x \ne 0$$
$$f'(0) = \mathop {\lim }\limits_{x \to 0} {{x^2 \sin ({1 \over x}) - 0} \over {x - 0}} = 0$$

 

[tiny-tim]

what now?
$$f(x)\; = \;x^2 \sin ({1 \over x}),x \ne 0$$
$$f'(0) = \mathop {\lim }\limits_{x \to 0} {{x^2 \sin ({1 \over x}) - 0} \over {x - 0}} = 0$$

uhh?

use the product rule …

f'(x) = … ?​

 

[HallsofIvy]

what now?
$$f(x)\; = \;x^2 \sin ({1 \over x}),x \ne 0$$
$$f'(0) = \mathop {\lim }\limits_{x \to 0} {{x^2 \sin ({1 \over x}) - 0} \over {x - 0}} = 0$$

Yes, that's correct. But tiny-tim's point, I think, is that the derivative for for x not 0 (which he didn't actually ask) is 2x sin(1/x)- cos(1/x) so f '(x) is not continuous at x= 0 and so does not have a second derivative there.

It is NOT true that "we can differentiate every function enless times" though that is certainly true for polynomials.

A simpler example is
$f(x)= (1/2)x^2$ if $x\ge 0$
$f(x)=-(1/2)X^2$ if x< 0

Then f'(x)= x if $s\ge 0$ and f'(x)= -x if x< 0: in other words f'(x)= |x| and that is not differentiable at x= 0.

I got that example, of course, by integrating |x|. Integrate again, to get
$f(x)= (1/6)x^3$ if $x\ge 0$
$f(x)= -(1/6)x^3$ if x< 0
and you get f having first and second deriviatives but not a third derivative. You can use that to get example of functions that are differentiable n times but not n+1 times for any integer n.

 

[transgalactic]

i assumed that in the case of prooving the second derivative point x=0 the function turns to 0 too.
$$f(x) = \ x^{2n} \sin (\frac{1}{x})$$
$$f'(x)=2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})$$

$$f''(x)=\lim_{x->0+}\frac{2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})-0}{x-0}=\\$$
$$f''(x)=\lim_{x->0+}\frac{2nx^{2n-2}\sin (\frac{1}{x})-x^{2n-1}\frac{1}{x^2}\cos(\frac{1}{x})-0}{1}$$

so in the plus side i will get 0
and on the minus side
but i can't keep doing this derivatives till i get two different limits
there must be an easier way
??

 

[tiny-tim]

… $$f''(x)=\lim_{x->0+}\frac{2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})-0}{x-0}=\\$$
$$f''(x)=\lim_{x->0+}\frac{2nx^{2n-2}\sin (\frac{1}{x})-x^{2n-1}\frac{1}{x^2}\cos(\frac{1}{x})-0}{1}$$

Why are you using this lim notation?

Just use the product rule and chain rule, and then see whether the result converges to a limit as x -> 0.

In particular:

does sin(1/x) converge as x -> 0?
does cos(1/x) converge as x -> 0?
does xsin(1/x) converge as x -> 0?
does xcos(1/x) converge as x -> 0?​

so in the plus side i will get 0
and on the minus side

sorry, no idea what you mean

 

[transgalactic]

here is the first derivative
$$f'(x)=2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})$$so i need to do limit for positive side of zero
equals limit of negative side of zero equals f(0)but its not a solutions
i can't keep doing derivatives
??

 

[tiny-tim]

so i need to do limit for positive side of zero
equals limit of negative side of zero equals f(0)

oh i see now …

you're talking about lim x -> 0+ and lim x -> 0-

it doesn't matter in this case …

either they both exist, or neither does.

here is the first derivative
$$f'(x)=2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})$$ …

transgalactic, when someone at PF helps by asking you a question, answer it!

even if you can't see the point …

I'll ask again: does sin(1/x) converge as x -> 0?

 

[transgalactic]

no
the function is bounded and Non convergent

so is the cos (1/x)

but x*sin(1/x) and x*cos(1/x) are convergent because lim of 0*bounded =0

 

[tiny-tim]

no
the function is bounded and Non convergent

so is the cos (1/x)

but x*sin(1/x) and x*cos(1/x) are convergent because lim of 0*bounded =0

That's right …

sin(1/x) and cos(1/x) oscillate between -1 and +1,

but xⁿsin(1/x) and xⁿcos(1/x) oscillate between -xⁿ and +xⁿ, and so converge to 0, for n ≥ 1.

i can't keep doing derivatives

Maybe, but you don't need to calculate the full derivative (with the correct factors), you only need to know the powers of x that are involved.

 

[transgalactic]

how is that??
how the power of X can affect if the derivative converge or not
??

 

[transgalactic]

That's right …

sin(1/x) and cos(1/x) oscillate between -1 and +1,

but xⁿsin(1/x) and xⁿcos(1/x) oscillate between -xⁿ and +xⁿ, and so converge to 0, for n ≥ 1.


Maybe, but you don't need to calculate the full derivative (with the correct factors), you only need to know the powers of x that are involved.

i don't know what you mean regarding the power things

can you explain that?

 

[transgalactic]

for n=0
we can't differentiate it

for n=1
we can differentiate it once
whats the general rule??

 

[tiny-tim]

for n=0
we can't differentiate it

for n=1
we can differentiate it once
whats the general rule??

That's better!

Try it for n= 2 …

how many times does it work? …

when it stops working, what's the spanner-in-the-works stopping it? …

when does that happen for larger n?

 

[transgalactic]

it can go forever
for every n>0 i get 0*bounded
there is no limitation
$$f(x) = x^{4} \sin (\frac{1}{x})\\$$
$$f'(x)=lim_{x->0}\frac{x^{4} \sin (\frac{1}{x})}{x}=lim_{x->0}{x^{3} \sin (\frac{1}{x})}=0$$

 

[tiny-tim]

it can go forever
for every n>0 i get 0*bounded
there is no limitation
$$f(x) = x^{4} \sin (\frac{1}{x})\\$$
$$f'(x)=lim_{x->0}\frac{x^{4} \sin (\frac{1}{x})}{x}=lim_{x->0}{x^{3} \sin (\frac{1}{x})}=0$$

uhh? none of that makes sense

try again, and write it out properly​

 

[transgalactic]

i was given this expression
$$\dfrac{\mathrm{d}^r}{\mathrm{d}x^r}\left(x^{2n}\sin\left(\frac{1}{x}\right)\right) &=\left((-1)^r x^{2(n-r)}+x^{2(n-r)+2}P_{n,r}(x)\right)\sin\left(\frac{1}{x}+r\frac{\pi}{2}\right)\\ &\quad+x^{2(n-r)+1}Q_{n,r}(x)\sin\left(\frac{1}{x}+(r-1)\frac{\pi}{2}\right) \end{align*}$$

is it in any use??

 

[transgalactic]

uhh? none of that makes sense

try again, and write it out properly​

for n=2 it differentiates
$$f'(x)=lim_{x->0}\frac{x^{4} \sin (\frac{1}{x})-0}{x-0}=lim_{x->0}{x^{3} \sin (\frac{1}{x})}=0$$

 

[tiny-tim]

i was given this expression
$$\dfrac{\mathrm{d}^r}{\mathrm{d}x^r}\left(x^{2n}\sin\left(\frac{1}{x}\right)\right) &=\left((-1)^r x^{2(n-r)}+x^{2(n-r)+2}P_{n,r}(x)\right)\sin\left(\frac{1}{x}+r\frac{\pi}{2}\right)\\ &\quad+x^{2(n-r)+1}Q_{n,r}(x)\sin\left(\frac{1}{x}+(r-1)\frac{\pi}{2}\right) \end{align*}$$

is it in any use??

did you get that from another forum?

it may be right (i haven't checked it) …

but it's useless to you if you don't know how to get it, isn't it? ​

for n=2 it differentiates
$$f'(x)=lim_{x->0}\frac{x^{4} \sin (\frac{1}{x})-0}{x-0}=lim_{x->0}{x^{3} \sin (\frac{1}{x})}=0$$

uhh? still makes no sense …

you haven't even used the product rule