How many times can we differentiate this function?

In summary, the function f(x) = x^{4} \sin (\frac{1}{x}) can be differentiated an infinite number of times, as long as n > 0. The general rule is that for n=0, the function cannot be differentiated, and for n>0, the function can be differentiated n times. The limitation occurs when n=0 and the function cannot be differentiated.
  • #1
transgalactic
1,395
0
how many times can we differentiate this function??

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

[/itex]
 
Last edited:
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  • #2
Hint: for what values of n can we differentiate it once? :smile:
 
  • #3


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..
 
  • #4
really?

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

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



[/itex]

{ 0,x = 0​
 
Last edited:
  • #5


what now?
[tex]
f(x)\; = \;x^2 \sin ({1 \over x}),x \ne 0
[/tex]
[tex]
f'(0) = \mathop {\lim }\limits_{x \to 0} {{x^2 \sin ({1 \over x}) - 0} \over {x - 0}} = 0
[/tex]
 
Last edited:
  • #6
transgalactic said:
what now?
[tex]
f(x)\; = \;x^2 \sin ({1 \over x}),x \ne 0
[/tex]
[tex]
f'(0) = \mathop {\lim }\limits_{x \to 0} {{x^2 \sin ({1 \over x}) - 0} \over {x - 0}} = 0
[/tex]

uhh? :confused:

use the product rule

f'(x) = … ?​
 
  • #7


transgalactic said:
what now?
[tex]
f(x)\; = \;x^2 \sin ({1 \over x}),x \ne 0
[/tex]
[tex]
f'(0) = \mathop {\lim }\limits_{x \to 0} {{x^2 \sin ({1 \over x}) - 0} \over {x - 0}} = 0
[/tex]

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
[itex]f(x)= (1/2)x^2[/itex] if [itex]x\ge 0[/itex]
[itex]f(x)=-(1/2)X^2[/itex] if x< 0

Then f'(x)= x if [itex]s\ge 0[/itex] 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
[itex]f(x)= (1/6)x^3[/itex] if [itex]x\ge 0[/itex]
[itex]f(x)= -(1/6)x^3[/itex] 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.
 
  • #8


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

[tex]
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}=\\

[/tex]
[tex]
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}
[/tex]

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
??
 
  • #9
transgalactic said:
… [tex]
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}=\\

[/tex]
[tex]
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}
[/tex]

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 :confused:
 
  • #10


here is the first derivative
[tex]
f'(x)=2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})
[/tex]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
??
 
  • #11
transgalactic said:
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
[tex]
f'(x)=2nx^{2n-1}\sin (\frac{1}{x})-x^{2n}\frac{1}{x^2}\cos(\frac{1}{x})
[/tex] …

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?
 
  • #12


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
 
  • #13
transgalactic said:
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 :smile:

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. :wink:
 
  • #14


how is that??
how the power of X can affect if the derivative converge or not
??
 
  • #15


tiny-tim said:
That's right :smile:

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

can you explain that?
 
  • #16


for n=0
we can't differentiate it

for n=1
we can differentiate it once
whats the general rule??
 
  • #17
transgalactic said:
for n=0
we can't differentiate it

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

That's better! :biggrin:

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? :wink:
 
  • #18


it can go forever
for every n>0 i get 0*bounded
there is no limitation
[tex]
f(x) = x^{4} \sin (\frac{1}{x})\\
[/tex]
[tex]
f'(x)=lim_{x->0}\frac{x^{4} \sin (\frac{1}{x})}{x}=lim_{x->0}{x^{3} \sin (\frac{1}{x})}=0
[/tex]
 
  • #19
transgalactic said:
it can go forever
for every n>0 i get 0*bounded
there is no limitation
[tex]
f(x) = x^{4} \sin (\frac{1}{x})\\
[/tex]
[tex]
f'(x)=lim_{x->0}\frac{x^{4} \sin (\frac{1}{x})}{x}=lim_{x->0}{x^{3} \sin (\frac{1}{x})}=0
[/tex]

uhh? none of that makes sense :confused:

try again, and write it out properly
 
  • #20


i was given this expression
[tex]
\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*}
[/tex]

is it in any use??
 
  • #21


tiny-tim said:
uhh? none of that makes sense :confused:

try again, and write it out properly

for n=2 it differentiates
[tex]
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
[/tex]
 
  • #22
transgalactic said:
i was given this expression
[tex]
\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*}
[/tex]

is it in any use??

did you get that from another forum? :rolleyes:

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? :smile:
transgalactic said:
for n=2 it differentiates
[tex]
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
[/tex]

uhh? still makes no sense …

you haven't even used the product rule
 

Related to How many times can we differentiate this function?

1. How do you know how many times a function can be differentiated?

The number of times a function can be differentiated is determined by the degree of the function. A function of degree n can be differentiated n times.

2. Can any function be differentiated an infinite number of times?

No, not all functions can be differentiated infinitely. The function must be continuous and have a defined derivative at each point in order for it to be differentiated.

3. How do I find the maximum number of times a function can be differentiated?

The maximum number of times a function can be differentiated is determined by the degree of the function. To find the degree, look at the highest power of the variable in the function.

4. Is there a limit to how many times a function can be differentiated?

Yes, there is a limit to how many times a function can be differentiated. The maximum number of times a function can be differentiated is equal to its degree.

5. Can the number of times a function can be differentiated change?

Yes, the number of times a function can be differentiated can change depending on the function itself. For example, if the function has a variable in the denominator, the number of times it can be differentiated may decrease.

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