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Homework Help: How many times can we differentiate this function?

  1. Feb 2, 2009 #1
    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: Feb 3, 2009
  2. jcsd
  3. Feb 2, 2009 #2

    tiny-tim

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    Hint: for what values of n can we differentiate it once? :smile:
     
  4. Feb 4, 2009 #3
    Re: how many times can we differentiate this function??

    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..
     
  5. Feb 4, 2009 #4

    tiny-tim

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    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: Feb 4, 2009
  6. Feb 6, 2009 #5
    Re: how many times can we differentiate this function??

    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: Feb 6, 2009
  7. Feb 6, 2009 #6

    tiny-tim

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    uhh? :confused:

    use the product rule

    f'(x) = … ?​
     
  8. Feb 6, 2009 #7

    HallsofIvy

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    Re: how many times can we differentiate this function??

    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.
     
  9. Feb 8, 2009 #8
    Re: how many times can we differentiate this function??

    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 cant keep doing this derivatives till i get two different limits
    there must be an easier way
    ??
     
  10. Feb 8, 2009 #9

    tiny-tim

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    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?​
    sorry, no idea what you mean :confused:
     
  11. Feb 8, 2009 #10
    Re: how many times can we differentiate this function??

    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 cant keep doing derivatives
    ??
     
  12. Feb 8, 2009 #11

    tiny-tim

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    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.
    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?
     
  13. Feb 8, 2009 #12
    Re: how many times can we differentiate this function??

    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
     
  14. Feb 8, 2009 #13

    tiny-tim

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    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:
     
  15. Feb 9, 2009 #14
    Re: how many times can we differentiate this function??

    how is that??
    how the power of X can affect if the derivative converge or not
    ??
     
  16. Feb 10, 2009 #15
    Re: how many times can we differentiate this function??

    i dont know what you mean regarding the power things

    can you explain that?
     
  17. Feb 10, 2009 #16
    Re: how many times can we differentiate this function??

    for n=0
    we cant differentiate it

    for n=1
    we can differentiate it once
    whats the general rule??
     
  18. Feb 10, 2009 #17

    tiny-tim

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    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:
     
  19. Feb 10, 2009 #18
    Re: how many times can we differentiate this function??

    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]
     
  20. Feb 10, 2009 #19

    tiny-tim

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    uhh? none of that makes sense :confused:

    try again, and write it out properly
     
  21. Feb 10, 2009 #20
    Re: how many times can we differentiate this function??

    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??
     
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