Results 1 to 5 of 5

Math Help - Correctly shown?

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    59

    Correctly shown?

    Just checking this now :\ I had to show that f is not differentiable at 0 so:

    Potential Proof:

    |\frac{f(0+h) - f(0)}{h} - 0 | = |\frac{f(0+h) - f(0)}{h} |

    =\fraq{{|f(h)|}} {{|h|}}

    \le \fraq{{h}} {{|h|}}

    = 1

    I think I have made a big booboo and gone completely wrong, here is the whole of the question (its not an exam question, but a simple question from a group that our lecturer has given us to try and do over summer before anyone asks)



    PS: It is spose to show f(h) over the modulus of h and in the line below, h over modulus of h
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by ramdrop View Post
    Just checking this now :\ I had to show that f is not differentiable at 0 so:

    Potential Proof:

    |\frac{f(0+h) - f(0)}{h} - 0 | = |\frac{f(0+h) - f(0)}{h} |

    =\fraq{{|f(h)|}} {{|h|}}

    \le \fraq{{h}} {{|h|}}

    = 1

    I think I have made a big booboo and gone completely wrong, here is the whole of the question (its not an exam question, but a simple question from a group that our lecturer has given us to try and do over summer before anyone asks)



    PS: It is spose to show f(h) over the modulus of h and in the line below, h over modulus of h
    Similar to this problem on another forum.

    Show f(x) = { x/2 if x rational , x if x irrational is not differentiable at 0

    Basic idea is that there are two possible (secant) slopes: 0 and 1, and you can always find values of x satisfying both in any (-delta, delta) \ {0}, delta > 0.

    I don't entirely follow your potential proof. Shouldn't you be subtracting L instead of 0 inside the first absolute value?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2010
    Posts
    59
    Ah looked through my proof, it is very wrong so, I looked at the other forum and got a few ideas and tried it that way and reached 2 inequalities that contradicted each other, now starting with part A (seeing as I did b first, d'oh)

    I would try:

    Let x'n = sequence of irrational numbers tending to x

    f(xn) = 0 -> 0 \ne x

    But that just goes wrong, maybe I have to set it up differently, maybe hmm, well i dunno guess I must look harder
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2010
    Posts
    5
    Quote Originally Posted by ramdrop View Post
    Just checking this now :\ I had to show that f is not differentiable at 0 so:

    Potential Proof:

    |\frac{f(0+h) - f(0)}{h} - 0 | = |\frac{f(0+h) - f(0)}{h} |

    =\fraq{{|f(h)|}} {{|h|}}

    \le \fraq{{h}} {{|h|}}

    = 1

    I think I have made a big booboo and gone completely wrong, here is the whole of the question (its not an exam question, but a simple question from a group that our lecturer has given us to try and do over summer before anyone asks)



    PS: It is spose to show f(h) over the modulus of h and in the line below, h over modulus of h
    Proof.
    a) Let \{x_n\} be a sequence such that x_n\rightarrow 0 (n\rightarrow \infty). To show f(x_n)\rightarrow f(0) (n\rightarrow \infty).

    |f(x_n)-f(0)|=|f(x_n)|\leq |x_n|\rightarrow 0 (n\rightarrow \infty)

    b)Aim to show \lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0} doesn't exist, i.e., \lim_{x\rightarrow 0}\frac{f(x)}{x} doesn't exist. From the definition of LIMIT, let a be a real number. To show

    \exists \varepsilon_0>0, \forall \delta>0, \exists |x|<\delta such that |\frac{f(x)}{x}-a| \geq \varepsilon_0.

    Actually, it is easy to find such \varepsilon_0:just set it to be \frac{a}{2}, noting that x should be irrational.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2010
    Posts
    59
    Thanks man, is that like the full proof seems really short
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. For the tringle shown what are the :
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 31st 2010, 10:59 AM
  2. Replies: 3
    Last Post: March 21st 2010, 04:40 AM
  3. Work shown: Solve the triangle a=3, c=1, B=100?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 4th 2010, 06:32 PM
  4. Calculate XZ in the diagram shown
    Posted in the Geometry Forum
    Replies: 1
    Last Post: August 22nd 2009, 03:07 PM

Search Tags


/mathhelpforum @mathhelpforum