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Prove that Any Cubed Function is Differentiable (delta-epsilon method)

  1. Dec 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that:

    Any function [itex] f [/itex] such that [itex] f(x)=x^3 [/itex] for any [itex] x \in [/itex] R is differentiable.

    2. Relevant equations

    Skip.

    3. The attempt at a solution

    Okay! So, to conclude, it must be shown that, for any [itex] a [/itex] in the domain of [itex] f [/itex],

    [itex] \displaystyle \exists \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} [/itex].

    Quickly,

    [itex] \displaystyle \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} [/itex]
    [itex] {} = \displaystyle \lim_{h \rightarrow 0} \left( 3a^2 + 3ah + h^2 \right) [/itex].

    Now, the latter limit must be proved (via delta-epsilon method) to equal [itex] 3a^2 [/itex]. (This is the part I am having most trouble with, where delta-epsilon proofs are still sort of a mystery to me.)

    Suppose [itex] \varepsilon > 0 [/itex] is given. So, we should be able to show that if [itex] 0 < \left| h \right| < 0 [/itex] for some [itex] \delta > 0 [/itex], then we have [itex] \left| \left( 3a^2 + 3ah + h^2 \right) - 3a^2 \right| = \left| h \right| \cdot \left| 3a + h \right| < \varepsilon [/itex]. Well, we could choose [itex] \delta = \varepsilon / \left| 3a + h \right| [/itex], however we must limit the size of [itex] \left| 3a + h \right| [/itex] in terms of our [itex] 0 < \left| h \right| < 0 [/itex] statement (why must we do this?). With that said, suppose [itex] \delta = 1 [/itex]. Then, [itex] \left| h \right| < 1 [/itex]. Yet, this means [itex] \left| 3a + h \right| \leq \left| 3a \right| + \left| h \right| < 3 \left| a \right| + 1 [/itex]. Therefore, the claim (hopefully!) follows if we choose [itex] \delta = \min \left\{ 1, \varepsilon / \left( 3 \left| a \right| + 1 \right) \right\} [/itex].

    QED


    (Thanks for any help and or advice!)
     
  2. jcsd
  3. Dec 27, 2011 #2

    SammyS

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    Did you intend to say (I hope):
    "... So, we should be able to show that if [itex] 0 < \left| h \right| < \delta [/itex] for some [itex] \delta > 0 [/itex], ..."​

    ... and similar correction in the other place that you wrote:
    [itex] 0 < \left| h \right| < 0 [/itex]​
    should also be
    [itex] 0 < \left| h \right| < \delta [/itex]​

    Otherwise, it's impossible for there to be such a number, h .

    Added in Edit:
    Your δ looks correct.

    Your proof in somewhat in the reverse order of what it needs to be.

    You have essentially shown the "scratch work" needed to find what δ you need in terms of ε. In your "final proof ", the one you turn in for credit, you need to show:
    If 0 < |h| < δ , (i.e. [itex]\displaystyle 0<|h|<\min\left(1,\ \frac{\varepsilon}{3|a|+1}\right) [/itex]) then [itex]|3ah+h^2|<\varepsilon[/itex]​
     
    Last edited: Dec 27, 2011
  4. Dec 27, 2011 #3

    Ray Vickson

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    Were you told that you must use the epsilon-delta method *directly*, or are you allowed to use and apply _properties_ of limits? If the latter, here is what facts I would use (and you COULD prove these by an epsilon-delta method!):

    (1) If lim_{h->0} g_1(h) = a_1 and lim_{h->0} g_2(h) = a_2 exist, then lim_{h->0} [g_1(h) + g_2(h)] exists and it equals a_1 + a_2 (In other words, the limit of a sum is the sum of the limits).
    (2) If c is a constant, then lim_{h->0} c*g(h) = c* lim_{h->0} g(h).
    (3) if g_1 and g_2 have limits, then lim g_1(h)*g_2(h) = lim g_1(h) * lim g_2(h) (that is, the limit of a product is the product of the limits).

    Now lim (3a^2 + 3ah + h^2) = lim(3a^2) + lim(3ah) + lim(h^2), by (1); lim(3ah) = 3a lim(h) = 0 (by (2); and lim(h^2) = (lim h)^2 (by (3)).

    RGV
     
  5. Dec 27, 2011 #4
    Hmmm... I would imagine I could use such properties of limits! For example, those properties were indeed proved in the textbook (via delta-epsilon method, too!), and we are allowed to use anything proved or stated by the author in proofs for problems (from the textbook). Thanks! I kind of forgot that I was (at least, I think) allowed to do that, haha.

    However, if you don't mind, I would greatly appreciate any commentary on my delta-epsilon proof, as I am sure I will be required to do one in the near future.

    EDIT: Oh, sorry! I didn't notice the first comment that indeed made comments on it. So, only if you have time, then, as that would be great!


    ---
    Much appreciation,

    5hassay
     
  6. Dec 27, 2011 #5
    Oh, dearest me, hahaha! Yes, I did mean what you speculated myself to mean (sorry!).

    And, about the order, I also seemed to forget about that, as well.

    Okay! Well, that's great, then! Progress here is definitely quite wonderful, as I would like to move on to the calculus part of Michael Spivak's, "Calculus," without a poor understanding in limit proofs slowing me down.


    ---
    Much appreciation,

    Lucas
     
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