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Homework Help: Need help with these Derivatives

  1. Jul 29, 2010 #1

    jgens

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    1. The problem statement, all variables and given/known data

    If [itex]f[/itex] is differentiable at [itex]a[/itex], prove the following:
    [tex]\lim_{h,k \to 0^+} \frac{f(a+h)-f(a-k)}{h+k} = f'(a)[/tex]

    2. Relevant equations

    N/A

    3. The attempt at a solution

    At the moment, I don't have a complete proof worked out, but I was wondering if someone could comment on the validity of this reasoning . . .

    Clearly, for every [itex]\varepsilon > 0[/itex], there exists a [itex]\delta_1 > 0[/itex] such that if [itex]0 < k < \delta_1[/itex] then . . .

    [tex]\left|\frac{f(a+h)-f(a-k)}{h+k} - \frac{f(a+h) - f(a)}{h} \right| < \frac{\varepsilon}{2}[/tex]*​

    Moreover, for this same [itex]\varepsilon[/itex], there must be some other number [itex]\delta_2 > 0[/itex] such that whenever [itex]0 < h < \delta_2[/itex], it follows that . . .

    [tex]\left|\frac{f(a+h)-f(a)}{h} - f'(a) \right| < \frac{\varepsilon}{2}[/tex]​

    From this, so long as [itex]0 < k < \delta_1[/itex] and [itex]0 < h < \delta_2[/itex], we have

    [tex]\left|\frac{f(a+h)-f(a-k)}{h+k} - f'(a) \right| < \varepsilon[/tex]​

    as desired.

    *I realize that this is the point that really needs some work, but I think that it should be a trivial (albeit potentially long-winded) exercise to find the proper [itex]\delta[/itex].
     
  2. jcsd
  3. Jul 30, 2010 #2
    Re: Derivatives

    Looks fine, just remamber that the first inequality with given [tex]\delta_1[/tex] must hold for all [tex]0 < h <\delta_2[/tex].
     
  4. Jul 30, 2010 #3

    jgens

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    Re: Derivatives

    Well, the first inequality should hold for any arbitrary h > 0, so I should have my bases covered. Thanks! I'll work on finding a proper [itex]\delta[/itex] now.
     
  5. Jul 30, 2010 #4
    Re: Derivatives

    Since we don't have any global assumptions on f, I don't think it's possible to find [tex]\delta[/tex] for arbitrary h. If, for example, for some h>0 function
    [tex]\phi_h(k)=\frac{f(a+h)-f(a-k)}{h+k}[/tex]
    is discontinuous at 0, it's hopeless.
     
  6. Jul 30, 2010 #5

    jgens

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    Re: Derivatives

    Well, I'm working on my phone right now, so I'm not sure if this is correct, but here are my thoughts. For every [itex]\varepsilon > 0[/itex], it's clearly possible to find a [itex]\delta_1 > 0[/itex] such that if [itex]0 < h < \delta_1[/itex], then the second inequality holds. Based on this choice of [itex]h[/itex] and the continuity of [itex]f[/itex] at [itex]a[/itex], it's then possible to choose a [itex]\delta_2 > 0[/itex] such that if [itex]0 < k < \delta_2[/itex], then the second inequality holds. This would show that for every number [itex]\varepsilon > 0[/itex], it's possible to find the necessary two numbers [itex]\delta > 0[/itex] for the final inequality to hold. However, looking through this, the dependence of [itex]\delta_2[/itex] on h seems like it might be problematic. Ugh :(

    Edit: I'm fairly certain that it's possible to find the necessary [itex]\delta[/itex]s that I allude to in this post; it's just is the second one's reliance on h problematic or is it a non-issue.
     
  7. Jul 30, 2010 #6

    jgens

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    Re: Derivatives

    Thanks! I'll try to post a fully worked out proof later tonight when I have access to a computer. If you don't mind, could you explain why the delta with dependence on h is problematic? I understand why it's certainly problematic if the delta is dependent on the same variable it's placing a bound on, but I can't figure it out exactly when it's dependent on a different variable. Thanks again!
     
  8. Jul 30, 2010 #7

    jgens

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    Re: Derivatives

    I would probably be able to replace delta 2's dependence on h with a dependence on delta 1. I'm not sure if this makes the situation any better, but I could really use some help here. Thanks.
     
  9. Jul 30, 2010 #8
    Re: Derivatives

    Sorry for deleting that post. Dependence is indeed problematic because of the definition of double limit, but I realised I made a critical mistake in my solution. I'll think about and post if I find anything useful. Good luck!
     
  10. Jul 30, 2010 #9

    Dick

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    Re: Derivatives

    Try this. Can you show min(L1,L2)<=(h*L1+k*L2)/(h+k)<=max(L1,L2) for h and k positive? Can you think how you might use that?
     
  11. Jul 30, 2010 #10

    jgens

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    Re: Derivatives

    Thanks for the help Dick! First, we need only note that ...

    [tex]h\min{(L_1,L_2)} + k\min{(L_1,L_2)} \leq h \dot L_1 + k \dot L_2 \leq h\max{(L_1,L_2)} + k\max{(L_1,L_2)}[/tex]​

    From which the inequality that you posted follows immediately. Next, we define [itex]L_1[/itex] and [itex]L_2[/itex] such that ...

    [tex]L_1 = \frac{f(a+h)-f(a)}{h}[/tex]
    [tex]L_2 = \frac{f(a)-f(a-k)}{k}[/tex]

    Using these definitions, we have that ...

    [tex]\min{(L_1,L_2)} \leq \frac{f(a+h)-f(a)+f(a)-f(a-k)}{h+k} = \frac{f(a+h)-f(a-k)}{h+k} \leq \max{(L_1,L_2)}[/tex]​

    And then using the squeeze theorem or explicitly writing out the [itex]\varepsilon-\delta[/itex] proof, we prove the desired result. Is this what you were getting at Dick?
     
  12. Jul 30, 2010 #11

    Dick

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    Re: Derivatives

    Sure, it's a squeeze theorem. And you know how to apply it. So why is min(L1,L2)<=(h*L1+k*L2)/(h+k)<=max(L1,L2) for h,k>0? Do you get that part?
     
  13. Jul 30, 2010 #12

    jgens

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    Re: Derivatives

    The first part of my post was meant to address that, I'm sorry if it wasn't clear. To outline it again without that pain of the maxs/mins, I'll assume (without loss of generality) that [itex]L_1 \leq L_2[/itex]. Clearly ...

    [tex]L_1(h+k) \leq h \dot L_1 + k \dot L_2 \leq L_2(h+k)[/tex]​

    Dividing through by [itex]h+k[/itex] we get ...

    [tex]L_1 \leq \frac{h \dot L_1 + k \dot L_2}{h+k} \leq L_2[/tex]​

    as desired.
     
  14. Jul 30, 2010 #13

    jgens

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    Re: Derivatives

    Thanks for your help anyway. I'm not familiar at all with the definition of a limit involving two variables (single variable calc. course here) so that makes sense why I was having such a difficult time explicitly understanding why the dependence was so problematic; after looking up the correct definition, it makes complete sense.
     
  15. Jul 30, 2010 #14

    Dick

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    Re: Derivatives

    That's a bit of circular reasoning and why are you putting dots on the L's? I know L1<=(h*L1+k*L2)/(h+k)<=L2 is true for L1<=L2 and h,k>0. But do you know why? I'll give you a hint. It's the same reason why L1<=t*L1+(1-t)*L2<=L2 for t in [0,1].
     
  16. Jul 30, 2010 #15

    jgens

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    Re: Derivatives

    Sorry about the dots, they were supposed to be multiplication signs. And what's particularly circular about my reasoning? If I choose arbitrary [itex]h,k > 0[/itex] and assume that [itex]L_1 \leq L_2[/itex], then [itex]hL_1 + kL_1 \leq hL_1 + kL_2[/itex] since [itex]kL_1 \leq kL_2[/itex]. The other inequality would follow from similar reasoning.

    I'm off to bed now, so I'll follow up on your last idea tomorrow morning. Thanks again Dick!
     
  17. Jul 30, 2010 #16

    Dick

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    Re: Derivatives

    Oh, nothing circular about your thinking. More of a problem with my following what you are doing. I was focusing on the dots. Don't worry about it.
     
  18. Jul 31, 2010 #17
    Re: Derivatives

    We can solve it similarly without any min/max reasoning. Observe that
    [tex]\left|\frac{f(a+h)-f(a-k)}{h+k}-f'(a) \right|
    =
    \left|\frac{f(a+h)-f(a) + f(a)-f(a-k)}{h+k}-f'(a)\right|
    =
    [/tex]
    [tex]
    \left|\frac{f(a+h)-f(a) -hf'(a)+ f(a)-f(a-k)-kf'(a)}{h+k}\right|
    \leq
    [/tex]
    [tex]
    \left|\frac{f(a+h)-f(a) -hf'(a)}{h+k}+\frac{f(a) - f(a-k)-kf'(a)}{h+k}\right|
    \leq
    [/tex]
    [tex]
    \frac{h}{h+k} \left|\frac{f(a+h)-f(a)}{h} -f'(a)\right|
    +\frac{k}{h+k} \left|\frac{f(a-k)-f(a)}{-k} -f'(a)\right| \leq
    [/tex]
    [tex]
    \left|\frac{f(a+h)-f(a)}{h} -f'(a)\right|+\left|\frac{f(a-k)-f(a)}{-k} -f'(a)\right|[/tex]
    Sorry for not posting this earlier, I wanted to complete you original idea.
     
    Last edited: Jul 31, 2010
  19. Jul 31, 2010 #18

    jgens

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    Re: Derivatives

    Thanks again Dick! And nice proof losiu99! I really appreciate all the help that you've given me.
     
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