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Epsilon and delta

  1. Jun 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Snap1.jpg

    2. Relevant equations
    Continuity:
    $$\vert x-c \vert < \delta~\Rightarrow~\vert f(x)-f(c) \vert < \epsilon$$
    $$\delta=\delta(c,\epsilon)$$

    3. The attempt at a solution
    $$\vert f(x)-f(c) \vert <\frac{1}{2}f(c)~\Rightarrow~\vert x-c \vert < \delta_1$$
    So i have this δ1 but what do i do with it?
    And ε=½f(c) is big, maybe it will be in the negative zone.
    Maybe i have to find a δ such that ##~\vert f(x)-f(c) \vert =0~##?
    There is such a δ, so why was advised to take such a large ε?
     
  2. jcsd
  3. Jun 2, 2017 #2

    PeroK

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    You are still starting these proofs the wrong way round. Somehow you have to train yourself to stop writing things like:

    $$\vert f(x)-f(c) \vert <\frac{1}{2}f(c)~\Rightarrow~\vert x-c \vert < \delta_1$$

    You must, must, must stop yourself from doing this.

    For this problem I would first try to "prove" it using a graph of the function and a geometric argument.
     
  4. Jun 2, 2017 #3

    PeroK

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    The above is continuity.

    And this is what you write. You must see the difference. Every time you turn it round the wrong way.
     
  5. Jun 2, 2017 #4
    I was wrong at the definition of continuity:
    $$\vert f(x)-f(c) \vert < \epsilon~\Rightarrow~\vert x-c \vert < \delta$$
    The ε is to the intersection with x
    Snap4.jpg
     
  6. Jun 2, 2017 #5

    Mark44

    Staff: Mentor

    No, what you have above is backwards. The implication you showed in post 1 has the implication in the right order.
    In words, "If x is close to c, then f(x) will be close to f(c)"
    The delta and epsilon quantify the "close to" terms.
    ??
    In your drawing, where is x? Where is c? Is the circled point on the curve (c, f(c))?
     
  7. Jun 2, 2017 #6

    PeroK

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    :headbang:
     
  8. Jun 2, 2017 #7
    I had a friend, a fraternity brother actually, who always said that his ambition in life was to be an epsilon and delta picker.
     
  9. Jun 2, 2017 #8
  10. Jun 3, 2017 #9
    If f(c)>0 i can take ε small and it will still be ##~\vert f(x)-f(c) \vert >0~## and find a δ because of continuity, so why do i need the ##~\epsilon=\frac{1}{2}f(c)~##?
     
  11. Jun 3, 2017 #10

    Mark44

    Staff: Mentor

    In your drawing in post #8, you have ##\epsilon = f(c)##, which isn't what the hint is saying.
     
  12. Jun 3, 2017 #11
    Snap5.jpg Thank you Mark, Dr.D and PeroK
     
  13. Jun 3, 2017 #12

    Mark44

    Staff: Mentor

    @Karol, did you actually prove the theorem? The problem asks you to prove that statement, and illustrate with a sketch.
     
  14. Jun 3, 2017 #13
    Because of continuity i can find a δ for ##~\epsilon=\frac{1}{2}f(c)~##, so in this interval: ##~c-\delta<x<c+\delta~##, f(x)>0
     
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