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Epsilon-Delta Proof with 2-Variable Polynomial

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a specific number δ>0 such that if x2 + y2 = δ2, then |x2+y2+3xy+180xy5 < 1/10 000.

    Answer: Choose δ < 0.002
    2. Relevant equations
    ε-δ def'n of limit: lim (x,y) → (a, b) f(x) = L if for every ε > 0 there exists a δ > 0 such that 0 < √(x-a)2+(y-b)2, |f(x) - L| < ε.

    3. The attempt at a solution
    So, f(x) = |x2+y2+3xy+180xy5 and because x2 + y2 = δ2, the limit is being taken as (x, y) → (0, 0) and L = 0.
    I don't understand how I would start this problem because it's not rational, like I'm used to. I can do this:
    (x) = |x2+y2+3xy+180xy5 = x2 + y2 + 3|x||y| + 180|x||y5|, but now I am stuck. Any help is greatly appreciated.
     
  2. jcsd
  3. Sep 23, 2013 #2

    jbunniii

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    Well, this equation is certainly not true, but if you replace the ##=## with ##\leq##, then it is true, because of the triangle inequality. So this gives you
    $$|x^2 + y^2 + 3xy + 180xy^5| \leq |x^2 + y^2| + |3xy| + |180xy^5|$$
    Assuming ##x## and ##y## are real, it's true that ##|x^2 + y^2| = x^2 + y^2##. And the absolute value of a product equals the product of absolute values. So we can rewrite the right hand side as
    $$x^2 + y^2 + 3|x|\cdot |y| + 180 |x| \cdot |y|^5$$
    You know that ##x^2 + y^2 = \delta^2##. Can you find a way to bound the other terms in a way that depends on ##\delta##?
     
  4. Sep 23, 2013 #3
    Sorry, I meant to say that x2+y2 < δ^2

    I am still very confused.

    Can I do something like this?

    0 ≤|x2 + y2 + 180xy5| ≤ x2 + y2 + |3xy| + |180xy5| ≤ |3x2y2|+|180x2y5| = |3x2y2(|1+60y3|)
     
  5. Sep 23, 2013 #4

    jbunniii

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    It isn't necessarily true that ##|3xy| \leq |3x^2 y^2|## or that ##|180xy^5| \leq |180x^2 y^5|##. This is true if ##x \geq 1## and ##y \geq 1## but not in general.

    Here is a hint. How does ##|x|## compare to ##\sqrt{x^2 + y^2}##?
     
  6. Sep 23, 2013 #5

    Mark44

    Staff: Mentor

    No. You can't turn a sum into a product like that.

    IOW, you can't do this: x2 + y2 + |3xy| + |180xy5| ≤ |3x2y2|+|180x2y5|

    And you can't do what you did in the last expression, either.
     
  7. Sep 23, 2013 #6
    Okay. So,
    |x| ≤ [itex]\sqrt{}x2+y2[/itex]< δ
    and
    |y| ≤ [itex]\sqrt{}x2+y2[/itex]< δ

    Then,
    |x2+y2+3xy+180xy5| < δ+δ+3δ2+180δ6

    Right?

    Also, thank you so much!
     
  8. Sep 23, 2013 #7

    jbunniii

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    So far so good.

    That's not quite right. Try writing out the intermediate steps:
    $$\begin{align}
    |x^2 + y^2 + 3xy + 180xy^5| & \leq x^2 + y^2 + 3|x| |y| + 180 |x| |y|^5 \\
    & \leq \delta^2 + 3|x| |y| + 180 |x| |y|^5 \\
    &\leq ???\end{align}$$
     
  9. Sep 23, 2013 #8
    Isn't |x2+2+3xy+180xy5| ≤ δ2
    ie. it doesn't matter what you add to the right side because it will always make δ2 larger, in this case?
     
  10. Sep 23, 2013 #9

    jbunniii

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    You're trying to solve for ##\delta## which will satisfy the requirement in your problem statement. You need to find a bound for ##|x^2 + y^2 + 3xy + 180xy^5|## that depends only on ##\delta##, not ##x## or ##y##. You have the constraint ##x^2 + y^2 < \delta^2##. From this, in post #6 you found some bounds for ##|x|## and ##|y|##. Now apply those to the right hand side in post #7.
     
  11. Sep 23, 2013 #10

    jbunniii

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    To address this question directly, no, it's not necessarily true. All you know is that ##x^2 + y^2 < \delta^2##. But ##|x^2 + y^2 + 3xy + 180xy^5|## may be bigger than ##x^2 + y^2##, so it isn't necessarily bounded by ##\delta^2##. Try plugging in ##x=y=1## and ##\delta = 2## to see this.
     
  12. Sep 23, 2013 #11
    Waiiiiit. 4δ2 + 180δ6
     
  13. Sep 23, 2013 #12
    would be the right side?
     
  14. Sep 23, 2013 #13

    jbunniii

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    Yes, that looks right:

    ##x^2 + y^2 < \delta^2##
    ##3|x||y| < 3\delta^2##
    ##180|x||y|^5 < 180\delta^6##

    so

    ##x^2 + y^2 + 3|x||y| + 180|x||y|^5 < 4\delta^2 + 180\delta^6##
     
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