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Homework Help: Preparing for test

  1. Aug 3, 2005 #1
    I have two questions i hope you can help me with:

    The first one is about series of positive terms like, for example [tex]\sum _{i=0}^{inf}(n^{1/n}-1)[/tex].
    Can i say that as the square of this series of positive terms diverges by the "nth term test", the original series also diverges.
    If i cant use this, could you please point me on the right direction in order to do determine the convergence of this series?.

    My second question is about a limit by definition(lambda-epsilon proof). I have to find out if the following function is continuous and/or differentiable:
    [tex]f(x,y) =\left\{\begin{array}{cc}\frac {x*y} {\sqrt x -\sqrt y},&\mbox{ if }x\neq y\\0, & \mbox{x=y}\end{array}\right[/tex]

    Everything i tried indicates that this function is continuous, but i cant prove that the limit when (x,y)->(0,0) of f is in fact 0.
    I you want me to explain a little better what i have done just ask. I would really appreciate any help you could give me.
    Many Thanks, Paul.
     
  2. jcsd
  3. Aug 3, 2005 #2

    Galileo

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    A general hint for these kind of limit problems. When you want to prove such a limit exists and is zero, you usually need to find an upper-bound for that function which also tends to zero. This is basically the squeeze theorem.
    There are probably other ways to do it, but playing around I found:

    [tex]\left| \frac{xy}{\sqrt{x}-\sqrt{y}}\right| \leq \frac{|xy|}{\sqrt{x}+\sqrt{y}}= |x|\cdot \left| \frac{y}{\sqrt{x}+\sqrt{y}}\right| \leq |x|\sqrt{y}[/tex]
    The first inequality uses the triangle inequality on the denominator. The second is obvious.
    The right side clearly goes to zero.
     
  4. Aug 3, 2005 #3
    thanks

    Ok, I understand what you did, and i was thinking to do something very similar, but the inequalities didnt came up to my mind. I now have to find out if it is differentiable or not. Would i be on the right track if i use the theorem that says that if the partial derivatives are continuous in an "interval", then the function is also differentiable on that "interval"???? I think i should use this instead of the differentiability limit, because i dont know which are the points to test for differentiability. Am i right????

    Well i found out the partial derivatives and i realized that they are continuous everywhere except on (0,0) (I forgot to tell on the initial problem statement that the function is defined only for x and y postive or equal to 0), so i used the differentiability limit and i found that it was differentiable on (0,0).
    Is my answer and PROCEDURE correct????

    Do you have any hints for the first problem????

    Thankfully, Paul.
     
  5. Aug 3, 2005 #4

    Galileo

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    This is a mistake. The right side is smaller than the left side, since [itex]\sqrt{x}+\sqrt{y} \geq |\sqrt{x}-\sqrt{y}|[/itex]
     
  6. Aug 3, 2005 #5
    checking

    OOOUch, i didnt check your work correctly the first time, but now i see it was wrong. So then how could i prove this limit????
    Assuming i could prove the continuity limit, i am pretty sure i could prove the differntiability also.
    Grateful, Paul.
     
  7. Aug 3, 2005 #6
    Polar Coordinates.
     
  8. Aug 4, 2005 #7

    Galileo

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    Actually, it's pretty easy to see the limit doesn't exist (so no need to check for differentiability).

    Note that the line y=x (where x>0) doesn't lie in the domain of f. The denominator can gets arbitrality small in any neighbourhood around (0,0). Therefore, if you choose an [itex]\epsilon[/itex]-neighbourhood around (0,0), no matter how small, you can approach the line y=x from two sides inside that neighbourhood. Take a circle of radius [itex]r<\epsilon[/itex] and approach y=x from two sides. The product xy is nonzero near that line, but [itex]\sqrt{x}-\sqrt{y}[/itex] approaches [itex]0^+[/itex] and [itex]0^-[/itex] respectively.

    About the sum, you use 'i' as your index, but have 'n' in the expression, did you mean:

    [tex]\sum_{n=0}^{\infty}(n^{1/n}-1)[/tex]

    I`m not familiar with the 'n'th-term test'.
     
  9. Aug 4, 2005 #8
    ok

    ok, so problem solved for the second one.
    About the first one, it is as you said:
    [tex]\sum_{n=0}^{\infty}(n^{1/n}-1)[/tex]

    And what i called the "n-th term root" is just the propertie that says that if a series [tex]\sum_{n=0}^{\infty}(A_n)[/tex] converges, then [tex]\lim_{n->\infty}A_n=0[/tex]

    I am sorry, but i didnt know how this property was called in english.

    I found another way to solve this exercise but again i am not sure if what i did is correct:
    I divided [tex]n^{1/n}-1[/tex] by [tex]n^{1/n}[/tex] and took the limit of this when n tends to infinite. It gave me 0, so i concluded that the series:
    [tex]\sum_{n=0}^{\infty}(n^{1/n}-1)[/tex] and [tex]\sum_{n=0}^{\infty}(n^{1/n})[/tex] behave the same way(i think it is because the limit comparison test right??).
    As the second series diverges, then the original series does also diverges.

    I would really appreciate if you could tell me if any of the two ways i tried are correct.
    I have my doubts about the second one specially, because i dont remember if i could use the limit comparison test when the limit A_n/B_n is equal to 0 with n tending to infinite.

    Thank you, Paul.
     
    Last edited by a moderator: Aug 4, 2005
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