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Grad School

  1. Aug 27, 2014 #1
    I started graduate school this week after getting my BS in May of 2012. I'm having to review a lot of material. Heh. I've finished most of the homework, but I'm a bit stumped on two easy problems.

    The induction proof is no problem; I just haven't been able to find a formula. The nth partial sums are like 2, 8, 12, 20, 40, etc.

    I've played with the Alternate Triangle Inequality a bit and a Theorem from the book:

    |x| - |y| < |x - y| < ε

    |x| < |y| < ε
     
  2. jcsd
  3. Aug 27, 2014 #2

    Fredrik

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    I've been trying for ten minutes now, and I still don't see it. If I don't see it within the next few minutes, I think someone else will have to take this one. (Ignore what I wrote here before I edited. The tip I gave you was useless).

    No need for that sort of thing. |x-y|=0 implies x=y, so it's sufficient to prove that |x-y|=0. This reduces your problem to the following: If ##0\leq x<\varepsilon## for all ##\varepsilon>0##, then x=0.
     
    Last edited: Aug 27, 2014
  4. Aug 27, 2014 #3
    No problem. I tried to find a formula online but couldn't. From seeing kind of similar summations, my guess is that it's maybe a quadratic.

    Eh. I'm not quite sure how that leads to me to show that x=y. I'm not yet used to working full-time and then going home to do homeWORK. :zzz:
     
  5. Aug 27, 2014 #4

    Fredrik

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    The absolute value is defined by
    $$|x|=\begin{cases}x & \text{if }x\geq 0\\ -x & \text{if }x<0\end{cases}$$ It follows from this definition that 0 is the only real number with absolute value 0.
     
  6. Aug 27, 2014 #5
    ##\sum k(k+1)=\sum k^2+\sum k##
     
  7. Aug 27, 2014 #6
    I guess it seems (but I know that you're not***) like you're using the conclusion in the proof. However, it seems that a proof by contradiction might be easiest. If x ≠ y, then it would imply that there's some ε > 0 such that |x - y| > ε.

    I'm simply trying to start and use the given relation. Is your approach akin to

    0 ≤ |x - y| < ε for all ε > 0

    and inferring that |x - y| = 0?
     
    Last edited: Aug 27, 2014
  8. Aug 27, 2014 #7
    I think he wants a formula involving n such as

    1 + 2 + ... + n = (1/2)n(n+1)
     
  9. Aug 27, 2014 #8
    I understand that. Do you think you could maybe use the known formulas for ##\sum\limits_{k=1}^nk## and ##\sum\limits_{k=1}^nk^2## along with the hint I gave to find a formula for ##\sum\limits_{k=1}^nk(k+1)##?
     
  10. Aug 27, 2014 #9
    Ah. Yes. Let me look into that.

    =(1/6)n(n+1)(2n+4)
     
  11. Aug 27, 2014 #10

    Fredrik

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    For all x>0, we have |x|=x>0. For all x<0, we have |x|=-x>0. These results (which follow from the definition of the absolute value...and what I'm saying in the next paragraph) rule out the possibility that a non-zero real number could have absolute value 0.

    The result that -x is positive when x is negative isn't entirely trivial. It follows from the axiom (that's part of the definition of real numbers) that says that you can add any real number to both sides of any inequality. So if ##x<0##, we have ##x+(-x)<0+(-x)## and therefore ##0=x+(-x)<0+(-x)=-x##.

    That works, but I prefer to ignore x and y solve the simpler problem: If ##0\leq x<\varepsilon## for all ##\varepsilon>0##, then what is x? (Proof by contradiction sounds like a good idea. If x≠0, then...).
     
    Last edited: Aug 27, 2014
  12. Aug 27, 2014 #11

    Zondrina

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    Hint:

    $$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$

    You can show this by induction. Similarly by induction and the above formula you can show:

    $$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$
     
  13. Aug 27, 2014 #12
    gopher_p and Zondrina, thank you for the hint. I successfully worked the problem; I knew that I was missing something simple. I'm slowly getting back into things.
     
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