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Showing exp is positive

  1. Nov 23, 2007 #1

    Päällikkö

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    [SOLVED] Showing exp is positive...

    ... well not quite. This isn't actually homework, but here's what I'd like to prove:
    [tex]\forall n \in \mathbb{N} : \sum_{k=0}^{2n}\frac{x^k}{k!} \ge 0[/tex].

    I tried going about it through induction, but got into trouble quite early on: I couldn't prove that
    [tex]\sum_{k=0}^{2(n+1)}\frac{x^k}{k!} = \sum_{k=0}^{2n}\frac{x^k}{k!} + \frac{x^{2n+1}}{(2n+1)!} + \frac{x^{2n+2}}{(2n+2)!} \ge 0[/tex],
    as using
    [tex]\sum_{k=0}^{2n}\frac{x^k}{k!} \ge 0[/tex]
    here I'd get
    [tex]\frac{x^{2n+1}}{(2n+1)!} + \frac{x^{2n+2}}{(2n+2)!} \ge 0[/tex],
    which cannot be proven as it's false.

    So obviously I'm in need of a different viewpoint. My brain is all tangled up, I do hope the problem isn't as easy as it looks at a first glance.
     
  2. jcsd
  3. Nov 23, 2007 #2

    rock.freak667

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    Well try bringing the 2 fractions to the same base by saying that (2n+2)!=(2n+2)(2n+1)!
     
  4. Nov 23, 2007 #3

    Päällikkö

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    Which equation are you talking about? I've tried that plenty before, to no avail.

    I forgot to mention in the first post that I managed to show that
    [tex]\frac{x^{2n+1}}{(2n+1)!} + \frac{x^{2n+2}}{(2n+2)!} \ge -\frac{(2n+1)^{2n+1}}{(2n+2)!}[/tex],
    but that lead to nowhere as the expression
    [tex]\sum_{k=0}^{2n}\frac{x^k}{k!} \ge \frac{(2n+1)^{2n+1}}{(2n+2)!}[/tex]
    does not hold.
     
    Last edited: Nov 23, 2007
  5. Nov 23, 2007 #4

    Dick

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    Try this. The Lagrange form of the Taylor series remainder is R(2n)(x)=exp(y)*x^(2n+1)/(2n+1)! for some y between 0 and x. So if I call your series P(2n)(x), then P(2n)(x)+R(2n)(x)=exp(x). Clearly you have nothing to worry about if x>0. If x<0 then R(2n)(x)<0, but exp(x)>0. Reach a conclusion about P(2n)(x).
     
  6. Nov 23, 2007 #5

    EnumaElish

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    I'd start with S(1) = [itex]\sum_{k=0}^{2}{x^k}/{k!} = 1 + x + x^2/2[/itex]. Assume x < 0, then show S(1) > 0 because the first and the third terms trump the second. Then generalize to S(n).
     
  7. Nov 23, 2007 #6

    Päällikkö

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    My thinking in the first post works for x > 0, but runs into trouble when x < 0 (More precisely, everything except -2n - 2 < x < 0 should work out directly (I haven't actually tried to prove that the assertion holds in the mentioned region separately, maybe that'll work)). You confronted this difficulty with a quite involved theorem about Lagrange remainders and the properties of exp. Although correct, I'm not completely satisfied: I'd rather have the proof not relying on the properties of exp.

    This is indeed exactly what I've done in the first post. As I pointed out, the difficulty started when trying to prove the implication S(n) => S(n+1).
     
  8. Nov 23, 2007 #7

    Dick

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    I don't consider Taylor series remainder terms to be be rocket science, but maybe there is a more direct approach. I wish you luck in finding it. I couldn't come up with anything.
     
  9. Nov 23, 2007 #8

    EnumaElish

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    That must be because any "excess positivity" in S(n) actually gets to be used in S(n+1) to soak up (cancel out) the additional negativity introduced (together with some additional positivity) while one goes from n to n+1.

    I.e. you have to assume not only that S(n) > 0 (which is what your attempt at induction assumes), but actually S(n) > M(n) > 0 so that S(n+1) > 0.
     
    Last edited: Nov 23, 2007
  10. Nov 24, 2007 #9

    Päällikkö

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    If you'd read my posts, you would have noticed that that's what I've tried. To recap: there exists an x such that
    [tex]\frac{x^{2n+1}}{(2n+1)!} + \frac{x^{2n+2}}{(2n+2)!} = -\frac{(2n+1)^{2n+1}}{(2n+2)!}[/tex],
    but
    [tex]\sum_{k=0}^{2n}\frac{x^k}{k!} \ge \frac{(2n+1)^{2n+1}}{(2n+2)!}[/tex]
    does not hold for all x. That is, one cannot find an M(n) independent of x (of the kind that would help with the proof). This makes the proof trickier.

    EDIT: Got it.
     
    Last edited: Nov 24, 2007
  11. Nov 25, 2007 #10
    Can you show the proof; I wanna check whether my proof is same or not.
     
  12. Nov 25, 2007 #11

    Päällikkö

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    It relies on the fact that:
    [tex]\frac{d^2}{dx^2}\sum_{k=0}^{2(n+2)}\frac{x^k}{k!} = \sum_{k=0}^{2n}\frac{x^k}{k!}[/tex]

    The rest should easily follow.
     
  13. Nov 25, 2007 #12

    arildno

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    The functional properties of the exponential function, that Exp(x+y)=Exp(x)*Exp(y) and Exp(0)=1 are sufficient to prove that Exp(x) are greater than 0 for ALL x, since Exp(-x)=Exp(0)/Exp(x) for arbitrary positive x.
     
  14. Nov 25, 2007 #13
    Mine is same.
     
  15. Nov 25, 2007 #14

    Dick

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    So to let the rest of us catch up, how did you do it using only that?
     
  16. Nov 26, 2007 #15

    Päällikkö

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    [tex]\forall n \in \mathbb{N} : \sum_{k=0}^{2n}\frac{x^k}{k!} > 0 \quad (1)[/tex]
    I changed the greater than or equal sign to a greater than (might as well go for the stronger version as the proof doesn't differ one bit).
    So :
    [tex]\sum_{k=0}^{2}\frac{x^k}{k!} = 1 + x + \frac{x^2}{2} > 0[/tex].

    Therefore we may assume that (1) holds for some n. Now
    [tex]\frac{d^2}{dx^2}\sum_{k=0}^{2(n+1)}\frac{x^k}{k!} = \sum_{k=0}^{2n}\frac{x^k}{k!} > 0[/tex]
    (Apparently I have a typo in my last post in the indices, that might be confusing)

    It follows that the function has exactly one minimum, at the point x0:
    [tex]\frac{d}{dx}\sum_{k=0}^{2(n+1)}\frac{x_0^k}{k!} = 0[/tex]

    With the above equation, we can drop a whole lot of terms from the original:
    [tex]\sum_{k=0}^{2(n+1)}\frac{x^k}{k!} > \sum_{k=0}^{2(n+1)}\frac{x_0^k}{k!} = \frac{x_0^{2n+2}}{(2n+2)!} > 0[/tex]

    This holds for all x.
     
    Last edited: Nov 26, 2007
  17. Nov 26, 2007 #16

    Dick

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    Thanks.
     
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