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Fun math proof

  1. Nov 12, 2005 #1
    Any nice proofs for this?

    [tex]2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}
    [/tex]




    I hope the tex came out alright. have fun!

    ps. n is any natural number.
     
    Last edited: Nov 12, 2005
  2. jcsd
  3. Nov 12, 2005 #2
    ill post solution in the morning.
     
  4. Nov 12, 2005 #3

    daniel_i_l

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    Gold Member

    I doesn't look that hard if you prove each half by induction, and if first you multiply everything by sqrt(n)/sqrt(n). But then again there could be a catch somewere. If I have time I'll try it.
     
  5. Nov 12, 2005 #4
    [tex]
    \begin{align*}
    2\sqrt{n+1} - 2\sqrt{n} &= 2(\sqrt{n+1} - \sqrt{n}) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} \\
    &= \frac{2}{\sqrt{n+1} + \sqrt{n}} \\
    &< \frac{2}{\sqrt{n} + \sqrt{n}} \\
    &= \frac{1}{\sqrt{n}}
    \end{align*}
    [/tex]


    Alternatively, let [itex]f(x) = \sqrt{x}[/itex], then by the MVT on [n, n+1]:

    [tex]
    \begin{align*}
    \frac{\sqrt{n+1} - \sqrt{n}}{1} &= f'(x) \\
    &= \frac{1}{2\sqrt{x}}, \text{ with x} \in (n, n+1) \\
    &< \frac{1}{2\sqrt{n}}
    \end{align*}
    [/tex]

    Second one should be similar.
     
    Last edited: Nov 12, 2005
  6. Nov 13, 2005 #5
    Yes, my solution ressembled the first one. Although, i was very amused at seeing the MVT making such short work of it. nice job. heres the pickle. i forgot the solution to this, ill think about it over the next few days, but have a try:
    Using this inequality, evaluate to the [tex]\pm 1[/tex] the integer part of the sum
    [tex] 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{1000000}}[/tex]
    any progress, you let me know.
     
  7. Nov 13, 2005 #6
    hehe, alright, it came back to me, but take a try at it anyways.
     
  8. Nov 15, 2005 #7
    Finally, heres the cherry on top. Prove that the integer part of the following expression is odd.

    [tex](\sqrt{n}+\sqrt{n+1})^2[/tex]

    have fun! im interested in seeing some original solutions as i think there are better ones than the ones i have seen.
     
    Last edited: Nov 15, 2005
  9. Nov 18, 2005 #8
    no takers? i post solutions tomorrow.
     
  10. Nov 18, 2005 #9
    I think some appropriate definite integration of 1/sqrt(x) will give the desired approximation.
     
  11. Nov 18, 2005 #10
    This follows from the following 2 observations:
    1.
    [tex](\sqrt{n}+\sqrt{n+1})^2 + (\sqrt{n}-\sqrt{n+1})^2 = 4n+2 [/tex]
    2.
    [tex](\sqrt{n}-\sqrt{n+1})^2[/tex] is really small.
     
  12. Nov 19, 2005 #11

    CRGreathouse

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    Science Advisor
    Homework Helper

    [tex]2\sum_{k=a}^b\left(\sqrt{k+1}-\sqrt{k}\right)<\sum_{k=a}^b\frac{1}{\sqrt k}<2\sum_{k=a}^b\left(\sqrt{k}-\sqrt{k-1}\right)[/tex]

    [tex]2\left(\sqrt{b+1}-\sqrt{a}\right)<\sum_{k=a}^b\frac{1}{\sqrt k}<2\left(\sqrt{b}-\sqrt{a-1}\right)[/tex]

    [tex]2\left(\sqrt{1000001}-\sqrt{1}\right)<\sum_{k=1}^{1000000}\frac{1}{\sqrt k}<2\left(\sqrt{1000000}-\sqrt{0}\right)[/tex]

    [tex]1998<\sum_{k=1}^{1000000}\frac{1}{\sqrt k}<2000[/tex]

    By direct calculation
    [tex]1998.5401<\sum_{k=1}^{1000000}\frac{1}{\sqrt k}<1998.5402[/tex]
     
    Last edited: Nov 19, 2005
  13. Nov 19, 2005 #12
    well done, youve saved me a bit of typing! yes thats the solution i have as well.
     
  14. Nov 19, 2005 #13
    interesting, but i find your second premise a little weak, as in... not justified. it definitely is true though. i could be wrong though, its just what if n is very large?
     
  15. Nov 19, 2005 #14

    CRGreathouse

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    Homework Helper

    It decreases strictly as n increases. In particular, if n>0, [tex]-1<\sqrt n-\sqrt{n+1}<0[/tex]. This gives the desired result.
     
  16. Nov 19, 2005 #15
    interesting! im curious, is there a nice proof of this inequality?
     
  17. Nov 19, 2005 #16

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    The original problem looks like an easy application of Taylor series! The 0-th order approximation is:

    [tex]
    f(x + e) = f(x) + f'(x^*) e
    [/tex]

    for some [itex]x^*[/itex] between x and x+e inclusive.

    Letting f be the square root function and e be 1, we have:

    [tex]
    \sqrt{x + 1} - \sqrt{x} = \frac{1}{2 \sqrt{x^*}} \leq \frac{1}{2 \sqrt{x}}
    [/tex]

    because the rightmost term is an upper bound for the middle term. Letting e = -1, we have:

    [tex]
    \sqrt{x - 1} - \sqrt{x} = - \frac{1}{2 \sqrt{x^*}} \geq -\frac{1}{2 \sqrt{x}}
    [/tex]

    because the rightmost term is a lower bound for the middle term. (Note the signs!)

    Therefore,

    [tex]\sqrt{x+1} - \sqrt{x} \leq \frac{1}{2 \sqrt{x}} \leq \sqrt{x} - \sqrt{x - 1}[/tex]

    or

    [tex]2\sqrt{x+1} - 2\sqrt{x} \leq \frac{1}{\sqrt{x}} \leq 2\sqrt{x} - 2\sqrt{x - 1}[/tex]
     
    Last edited: Nov 19, 2005
  18. Nov 19, 2005 #17
    Yes. It's actually quite easy to see why this is true, since the terms in the middle are consecutive square roots. A formal proof goes something like this:

    [tex]\sqrt{n+1} - \sqrt{n} > 0 \text{ is obviously true for } n \in \mathbb{N}[/tex]

    It remains to show that:

    [tex]\sqrt{n+1} - \sqrt{n} < 1[/tex]

    Squaring both sides and playing around, we see that this follows from [itex]n < n+1[/itex].

    [tex]
    n < n+1
    \Rightarrow n^2 < n(n+1)
    \Rightarrow n < \sqrt{n(n+1)}
    \Rightarrow 2n - 2\sqrt{n(n+1)} < 0
    \Rightarrow n + 1 - 2\sqrt{n(n+1)} + n < 1
    \Rightarrow (\sqrt{n+1} + \sqrt{n})^2 < 1
    [/tex]

    So we have:

    [tex]
    0 < \sqrt{n+1} - \sqrt{n} < 1
    \Rightarrow -1 < \sqrt{n} - \sqrt{n+1} < 0
    [/tex]
     
  19. Nov 19, 2005 #18
    If you say so, i havent finished my cal 2 yet unfortunately, hehe. but ill keep it in mind!
     
  20. Nov 19, 2005 #19
    oustanding, thanks for that.
     
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