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Olympiad problem -- Sum involving many square roots...

  1. Jun 11, 2017 #1
    √(2-√(2^(2)-1))+√(4-√(4^(2)-1))+√(6-√(6^(2)-1))+...+√(80-√(80^(2)-1))
    How the find it's value
     
  2. jcsd
  3. Jun 11, 2017 #2

    Nidum

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    Do you have any ideas yourself ?

    Just out of curiosity are there any rules about how these problems have to be solved ? Do the solutions always have to be analytic ones ?

    Personally if I wanted an answer to this problem for some practical purpose I would just write a program to sum all the terms numerically - it would only be a few lines of coding .
     
  4. Jun 11, 2017 #3
    Well you are not supposed to use a calculator and the answer should be rounded of to nearest integer if needed.
     
  5. Jun 11, 2017 #4
    Well i have an idea and here it is:
    So if y=2-√(2^(2)-1)
    y=2-√3
    y(2+√3)=1
    And this works for all the terms.
     
  6. Jun 11, 2017 #5
    But I don't know how to connect it to the equation
     
  7. Jun 11, 2017 #6

    mfb

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    I moved the thread to our homework section.

    Or y=1/(2+√3).
    If you look at larger values like 1/(78+√(782-1)) or 1/(80+√(802-1)), do you see how you could approximate that?
    Does this lead to a familiar series?
     
  8. Jun 11, 2017 #7
    Convering series? But how can I estimate the final and?
     
  9. Jun 11, 2017 #8

    mfb

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    With an approximation for this series (well, the finite partial sum, to be more precise).
     
  10. Jun 11, 2017 #9
    Can I say this

    40
    Summation (√2n-√(2n)^(2)-1))
    n=1
    And how can I find the partial sum for this
     
  11. Jun 11, 2017 #10

    mfb

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    This is just your original problem written in a different way. Sure.
    Find a suitable approximation, see above.
     
  12. Jun 12, 2017 #11
    I'm posting my solution because this thread wasn't originally in homework section and doesn't appear to be homework.
    After some algebra I write the problem as$$\sqrt{2}\sum_{n=1}^{40}\sqrt{n-\sqrt{n^2-.25}}$$
    I multiply the sum by,
    $$\frac{\sqrt{n+\sqrt{n^2-.25}}}{\sqrt{n+\sqrt{n^2-.25}}}$$
    I approximate$$\sqrt{n^2-.25}\approx n$$
    And after a little more algebra the sum becomes approximately$$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}}$$
    I approximate ##\sqrt{2}\approx 1.5 ## ## \sqrt{3}\approx 2## etc. I note the first two terms ##\approx \frac{1.75}{2}##, the next five terms##\approx 5\frac{1}{2}\frac{1}{2}##, the next five terms ##5\frac{1}{2}\frac{1}{3}##, the next ten terms ##\approx 10\frac{1}{2}\frac{1}{4}##, and the last eighteen terms ##\approx 18\frac{1}{2}\frac{1}{6}##. Adding it all up I get 5.2.
     
  13. Jun 12, 2017 #12

    fresh_42

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    Hmm, the program (formula from post #9) ended with ##5.65685 ...##
     
  14. Jun 12, 2017 #13

    mfb

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    Quite good for an approximation. If we round we are off by 1.

    Replace the sum by an integral:
    $$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}} \approx \frac 1 2 \int_{0.7}^{40.5} \frac{dx}{\sqrt{x}} = [\sqrt{x}]_{0.7}^{40} = \sqrt{40}-\sqrt{0.7} \approx 6.3 - 0.85 = 5.45$$
    0.7 is black magic based on intuition, the square roots were estimated without calculator.

    Compare it to the precise value of the sum:
    $$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}} \approx 5.6338$$

    If we round that, we are still off by 1, but now the estimate is within 0.2 of the precise value of the original problem.

    We get a better result with a numerical evaluation of the first two terms:
    $$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}} \approx \frac 1 2 \left( 1+\frac{1}{\sqrt 2} + \int_{2.5}^{40.5} \frac{dx}{\sqrt{x}} \right) = 0.5 + 0.35 + \sqrt{40}-\sqrt{2.5} \approx 0.85 + 6.33 - 1.59 = 5.59$$
    Again without calculator (and without using the result from above...), but with a bit more thought about the square roots.
     
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