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Math Help - Approximate Sqrt(2)

  1. #1
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    Approximate Sqrt(2)

    Approximate \sqrt{2}
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  2. #2
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    \sqrt2\approx1.414
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    Approximate \sqrt{2} to #.## and show how please.
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    The Babylonian Method works well here. It makes use of the bisection method and the fact that if you divide a number by its square root, you get the square root.

    You need to make an initial guess.

    Since \sqrt{1} < \sqrt{2} < \sqrt{4} that means 1 < \sqrt{2} < 2.

    Average these endpoints to get 1.5 is a good starting guess.


    \frac{2}{1.5} = 1.33333333.

    So that means 1.33333333 < \sqrt{2} < 1.5.

    Average these out and you get a new guess.

    Follow this process until you reach your desired level of accuracy.
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  5. #5
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    Quote Originally Posted by razemsoft21 View Post
    Approximate \sqrt{2}
    It helps if you post all of the information relevant to the question. Approximate using what method? To how many decimal places? etc.
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    Quote Originally Posted by mr fantastic View Post
    It helps if you post all of the information relevant to the question. Approximate using what method? To how many decimal places? etc.
    Approximate using what method? ----> If I know the method, no need to ask the question !
    To how many decimal places? ----> already mentioned in post #3 (#.## means to 2 decimal places.

    I think the method of Prove It works
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  7. #7
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    Quote Originally Posted by razemsoft21 View Post
    Approximate using what method? ----> If I know the method, no need to ask the question !
    To how many decimal places? ----> already mentioned in post #3 (#.## means to 2 decimal places.

    razemsoft21, it helps if you tell us the full details in the first post. Saying Approximate \sqrt2 isn't enough.

    Also, there are many methods of solving squareroots. If you don't know how to solve it, then maybe asking for someone to tell you a method of solving squareroots.

    Doing research would help too: Computing Square roots - Wikipedia. Here, it shows you many methods of how to solve a square root of a number.
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    Hello, razemsoft21!

    Where did this problem come from?
    It's unlikely that you would be assigned this problem
    , , without being taught any approximation methods.


    Approximate \sqrt{2}

    The very worst you could do is guess-and-adjust.

    Since \sqrt{2} is between 1 and 2 . . .

    \begin{array}{ccccccc}<br />
\text{Try 1.5:} & 1.5^2 &=& 2.25 & \text{too big} \\<br />
\text{Try 1.4:} & 1.5^2 &=& 2.25 & \text{too small} \\<br />
\text{Try 1.45:} & 1.45^2 &=& 2.1025 & \text{too big} \\<br />
\text{Try 1.42:} & 1.42^2 &=& 2.0164 & \text{too big} \\<br />
\text{Try 1.42:} & 1.41^2 &=& 1.9881 & \text{too small}<br />
\end{array}

    We have the two decimal places.
    Which is the "better" answer: 1.41 or 1.42 ?

    1.41 gives an error of: . 2 - 1.9881 \:=\:0.0119

    1.42 gives an error of: . 2.0164 - 2 \:=\:0.0164


    Since 1.41 has the smaller error: . \sqrt{2}\:\approx\:1.41
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  10. #10
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    . . . An elementary approximation of \sqrt{N}


    Let a_1 be your first approximation to \sqrt{N}.

    Substitute into: . \dfrac{N+a_1^2}{2a_1} \;=\;a_2 .[1]

    This produces a_2, a better approximation.


    For more accuracy, substitute a_2 into [1] . . . and so on.


    Example: . \sqrt{2},\;\;N = 2


    Let a_1 = 1.5

    We have: . a_2 \:=\:\dfrac{2 + 1.5^2}{2(1.5)} \;=\;1.41666\hdots

    . . Check: . (1.41666\hdots)^2 \:=\:2.006944443


    Then: . a_3 \:=\;\dfrac{2 + (1.4166\hdots)^2}{2(1.41666\hdots)} \:=\:1.414215686

    . . Check: . 1.414256862^2 \:=\:2.000007749


    Then: . a_4 \:=\:\dfrac{2 + 1.41425686^2}{2(1.414256862)} \:=\:1.414213562

    . . Check: . 1.414213562^2 \:=\:1.999999999


    Therefore: . \sqrt{2} \;\approx\;1.424213562 .to 9 decimal places.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Why does [1] provide better and better approximations?


    We want an approximation of \sqrt{N}.

    Our first guess is a_1.

    We have factored N into two factors: . a_1\,\text{ and }\,\dfrac{N}{a_1}

    If we are extremely lucky, the two factors will be equal.
    . . (Think about it!)

    Since it is unlikely that the two factors are equal,
    . . a better approximation is the average of the two factors.
    . . (The number exactly halfway between them.)

    So we use: . a_2 \;=\;\dfrac{a_1 + \frac{N}{a_1}}{2} <br />

    . . which simplfies to: . a_2 \;=\;\dfrac{N+a_1^2}{2a_1} . . . . . ta-DAA!
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  11. #11
    ggn
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    this is more accurate

    1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799
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  12. #12
    ggn
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    we must appreciate the works of Robert Nemiroff he calculated to 1 million digits just check this out!

    The Square Root of Two to 1 Million Digits
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  13. #13
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    Thank you for help and especial THANKS for:
    Prove It , Educated , Soroban
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  14. #14
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    Quote Originally Posted by razemsoft21 View Post
    Approximate \sqrt{2} to #.## and show how please.
    Here's a perfectly valid method: a calculator (for example, the one included with "Windows") gives \sqrt{2} as 1.4142135623730950488016887242097. To two decimal places, that is 1.41.
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