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Estimation of Square roots

  1. Jan 13, 2005 #1
    How can I estimate the value of sqrt. of a particular no. which is not a perfect square(e.g. 125) without using the calculator?
    Secondly how does the Calculator solve it? Does it use logrithms? If it does how does it solve them?
     
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  3. Jan 13, 2005 #2

    Integral

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  4. Jan 13, 2005 #3
    Thanks Integral but I'm still confused... I'm just in 10th grade and I think the method there is for much senior students. Isnt there any simpler form of this method or any other simple method to find it?
    Any help will be appreciated.
     
  5. Jan 13, 2005 #4
    125=25x5. [tex]\sqrt{5^2x5} =5\sqrt{5}=5\sqrt{2^2+1}=5(2+1/4-1/64+-+)[/tex]

    The second part comes from using the method of Newton for fractional binominal expansion: [tex]\sqrt{4+1}=4^{1/2}+(1/2)4^{-1/2}+\frac{(1/2)(-1/2)}{2!}(4^{-3/2})[/tex]

    That is a very useful method, particularly when one of the two terms under the square root is just 1. (In that case we have 1^0, 1^1, 1^2, where each successive term is multiplied to each new term, and so is irrelevant.)

    However, we could also look at this:
    [tex]\sqrt{121+4} =11 +(1/2)121^{-1/2}(4^1) =11+2/11-+-[/tex]

    Which for two terms is just 11 +2/11. Now 1/11 = .090909..., so 2/11 = .18181818....giving 11.18 as a quick approximation, and we know the next term in the expansion would have been negative, so the real answer is a little less than 11.18181818...., but for two terms, we are completely correct!
     
    Last edited: Jan 13, 2005
  6. Jan 13, 2005 #5

    NateTG

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    Binary Search method:
    This method may seen simple, but it's really quite fast and easy to program.
    You're trying to find the square root of [itex]n[/itex].
    Start with some [itex]x_0=n[/itex] if [itex]n \geq 1[/itex] and [itex]x_0=1[/itex] otherwise (so [itex]x_0^2>n[/itex]).
    Now, generate [tex]x_n[/tex] using the following rules
    If
    [tex]x_{i}^2 > n[/tex]
    then
    [tex]x_{i+1}=x_{i} - \frac{x_0}{2^i}[/tex]
    If
    [tex]x_{i}^2 <i[/tex]
    then
    [tex]x_{i+1}=x_{i} + \frac{x_0}{2^i}[/tex]
    If
    [tex]x_{i}^2=i[/tex]
    then you've got the exact square root.
     
    Last edited: Jan 13, 2005
  7. Jan 13, 2005 #6

    dextercioby

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    I'm surprised that noone suggested the old-fashoined method i learnt in the 6-th grade,10 years ago.The algorithm of extracting sqrt.Take the number and write it in decimal notation and then extract sqrt with arbitrary approximation using the algorthm.

    Take 7.Write it 7.00 00 00 00 .It's still 7,right??The fact that u added 8 zeros after the dot means u'll be getting the 4 decimals from [itex] \sqrt{7} [/itex].Now apply to it the algorithm described here

    Daniel.

    PS.I still reccomend logarithm tables in the absence of computers.For example,u might need to extract root of order [itex] \pi\sqrt{4.5} [/itex].There's no algorithm in that case.
     
  8. Jan 13, 2005 #7
    A simple method to approximate a square root is this:
    example:
    [tex] \sqrt{42}[/tex]
    pick any number lower than 42 and divide 42 by this number:
    [tex] x = \frac{42}{6} = 7[/tex]
    take the average of both divisors:
    [tex]\frac{6+7}{2} = 6.50[/tex]
    divide 42 by this average:
    [tex]x = \frac{42}{6.50} = 6.46[/tex]
    take the average:
    [tex]\frac{6.5 + 6.46}{2} = 6.48[/tex]
    divide 42 by this average:
    [tex]x = \frac{42}{6.48} = 6.48[/tex]

    [tex] \sqrt{42} \approx 6.48[/tex]

    you can of course use more decimals and repeat dividing and taking averages more times to get a closer approximation
     
  9. Jan 13, 2005 #8

    BobG

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    Learn to square numbers really fast.

    https://www.physicsforums.com/showthread.php?t=53229&page=2&pp=15
    There's other short cuts you could learn, too.

    If you get to the square of the nearest 0 or 5 (20^2 = 400, 25^2 = 625, etc) and understand the rate that the square is increasing (or decreasing) using the fourth and fifth term, you can get pretty close.

    If you really understand the rate that the square is increasing, you can interpolate between the two closest integers a lot more accurately than you'd think (you round off your fractions to avoid unwieldy numbers and kind of get a feel for how you have to adjust). When I had to solve a lot of problems, I could pretty well count on being at least to the nearest hundredth.

    Of course, when I was in high school, electronic calculators were one of those new-fangled technological inventions that only professionals or rich kids could afford. My calculator was a Post Versalog 1460, made of the finest Hemmi bamboo. Actually, that was pretty high end for high school - most kids had the cheap Picketts. I was lucky - my dad needed a pretty high end calculator for his job, so I got his Post when he switched over to an electronic calculator. Since you had to keep track of significant digits yourself, being able to get a fairly accurate estimate was always an advantage (okay, my estimates were usually accurate to more significant digits than my slide rule could give me, but that's beside the point).
     
  10. Jan 13, 2005 #9
    Lets take a number, say..............., 154372. How would I find the square root of that because Dex's way was working until I found my left number was too large and BobG's method has gone over my head.

    The Bob (2004 ©)
     
  11. Jan 13, 2005 #10

    NateTG

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    Newton's method:
    Let's start by guessing 500.
    154372/500=308.744
    The average is 404.372
    154372/404.372=381.75...
    The average is 393.06
    154372/393.06=392.744
    and so on.

    Binary search:
    Start with 512 (starting with apower of two makes this work a bit better...)
    512*512 is to big subtract 256 -> 256
    256*256 is too small add 128 -> 384
    384*384=147456 is too small, add 64 -> 448
    448*448 is too big subtract 32 -> 416
    416*416 is too big subtract 16 -> 400
    400*400 is too big subtract 8 -> 392
    392*392 is too small, add 4 -> 396
    396*396 is too big subtract 2 -> 394
    394*394 is too big subtract 1 -> 393
    393*383=154449 is too big, subtract 0.5 ->392.5
    and so on.
     
  12. Jan 13, 2005 #11
    dextercioby Homework Helper I'm surprised that noone suggested the old-fashoined method i learnt in the 6-th grade,10 years ago.The algorithm of extracting sqrt.Take the number and write it in decimal notation and then extract sqrt with arbitrary approximation using the algorthm.

    Boy! You guys are slow to take a hint. He even tells you where to find the method: http://www.geocities.com/cnowlen/Cathy/Emat4680/Squareroot.htm

    In the case above you seperate two digits at a time: 15,43,72. The first step is to find a square, the largest and less than 15, that's 3, then you subtract 9 and bring down two more digits to get 643. (I don't have a better way of showing this.) Now that you have 3 above the number, you double this and make it 6 on the left, then you look at (10x6 +?)(?) is the largest answer less than 643. In this case it is 69, so 69x9 = 621, which you subtract from 643 and add on the last two digits, which are 72, to get 2272.

    Now you have two digits above the number 39, so you double this getting 78 and look at (10x78+?)(?) is less than 2272. Etc, etc.... Check the example he suggests.

    When I went to school before they had calculators, that was the standard method, you know. Today, it seems mostly a lost art! Obviously it is a direct method.
     
    Last edited: Jan 13, 2005
  13. Jan 13, 2005 #12

    BobG

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    Change it to 1543.72 x10^2 (I can't do square roots for numbers greater than 10000). It's between 1225 (35^2) and 1600 (40^2) and closer to 1600. Dropping to 39 means I need to subtract 80, then add 1, which drops me to 1521. The square root is between 39 and 40, around 39.25 (I don't have as good a feel for the interpolation as I did before I could use a calculator).

    Edit: Make that about 392.5 (I forgot my powers of ten. Divide those by 2 for the square root).

    (In fact, not near as good a feel, since the answer is much closer to 392.9)
     
    Last edited: Jan 13, 2005
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