Approximating a Square Root

Поделиться
HTML-код
  • Опубликовано: 4 фев 2025

Комментарии • 12

  • @ali8800
    @ali8800 3 месяца назад +4

    Wow! Really informative and concise.

  • @ingiford175
    @ingiford175 23 дня назад

    FYI the doubling the value and using that as the denominator, is just taking the second value of the continued fraction. Also can be found via newton's method.

  • @cyruschang1904
    @cyruschang1904 23 дня назад

    4.1 x 4.1 = 16.81
    4.2 x 4.2 = 17.64
    4.125 x 4.125 =17.0156
    4.1245 x 4.1245 = 17.0115
    4.123 x 4.123 = 16.999
    √17 is about 4.123

  • @robertveith6383
    @robertveith6383 3 месяца назад +1

    @Zach's Math Zone --

  • @markryabenkiy4080
    @markryabenkiy4080 3 месяца назад +4

    square root of 68 is just 2sqrt(17)

    • @frankoptis
      @frankoptis 3 месяца назад

      sqrt(68) = sqrt( 4 * 17) = sqrt(4) * sqrt(17) = 2 * sqrt(17) - good find! The only thing I struggle with is how he calculates the fraction to be "double", e.g. from 8^2 to the approximation 2 * 8 as the fraction part.

    • @dmdeemer
      @dmdeemer 3 месяца назад +3

      @@frankoptis It is because the derivative of sqrt(x) is 1/(2*sqrt(x)). This approximation finds the tangent to a known point on the sqrt curve A: f(x) = sqrt(A) + (x-A) sqrt'(A) = sqrt(A) + (x-A)/(s*sqrt(A)). As such it works well for the examples given in the video that are just over a square number, and less well for values that are just under a square number (like 24 or 35). Those will have quite a bit more error.

  • @jeffdege4786
    @jeffdege4786 26 дней назад

    Isn't this just a single pass of Newton"s method?

  • @jimurban5367
    @jimurban5367 3 месяца назад +3

    If you’re going to calculate the square root by hand anyway, just use the square root algorithm.

    • @slamopfpnoobneverunsub5362
      @slamopfpnoobneverunsub5362 3 месяца назад +2

      is the square root algorithm involve those fraction a+b and a+2b and stuff? Not sure which one are you talking about.