Approximating a Square Root

[@ali8800]

Wow! Really informative and concise.

[@ZachsMathZone]

Thanks for watching!

[@markryabenkiy4080]

square root of 68 is just 2sqrt(17)

[@frankoptis]

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]

@@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.

[@robertveith6383]

@Zach's Math Zone --

[@jimurban5367]

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

[@slamopfpnoobneverunsub5362]

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