Friday, June 1, 2012

Finding the square root of an integer using Calculus

Have you ever been in a situation where you needed to find the square root of an integer (especially imperfect squares), but had no calculator close by? I have. I however remember learning a Calculus trick (in senior high) that has always come to my aid in such situations. Grab a piece of paper and a pen, and come along as I share this trick. The trick requires that you identify some things:
  • The integer (Obviously!)
  • A perfect square that can be got from the integer
  • The remainder in subtracting the perfect square from the integer
The steps involved in deriving the formula for finding the square root of an integer is shown below:


As an example, let's find the square root of the integer 7:

PS: You will notice (after a number of practices) that the values got using this method differ slightly from the ones got from the calculator. Do not be alarmed, the values are still correct nonetheless.

I do hope this is helpful and would come in handy when we need it.
Please let me know what your thoughts on any other trick you know. Waiting for your comments. ;-)


  1. Brilliant! Never seen this before. Thanks for sharing...

    1. Thanks. I hope it comes in handy...

  2. I don't know calculus but it appears you're talking about something like this?

    1. Yes Komla. Same idea, different approaches. ;-)

  3. Manual square root analogous to long division:

    1. Better link:

      and am I the only person to remember this method?

  4. Another approach, without calculus, but that is basically the same is this:

    0) looking for sqrt(X) = Y => X = Y^2

    1) nearest integral square: X0 = Y0^2

    1.1) Y0 may be greater or less than, i.e. next integer up or down from, Y

    2) deltaY = Y - Y0; Y = Y0 + deltaY

    3) deltaX = X - X0

    4) Y^2 = (Y0 + deltaY)^2 = Y0^2 + 2 Y0 deltaY + deltaY^2

    5) approximation: drop deltaY^2; less than 1/4 if Y0 is closest integer to Y

    5.1) this is the same as the [deltaY/deltaX = dy/dx (partial)] in the post above

    6) Y^2 ~ Y0^2 + 2 Y0 deltaY

    7) solve for deltaY:

    7.1) deltaY ~ (Y^2 - Y0^2) / (2 Y0)

    7.2) deltaY ~ (X - X0) / (2 Y0)

    7.2) deltaY ~ deltaX / (2 sqrt(X0)) QED

    8) Finally, error in Y calculated by this
    method, assuming deltaY < 1/2, will be
    less than 1/(8 X0)