Approximation of Square Roots

Square root is common function in mathematics. It has a wide range of applications from the field of mathematics to physics. Sometimes it gets hard to calculate square root of a number, especially the one which are not actually square of a number. Often the method we employ are to tedious work with decimals. Here is a guide to find square root or rather their approximates.

Let $n$ be the number whose square root we need to calculate. Let $n$ can be written as $p+q$ where $p$ the largest perfect square less than $n$ and $q$ be any positive real number. Then,

$\sqrt{n} = \sqrt{p+q} ≈ \sqrt{p} + \frac{q}{2\sqrt{p}+1}$

Approximate the square root of 968.

---

Let us first find the perfect square less than $968$. To do this we would be comparing $968$ with perfect squares that are easy to figure out like $30^2=900$.

Clearly,

$$900<968$$

Hence,

$$30<\sqrt{968}$$

Now try the square of another number greater than $30$ like $31^2=961$ and $32^2=1024$.

Clearly,

$$961<968<1024$$

$$31<\sqrt{968}<32$$

Now applying my formula,

$$\sqrt{968} = \sqrt{961+7}≈ \sqrt{961} + \frac{7}{2\sqrt{961}+1}$$

$$\sqrt{968} ≈ 31+\frac{7}{2(31)+1}$$

$$\sqrt{968} ≈ 31+\frac{7}{63}$$

$$\sqrt{968} ≈ 31+\frac{1}{9}$$

$$\sqrt{968} ≈ \boxed{31.11}$$

You can cross check by squaring the answer $(31.11)^2=967.8321$.

That is indeed a good approximation.

Here is the proof of the formula mentioned above.

The proof of the formula lies inside the formula itself. The following formula is based on assumption that roots between two perfect squares are uniformly distributed.

To understand it better we can use a graph-

We know that $2\sqrt{p}+1$ integers exist between a perfect square $p$ and $(\sqrt{p}+1)²$.

Let,

$n$ be any real number that exists between $p$ and $(\sqrt{p}+1)²$. Also, $n-p=q$.

So,

$\sqrt{n}$ will exist between $\sqrt{p}$ and $\sqrt{p}+1$.

Clearly the difference between $\sqrt{p}$ and $\sqrt{p}+1$ is $1$, now $2\sqrt{p}+1$ roots can be uniformly distributed between $\sqrt{p}$ and $\sqrt{p}+1$, if each root occupies $\frac{1}{2\sqrt{p}+1}$ space.

The $q^{th}$ number will occupy $\frac{q}{2\sqrt{p}+1}$ space. Hence the formula turns out to be

$\sqrt{n} = \sqrt{p+q} ≈ \sqrt{p} + \frac{q}{2\sqrt{p}+1}$

This section deals with conventional method of square root-finding. This method is more accurate, however it is a tedious one too.