# Calculating the root of a number by hand

• I
• NODARman
NODARman
TL;DR Summary
.
Hi, is it possible, is there any formula that can help me to take root from (for example) 1,2 without a calculator (by hand)?
For example, there is a cos(x) formula that can be calculated on the paper:
$$\cos x=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n}}{(2 n) !}$$

There is the Babylonian method for roots, but it's not as accurate as the cos(x) formula.
$$x^{\prime}=\frac{1}{2}\left(x+\frac{n}{x}\right)$$

## Answers and Replies

Mentor
Hi, is it possible, is there any formula that can help me to take root from (for example) 1,2 without a calculator (by hand)?
It's not clear to me what you are asking here.
Which root -- square root, cube root, etc.?
In your example are you asking about the square root of 1.2; i.e. ##\sqrt{1.2}##?
NODARman said:
For example, there is a cos(x) formula that can be calculated on the paper:
$$\cos x=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n}}{(2 n) !}$$
This formula has nothing to do with roots. It is the Maclaurin series representation for the cosine function. There's a corresponding series for the sine function and many other functions.

NODARman said:
There is the Babylonian method for roots, but it's not as accurate as the cos(x) formula.
$$x^{\prime}=\frac{1}{2}\left(x+\frac{n}{x}\right)$$
I don't believe this derives from the Babylonians, since they knew nothing about derivatives. This formula derives from a technique called the Newton (or Newton-Raphson) method.

If you're interested in calculating square roots by hand, I was taught a technique back when I was in the 8th grade, a long time ago. The technique is somewhat akin to long division. As far as I know, it's no longer taught. Here's a link to a youtube video:

Last edited:
FactChecker, PeroK, DrClaude and 3 others
Homework Helper
Gold Member
I don't believe this derives from the Babylonians, since they knew nothing about derivatives.
You don't need to know anything about derivatives to derive this method, you simply need to realise that if ## x ## is an understimate of ## \sqrt n ## then ## \frac n x ## is an overestimate (and vice versa) and therefore the midpoint ## x^{\prime}=\frac{1}{2}\left(x+\frac{n}{x}\right) ## is a better estimate.

The term "Babylonian method" is commonly used, although I am not aware of a confirmed source. See e.g. https://demonstrations.wolfram.com/BabylonianAlgorithmForComputingSquareRoots/.

Edit: https://www.sciencedirect.com/science/article/pii/S0315086098922091 seems to provide a source.

DrClaude, PeroK and NODARman
NODARman
It's not clear to me what you are asking here.
Which root -- square root, cube root, etc.?
In your example are you asking about the square root of 1.2; i.e. ##\sqrt{1.2}##?

This formula has nothing to do with roots. It is the Maclaurin series representation for the cosine function. There's a corresponding series for the sine function and many other functions.

I don't believe this derives from the Babylonians, since they knew nothing about derivatives. This formula derives from a technique called the Newton (or Newton-Raphson) method.

If you're interested in calculating square roots by hand, I was taught a technique back when I was in the 8th grade, a long time ago. The technique is somewhat akin to long division. As far as I know, it's no longer taught. Here's a link to a youtube video:

x' is not a derivative of x, it means the new result of x.

Mentor
x' is not a derivative of x, it means the new result of x.
Without any explanatory context, a "primed" variable would ordinarily be interpreted to mean the derivative of that variable.

MidgetDwarf
These algorithms fall into something we call algorithms in mathematics. A numerical analysis book, has many of these types of solutions, not just for roots.

To give you a better answer. What are you trying to find the root of? a square root, cubic, function? A particular example would help.

Gold Member