Calculate any square root without a calculator!

A simple formula will be derived that gives the power to calculate any square root without the use of a calculator.

[ChrT]

This is EXACTLY the same as one step of the Newton algorithm to find the zero of the function f(x)=x^2-a. (i) guess the root x (a nearby number) and then (ii) perform one Newton step to improve the guess x, i.e. subtract f(x)/f'(x), which leads to x-(x^2-a)/2x = (x^2+a)/2x. Since the Newton algorithm converges extremely fast, this approximation is so good. Fact is, if you replace x by the new approximation and again subtract f(x)/f'(x), each step will about double the number of correct digits. (quadratic convergence)

[Bromon]

“No calculator required” you overestimate my ability to divide.

[ARBB]

This has always been a very useful fact I've used to mentally calculate roots. Never seen the proof tho, nice!

[duongbinh]

Something like the AM-GM (Arithmetic Mean-Geometric Mean) inequality.

The closer x^2 is to a, the higher the accuracy of the estimate.

[edmondscott]

These are the first 2 terms of the binomial expansion of (x^2 + a) ^1÷2, where x =8 and a = 2.

[DFeitoza]

That’s a quite beautiful equation. I use basic math to demonstrate how it works and calculate the error.

N=n² Q=q²

-> (n-q)² = n²+q²-2nq

-> 2nq = n²+q² - (n-q)²

-> n = (n²+q²)/2q - (n-q)²/2q

-> n≈(n²+q²)/2q

n -> square root of interest q -> perfect square closest to n

-> Err=(n-q)²/2q

Err -> Error We can see that the error Err is proportional to the difference between n and q. And inversely proportional to q.

So, when q is larger, the error gets smaller.

That explains the shape of that plot.

[aeromoe]

Black ink (more contrast) and bigger hand written text and printed examples would be much appreciated and more helpful.

[Ujjwal_Blunders]

Loved the lesson,
Btw can you mention what pen you are using, it looks so cool !!

[tunneloflight]

Use a calculator or spreadsheet. They are ubiquitous. If you don't have one, use the long division method to extract as many digits as you wish with no error at all, save what ever you didn't calculate. No need for approximation.

[tomdekler]

This IS a very neat tool! I can't say I fully understand the video as I have no idea about what taylor expansions are, but seeing the formula work in action has been fairly magic.

That having been said, for me this formula is a lot of unnecessary arithmetic and I encourage everyone to instead learn the derived formula:
√a ≈ x + (a - x²)/2x

The first step, looking for x, is essentially looking for everything that's gonna be in front of the decimal point.*
So you already know this bit. The answer is going to be x plus an approximate of the remainder, which is (a - x²)/2x

These steps feel very intuitive to me.

In the formula as you wrote it down, you're taking x, squaring it into a big number that you add to another big number, and then dividing that very big number by 2x, which you then have to divide into n2x (where n will always* turn out to be equal to x) + a remainder. It leads to a nice single fraction, sure, but not an easy one to calculate.

The derived formula I propose makes this easier.

So for instance sqrt(137) = 11 + (137-121)/22 = 11 + 16/22. In one glance I can see that 16/22 is 8/11, which is pretty close to 7/10 so my rough approximation would be 11.7, which is very close indeed.

Your formula would have me calculate 258/22, which isn't _too_ hard to work out but I much prefer just instantly having the 11.something answer right from the getgo.

*note: The only exception would be if a - x² > 2x, where the remainder will be larger than 1. Since the difference between two squares is 2x+1, this only happens when _a_ is a non-integer number between x² + 2x and x² + 2x + 1.
Even then, I would rather calculate something like 13 + 26.5/26 ≈ 14 over having to calculate 364.5/26

[Malaphor]

Return of the Square format for the Square root.

[mstarsup]

1: nice approximation, i like that.
2: no need for taylor approximations, just developping (x-sqrt(a))^2 leads to the result, as when isolating sqrt(a), the additional term (x-sqrt(a))^2/(2x) goes to 0 as a goes to infty.
3: would have been nice to use a logarithmic scale for the x-axis for the graphs at 1:50 - to show the remainder vanishes as a grows large.

[gigachadkartik]

EASY PROOF:

Assume x ≈ √a,
then (x-√a) ≈ 0
or, (x-√a)² ≈ 0
or, x² + a -2x√a ≈ 0
or, x² + a ≈ 2x√a
or, √a ≈ x² + a /2x (PROVED)

The closer our assumption is the more accurate the formula is that means
closer value of a is to a perfect square number (like 24, 26, 48, 51, 80, 82, etc...) the more accurate we are.

[AikaterineG]

This reminds me of Julia's regression square root formula (Don't confuse it with the package of the Julia programming language)

[eeetube]

How to get x? You have to remember the closest integer square root to use this method.

[berndmayer]

or you break off the binomial series after the linear term

[timetravellingblockhead]

You know a maths video is going to be good when it's a 4:3 ratio video, with a fountain pen, by an obscure YouTube channel, where the video has become inexplicably popular

P.S Is your grip on the fountain pen comfortable?

[CliffSedge-nufv]

I prefer just using the Taylor series to first or second order. This is essentially the same thing, but seems to over complicate things.

[DotaMobaUnionRu]

But you can truncate the Taylor series at the later terms, as well.

[thomaskember]

When I first looked at this formula, I thought well what is x? It turns out x is the nearest perfect square or square number it is also called nearest to a. This is easy to find if a is less than 100. But what if a is in the thousands or millions? This is not so easy.