(para espanol: Aqui por favor.)

As part of my ongoing quest to emulate Richard Feynman, I’ve taken an interest in mental math. Particularly, trying to get a “sense” of different calculations. I thought it would be cool if I had a decent method of calculating square roots in my head. They come up often enough, any time you’re trying to find the length of the Hypotenuse of a right triangle, for instance, or figuring out what the dimensions of a square-ish room might be, if you know the square footage. I came up with a decent method, which can be expressed as follows:

*(The φ _{n} part is a magic function I made up for the sake of notation. It equals the closest perfect square less than n. So φ_{32} = 25, and φ_{80} = 64. The part to the right of the “+” gives the whole number part of the answer. The fraction to the left gives the decimal part of the answer.)*

That’s mostly for show, and is probably the most complicated possible way to express the method. I have it there because fancy symbols make me feel important. But it does explain exactly what my method is. The idea of it is, that if you memorize the perfect squares, then you can make a pretty good guess as to the square root of a number that falls between them.

X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

X^{2} |
1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 |

If you’re given the number 47, you can tell right away that the square root is between 6 and 7. To find the next decimal place, the key is that the closer the original number is to the next perfect square, the closer the square root is to the next value. There are 13 numbers between 36 and 49 ( 7^{2} – 6^{2}), and 47 is 11 parts of the way there. So, we estimate that √47 is 6 and 11/13th’s, which works out to 6.846… which is actually 99.86% of the true answer (6.8556…). – Try again with 27: 5^{2} is 25, 6^{2} is 36. So the answer is between 5 and 6. There are 11 numbers between 25 and 36, and 27 is 2/11ths of the way there, so 5 + 2/11 = 5.1818… = 99.72% of the true answer: 5.19615…

So how accurate is this method? For numbers greater than 10, you’ll always be within 2% (and < 1% in nearly all cases). The higher the number, the more accurate the method gets. Here’s a cool chart. For each number “n”, it tells how far off your approximation of √n will be:

Doing this mentally is a bit cumbersome at first, but gets quicker and quicker with moderate amounts of practice. I’ve started finding little shortcuts for Weird fractions like 7/13ths- approximations upon approximations, but that’s what I was after I suppose.

One further note: I tried doing a string of 10 random square root problems using the calculator on my cell-phone. The buttons were awkward enough, that it took me about the same amount of time as it did to do it in my head. Of course, the cell phone was much more accurate. Still though, it could make a good bar wager, to race someone in square roots, mentally vs. calculator, to one decimal point.

Very Interesting post. I wish I could do math like that but alas I’m still stuck at trig.

Glad you liked my post. As for the Math involved, I think the real key is practice. It’s mostly just doing division, and it works reasonably well even if you fudge the numbers a bit. And there’s nothing wrong with being on Trig! You might want to check out this article I wrote last week that deals with some cool Trig applications.

Ah, nostalgia. Mr. Blomquist, my highschool calc teacher, taught me this handy trick. He was also a big fan of ‘mental math’. Bravo sir.

GBHS?

Cool!

There’s a possible simplification of your formula that you may already know about due to the fact that

(n+1)^2 – n^2 = n + (n+1)

eg. 8^2 – 7^2 = 15 = 8 + 7

So the calculation for n = 60 can be computed as

( (60 – 49) / (7 + 8) ) + 7

which may be easier to perform in your head than

( (60 – 49) / (64 – 49) ) + 7

I’m just quite dumb at math. This helps me a bit. Thanks. lol

Nice post, iv been trying to do a couple, it is a little combersome but with enough practice should work very well. Thanks!

Sigh. Maybe you coulda started off the article saying something like “This post describes how to approximate a function using linear interpolation of known function values. This is commonly done in every math/science/engineering professional’s head every day. Here I’ll give an example using the square-root function…

Congratulations on discovering linear interpolation!

The approximation is similar to what I use, but instead of the ‘range’ between the two roots, I use the double of the lower root. In the case of your example (27), it would be 5 + 2/10 instead. For 47, 6 and 11/14 instead.

I don’t know how accurate this is, but I find it easier to calculate.

11/12, oups.

Using the formula in the article is guaranteed to give you a result that is less than the true value.

Conversely, this formula is guaranteed to exceed the true value.

So you know even more than you thought!

eg. for n = 95

9 + (95 – 81)/(9 + 10) = 9.7368… sqrt(95)

er, that is

9 + (95 – 81)/(9 + 10) = just under sqrt(95), and

9 + (95 – 81)/(9 + 9) = just over sqrt(95)

What about for n > 12 where people start becoming lazy to memorize the roots of perfect squares? I guess for numbers divisible by 100, 10000, etc its fairly easy, but what about the rest?

Yes, you’ve inadvertently rediscovered linear interpolation via a wholly unique path and application. Awesome! Seriously, that’s the fun in mathematics and science, discovering that multiple paths can lead to a similar technique. Its speaks more about the underlying universality of algorithms than anything else, and that’s the magic. Feynman would approve. Anyone commenting otherwise has lost the joy of self-derived discovery, I sigh for them instead.

As others have mentioned, this is simply a linear interpolation between the closest pair of perfect squares.

A small adjustment: instead of (\sqrt{\varphi_n}+1)^2 in the denominator there why don’t you say \varphi_{n+1}?

according to his definition, \varphi_{n+1} does not equal ( \sqrt(\varphi_n) + 1 )^2

for example, varphi_5 = varphi_6 = 4 since 4 is the largest perfect square less than 5 or 6.

Are you aware of the curse of 47? lol google it.

That’s great. Ill try to keep that method in my head. Hey, did anyone else notice that the graph looks like a bouncing ball?

I wonder what the connection is between square roots and ‘bounce decay’?

Jack has nailed your solution perfectly — it’s a rather roundabout way of doing (and explaining) linear interpolation.

I’d suggesting a more Feynman-like approach would involve using more memory and more heuristics. First of all — memorize a few square roots. Most of us know the square roots of the numbers from 1 to 10 to at least two significant figures, right? Now observe that sqrt(ab) is sqrt(a) x sqrt(b) and you have a huge capacity to estimate square roots.

Second, you can easily get a midpoint to your linear interpolation using the handy rule (x + 1/2) ^ 2 = x x (x + 1) + 1/4. E.g. 2.5 squared is 2 x 3 + 0.25 = 6.25. The errors in your linear interpolation drop hugely with that extra data point between each integer value.

Next, calculating fractions in your head tends to be pretty horrible (for me at least). Rather than try to remember what 4/11 looks like, take the smallest power of 10 greater than the increment you’re working with — e.g. for squares between 1^2 and 2^2 that would be 10, divide that by the difference (3) to get 3.3 (or 0.33) and multiple by that directly to get your estimated square roots — 1.33, 1.67. For any fractions more complicated than quarters this will be much less painful than trying to remember what 13/17 is.

Next, memorize a few logarithms base 10 (the numbers from 1 to 10 are a good start). Now recall that the log of sqrt(x) is half the log of x — this along with other simple rules (log of 10x is log x + 1) gets you lots of precision for guesstimating not just square roots but all kinds of ugly multiplications and divisions.

Look up “Dead Reckoning” by Ronald Doerfler. Goes into arbitrary roots, trig, logarithms and their inverses. All done mentally.

this is actually pretty important to do on the MCAT (medical admissions test). I wonder if you would be able to apply this same thought process to a slightly more difficult problem. For example if you needed to estimate the square root of 6 x 10^-5 what would the answer be?

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