Tag Archives: party tricks

How I won the talent show

So I don’t want to brag or anything but I won the neighborhood talent show tonight.  Look, here’s my trophy:

My act was “The Great Squarerootio.”  I drank beer and computed square roots in my head.  So I thought it might be fun to explain how to do this!  Old hat for my mathematician readers, but this is elementary enough for everyone.

Here’s how it works.  Somebody in the audience said “752.”  First of all, you need to know your squares.  I think I know them all up to 36^2 = 1296.  In this case, I can see that 752 is between 27^2 = 729 and 28^2 = 784, a little closer to 729.  So my first estimate is 27.  This is not too bad:  27^2 = 729 is 23 away from 752, so my error is 23.

Now here’s the rule that makes things go:

new estimate = old estimate + error/(2*old estimate).

In this case, the error is 23 and my old estimate is 27, so I should calculate 23/2*27; well, 23 is just a bit less than 27, between 10 and 20% less, so 23/2*27 is a bit less than 1/2, but should be bigger than 0.4.

So our new estimate is “27.4 something.”

Actual answer: 27.4226…

Actual value of 27 + 23/54: 27.4259…

so I probably would have gotten pretty close on the hundredth place if I’d taken more care with estimating the fraction.  Of course, another way to get even closer is to take 27.4 as your “old estimate” and repeat the process!  But then I’d have to know the square of 27.4, which I don’t by heart, and computing it mentally would give me some trouble.

Why does this work?  The two-word answer is “Newton’s method.”  But let me give a more elementary answer.

Suppose x is our original estimate, and n is the number we’re trying to find the square root of, and e is the error; that is, n = x^2 + e.

We want to know how much we should change x in order to get a better estimate.  So let’s say we modify x by d.  Then

(x+d)^2 = x^2 + 2xd + d^2

Now let’s say we’ve wisely chosen x to be the closest integer to the square root of n, so d should be pretty small, less than 1 at any rate; then d^2 is really small.  So small I will decide to ignore it, and estimate

(x+d)^2 = x^2 + 2xd.

What we want is

(x+d)^2 = n = x^2 +e.

For this to be the case, we need e = 2xd, which tells us we should take d = e/2x; that’s our rule above!

Here’s another way to think about it.  We’re trying to compute 752^(1/2), right?  And 752 is 729+23.  So what is (729+23)^(1/2)?

If you remember the binomial theorem from Algebra 2, you might remember that it tells you how to compute (x+y)^n for any n.  Well, any positive whole number n, right?  That’s the thing!  No!  Any n!  It works for fractions too!  Your Algebra 2 teacher may have concealed this awesome fact from you!  I will not be so tight-lipped.

The binomial theorem says

(x+y)^n = x^n + {n \choose 1} x^{n-1} y + {n \choose 2} x^{n-2} y^2 + \ldots +

Plugging in x=729,y=23,n=1/2 we get

(729+23)^{1/2} = 729^{1/2} + {1/2 \choose 1} 729^{-1/2} \cdot 23 + {1/2 \choose 2} 729^{-3/2} \cdot 23^2 + \ldots

Now 729^{1/2} we know; that’s 27.  What is the binomial coefficient 1/2 choose 1?  Is it the number of ways to choose 1 item out of 1/2 choices?  No, because that makes no sense.  Rather:  by “n choose 1” we just mean n, by “n choose 2” we just mean (1/2)n(n-1), etc.

So we get

(729+23)^{1/2} = 27 + (1/2) \cdot 23 / 27 + (-1/8) \cdot 23^2 / 27^3 + \ldots

And look, that second term is 23/54 again!  So the Great Squarerootio algorithm is really just “use the first two terms in the binomial expansion.”

To sum up:  if you know your squares up to 1000, and you can estimate fractions reasonably fast, you can get a pretty good mental approximation to the square root of any three-digit number.  Even while drinking beer!  You might even be able to beat out cute kids in your neighborhood and win a big honking cup!

 

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