Mathematics in School

30 Mathematics in School, March 2020 The MA website www.m-a.org.uk It was the first class of Year 12. Mr Rowley, writing on the blackboard, said and the entire Further Maths class reached for a calculator. (This was in the olden days, like, 1994 or so.) The very quickest of us had pressed the ‘on’ button before he tutted and said “4 and a bit. 4.04.” I don’t think he’d set out to impress us – as the course went on, it became clear that estimating numerical answers was pretty much a compulsive habit for him – but we were eating out of his hand from that moment on. We reasoned, if someone was that good with numbers, they must be that good at maths. And that’s the premise of this article: if you can learn a few number tricks that look impressive but turn out to be straightforward, you can take a huge shortcut on the road to convincing your students you know what you’re doing. Obviously, you don’t need to learn any of these tricks, but I promise: the look of absolute terror and disbelief when you reel off an accurate answer in your head is well worth the effort. So, how did Mr Rowley do it? I don’t know for sure, but I have some good guesses. I surmise that he had two relevant facts at his disposal: • The square root of 3 is about ; and • That estimate is about 1% too high. Seven divided by 7/4 is four. If we reduce the denominator by 1%, that’s roughly the same thing as increasing the numerator by 1% - and increasing the answer by 1% gives 4.04. The maths here isn’t at all difficult! The only tricky things are knowing the numbers and having the confidence to trot themout quickly. But who has time tomemorize great big lists of square roots? Certainly not busy teachers. Fortunately, estimating square roots is also fairly quick and simple. Estimating square roots Suppose you had to estimate 18 , and for some reason 3 2 wasn’t an acceptable answer. (Of course, if you know that 2 is on the high side of 1.41, you can say 3 2≈4.24 immediately. But let’s pretend you don’t.) I have two words for you. Word # 1: Newton. Word # 2: Raphson. Wait! Come back! You don’t have to do the whole differentiation thing in your head every time – in fact, once we’ve gone through the theory here, you can just learn the trick and forget about the evidence. I’ll recap the method quickly: if x 0 is a good estimate for a solution to f ( x ) = 0, then is usually a better estimate. The function we’re looking at here is f ( x ) = x 2 – 18, which has a zero when x = 18. The derivative is ′ f x ( ) = 2 x . And we can readily ballpark the number: the square root of 18 is between 4 and 5, and towards the lower end of that interval. Let’s use x 0 = 4 Then That’s not too shabby! When you look at the calculation, you might notice that the top of the fraction is the difference between the square you’ve picked and the target number; the bottom is just double your estimate. So, if we anticipate that we’re subtracting a negative, the simplified Newton-Raphson recipe for square roots is: • Pick the nearest square number whose square root you know (in this example, that’s 16 = 4 2 .) • Work out the signed difference between your target and this square (18 is 2 more than 16). • Divide this by double the perfect square root (2/8 = 1/4). • Add this on to the initial guess (the final answer is 4 + 1/4 = 4.25). With a little practice, this becomes routine. The square root of 34? It’s a bit short of 6. Two-twelfths short, so it’s about five and five-sixths, or 5.83. For the more algebraically-minded: when b is much smaller than a , because The last term is, by assumption, comparatively small. “ 7 3 , which is … ”, 7 4 x 1 = x 0 − f x 0 ( ) ′ f x 0 ( ) x 1 = 4 − 16 − 18 2 × 4 =4.25. a 2 + b 2 ≈ a + b 2 a , a + b 2 a ⎛ ⎝⎜ ⎞ ⎠⎟ 2 = a 2 + b + b 2 4 a 2 . Becoming a Mathematical Ninja by Colin Beveridge