The Mathematical Ninja takes a square root

"So," said the Mathematical Ninja, "we meet again."

"In fairness," said the student, "this is our regularly-scheduled appointment."

The Mathematical Ninja was unable to deny this. Instead, it was time for a demand: "Tell me the square root of 22."

"Gosh," said the student. "Between four-and-a-half and five, definitely. 4.7 or so?"

The Mathematical Ninja nodded. "Four and seven tenths is an excellent initial approximation. Allow me to demonstrate a better one."

The student allowed.

"You have correctly - I presume - made your initial estimate by noting that 22 is 3 smaller than $5^2$, so $5 - \frac{3}{2\times 5}$ is close to the root."

"Of course, sensei," said the student, who had done nothing of the sort. He had interpolated quickly.

"A better second estimate is $5 - \frac{30}{97}$. One gets that - in this case, with the key numbers 3 and 10 and a previous estimate of $\frac{a}{b}$, as $\frac{3b}{10b+a}$."

"Thirty... over 100 + 3. Wai... wah!"

"$a$ is -3, remember."

The student was unlikely to forget that now. "So, I could apply the same process again to get a correction of..." (pause for thought) " $\frac{291}{940}$?"

The Mathematical Ninja nodded again. "You could continue the process for as long as you felt the need - each iteration gets you a slightly better estimate. $5-\frac{291}{940}$ is correct to one part in half a million."

The student felt that there was little need to make it any more accurate than that - although he'd never tell the Mathematical Ninja that. He had developed a little sense.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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2 comments on “The Mathematical Ninja takes a square root

  • elmer

    I don’t get this part ‘so 5 – 3/2×5 is close to the root’

    • Colin

      That’s one of the Mathematical Ninja’s favourite tricks!

      You start from $\sqrt{22} = \sqrt{25-3} = 5 \sqrt{1 – \frac{3}{25}}$, then apply the binomial expansion to the square root: $\left(1 + x\right)^{\frac{1}{2}} \approx 1 + \frac{1}{2}x + …$. That gives $5 \left(1 – \frac{3}{2\times 25}\right) = 5 – \frac{3}{2 \times 5}$.

      Generally, a fairly good approximation for a square root in the form $\sqrt{a^2 + b}$, with $b < a$, is $a + \frac{b}{2a}$. Hope that helps!

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