PROBLEM: Given the fact that seven tenths is approximately equal to the square root of a half, use a binomial expansion to produce an infinite series converging to the square root of two.

SOLUTION:

7/10≃√½, then 10/7≃√2 and (10/7)²=100/49. But 100/49-2/49=98/49=2.

Therefore, √(100/49-2/49)=√2, and 100/49-2/49 can be written (100/49)(1-0.02) and the square root can be written: (10/7)√(1-0.02) or (10/7)(1-0.02)^½.

Now we can expand the binomial:

(10/7)[1-½(0.02)+(1)(3)(0.01)²/2!-(1)(3)(5)(0.01)³/3!+...+(-1)ⁿ(1)(3)...(2n-1)(0.01)ⁿ/n!+...].

We now have an infinite series for √2.

COROLLARY

By simply changing 10/7 to 5/7 we get a series for √½, because √½=√2/2. By changing 10/7 to 20/7 we get a series for √8, and so on for other multiples.