[seqfan] Mnemonic for square root of 2 (and of 0.5)

[Robert Munafo]

On 9/24/12, Alonso Del Arte <alonso.delarte at gmail.com> wrote:
> [...]
> Only problem was that I could only remember 0.707. Nicole used her cell
> phone's "scientific" calculator but that only gave it to five decimal
> places, so that's what I went with. (I think it would have been too much of
> an imposition to have her incur data charges to connect to the OEIS on her
> Sprint phone).

If you can remember sqrt(2), then you can divide it by 2 to get
sqrt(0.5) ... and to 8 digits, sqrt(2) is:

  1.4 14 21 35 ...

all multiples of 7, specifically 7 times the sequence 2, 2, 3, 5...
This is not a coincidence, it is a pattern shared with certain other
square roots with repeating decimals (such as sqrt(62)=7.874007874...)

The reasons oddly have to do with the fact that 2+2*7^2 = 10^2, and
the Taylor series of sqrt(1/(1-2x)). The coefficients are
(2*n)!/(2^n*(n!)^2), or you can use OEIS sequences A001790 and A060818
-- but you don't have to remember any of that if you can remember "7
times the sequence 2, 2, 3, 5".

For rough approximation you could add "9" to the sequence and be
pretty close: 1.4 14 21 35 63 is very close to sqrt(2). Taking each
term and dividing by 2, sqrt(0.5) is approximately:

  0.7
+0.007
+0.000105
+0.00000175
+0.0000000315
=0.7071067815

which is correct except for the last digit.

-- 
  Robert Munafo  --  mrob.com
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