On Sun, Feb 5, 2012 at 10:20, David Wilson <davidwwilson@comcast.net> wrote:
What I thought was notable about this solution was: - It arises naturally from a common length conversion. [...] [and]
sqrt(62) = 7.874 007874 ...
I will guess that small integer square roots with pretty blocks like this are rare in any base.
Well, of course there's 2*7^2 = 98, which gives us the pretty digit pattern: sqrt(2) = 1.4 14 ... More on this later, plus a mnemonic for the coefficients of Series[sqrt(1/(1-2x))] In base 5, because 111^2*3963+2 = 5^11, the square root of 3963*5 = 19815_10 = 1113230_5 = 1030.34033 00103034033 00132103324... where the digit-groups are 11 digits long. In base 6, because 46*13^2+2 = 6^5, the square root of 46*6 = 276_10 = 1140_6 = 24.34 02434 04053... where the digit-groups are 5 digits long. In base 7, because 977423*17^2+2 = 7^10, the square root of 977423_10 = 11210426_7 = 2611.43464 0261143464 0421565362... where the digit-groups are 10 digits long. In each of these examples, the third and 4th pretty digit-groups are 3/2 and 5/2 times the first two digit groups. For example in the familiar square root of 2, the first 8 digits (in groups of 2) are 14, 14, 21, 35 and 14*1.5 = 21, 14*5/2=35. (You can check the other pretty digit examples above, but remember I showed each in its native base and in odd bases 1/2 is an infinite "repeating decimal". So 261143464 times 3/2 in base 7 is 421565362.3333... where the "0.3333..." is 1/2 in base 7. These 3's overrun the following digit-group which hides the 5/2 term.) So there are 4 examples (-: The digits of the square root of 2, taken in pairs and divided by 14, are one way of remembering the coefficients of the series expansion of sqrt(1/(1-2x)), if you ever find yourself in the need and are stuck with just a pocket calculator. (-: The "62" ought to be 61.25, but it's overrun by the leading "1" of the following term which is 14*63/8=110.25) (And yes, I think these are cool, thank you David. I'll add a bit about it to my numbers pages.) - - - Related to this (in my mind at least) is the Catalan numbers and (1-sqrt(1-4x))/2... Folks who experiment with nice-looking digits might eventually play around with things like: 1.001^2 = 1.002001 1.002001^2 = 1.004006004001 How about a sequence of digits where you can square it and get almost the same sequence? 0.00010001000a000b000c...^2 = 0.00000001000a000b000c... where the two sides differ only in one digit. This can be worked out iteratively: A_0 = 0 A_1 = A_0^2 + .0001 = .0001 A_2 = A_1^2 + .0001 = .00010001 A_3 = A_2^2 + .0001 = .000100010002... A_4 = A_3^2 + .0001 = .0001000100020005... A_5 = A_4^2 + .0001 = .00010001000200050014... and so on. In the limit you get (1-sqrt(1-0.0004))/2 = .000100010002000500140042013204291430... which has the same digits as the easier-to-remember 50-sqrt(49*51) = .0100010002000500140042013204291430... On 2/5/2012 3:09 AM, Robert Munafo wrote:
The other of David's properties, regarding the geometric mean, also comes from (10^N-2) / 10^K, as you can see here:
1/sqrt(99.9998) = .**10000010000015000025000043750... [...] These all have the same pattern, related to the series expansion of sqrt(1/(1-2x)), which you can get by "Series[sqrt(1/(1-2*x)),{x,0,**8}]" in Mathematica or Wolfram Alpha. The coefficients are: 1, 1, 3/2, 5/2, 35/8, 63/8, 231/16, 429/16, 6435/128, ... The numerators are https://oeis.org/A001790, which are also the numerators of the sqrt(1/(1-x)) series expansion because only the denominators change.
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