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Talk:Essay:Best New Conservative Words (view source)
Revision as of 16:11, June 22, 2011
, 16:11, June 22, 2011→Graph of words over time: @jcw, Aschlafly
:::::::August, I have an open mind about this. I don't see how we could so easily find [[conservative]] words that double by century if the underlying pattern were not there. But please explain if you think that is in error.--[[User:Aschlafly|Andy Schlafly]] 22:27, 21 June 2011 (EDT)
:::::::August, I have an open mind about this. I don't see how we could so easily find [[conservative]] words that double by century if the underlying pattern were not there. But please explain if you think that is in error.--[[User:Aschlafly|Andy Schlafly]] 22:27, 21 June 2011 (EDT)
I hope we can agree on the function ''F<sub>theo</sub>'' - it should be uncontroversial:
*one layer exists from 1 - 2 - 4 - 8 words, i.e., 2<sup>0</sup>, 2<sup>1</sup>,2<sup>2</sup> and 2<sup>3</sup>
*the partial sums are 1 - 3 - 7 - 15, i.e., 2<sup>1</sup>-1, 2<sup>2</sup>-1, 2<sup>3</sup>-1, and 2<sup>4</sup>-1
*so, with the turn of the n-th century, there should be 2<sup>n+1</sup>-1 words
*for K layers, the number is K * (2<sup>n+1</sup>-1). Each layer has 15 words, thus, if there are '''N''' feasible words, the number of layers is '''N'''/15
*now adjust for years instead of centuries, and don't start with the first, but with the 17th, and you get the formula <math>F_{theo}(t) = \frac{\#words }{15}</math><math>(2^{\frac{t-1599}{100}}-1)</math>.
'''@jcw: ''' ''Another editor has attempted a similar argument above, but without success.'' I read the sections above, and I ''was'' convinced by the argument.
'''@Aschlafly:''' ''I don't see how we could so easily find [[conservative]] words that double by century if the underlying pattern were not there.'' The effects of the miscount ([[Talk:Essay:Best_New_Conservative_Words#PERFECTION: 20-40-80-160 BY CENTURY]]) have shown that you are able to match any pattern you were looking for.
[[User:AugustO|AugustO]] 12:11, 22 June 2011 (EDT)
== Americanadians ==
== Americanadians ==