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User:Natsuhata/Draft: Difference between revisions

Revision as of 12:28, 16 June 2024

average age by percentage of daily left stew

stew left in bowl (${\textstyle p}$)	average age (${\textstyle A_{p}}$)
0 %	0 seconds
0.1 %	87 seconds
1 %	15 minutes
10 %	3.0 hours
20 %	7.5 hours
30 %	15 hours
38 %	24 hours
40 %	27 hours
50 %	2 days
56.5 %	3 days
60 %	3.8 days
70 %	7.8 days
72.99 %	10 days
80 %	20 days
90 %	90 days
94.9009 %	365 days
95 %	380 days
99 %	27 years
99.9 %	2,735 years
99.99 %	273,753 years

Let ${\textstyle D\in \mathbb {N} }$ be the age of the perpetual stew in days and let ${\displaystyle p\in [0,1)}$ the percentage (where ${\displaystyle p=0.17}$ equals to 17 %) of stew left in the pot after every day. The stew is filled with fresh ingredients and stirred thoroughly at the beginning of each day. Then the average age (in days) of the stew at the time of the refilling is given by the partial sum $${\displaystyle A_{p,D}=\sum _{d=0}^{D}d\,p^{d}={\frac {D\,p^{D+2}-(D+1)\,p^{D+1}+p}{(1-p)^{2}}},}$$whose age limit ${\textstyle A_{p}}$ with respect to ${\textstyle D}$ is given as the series$${\displaystyle A_{p}=\lim _{D\to \infty }A_{p,D}=\sum _{d=0}^{\infty }d\,p^{d}={\frac {p}{(1-p)^{2}}}.}$$The partial sum ${\textstyle A_{p,D}}$ consists of nonnegative summands, hence increases as ${\textstyle D}$or${\textstyle p}$ is increasing, and ${\textstyle A_{p}}$ is the limit and an upper bound for ${\textstyle D\to \infty .}$ The limit ${\textstyle \lim _{p\to 1}A_{p}=\infty }$ tends to infinity and behaves as naively expected. Naturally, ${\textstyle A_{p,D}}$ and ${\textstyle A_{p}}$ describe upper bounds if ${\displaystyle p\in [0,1)}$ is an upper bound for the amount of stew left in the pot after every day. Trivially the continuation is ${\textstyle A_{1,D}=D.}$