Editing Talk:2435: Geothmetic Meandian
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:How can this be formulated as a PDE when F isn't even differentiable? | :How can this be formulated as a PDE when F isn't even differentiable? | ||
:Besides, R(Fn+1) < R(Fn) does not imply limit R(Fn) = 0 (Think R(n) := 1+1/n). -- [[User:Xorg|Xorg]] ([[User talk:Xorg|talk]]) 02:50, 12 March 2021 (UTC) | :Besides, R(Fn+1) < R(Fn) does not imply limit R(Fn) = 0 (Think R(n) := 1+1/n). -- [[User:Xorg|Xorg]] ([[User talk:Xorg|talk]]) 02:50, 12 March 2021 (UTC) | ||
− | Agreed. F is not differentiable due to median. For arbitrary R(n) such as R(n)=1+1/n then limit R(Fn) != 0, however I do not define R(n) arbitrarily but define it as R(n)=max(An,Bn,Cn)-min(An,Bn,Cn) | + | Agreed. F is not differentiable due to median. For arbitrary R(n) such as R(n)=1+1/n then limit R(Fn) != 0, however I do not define R(n) arbitrarily but define it as R(n)=max(An,Bn,Cn)-min(An,Bn,Cn) --Ramakarl<br> |
[[User:snark]] | [[User:snark]] | ||
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and show that is is less than the first distance. | and show that is is less than the first distance. | ||
− | Agreed. I've removed reference to PDE or Heat equation except for the indirect similarity. The new suggestion for proof, which is incomplete, is inductive based on the observed alternation of the min/max between the geomean and arithmean. I believe it can be shown that for some N, there exists F(N)=k, where k=min(N)=max(N)=arithmean(N)=geomean(N)=median(N) within some epsilon, and k is the fixed point. This is because the min(n+1), median(n+1) and max(n+1) alternate between arithmean(n) and geomean(n), which are strictly inside the open interval ( min(n), max(n) ). | + | Agreed. I've removed reference to PDE or Heat equation except for the indirect similarity. The new suggestion for proof, which is incomplete, is inductive based on the observed alternation of the min/max between the geomean and arithmean. I believe it can be shown that for some N, there exists F(N)=k, where k=min(N)=max(N)=arithmean(N)=geomean(N)=median(N) within some epsilon, and k is the fixed point. This is because the min(n+1), median(n+1) and max(n+1) alternate between arithmean(n) and geomean(n), which are strictly inside the open interval ( min(n), max(n) ). --Ramakarl |
I believe we can produce a simpler, rigorous proof. Assuming a set of three is given, we can show that after every 2 iterations, the range is reduced by at least 1/3 of its original value, and therefore it converges exponentially to 0. We use the fact that each iteration, none of the three values will lie outside the range of the previous iteration. In addition, it can be shown that the arithmean lies at least 1/3 of the previous range away from the highest and lowest values of the previous iteration. | I believe we can produce a simpler, rigorous proof. Assuming a set of three is given, we can show that after every 2 iterations, the range is reduced by at least 1/3 of its original value, and therefore it converges exponentially to 0. We use the fact that each iteration, none of the three values will lie outside the range of the previous iteration. In addition, it can be shown that the arithmean lies at least 1/3 of the previous range away from the highest and lowest values of the previous iteration. |