17/11/14 22:00:12.64 agSxZaXK.net
>>159 追加
ところで、おまえ、下記のThomae's functionの「f is not differentiable at all irrational numbers.」証明を読んでないみたいだから、引用しておくよ（＾＾
https:／／en.wikipedia.org／wiki／Thomae%27s_function
Thomae's function
(抜粋)
f(x)= 有理数ｒが既約分数p／qで表されるとき、1／q 無理数で0 （注：>>83同様）
f is not differentiable at all irrational numbers.
All sequences of irrational numbers (ai≠ x0)_（i=1～∞ ） converging to the irrational point x0 imply a constant sequence (f(ai)=0)_（i=1～∞ ）,
identical to 0,
and so lim _（i→ ∞ ）｜ (f(ai)-f(x0)）／（ai-x0））｜=0.
According to Hurwitz's theorem, there also exists a sequence of rational numbers (bi=ki／i)_（i=1～∞ ）,
converging to x0, with (ki∈ Z ,i∈ N )） (ki∈ Z ,i∈ N )） coprime and |ki／i-x0|＜ 1／（√ 5）* i^2）.
Thus for all i,: ｜（f(bi)-f(x0)）／ （bi-x0）｜ ＞ （1／i - 0）／（1／(（√ 5）* i^2)））= √ 5* i ≠ 0 and so f is not differentiable at all irrational x0.
(引用終り)