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7

The author of the question you linked has previously posted a question here, on the $18^{\text{th}}$ October. That question was roomba'd today, approximately an hour and a half after the question you linked was posted. Thus at the time they posted the linked question, the "new contributor" criteria were not satisfied. They are now, but if I understand the ...


8

This seems to work. (Instead of <br> you can simply use two spaces, if you prefer.) >! If $K=1-\frac{K_1}{2K_2}$ then clearly $K<1$.<br> >! Let $\lambda=\frac1{2K_2}$ so $K=1-\lambda K_1$.<br> >! Then $0\le\frac12=1-\frac12=1-\lambda K_2\le1-\lambda F'(c)\leq1-\lambda K_1=K<1$.<br> This was the suggestion from an ...


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