– peter shor hours ago | show more comments answers active oldest votes the name version is: to be on a first name basis (with...) share | improve this answer edited hours ago answered hours ago silverface silverface silver badges bronze badges comparing british and american ngram charts, i'm quite surprised...
bronze badges $\endgroup$ add a comment | answer active oldest votes $egingroup$ probably you know how to finish this proof but i will try to write it out for anyone who might need it. as you have noted, we must find a subset $s$ that contradicts the hypotheses so that the theorem is proved by contradiction...
thanks @j.w.tanner ) $$p_n=\dfrac -\dfrac \left(-\dfrac ight)^n$$ the solution for "$n$ is even" case is the same up to heads $\leftrightarrow$ tails swap; since $p_{heads}=p_{tails}= / $ it does not matter. share | cite | improve this answer edited hours ago answered hours ago kludg kludg silver badges...
extit{arbitrary}$ point in $[x_k,x_{k- }]$ can always be chosen to equal zero because in every interval there is a point $x$ such that $f(x)= $. it follows that $\int^ _ f(x)dx= $ if it exists. share | cite | improve this answer edited hours ago answered hours ago matematleta matematleta . k gold badges...
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