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Oct
12
reviewed Approve Prime numbers qns
Oct
12
reviewed No Action Needed Trigonometry Question - find the width of the river and height of the post
Oct
12
reviewed Looks OK Correct notation for “for all positive real $c$”
Oct
12
reviewed Looks OK Taylor expansion for $\arcsin^2{x}$
Oct
12
reviewed Looks OK Is differentiating on both sides of an equation allowed?
Oct
9
reviewed Leave Closed Sum of reversed numbers?
Oct
9
reviewed Reopen If some of your boomerangs don't come back, how many throws will you get?
Oct
9
reviewed Reopen Order of the sum of elements of the inverse of a matrix
Oct
9
reviewed Reject
Oct
9
reviewed Reject
Oct
7
comment minimum number of times to change tyres.
To be honest, I do not even know how to prove the above solution is optimal. I cannot find any better solution does not constitute a proof!
Oct
7
asked minimum number of times to change tyres.
Oct
5
reviewed Edit definition of a sufficient statistic
Oct
5
comment Holder continuity constant and dervative.
I also know my fn are monotonically decreasing. Somehow my set of conditions allows to prove something, but it is hard to write down all of them
Oct
5
comment Holder continuity constant and dervative.
no, i have proved it before, but i forgot how it was done now...
Oct
5
comment Holder continuity constant and dervative.
Yes, also i know the derivatives of fn are bounded. I know f has derivatives everywhere except at one point, where it still admits left and right derivative, but not sure if they agree. Also fn and f are all convex.
Oct
5
asked Holder continuity constant and dervative.
Oct
3
comment If $f$ is continuously differentiable in $[a,b]$, $f(a)=f(b)$, and $f'(a)=f'(b)$, then there exist $a<x_1<x_2<b$ such that $f'(x_1) = f'(x_2)$.
i dont understand the down votes, because the arguments seems fine to me.
Oct
3
comment Variation of Coupon Collector's Problem
However, I also have to factor in the fact that the boxes themselves also have a probability to dispense coupons. -- not concise, in my opinion.
Oct
3
reviewed Looks OK if I know $f(x+1) = 2f(x) + 1$, how do I solve f(x)