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0

I would say it's last image A -> B = 90 degree B -> C = 180 degree C -> D = 270 degree D -> E = 360 degree E -> new one = 450 degree


4

(This answer begins with $\,m=-2\,$ and provides only general expressions for nonnegative $m$) Let's start with the well known : $$\tag{1}F(x):=2\,\arcsin(x)^2=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2\binom{2n}{n}}$$ From this we deduce : \begin{align} F'(x)&=4\,\frac{\arcsin(x)}{\sqrt{1-x^2}}&=4\sum_{n=1}^\infty \frac{(2x)^{2n-1}}{n\binom{2n}{n}}\\ ...


0

This sounds like a Data Mining problem; specifically, the problem of Frequent Pattern Mining. If you want to find various frequent itemsets given a transaction database, then you'll probably be interested in algorithms such as Apriori and FP-Growth.


1

$N-1\;\; x's$ of length $2 [1-2\;\;thru\;\; (N-1)-N]$ $N-2\;\; x's$ of length $3 [1-2\;\;thru\;\; (N-2)-N]$ ..... ..... $ 2\;\; x's$ of length $N-1 [1-(N-1)\;\;thru\;\; (N-1)-N]$ sum $ = [1+2+3+........(N-1)] - 1$ You must be knowing the formula for the sum of terms within the brackets $[\;\;\;]$ Added: a combinatorial approach Consider a string of ...


1

For $N$ x in a row, we have $N$ x, we have $N-1$ of 2 x's in a row, N-2 of 3 x's in a row. In general we have $N-k+1$ of $k$ x's in a row. To see why is that so, construct a window of length $k$ and move it.



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