test3

 as: Be yourself, trust yourself, accept yourself

P(i, 2j) &=& \sin\Big(\frac{i}{n^{2i/d}}\Big)\P(i, 2j+1) &=& \cos\Big(\frac{i}{n^{2i/d}}\Big)\end{eqnarray}

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	be	yourself	accept	trust	positive	good
Bag of words (binary)	1	1	1	1	0	0
Bag of words (count)	1	3	1	1	0	0
TF	1/6	3/6	1/6	1/6	0	0
IDF	$$	$$	log(2/1)	log(2/1)
IDF in Keras	log(1+2/3)	log(1+2/2)	log(1+2/2)	log(1+2/2)	log(1+2/2)	log(1+2/2)
TF-IDF in Keras	(1+log(1)) * log(1+2/3)	(1+log(3)) * log(1+2/2)	(1+log(1)) * log(1+2/2)	(1+log(1)) * log(1+2/2)	0	0

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