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Nov 2, 2016 at 17:04 vote accept Srishti M
Apr 6, 2015 at 18:34 comment added Yuri Nenakhov You can assume $x_n = 0, 1,\ldots N-1$. After you find the approximation $y_n = t_d \cdot x_n^d$, you may find $x_n' = \frac{x_n}{t_d}$. This will give you $\theta = {x'}^d$.
Apr 6, 2015 at 18:04 history edited Yuri Nenakhov CC BY-SA 3.0
added 88 characters in body
Apr 6, 2015 at 17:56 history answered Yuri Nenakhov CC BY-SA 3.0