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Mar 18, 2013 at 18:23 comment added Aaron Meyerowitz Also, think of training your universal function to answer $x=n$ given $x=n$. Then you have $10$ data points indicating the identity function. Now get it to answer $x=n$ given $n=x$ so you have 10 more data points $(2^{n+7}3^55^1,2^13^55^{n+7})$ This is not something a polynomial does well and your other equations are all broken until you go to a degree $19$ polynomial. Going up to two digits only increases the complexity. Consider $x-0=n$ if you wish.
Mar 18, 2013 at 16:59 comment added Reuben Thank you for a concrete explanation, it has improved my understanding of the issue much.
Mar 18, 2013 at 14:55 history answered Aaron Meyerowitz CC BY-SA 3.0