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A number of assumptions seem to be missing from this article. Since the author is using the sigmoid function which is smooth, this argument actually only applie
by qwerty1793 5y ago
A number of assumptions seem to be missing from this article. Since the author is using the sigmoid function which is smooth, this argument actually only applies to approximating smooth functions. That is, you dont just need f to be continuous, you need all of its derivatives to exist and all be continuous. Also, since we are only able to have finitely many neurons, we need to be able to approximate f using step functions with definitely many pieces. So this argument can only be used if f is constant outside of a compact region.
- hprotagonist 5y agoat which point, a fourier transform is a hell of a lot cheaper ;)
- joppy 5y agoWhy does the function you are approximating need to be smooth? From the paper cited in the article, all you need is for f to be continuous on a compact subset of R^n.