Question about Nearest Neighbor Exponential Blowup - Week 6b
The second term in the bound for the error of the 1-NN classifier is \(4c \sqrt{d} N^{-\frac{1}{d+1}} \)
When we fix \(N\) and let \(d\) tend to \(\infty\), how does this term blow up? If we ignore \(\sqrt{d}\), we see that if \(d\) grows, the power tends to 0, and 1/(anything raised to 0) is 1, so the term tends to 1?
"If you double d then you will have to square your number of samples to get a constant term. This is why we say that it is growing exponentially with d."
Question about Nearest Neighbor Exponential Blowup - Week 6b
The second term in the bound for the error of the 1-NN classifier is \(4c \sqrt{d} N^{-\frac{1}{d+1}} \)
When we fix \(N\) and let \(d\) tend to \(\infty\), how does this term blow up? If we ignore \(\sqrt{d}\), we see that if \(d\) grows, the power tends to 0, and 1/(anything raised to 0) is 1, so the term tends to 1?
1
I think that it could help you.
https://www.oknoname.com/EPFL/CS433/topic/2500/lecture-6b-knn-upper-bound/
"If you double d then you will have to square your number of samples to get a constant term. This is why we say that it is growing exponentially with d."
1
Thanks! Missed that thread.
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