the mathematics that support AI, ie. Linear Algebra and Statistics, etc.
https://link.springer.com/article/10.1007/s10994-018-5742-0
After listening to Lex's interview (on AI gitter) I have seen a video from NYU, but got lost in the math quickly.
But principle looks interesting. Better than SVM he claims, (i think).
I'm studying limits in calculus, epsilon delta, and so on.
But I ran into a problem with the proper definition of square root/taking the square root (and other possibly all even roots, and maybe also odd roots) where I was unsure of when +/- appears in equations.
I have not found a good complete and insightful (semi-rigorous) reference within my books or online on the topic of rational exponents.
Most websites get it wrong, but wolfram gets it totally correct:
From a book that doesn't explain why in a helpful way:
Question: simplify (x^12)^(1/4) where x is a real number (not just positive):
Answer: |x|^3 where |.| is absolute value
https://www.wolframalpha.com/input/?i=(x%5E12)%5E(1%2F4)
Google gives an answer with imaginary 'i', which I'm not targeting right now, unless is elucidates my abs() problem.
https://www.google.com/search?client=firefox-b-1-d&ei=jcE_Xb7VGMu7tgXYgKmgCA&q=(-3^12)^(1%2F4)
Can you offer any insights?
A technical point: When you are dealing with these exponents with variables, you might have to take account of the fact that you are sometimes taking even roots. Think about it: Suppose you start with the number –2. Then:
(−2)2=4=2≠−2\large{\sqrt{(-2)^2} = \sqrt{4} = 2 \neq -2}(−2)2
=4
=2≠−2
In other words, you put in a negative number, and got out a positive number! This is the official definition of absolute value:
∣x∣=x2\large{\lvert x \rvert = \sqrt{x^2}}∣x∣=x2
Yeah, I know: they never told you this, but they expect you to know somehow, so I'm telling you now.
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So if they give you, say, x3/6, then x had better not be negative, because x3 would still be negative, and you would be trying to take the sixth root of a negative number. If they give you x4/6, then a negative x becomes positive (because of the fourth power) and is then sixth-rooted, so it becomes | x |2/3 (by reducing the fractional power). On the other hand, if they give you something like x4/5, then you don't have to care whether x is positive or negative, because a fifth root doesn't have any problem with negatives. (By the way, these considerations are irrelevant if your book specifies that you should "assume all variables are non-negative".)