I hope that's not an answer you'd give in a live seminarWhat I'm really saying I suppose: I did not come down in the last shower.
Never had a reason to. In speakers, honesty is the best policy. Don't dig yourself into a hole.I hope that's not an answer you'd give in a live seminarWhat I'm really saying I suppose: I did not come down in the last shower.
Weird answer.The paper was discussing properties of minimas, not optimisation.
[$]x^ {1/x}=\exp\left[\frac{\log x}{x}\right][$] so ,maximize [$]\frac{\log x}{x}[$]Prove that [$]e[$] is the number that maximises [$]\sqrt[x] {x}[$] for [$]x > 0[$].
And if we continue with [$]log x/x < 1 [$] we conclude [$]\sqrt[x] {x} < e [$] for any [$]x>=0[$].[$]x^ {1/x}=\exp\left[\frac{\log x}{x}\right][$] so ,maximize [$]\frac{\log x}{x}[$]Prove that [$]e[$] is the number that maximises [$]\sqrt[x] {x}[$] for [$]x > 0[$].
he does seem to have made a mountain out of a molehill.Compare with Steiner's proof in Crelle's journal (overkill?)
http://www2.washjeff.edu/users/mwolterm ... rie/89.pdf
[$]e^x[$] is difficult with standard GD (bad answers) but [$]e^{-x}[$] is better. But we see how the learning rate is important.The only solution left is tie solve this problem using (a hand-crafted?) ML algorithm?
Can I just look at the diagram and give the answer?
"Das ist nicht nur nicht richtig; es ist nicht einmal falsch!".π⁴ + π⁵ ≈ e⁶
to seven significant figures
Ref
Analysis Fact @AnalysisFact
it's true but not necessarily useful. You want [$]e^{5}[$] and this is a formula for [$]e^{6}[$]"Das ist nicht nur nicht richtig; es ist nicht einmal falsch!".π⁴ + π⁵ ≈ e⁶
to seven significant figures
Ref
Analysis Fact @AnalysisFact