Peter Humphries October 7, 2019 at 4:03 am on Notes from a talk at the Maine-Quebec Number Theory ConferenceOmega results of this form using this method are presented quite nicely in Chapter 15 of Montgomery and Vaughan; they give applications towards sign changes of the Chebyshev psi function and of partial sums of the Mobius function.

Donna ebert September 27, 2018 at 3:55 pm on AboutI appreciate your taking the time to reply. The version of FreeCell I play online does count taking a card from the tableau (and the Freecells) as a move when adding them to the foundation. I never realized others did not.

David Lowry-Duda (mixedmath) September 27, 2018 at 8:20 am on AboutI believe you are referring to: this post on MathSE. Yes, that's right. In the typical language, a "move" consists of one mouse-click-and-drag action. When there are open freecells, it is possible to move multiple cards using one mouse operation (as a shortcut to doing all the card moves individually). As an example, if all 4 freecells are open, you ---

Donna September 27, 2018 at 1:38 am on AboutI see nowhere to submit an email so I am asking my question here. On another site you answered a question about the average number of moves to solve Freecell. You gave an in-depth answer about the studies that had been done (thank you) which was 45. What I cannot comprehend is how anyone can move 52 cards, one at ---

David Lowry-Duda (mixedmath) September 24, 2018 at 6:33 pm on An intuitive introduction to calculusThank you. You could also write $P(t) = ae^{kt + C_1}$. But this may give the impression that $a$ and $C_1$ carry different data --- but they don't. To see the equivalence, you can write $ae^{kt + C_1} = a e^{kt} e^{C_1} = (a e^{C_1}) e^{kt} = A e^{kt}$, where $A = a e^{C_1}$. There are two fundamental degrees of ---

Luis Alberto Torres Cruz September 23, 2018 at 9:50 am on An intuitive introduction to calculusGreat post! Thank you for taking the time to write this. Regarding the example on population growth, you state that if P'(t) = k*P(t), then P(t) = ae^(kt). But wouldn't a more general expression be: P(t) = ae^(kt+C1), where C1 is a constant?