# User Portlet Christos Papahristodoulou
Discussions
Jan, I agree with you. My question had to do with the selected methodology of "excess deaths" which can give absurd numbers, i.e. below the normal. When in an average year there are say 10,000 "normal deaths" and in the COVID19 year you observe...
Enrique, I am afraid you are right. I update my post and will be clear soon, with new data. Preliminary, the number of deaths in Sweden will be more than 6,200 by the end of July, where it seems to be the time when the pandemic is over. In addition,...
Thanks very much! I have a new document on Sweden, a country that follows its own "open" strategy and so far has paid a huge price! Please see ti here: https://community.wolfram.com/groups/-/m/t/1990972
As Mariusz formulated, you need to specify the value of parameter y, in order to get your x. But you can't invert that function. For instance with y = 0.5, you get x -> 1.5708, but if you set x = 1.5708, you get y = 0.5 and y = 1.5, as the two pair...
Check the three alternatives in the file. There are many solutions if z (the right part side) is positive (or Integer). &[Wolfram Notebook] : https://www.wolframcloud.com/obj/dca5e81e-12db-4182-b637-fe2009d147ef
What about if the Recovery equation changes to something like: equationR = r'[t] == (1 - w) k i[t], where w = a "death" parameter, such as w = 0.005? Then, DSolve[equationR, {r[t]}, {t}] gives: r[t] -> C + Inactive[Integrate][(k i[K] -...
Mycket bra Jan! Sverige är det enda landet i Europa som fortsätter öppet...Vi börjar bli oroliga här...
Thanks Gianluca! It has saved my afternoon so that I can watch Serie A matches later!
Hi, I think it is better if you "create" some data first, like below: datax = Table[{i, 60 - j + 1}, {i, 10}, {j, 15}]; datay = Table[{i, 60 - j + 1}, {i, 10}, {j, 15}]; streamData = {datax, datay}; ...
Nice Manipulate! You need to make the bounds of both Rho and Theta similar, i.e. make the Theta as you have it with Rho, { \[Theta], 0}, 0, 10]. Regards