User Portlet
| Discussions |
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| This is exactly what I wanted to enable me to see particle trajectories in 3D. Thanks. |
| Hi Eric, I have version 12.3 of mathematica on a MacBook Pro. I don't think the test is perfectly running but I can see results for every slider with a manually adapted saturation. Hedwig |
| A new special function: **Srivastava-Daoust** function for **N** variables. Srivastava-Daoust** You can use this function to express: 1.All four **Appell** function 2.**Horn** Function -**34** distinct convergent hypergeometric series of... |
| Kuba, The final line of your post solved my problem. Here is the test function that saves state for different names. test[name_String, init_Integer] := Block[{state}, If[NameQ[name "$"], (state = Symbol[name "$"];... |
| Eric, I hadn't seen your other thread. I just glanced at it. I can give a few comments on Mathematica's notation vs engineering notation. I think it is fine that you are creating a way to convert between the two. However, I don't have a problem... |
| Thanks, Szabolcs. Well, the story behind my question is that I'm developing an environment for working with Boolean algebra in an engineering context. I've been fascinated by Boolean algebra since I read Boole's *Laws of Thought* when I was... |
| Sorry, when I posted it the first time, I got a message that some further classification was needed, so I added another one and posted. But it looks like the first one when through anyway. Eric |
| Mathematica doesn't use the usual boolean notation with which circuit designers are familiar. The plan is to enter boolean expression as a c d + b d not a + a not b not c (a not c not d + b not c not d) and produce overbars on the... |
| Hi Harun, I tried every possible permutation trying to find that local maximum. No deal. But even in Mathematica itself, which can access Alpha, your command could not zero in on the local maximum. However, when I asked for `local maximum y... |
| If you include the equation for `a`: y=t-DT; DT=300 a/9.81; a=((B/2)^2/3)/((Tu/2)^2/3*(h-hv+tu)); hv=(Vi*d)/(4 B*(Tu/2)^2)^(1/3); B=Vi*g*d; g=(9.81 (t-x))/300; Vi=Q/l; In[34]:= y Out[34]= t-(0.0327 d^2... |