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CA Trevillian
CA Trevillian
3:00 AM
D
Is there anything like this for the Wolfram language? Apologies for the bad formatting, I’m on mobile and they changed the interface... linkedin.com/posts/…
 
 
2 hours later…
Nasser
Nasser
4:51 AM
@xzczd I just got the chance to try your example in Maple, and as expected, Maple did detect input type is not valid

inttrans:-laplace(sin(1),1,1)

Error, invalid input: `inttrans/laplace` expects its 2nd argument, t, to be of type name, but received 1
 
 
3 hours later…
P. Fonseca
P. Fonseca
7:51 AM
@kirma I guess that on the first time he will just do a sympathy smile. But if you insist, he will eventually direct you to one of his colleagues more specialized on brain issues... :)
 
 
3 hours later…
Szabolcs
Szabolcs
10:36 AM
@b3m2a1 MicroTest was really just for my own use and it's far, far from a fully fledged testing solution. There was a very simply motivation behind it (which you should be aware if you are designing a testing package). I was working with managed library expressions. These are LibraryLink data structures that get automatically destroyed when there are no more references to them. So, x = MakeFoo[]; creates something, and x=. is sufficient to ensure that your own destructor is run.
Except that MUnit tends to keep references to symbols it touches, or their past values. This prevents the destructor from running. MUnit does this is a non-transparent manner.
Now you have a choice: 1) try to hack something complicated that you do not fully understand and you have no control over its development 2) or just create a dead-simple transparent system that you have full control over.
@njpipeorgan Thanks for the notice! I was quite sick the last few days, and now I'm behind with work, so I won't be on this site too much for a while.
@ChrisK What's the context for this discussion? The article is paywalled. I was in the news the claims that "the data is too perfect", and I was extremely sceptical, but I was too sick to look up the source (just a bad flu)
 
Szabolcs
Szabolcs
11:02 AM
 
Szabolcs
Szabolcs
11:44 AM
I'm a physicist, not a statistician, and I am generally wary of goodness of fit measures like r-squared ... it's clear enough that they're abused way too often when explaining things. So I don't know what to say. But but ... 
In[15]:= data = Table[{x, Exp[3 x]}, {x, 0, 1, 0.05}];

In[16]:= fm = LinearModelFit[data, {1, x, x^2}, x];

In[17]:= fm["RSquared"]
Out[17]= 0.9917
There's 0.99 and that's an exponential (with plenty of curvature in it), not a quadratic
On the other hand, if I take the data from Wolfram, 
rd = ResourceData["Epidemic Data for Novel Coronavirus COVID-19"];
ts = rd[1, "Deaths"]; (* this is Hubei *)
This is new deaths (i.e. derivative of deaths) curve from Jan 22 to Feb 11, inclusive. Yes, I truncated the data to the "clean-looking" part.
That looks like a line, so the quadratic fit really is quite good
 
 
1 hour later…
kirma
kirma
12:51 PM
@P.Fonseca :)
@Szabolcs The context was my earlier quip on how perfect curve fits the official Chinese data seemed to have (not particularly scientific in my case).
 
Szabolcs
Szabolcs
1:18 PM
I can't fathom why they would fake that data. That the data is not even close to reality, that much was clear. But that the data would be made up and not based on measurement (even if very flawed measurement) ... that I cannot understand.
 
kirma
kirma
2:01 PM
@Szabolcs My thoughts exactly.
 
halirutan
halirutan
2:23 PM
I just leave this here. Got it from a friend, and I think it's hilarious
user image
7
 
 
3 hours later…
Mats Granvik
Mats Granvik
5:33 PM
(*start*)
Clear[T];
nnn = 10;
m = 3;
Table[
T = Tuples[Table[(Range[k] - 1)/k^0, {k, 1, nn}]];
a = Sort[Table[Total[T[[k]]], {k, 1, Length[T]}]];
Count[a, m], {nn, 0, nnn - 1}]
g1 = ListLinePlot[a];

Clear[T];
T[1, 1] = 1;
T[n_, 1] = 0;
T[n_, k_] := T[n, k] = Sum[T[n - i, k - 1], {i, 1, k - 1}];
Table[T[m + k, k], {k, 1, nnn}]
Show[g1]
(*end*)
Is this the error function?
The inverse error function, I meant.
 
 
1 hour later…
Mats Granvik
Mats Granvik
7:01 PM
I get the Mahonian numbers now:
(*start*)
Clear[T];
nnn = 10;
Table[
T = Tuples[Table[(Range[k] - 1)/k^0, {k, 1, nn}]];
a = Sort[Table[Total[T[[k]]], {k, 1, Length[T]}]];
Differences[Flatten[Position[Differences[a], 1]]], {nn, 0, nnn - 1}]
(*end*)
 
Szabolcs
Szabolcs
7:23 PM
@kirma Could it be that their data flow is extremely choppy and unreliable, but people expected daily updates, so they smoothed the data in some way (which means they needed to do prediction for some days, and then slowly correct following days to keep the cumulative count correct)? That still would not explain such a perfect quadratic growth.
 

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