Community RSS Feed
https://community.wolfram.com
RSS Feed for Wolfram Community showing any discussions in tag Algebra sorted by activeUnable to find roots for a simple equation
https://community.wolfram.com/groups/-/m/t/2455951
I tried to find a solution to this simple equation using Mathematica's Solve:
In[37]:= Solve[(2 Sqrt[2] V (m \[Epsilon]f)^(3/2))/(
3 \[Pi]^2 \[HBar]^3) == Ntot, \[Epsilon]f]
Out[37]= {{\[Epsilon]f ->
Root[-9 Ntot^2 \[Pi]^4 \[HBar]^6 + 8 m^3 V^2 #1^3 &, 1]}}
This does not give an expression for my fermi energy. So I tried reduced with some more result
In[46]:= Reduce[-9 Ntot^2 \[Pi]^4 \[HBar]^6 +
8 m^3 V^2 \[Epsilon]f^3 == 0, \[Epsilon]f] // FullSimplify
Out[46]= 2 m \[Epsilon]f ==
3^(2/3) \[Pi]^(4/3) (Ntot/V)^(2/3) \[HBar]^2
I don't understand why the 2m is in front of the fermi energy but at least I can read off the solution. But if even now Mathematica cannot handle my equation:
In[54]:= Solve[
2 m \[Epsilon]f ==
3^(2/3) \[Pi]^(4/3) (Ntot/V)^(2/3) \[HBar]^2, \[Epsilon]f]
Out[54]= {{\[Epsilon]f ->
Root[-9 Ntot^2 \[Pi]^4 \[HBar]^6 + 8 m^3 V^2 #1^3 &, 1]}
Why is it that it cannot find an explicit expression for \[Epsilon]f? And how to get it in another way?&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/3b5188c1-603a-4302-a07e-a94fab9fe750Willem Verheijen2022-01-26T07:45:36ZAn asymptotically closed loop of tetrahedra
https://community.wolfram.com/groups/-/m/t/2456465
*WOLFRAM MATERIALS for the ARTICLE:*
> Elgersma, M., Wagon, S. An Asymptotically Closed Loop of Tetrahedra.
> Math Intelligencer 39, 40–45 (2017).
> https://doi.org/10.1007/s00283-016-9696-4
![enter image description here][1]
&[Wolfram Notebook][2]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=sdafq34dfagasfsdc4.jpg&userId=11733
[2]: https://www.wolframcloud.com/obj/fa551709-69ef-4eac-82be-e93ecd7869d3Stan Wagon2022-01-26T17:55:36ZCompute the prime implicants of a Boolean function?
https://community.wolfram.com/groups/-/m/t/2454010
Hello. Is there a function in the Wolfram Language to compute the prime implicants of a Boolean function?
If yes, which one?
If no, can I find one in a function repository or elsewhere?
Many thanks, any help highly appreciated.
FranciscoFrancisco Gutierrez2022-01-23T23:24:26ZDiscovering symbolic square roots?
https://community.wolfram.com/groups/-/m/t/2453684
Is there any technology in Mathematica (or even as pure research), that would let me discover formulas for square root of a matrix specified as a formula?
For example for any vector with positive entries adding up to 1, the following factorization seems to hold
$$A=\text{diag}(p)-pp'=B^T B$$
where
$$B=\text{diag}(\sqrt{p}) - (\sqrt{p}) p^T$$
As you can check with the code below:
```
p = {1, 10, 3};
p = With[{q = Abs[p]}, q/Total[q]]; (* positive + normalized *)
A = DiagonalMatrix[p] - Outer[Times, p, p];
B = DiagonalMatrix[Sqrt[p]] - Outer[Times, Sqrt[p], p];
A == Transpose[B] . B
```
I got this formula by expanding the components and staring at them for a while. This obviously doesn't scale. This was a factorization of cross-entropy hessian, and there are at least 20 more [loss functions](https://pytorch.org/docs/stable/nn.html#loss-functions) I'd like to do .... can anyone think of a way such formulas could be discovered automatically?
This begs the question of what "formula" is....if we restrict attention to factorizations of Hessians, a symbolic way to derive them is [this](https://arxiv.org/abs/2010.03313) . You can see that if your original loss was specified in terms of pointwise ops and some tensor contractions, thenĀ its Hessian will consist of a sum of terms with some pointwise ops and some tensor contractions.
The author posted formula for 3rd derivative of cross-entropy loss on Mathematica [stack-exchange](https://mathematica.stackexchange.com/a/262193/217) so you can see that small formulas remain small after differentiation.Yaroslav Bulatov2022-01-23T20:08:55ZWolfram|Alpha doesn't understand query about multivariable problem
https://community.wolfram.com/groups/-/m/t/2454874
Hello. I'm new to Wolfram and I've been trying to run this multivariable problem.
Solve for b, B and k in terms of a, A, c, C, f. F, alpha:
b = lsin(alpha)+a,
B = lcos(alpha)+A,
d = a+k(b-a),
D = A+k(B-A),
e = b+k(c-b),
E = B+k(C-B),
f = e+k(d-e),
F = E+k(D-E)Joseph Choi2022-01-24T21:57:31Z