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| There are 5 variables seemingly $g1,g2,s,s2,m$ and three plotting directions available. What is meant with *all the variables*? Let Clear[G, m0, f, f1] G = 6.67*10^(-11); m0[s_, g1_, s2_, g2_, m_] := s^g1 s2^(-g2)/(g2 - g1)... |
| Standard procedure (after years everybody should know it by heart): Read in the manual, e.g. about [Assumptions][1], on this page see below [$Assumptions][2] and [Assuming][3]. [1]: https://reference.wolfram.com/language/ref/Assumptions.html ... |
| There is one more with wh: ![enter image description here][1] and the popular names show the importance of doing: ![enter image description here][2] [1]:... |
| Doing the `r` under assumptions is the first step In[2]:= Integrate[(r^2 Cos[A r Cos[x]] Exp[-r^2/2])/(a^2 Cos[x]^2 + b^2 Sin[x]^2), {r, 0, Infinity},(* {x,0,Pi/2} *) Assumptions -> {A > 0, a > 0, b > 0}] Out[2]=... |
| And here is the latest and greatest In[4]:= $Version Out[4]= "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" In[5]:= Integrate[BesselI[0, x] BesselK[1, x]/x^2, {x, 1/2, 3/2}] Out[5]= 2/9 (8 + \[Pi]^( 3/2)... |
| > But I don't know how to get the eigenvalues by the right boundary at > infinity. This is usually done by a so-called *Ansatz*, e.g. if $\lim_{x->\infty}f(x) = 0$ is the boundary condition, one sets $f(x) = g(x) \exp(-x)$; this can work only if... |
| It's the classic omission: `Evaluate[]` to resolve subexpressions before evaluation of intended main expression. You did not give all needed input, so that had to be created out of the sky to proof the point: In[60]:= Clear[list0] list0... |
| Do this again with In[18]:= $Version Out[18]= "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)" telling in ref/InverseFourierSequenceTransform > InverseFourierSequenceTransform is the same as FourierCoefficient: ... |
| IMHO Lie Groups and Algebras are not built-in into Mathematica, here is a glimpse from the internet - [Mathematica package for Lie algebra computations?][1] - [LieART A Mathematica Application for Lie Algebras and RepresentationTheory][2] -... |
| Can you use it to *p!nkify* a portrait? ![enter image description here][1] ... Rock singer P!nk wearing a fan gift ... (around or before 2015). [1]:... |