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4

You can use SymmetricReduction[] for this. In particular, if you want the elementary symmetric polynomials to be represented symbolically instead of explicitly, you can use the three-argument form like so: SymmetricReduction[q, Array[p, 5], Array[e, 5]] // Total 1 - 3 c + c^2 + (1 - 2 c) e[1] + (1 - c) e[2] + e[3] + e[4]


0

This may be of some use to you: HornerForm[-1 - x + 3 x^3 - x^4 + x^5 - 3 x^6 + 2 x^7 + x^9] -1 + x (-1 + x^2 (3 + x (-1 + x (1 + x (-3 + x (2 + x^2))))))


1

Try fulleq=Append[eqnsMain, eqnsAdd]; And then Eliminate[fulleq, c] Using the {} parenthesis to fuse them would require Flatten on the result, as Mathematica will combine the vectors, not their components.


1

eqns = {AA == 2 a + 4 b + 45 c + 5, BB == 4 a + 45 b + 31 c + 78, CC == 0.23 a + 0.4 b + 4.35 c + 0.12, DD == 0.73 a + 0.2 b + 43.455 c + 3.12}; Substituting for c eqns2 = eqns /. c -> 43.5 AA + 34 b + 32 (* {AA == 5 + 2 a + 4 b + 45 (32 + 43.5 AA + 34 b), BB == 78 + 4 a + 45 b + 31 (32 + 43.5 AA + 34 b), CC == 0.12 + 0.23 a + 0.4 b + ...


5

linearizeEquation[expr_, f_, fp_, order_] := Block[{e}, Expand@Normal@Series[expr /. f -> (fp + e dF[#] &), {e, 0, order}] /. e -> 1 ] linearizeEquation[f[x] + (1 - f[x]^2) + D[f[x], {x, 2}] + f[x] D[f[x], x], f, f0, 1] (* 1 + f0 - f0^2 + dF[x] - 2 f0 dF[x] + f0 dF'[x] + dF''[x] *)


1

Not very elegant, but this could work. Use Factor as follows: In[6]:= expr = 2 A + 4 B + 2 C + 6 Out[6]= 6 + 2 A + 4 B + 2 C In[7]:= Factor[expr][[2]] Out[7]= 3 + A + 2 B + C 


5

So long as it is only for output purposes, there is probably no real harm in doing this, and so it does not seem to be a matter of "force" in this case--defining an output form is a perfectly reasonable thing to do. The situation here is rather like in this recent question, and so I propose a similar solution. The most sensible (and least intrusive) way to ...


6

If you really insist on getting exactly this output within a standard evaluation, you can do the following: Unprotect[Power]; Format[Power[2, Rational[-1, 2]]] := HoldForm[Sqrt[2]/2]; Protect[Power]; This leads to the desired output: 1/Sqrt[2] $$\frac{\sqrt{2}}{2}$$ Sin[1/Sqrt[2]] $$\sin\left(\frac{\sqrt{2}}{2}\right)$$ N[%] ...


4

I would make my own SomethingForm function like this: SpecialForm[expr_] := expr /. { 1/Sqrt[2] -> HoldForm[Sqrt[2]/2] } SpecialForm[1/Sqrt[2]] SpecialForm[5 10 + 3^Sqrt[3] + 1/Sqrt[2]]


4

A slight variation of garej's comment: Divide @@@ HoldForm[{Sqrt[2], 2}]



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