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Questions on the analytic and numerical equation solving functions of Mathematica (Solve, Reduce, NSolve, FindRoot, DSolve, RSolve, etc.).

Might make sense to solve for the trig and hyperbolic values because that can be done as a polynomial system. Then use those and perhaps myltiple inverses to recover the actual values. To do this, ex …
answered Jan 24 '17 by Daniel Lichtblau
Here is something of a brute-force approach. poly = c*(-4*c + 2*d) + d*(2*c - 4*d + 2*e) + e*(2*d - 4*e + 2*f) + (-2*b + 2*f + 2*g - 4*h)*h + f*(2*e - 4*f + 2*h) + g*(-4*g + 2*h) + j*(126 …
answered Apr 25 '17 by Daniel Lichtblau
One can use the fact that the second term in the analytic form of the Fibonacci numbers shrinks geometrically. So the idea is to solve for the first term equal to whatever polynomial, round the result …
answered Nov 3 '17 by Daniel Lichtblau
It requires a transformation that is not generically valid. Also it is a bit hard to make it happen using Simplify due to the default complexity measure. Solve[ Simplify[((v k)^-n (k v^n - v k^n))/( …
answered Nov 11 '14 by Daniel Lichtblau
One can use GroebnerBasis to eliminate variables, and it is set up in a way that tends to be more efficient than Eliminate (which may be using some dated technology). polys = {-a - 2 a c - 3 …
answered May 21 '15 by Daniel Lichtblau
Set to zero all variables that correspond to zeros in the fixed vector. vec = {0, 0, 0, 1/Sqrt, 0, 1/Sqrt, 1/Sqrt, 0}; vars = Array[x, Length[vec]]; zvars = Pick[vars, vec /. { 0 -> True, _? …
answered Jan 12 '16 by Daniel Lichtblau
One approach would be to convert to algebraic expressions, do the numeric substitutions, then find a Groebner basis that in effect "triangularizes" it so that the variables of interest can be solved i …
answered Mar 9 '17 by Daniel Lichtblau
Not a good idea to start variable names with capitals, hence the renaming below. ll = 10; w = 0.02; h = 0.02; yy = 209*10^9; mi = w*(h^3)/12; ei = yy*mi; mu = 0.001; rho = 7800; aa = w*h; cn = 4*Pi*m …
answered Jan 26 '18 by Daniel Lichtblau
If the binomials are expanded via FunctionExpand then the result is rational functions, which can be handled by Solve. Solve[ FunctionExpand[{Binomial[x + 1, y]/Binomial[x, y + 1] == 6/5, Binomi …
answered Mar 1 '18 by Daniel Lichtblau
Sort of embarassed I forgot this method for a few days. Anyway, one can get a set of possible moduli that work by computing what is called a strong Groebner basis of the defining polynomials over the …
answered Feb 13 '18 by Daniel Lichtblau
In addition to what I wrote in the comment, you can get a result much faster if you avoid using the explicit eigenvalues. Their product is the constant term of the characteristic polynomial so use tha …
answered Sep 28 '14 by Daniel Lichtblau
There is a typo either in the formatted equations or in the code. Assuming the latter, the fourth equation should have a double derivative. Fixing that we move from rational functions to polynomials a …
answered Apr 2 by Daniel Lichtblau
This may give you some ideas. In:= all = {a, b, c, d, e}; mean = Mean[all]; a1 = Accumulate@Prepend[all, 0]; a2 = Accumulate@Append[Table[mean, {i, Length[all]}], 0]; a3 = a2 - a1; Your soluti …
answered Apr 12 '12 by Daniel Lichtblau
One can get candidate values by solving for derivatives equal to zero. eqs = {y9[t] == -0.0019 y1[t] - y13[t] (-0.25 y2[t] + 0.248 y7[t]) - 0.45 Derivative[y1][t] + 0.448` (D …
answered Dec 22 '15 by Daniel Lichtblau
This involves what is known as l_0 minimization, and that is in general a "hard" problem. I can show a common surrogate method which somethis works (not quite in this case though). It is to instead do …
answered Jan 26 '18 by Daniel Lichtblau

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