In Some Notes on Internal Implementation especially in Algebra and Calculus one finds interesting subtleties and differences between these two functions, e.g. The code for Solve and related functions ...

Maybe this ? ClearAll["Global`*"]

General remarks In General Relativity we work in a 4-dimentional Lorentzian manifold i.e. there is a metric tensor $g$ of signature $(+,-,-,-)$ or $(-,+,+,+)$. Theses signatures are mathematically ...

The integrand has two singular points: Solve[ 4z^2 + 4z + 3 == 0, z] {{z -> 1/2 (-1 - I Sqrt[2])}, {z -> 1/2 (-1 + I Sqrt[2])}} At infinity it becomes zero: Limit[ 1/Sqrt[ 4 z^2 + 4 z + 2], ...

A shorter introduction to working with Root objects is in the below answer. Solutions to algebraic or transcendental equations are expressed in terms of Root objects whenever it is impossible to find ...

In general, a typical root of a negative number is complex, so you need to get rid of most roots. A nice approach would be Root, e.g. Root[ x^3 + 8, #] & /@ Range[3] {-2, 1 - I Sqrt[3], 1 + I ...

Mathematica wouldn't be much helpful if one applied only formulae calculated by hand. Here we demonstrate how to calculate the desired geometric objects with the system having a definition of the ...

You can make use of the following options in Plot, e.g. : Plot[ Tooltip @ {x^2, x^3, x^4}, {x, -2, 2}, PlotStyle -> {Red, Green, Blue}, PlotRangePadding -> 1.1] /. {Tooltip[{_, ...

You can do this : s = Solve[y^2 == 13 x + 17 && y == 193 x + 29, {x, y}]; xx = s[[All, 1, 2]]; yy = s[[All, 2, 2]]; Now you can access solutions, this way xx[[1]], yy[[2]]. If you prefer to ...

Defining the function F and a subset of its domain : pts : F[z_] := (5 - I z)/(5^2 + z^2) pts = {-7, -2, 0, 2, 7}; the most straightforward way fulfilling the task is based on ParametricPlot and ...

You can plot in 3 dimensions only real and/or imaginary parts of a function. One can make use of Plot3D, but since there was a question how the sine function looks like on the unit circle, first I ...

If not assumed otherwise m and n can be whatever, so you can do e.g. this : Solve[ Prime[n] + Prime[m] == 100, {n, m}, Integers] {{n -> 2, m -> 25}, {n -> 5, m -> 24}, {n -> 7, m -&...

Absolutely unbeatable : Tuples[{{1997}, data1}] All other methods are much slower even Verbeia's {1997,#}&/@ data1 or dws' Thread[{1997, data1}] Tuples[{{1997}, data1}] === ({1997, #} & /@ ...

We define the function f and multiple constraint functions g1, g2: f[x_, y_, z_] := x y + y z g1[x_, y_] := x^2 + y^2 - 2 g2[x_, z_] := x^2 + z^2 - 2 then, in order to find necessary conditions for ...

Use Reduce[(1/x) Cosh[x/2] == Sqrt[2], x, Reals] or Solve[(1/x) Cosh[x/2] == Sqrt[2], x, Reals] the latter yields {{x -> Root[{-E^(-(#1/2)) - E^(#1/2) + 2 Sqrt[2] #1 &, 0....

The default value of $NumberMarks Automatic means that ` should by default be used in arbitrary-precision but not machine-precision numbers. Arbitrary-precision numbers can contain an arbitrary ...

The modern differential geometry is a vast subject and while not specified exactly what you need the question is a bit too general. I would rather point out a few references. If you are looking for ...

I recommend using Transpose twice since it is more efficient than other approaches. Moreover Plus has the Listable attribute, thus one need not map Plus over a list (vector). Transpose[v1 + ...

There is an especially useful function for this kind of task: FrobeniusSolve[{a, b, c}, d] for finding the list of all solutions to the equation a x + b y + c z == d, where a,b,c are given positive ...

We couldn't be really pleased if we didn't exploit existing Mathematica functionality to get exact solutions. Here we provide them with Reduce rewriting the given system to an exact one and using a ...

The way you could use ContourPlot here, assuming your variable f is complex (f == x + I y) : eqn[x_, y_] := (25 Pi ( x + I y) I)/(1 + 10 Pi ( x + I y) I) {ContourPlot[Re@eqn[x, y], {x, -1, 1}, {y, -...

This is indeed a serious and problematic issue. We know many similar problems with symbolic integration which provides Integrate. There were some improvments in newer versions of the system but also ...

General remarks These are are crucial aspects of solving equations symbolically: So far (in general) Mathematica cannot solve transcendental equations when two unknowns are involved, nevervetheless ...

Instead of doing some transformation on the original ContourPlot we can do an exponential rescaling of the original variables in the ContourPlot, so this is somewhat different approach to get roughly ...

There are many ways to proceed, the best one uses FrobeniusSolve : I Since we know, that a x + b == y /. Solve[{-4 a + b == 11, 16 a + b == -1}, {a, b}] // Simplify {3 x + 5 y == 43} we find ...

There are much better programming methods in Mathematica than loops. Here is an approach based on IntegerPartitions, it chooses 5 numbers that sum up to 35: RandomChoice[ IntegerPartitions[35, {5}]]...

There is no need to play around with ReplaceAll, Rule, Block, Module or whatever using D, since you have an oparator Derivative really fulfilling your needs while you need not bother if the arguments ...

This is a good way : DensityPlot[ Rescale[ Arg[Sin[-x - I y]], {-Pi, Pi}], {x, -Pi, Pi}, {y, -Pi, Pi}, MeshFunctions -> Function @@@ {{{x, y, z}, Re[Sin[x + I y]]}, ...

For this function: f[z_] := (1 - E^z + z)/(z^3 (z - 1)^2) there are no branch cuts in the complex plane therefore we simply use Cauchy integral theorem and the related formula of the complex residue,...

One can use Solve as well, e.g. s = Solve[ BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[ Sin[x]] == 0 && 0 < x < 10, x] Solve::incs: Warning: Solve was unable to prove that the solution ...