# Search Results

Results tagged with Search options answers only user 184
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Questions on the analytic and numerical equation solving functions of Mathematica (Solve, Reduce, NSolve, FindRoot, DSolve, RSolve, etc.).

If you are looking for exact solutions you should substitute machine precission numbers by exact numbers. Therefore instead of your numerical coefficient I'd rather use this one: (7/5) (30^5) 2^(9/2 …
answered May 19 '13 by Artes
The result we get is correct and one shouldn't expect it in a different form since there are infinitely many solutions and there is no way to enumerate them all. That's why Mathematica returns the res …
answered Sep 10 '12 by Artes
Reduce works fine for a slightly more sophisticated expression, e.g. : Reduce[ ForAll[ x, x ∈ Integers, Mod[ x, 1] == 0], x] True however there is a bug in Solve : Solve[ Mod[x, 1] == 0, x, In …
answered Aug 17 '12 by Artes
Confirming that Mathematica 9 can easily solve this system unlike ver. 7 & 8 I'm going to suggest how to deal with it in earlier versions. there are many many complex solutions so the restriction o …
answered Mar 16 '13 by Artes
Since all the functions are roots of the same polynomial, a recommended approach uses a definition of one function instead of defining them separately. Thus I'd rather proceed along this way: f[x_, …
answered May 11 '13 by Artes
Solve works correctly, when returning {} it means there are no solutions. To demonstrate it let's rewrite your system: system = Thread[{-((2 (4 θ0 + θ1))/σ^2) - (2 (4 θ0 + θ2))/σ^2, …
answered Feb 28 '14 by Artes
There are infinitely many solutions. Here we demonstrate only a small part of the x - solution space in the complex plane. Let's start from the begining. There are no real solutions: Reduce[{ x - …
answered Sep 12 '17 by Artes
The canonical way to solve algebraic equations is to use Solve, e.g. 2 - 4 x - x^2 + 2 x^3 == 0 // Solve[#, x] & {{x -> 1/2}, {x -> -Sqrt[2]}, {x -> Sqrt[2]}} The above eqn //Solve[#,x]& is …
answered Feb 22 '12 by Artes
One can ask what is a nice solution, should it be as simple as possible? Isn't it a bit vague concept? Whatever that means there is a plenty of simple (and nice) solutions, e.g. simply x == 1 for all …
answered Mar 15 '14 by Artes
FindInstance[x + y == 1, {x, y}, Integers, 4] {{x -> -168, y -> 169}, {x -> 66, y -> -65}, {x -> 134, y -> -133}, {x -> 199, y -> -198}} or Table[ Reduce[x + y == 1, {x, y}, Integers] /. {C[1] …
answered Aug 22 '12 by Artes
At first, one should appropriately define the system of equations. Instead of machine precission numbers we prefer exact numbers therefore we would define: {a, b} /. Solve[{ 9/5 a + b == 1/100, 2 a …
answered Jul 22 '13 by Artes
You can make a WolframAlpha query directly from Mathematica (shortcut ==) : Solve y''(t)+3y'(t)=2t^4 Then just click the show steps link.
answered Feb 26 '12 by Artes
Solve is a really powerful function and its capabilities are much more extensive than polynomial equations as the other answers suggest. It can deal with many transcendental equations, however it does …
answered Mar 6 '17 by Artes
There are two real solutions, not only one as the question suggests: Maple is not an ultimate oracle, we have to understand what the solution is. It can be easily verified substituting solutions …
answered Jun 23 '14 by Artes
When you put sols={}; then calling FindRoot[alpha x == 1, Drop[sols, Length[sols]] you get FindRoot::fdss: "Search specification {} should be a list with 1 to 5 elements." because the sec …
answered Mar 15 '12 by Artes

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