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RSS Feed for Wolfram Community showing any discussions in tag Calculus sorted by activeNo output in solving system of trigonometric equations
https://community.wolfram.com/groups/-/m/t/2243285
Hello everybody, I hope you are doing all well today.
I'm new to Mathematica, I have a system of equations that I want to solve using Mathematica: <br><br>
$\begin{equation}
\begin{cases}
sin(x)cos( \epsilon y)cosh(\epsilon z) + sin(\epsilon x)cos(y)cosh(z) - sin(x(1 +
\epsilon)) = 0 \\
sin(x)sin(\epsilon y)sinh(\epsilon z) + sin(\epsilon x)sin(y)sinh(z) = 0
\end{cases}
\end{equation}$ <br><br>
I'm only interested in {x, y, z} values in the domain [0, 2*Pi] <br>
$\epsilon$ is a parameter that can take any value between 1.8 to 3.
I tried solving the problem first by setting $\epsilon = 2$, I used methods like `Solve` and `Nsolve`, but it didn't work. it took too much time to execute without any result. after a long day I found out about `Reduce` function, i used it and it works it gave me the full answer. but that's only for the case of $\epsilon = 2$. When I change $\epsilon$ to 2.5 it didn't gave me any answer. I left it to run to about 10 hours or more without any result. <br>
This is my fifth day with this problem, and I'm still stuck.
can anyone provide me with any advices. I really appreciate your help. Thank you so much <br><br>
This is my code for the case of $\epsilon = 2$
Reduce[{
Sin[x]*Cos[2.5*y]*Cosh[2.5*z] + Sin[2.5*x]*Cos[y]*Cosh[z] -
Sin[3.5*x] == 0,
Sin[x]*Sin[2.5*y]*Sinh[2.5*z] + Sin[2.5*x]*Sin[y]*Sinh[z] == 0,
0 <= x <= 2*Pi, 0 <= y <= Pi, 0<= z <= 2*Pi}, {x, w, y}, Reals]Mohamed El Ghafiani2021-04-13T17:04:13ZAppending a property to a variable for use in derivations
https://community.wolfram.com/groups/-/m/t/2242983
Folks -- one of my students asked if it is possible to directly associate a mathematical property with a variable for use in derivations involving that variable. For example, can one associate the property that a variable a is a real number that is greater than zero when performing integrals and the like? I know that one can make this assumption in the Integrate command and in other commands via the option Assumptions -> a>0 , but it would be useful to be able to do this at the beginning of a series of derivations that involve the variable, without constantly having to repeat the Assumption at every step.
Thanks -- Dan Dubin, UCSDDan Dubin2021-04-13T21:14:41ZA formula for the n-th Laplacian of a Gaussian function
https://community.wolfram.com/groups/-/m/t/2242639
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/f966d372-af2b-4b74-88d7-97d40410310c
[Original]: https://www.wolframcloud.com/obj/rauan234/Published/gaussian_laplacian.nbRauan Kaldybaev2021-04-12T15:48:11ZPreserve arguments' patterns while taking a Derivative?
https://community.wolfram.com/groups/-/m/t/2242146
Hello!
I've encountered a problem with loss of pattern information after using "function-producing" `Derivative` function.
Assume I have a scalar-valued function defined on vectors:
H[r_?VectorQ] := r.r
(evidently, `r` is assumed to be a three-dimentional vector, but id does not matter)
And then I would like to solve a differential equation like *dr / dt == grad(H)*. Note that I would like to keep the vectorial notation i.e. the solution must be a vector-valued function. I've tried the following:
NDSolveValue[{D[r[t], t] == Table[Derivative[xspec][H][r[t]], {xspec, IdentityMatrix[3]}],
r[0] == {1, 1, 1}}, r, {t, 0, 100}]
This returns an interpolated function, but it does not handle the derivatives of `H` properly. For example, if in the output of `Derivative` appeared term like `#1[[1]]` then it accepts `r[t]` as argument and evaluates simply to `t` and if there's `#1[[2]]` it throws a message that this part does not exist. It seems like `Derivative` loses argument check and the returned function does not formally require vectorial input anymore.
So, the question is: can I somehow tell WL that this derivative is still a function of a vector?Nikolay Shilov2021-04-12T12:32:31ZMathematica and logic
https://community.wolfram.com/groups/-/m/t/2242679
Do logicians use Mathematica? For instance for proof trees in sequent calculus. General answers are welcome.Stephan Spahn2021-04-12T19:39:01ZKernel quits while using DSolve[ ]?
https://community.wolfram.com/groups/-/m/t/2241599
Why does my DSolve command does not solve the Non-linear ODEs?
Clear["Global`*"]
(*Parameters*)
length = 1.5;
width = 0.25;
height = 0.25;
area = width*height;
cforce = -400;
dforce = (500*(X1^2));
stress = 10*displacement'[X1] + 100000*displacement'[X1]^3;
DE1 = D[stress*area, X1] + dforce;
DE2 = 10*displacement'[X1] + 100000*displacement'[X1]^3 + 6400;
(* Solving for the Exact Displacement Function *)
solution =
DSolve[{DE1 == 0, displacement[0] == 0, DE2 == 0 /. X1 -> length},
displacement[X1], X1];
displacement = displacement[X1] /. solution[[1]];
Print["The Exact Displacement Function of the Bar is"]
Print[displacement];
Print[Plot[{displacement}, {X1, 0, 1.5}, AspectRatio -> 0.4,
AxesLabel -> {"Bar Length (m)", "Axial Displacement (m)"},
PlotLegends -> {"Exact Displacement of the Bar"}]];
Print[" "]Aslam Kittur2021-04-10T18:10:30ZNIntegrate command does not converge
https://community.wolfram.com/groups/-/m/t/2241579
sol3 = NIntegrate[IntFunc, {X1, 0, 3}, {X2, 0, 2}];
![The solution yieals a 6 x 6 matrix solved for 2 vaiables X1 & X2][1]
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=NIntegrate.PNG&userId=2213110Aslam Kittur2021-04-10T14:53:18ZPlot 2 parameter solution from ParametricNDSolve with Table
https://community.wolfram.com/groups/-/m/t/2242051
I have a 2-D differential equation which I solve with ```ParametricNDSolve``` with initial time given as a parameters. I solve the equation using
twoeqnpara = ParametricNDSolve[{x'[t]==-3x[t] - y[t],y'[t] == x[t],x[0]==a,
y[0]==b},{x,y},{t,0,100},{a,b}].
When I try to plot with
pp = ParametricPlot[Evaluate@Table[{x[a,b][t],y[a,b][t]}/.twoeqnpara,
{a,-2,2,0.5},{b,-2,2,0.5}],{t,0,40},PlotRange->All]
I don't get any plot. But when I do for single parameter
pp = ParametricPlot[Evaluate@Table[{x[a,a][t],y[a,a][t]}/.twoeqnpara,{a,-2,2,0.4}],
{t,0,40},PlotRange->All,PlotLegends->Range[-2,2,0.4]]
It works and give a plot.
What I think the issue is when I use the command
Table[ x+y, {x,-2,2,0.5},{y,-2,2,0.5}]
, what the command is taking a single value of x and all the values of y. What I need is all the pairs possible from the 2 arrays so that can be used as a initial condition for plotting.Sahil Goyal2021-04-11T17:01:59ZHacking a complex function with Mathematica
https://community.wolfram.com/groups/-/m/t/2241822
&[Wolfram Notebook][1]
[1]: https://www.wolframcloud.com/obj/wolfram-community/Published/mmaHack.nbRobert Rimmer2021-04-11T01:21:31ZDistance between a point and others is an integer
https://community.wolfram.com/groups/-/m/t/2241420
I have 4 points in a 2D plane
{{0, 0}, {20, 0}, {138/5, (24 Sqrt[6])/5}, {25, 10 Sqrt[6]}}
and wish to find a 5th point such that the distances to the 4 points is an integer.
I have tried solving for a point to the point {138/5, (24 Sqrt[6])/5}, specifying Integer elements and reals and cannot get any specific point as an answer, it's all conditionally based. Satisfying 4 points simultaneously seems unobtainable. My question is, is this solvable using Mathematica's built in functions? Or do I resort to a programmable approach?
Btw the pairwise distances between those 4 points are Integer.Paul Cleary2021-04-10T15:50:20ZEmpty plot using NDSolve[ ] ?
https://community.wolfram.com/groups/-/m/t/2239857
Hey guys,
I'm quite new here and hope i'm in the right section. In trying to solve a PDE system with NDSolve the plots I obtain in the end are empty. I read through quite a few posts regarding this issue but haven't found a solution that I could apply to my code.
The PDE system:
deqs = {D[m[x, t], t] == D[vm[x]/am[x] D[vm[x] m[x, t], x], x] + f[{m[x, t], c[x, t]}],
D[c[x, t], t] == D[vc[x]/ac[x]*D[vc[x]*c[x, t], x], x] - f[{m[x, t], c[x, t]}],
m[x, 0] == M, c[x, 0] == C0, m[0, t] == m[L, t] , c[0, t] == c[L, t]};
Most important code:
sol1 = NDSolve[deqs, {m, c}, {x, 0, L}, {t, 0, 10}]
Plot3D[m[x, t] /. sol1, {x, 0, L}, {t, 0, 10}]
I assume my mistake already is somewhere in these lines, f is a polynome, vc and vm are constants and ac and am are arctan(x) like functions.
To be precise:
f[{a_, b_}] = g*a^2*b - d*a
am[x_]:= (A(ArcTan[B(x-L/2)]+ Pi/2))^(-1);
ac[x_]:=(A (ArcTan[B(-x+L/2)]+Pi/2))^(-1);
vm[x_] := vM;
vc[x_] := vC;
Does anyone have an Idea why this doesn't work?
Thanks a lot in advance!Janis Köhler2021-04-08T13:39:37ZHow can I do NIntegrate of derivative of the function?
https://community.wolfram.com/groups/-/m/t/2239047
Hi everyone,
I define the function and define new function using the derivative and NIntegrate on it.
As an example, I made some simple case.
Clear[f,d]
f[x_?NumericQ, a_?NumericQ] := Sin[a x]
d[x_?NumericQ] := NIntegrate[D[f[x, a], x], {a, 0, 1}]
And try to get the result at some x.
With[{x = 2}, Evaluate@d[x]]
But it does not work...And I got error message
> General::ivar: 2 is not a valid variable.
> NIntegrate::inumr: The integrand \!\(\*SubscriptBox[\(\[PartialD]\), \(2\)]\(f[2, a]\)\) has evaluated to non-numerical values for all sampling points in the region with boundaries {{0,1}}.
How can I fix this?Sangshin Park2021-04-07T12:14:20Z