# Tag Info

1

ClearAll[sus]; D[sus[i, t], t] /. i -> 1 // InputForm (* Derivative[0, 1][sus][1, t] *) This is the first derivative with respect to the second argument, and with the first argument equal to 1. This is equivalent to D[sus[1, t], t] // InputForm (* Derivative[0, 1][sus][1, t] *) ClearAll[sus]; sus = {sus1[t], sus2[t], sus3[t]}; The derivative of a ...

0

The solution I get with Version 12.0.0 looks indeed inconsistent. I compare the solution rather close to the one shown on the documenation page for NDSolve in the section Possible Issues -> Partial Differential Equations with the example for the Laplace equation with initial values. For the partial differential equation system given and for the value set ...

2

It integrates if (1) we replace your exponent $3.5$ by the exact $7/2$, and (2) we change the order of integration: g[x1_, x2_, x3_, x11_, x22_, x33_] = -((3(4(x1-x11)^2-(x2-x22)^2-(x3-x33)^2)((x2-x22)^2+(x3-x33)^2))/ (8π((x1-x11)^2+(x2-x22)^2+(x3-x33)^2)^(7/2))); Integrate[g[x1, x2, x3, x11, x22, x33], x3, x33, x2, x22, x1, x11] (* lengthy output ...

1

A workaround and under certain assumptions we have: func= r*BesselJ[n, a r]*BesselJ[n, b r]; InverseMellinTransform[Integrate[MellinTransform[func, a, s], {r, 0, Infinity}, Assumptions -> {s > 1, b > 0, n \[Element] Integers, n >= 0, 2 + n > s}], s, a] (*DiracDelta[a - b]/b*) Maple 2020.2 Can deal:

1

I suggest you write a function, say f, that takes λ and r0 as arguments and makes a plot of a simplified version of the expression you call Dfigura over the domain {T0, T}. Like so: With[{α = 19*^-6, T = 300, T0 = 0, L = 5}, f[λ_, r0_] := Plot[2.44*^2 λ L/(r0 Sqrt[1 + 2 α t]), {t, T0, T}, PlotLabel -> Row[{"λ: ", N@λ, " r0: &...

1

See if the following code does what you seek. I chose the pitch over each range arbitrarily since you did not specify it. output is a list of triplet including $(\lambda, r_0, \%_{difference})$: ClearAll["Global*"] α = 19*10^-6; L = 5; rf = r0*Sqrt[1 + 2*α*ΔT]; Dabertura = 2*rf; Dfigura = (2*L*1.22*λ/Dabertura)*100; output = Flatten[ Table[ {...

1

To avoid parameter confusion I renamed m->mm and n->nn in the Wolfram Expression. Integrate[(t^(-1 + \[Alpha])*LaguerreL[mm, \[Lambda], p*t]* LaguerreL[nn, \[Beta], p*t])/E^(p*t), {t, 0, Infinity}] == (Gamma[\[Alpha]]*Gamma[1 + nn - \[Alpha] + \[Beta]]* Gamma[1 + mm + \[Lambda]]* HypergeometricPFQ[{-mm, \[Alpha], \[Alpha] - \[Beta]}, {-nn + \ \[...

0

As Michael E@ shows, there is no maximum -- there are choices of $h$ and $k$ such that $h/k^2$ is arbitrarily close to 0.1974. However, your question is explicitly about writing a particular constraint. It is possible to use domain specifications, for example Element[k,Integers]. Unfortunately, the list of domains doesn't include half-integers. But we ...

0

You can add a extra differential equation f'[x]==y2[x]*y1[x]^3* x^2 in your equations and solve f'[x] at the same time. so we need to set the initial value f[0.001]=c and c is a parameters,so we use ParametricNDSolve Clear[pde1, pde2, pde3, sol]; pde1 = -y1''[x] - (2*y1'[x])/x + ((y1[x])^3 + y2[x]) y1[x] == 0; pde2 = y2''[x] + (2 y2'[x])/x - (y1[x])^3 == 0; ...

5

To find the value of $h$ for a given $k$, introduce a new variable, $n=2h$ and use Maximize. As an example, for $k=5$ we can do this {f, cond}={h / k^2, {0 < h, h / k^2 > 0.1974}} /. h - > n/2; m = Maximize[{f, cond} /. k -> 5, n, Integers] (* {0.18, {n -> 9}} *) We can recover $h$ with h = n/2 /. Last[m]. We can generate the point $(h,k,... 5 The following produces an increasing sequence$(h_j/2)/k_j^2$, with$h_j,k_j$positive integers, that converges to 0.1974 = 987/5000: NestList[ # /. {x_ :> ( 2500 Numerator[x] + 2961 Denominator[x]/2)/(100^2 Denominator[x])} &, 1/8, n] Consequently, there is no maximum. The following produces an answer because of a numerical precision ... 2 Consider your integral in torootL $$\int_0^1 \frac{zl \cdot y^d}{\sqrt{\left(1 - (zl/zh)^{d + 1} y^{d + 1}\right) \left(1 + \frac{t^2}{1 - (zl/zh)^{d + 1}} - y^{2 d}\right)}} d y.$$ It was$d=3$and let$z=zl/zh$such that $$zl \cdot\int_0^1 \frac{ y^3}{\sqrt{\left(1 - z^{4} y^{4}\right) \left(1 + \frac{t^2}{1 - z^{4}} - y^{6}\right)}} d y.$$ Substitute$y^2=...

0

Denote the result with I. This logarithm has simple derivative which is 1/(a-x) with respect to a. then factor out the parameter a to get an integrant: f(x) = x^(N-1) * (1-x)^ ((N+1)-N-1) / (1-(1/a)x) / (1-x) If you integrate it from zero to one with respect to x you get a Lauricella FD function which is the same as an Appell F1 function, you choose which ...

4

Try NIntegrateinstead of Integrate because F[t] is numerical: F[t_] := t^2/(1 + (E^(2.13122 (-2.6 - t)) + E^(2.13122 (-2.6 + t))) (1/2 + 0.0739645 t^2)^1.00384); F1[t] doesn't depend on t: F1 = 12 NIntegrate[F[t], {t, 0, 10}] (*86.6488*) FF[q_?NumericQ] :=12/F1 NIntegrate[F[r] *Sin[q r]/(q r) r^2, {r, 0, 5}] Plot[FF[q], {q, .001, 20 }, PlotRange -> {{-...

0

F[r_] = 1/(1 + (E^(2.13122 (-2.6 - r)) + E^(2.13122 (-2.6 + r))) (1/2 + 0.0739645 r^2)^1.00384); g[t_] = t^2/F[t]; F1 = 12 NIntegrate[F[t], {t, 0, 10}]; FF[q_] := 12 NIntegrate[F[r]/F1*Sin[q r]/(q r) r^2, {r, 0, 5}] Plot[FF[q], {q, 2, 20}]

4

I think @Jagra has the right idea using PlotGrid. You can do it manually using Grid or GraphicsGrid as well, but you end up having to tweak quite a few parameters. If you want to get them to share an x-axis, you can change Jagra's answer like this: pg = ResourceFunction["PlotGrid"]; data = Table[{x, i PDF[NormalDistribution[], x]}, {i, 3}, {x, -5, ...

4

If I've understood, it seems like you just want a variation of the answer you received in Plot Figures in two columns. ResourceFunction["PlotGrid"][ { {Plot[x, {x, 0, 1}, Frame -> True, PlotLabel -> "Some Label"]}, {Plot[x, {x, 0, 1}, Frame -> True, PlotLabel -> "Some Label"]}, {Plot[x^2, {x, 0, 1}, ...

3

I think the only real problem keeping you from using the more modern Quantities framework is the fact that Integrate[a0, t, t] and a0 Integrate[1, t, t] are not equivalent if one or more accelerations in a0 are zero. In the first case we get {0 m/s^2, -4.9 t^2 m/s^2} and in the second case we get {t^2 (0 m/s^2), t^2 (-4.9 m/s^2)}. Since we're integrating ...

4

The following seems to reproduce what you have in MMA 12 using the Quantity framework. Not that I am a huge fan of its implementation, but one can make it work. ClearAll["Global*"] Setup: r0Vec = {Quantity[0, "Meters"], Quantity[0, "Meters"]}; v0 = Quantity[37, "Meters"/"Seconds"]; v0Vec = AngleVector[{v0, ...

1

Use the definition of arc length. $$\int_{0}^{10} |r'(t)|\,\mathrm{d}t$$ r[t_] := {Sin[t], Cos[t], t}; Integrate[r'[t] // Norm, {t, 0, 10}] Integrate[r'[t]^2 // Total // Sqrt, {t, 0, 10}] NIntegrate[r'[t]^2 // Total // Sqrt, {t, 0, 10}] 10 Sqrt 14.1421

1

Consider the trajectory $r(t) = (\sin(t), 2 \cos(t))$: ParametricPlot[ {Sin[t], 2 Cos[t]}, {t, 0, 10}, PlotStyle -> Directive[Thickness[0.02], Opacity[0.3, Black]] ] You can use ArcLength directly to calculate the length of the parametric path. If you use exact numbers, Mathematica will attempt to find a symbolic answer, which may take quite a while: ...

3

I read Your question carefully. (A) You give a value $q$ and your image gives values $q_k$ with $k={c,h}$. OK. That is cold and hot. But q changes from right to left an back. It is constant and has to be modelled. (B) All functions start parallel to the x-axis up to $T_{initial}$ somewhat below $T=400K$. It is not logical to have something on a temperature ...

5

When you use approximate floating-point numbers in an exact solver, you invite trouble. It's hard to figure out whether accumulated round-off error is causing failure or if the problem simply cannot be solved symbolically by Wolfram. Mathematica has improved over the years (imo) its ability to handle FP numbers in exact solvers, but round-off error is an ...

2

Graphics3D[ Table[{Hue[(3 - z)/3], Cylinder[{{2, z, z}, {2, z, z + .1}}, (3 - z)*2]}, {z, 0, 3, .1}] ] This doesn't help you, but I thought I'd play with stacked cylinders, following a simple offset path and with the radius being twice the quantity of 3-z. Since I tend to abuse my parameters, I just colored each cylinder based on that same parameter.

3

ClearAll[pisaF] pisaF[ρ_: 1/2, μ0_: {0, 0}, μ1_: {3, 3}, σ_: {1, 1}] := PDF[BinormalDistribution[(1 - #3) μ0 + #3 μ1, σ, ρ]][{#, #2}] &; Row[RegionPlot3D[z <= #[x, y, z], {x, -5, 5}, {y, -5, 5}, {z, 0, 1}, MeshFunctions -> {#3 &}, Mesh -> 10, PlotRange -> All, PlotLabel -> Style[#2, 16], ImageSize -> Medium, ...

7

Let $r : [0,1] \longrightarrow [0,\infty[$ be decreasing with $r(0)=\infty$ and $r(1)=0$. For $c\in [0,1[$ set $s:= c r$ then $$r'(t') < -|s'(t)|$$ and $$(x-c r(f(x,y)))^2 + y^2= r^2(f(x,y))$$ such that $$f(x,y)= r^{-1}\left(\frac{1}{c^2-1}\left(c x \pm \sqrt{x^2-(c^2-1)y^2}\right)\right).$$ $r^{-1}$ just has to be chosen such that $f$ is differentiable, ...

4

Reply the comment According my original answer,we can rewrite my equation to get the two functions s[t] and r[t]. s[t_]=t/2; r[t_]=Sqrt[-Log[t]] seems work toward your new edition. Plot[Sqrt[-Log[t]], {t, 0, 1}] Limit[Sqrt[-Log[t]], t -> 0] Original My approach just as flinty by use ContourPlot3D, and my idea is use the deformation of the Gauss function ...

7

The equation of a circle at offset $(x_0,y_0)$ and radius $r$ is $(x-x_0)^2+(y-y_0)^2=r^2$. Therefore we can make the tower if we make $x_0$, $y_0$, and $r$ a function of height, moving $(x_0,y_0)$ along a line and shrinking $r$ as we go: With[{a = 0.6, b = 0.3}, Show[ ContourPlot3D[(x - a*t)^2 + (y - b*t)^2 == 0.04/t^2 (1 - t) + t (1 - t), {x, -2, 2}...

5

(Extended comment, not an answer.) Imported the STL file from here: https://www.thingiverse.com/thing:2733780/files . One approach would be to use intersections of this STL-file mesh object with planes orthogonal to the X-axis and derive (pairs of) functions over a certain set of X values. rgSTLPisa = Quiet[Import[ "https://cdn.thingiverse.com/...

-2

If the task is abstract it is sensible to draw improvised scetches of what to work might be. This looks very much like the integration over some profile like, LorentzianFunction. This enters some terminology to handle such a problem. I run the given code and found as so very ofthen that introducing: MinRecursion -> 20, MaxRecursion -> 20, AccuracyGoal -...

5

Solution for one set of parameters with cooling is very simple, but for the case of heating we need some small modification of code. Code for cooling case: ah = 342496;(*J/mol*)R = 8.314;(*J/mol.K*)A = 7.6*10^-38;(*s*)b = 0.67; x = 0.49; q = -0.1/60;(*K/s*)T0 = 500;(*K*)Tfinal = 350;(*K*)dt = 300;(*s*)n = IntegerPart[Abs[(T0 - Tfinal)/dt/q]];(*number of ...

1

Get a graphical overview to see, the function has no minimum and is zero not at single points, but at a curve depending on parameters. int = -3/(96 Pi^2 z^4) Integrate[ Sqrt[x^2 - y^2] x^2 E^-x, {x, y, Infinity}, Assumptions -> y > 0] rest = 3/(16 Pi^2 (1 - r^2)^2 z^4) (1 + 1/3 (2 - w^2 + (3 - r^2)/(1 - r^2)) (1 - r^2) z^2 + w^2/6 (w^2 - 2) (1 ...

1

I managed to sort your data. Here is the time history. The data is in c1. ListLinePlot[c1] Now we take the Fourier transform and plot n = Round[Length[c1]/2]; ft = Fourier[c1, FourierParameters -> {-1, -1}]; ListLogLogPlot[Abs[ft[[1 ;; n]]]] Hope that helps. Edit A comment below suggests you want the power spectral density. You may want this but if ...

3

Let's define a symbolic rank 4 tensor (of dimension 3): MatrixForm[symbolicRank4=Array[Subscript[a,StringJoin[ToString/@{##}]]&,{3,3,3,3}]] We can symmetrize this manually using the permutations you suggest: MatrixForm[manualSymmetrization=Simplify[Mean[TensorTranspose[symbolicRank4,#]&/@Permutations[Range]]]] or using the built-in Symmetrize: ...

1

Develop the driving force Abs[t] into FourierSeries and solve diff equation for the general coefficients. FourierSeries[Abs[t], t, 3, FourierParameters -> {1, 1}] (* -((2 E^(-I t))/\[Pi]) - (2 E^(I t))/\[Pi] - (2 E^(-3 I t))/( 9 \[Pi]) - (2 E^(3 I t))/(9 \[Pi]) + \[Pi]/2 *) fc[n_] = FourierCoefficient[Abs[t], t, n, ...

3

Here we use Mathematica to verify that the -Grad satisfy the maximum decrease. the result1 according to the definition of derivative along a vector v. Now it is equal to result2 or result3 means that the direction derivative is equal to $\nabla T\bullet v$, the projection of the gradient of $T$ to $v$. After that, we use Maximize to find the direction v ...

1

Use Derivative instead of D: {#, intSL[.01, #, 1], Derivative[0, 1, 0][intSL][0.01, # , 1]} &[.3] (*{0.3, -0.00218296, 0.0149288}*) {#, intSL[.01, #, 1], Derivative[0, 1, 0][intSL][0.01, # , 1]} &[.6]; % //Chop (*{0.6, -0.000430944, 0.00225041}*) The evaluation takes some time and shows several messages "FindRoot::jsing: Encountered a singular ...

3

Clear[p, q, r, x, y] p = x + y; q = 2 x + y; r = p + q 3 x + 2 y D[r, x] 3 Or: Clear[p, q, r, x, y] r[x_, y_] := (p = x + y; q = 2 x + y; p + q) D[r[x, y], x] 3

1

As a product of visual inspection, taking data from $\approx 80$ to $120$ and using the model $$f(a,b,\sigma_1,\sigma_2,x_1,x_2,x)=a e^{-\left(\frac{x-x_1}{\sigma_1}\right)^2}+b e^{-\left(\frac{x-x_2}{\sigma_2}\right)^2}$$ data = Get["https://pastebin.com/raw/2jgDw4iQ"]; reddata = Take[data, {990, Length[data]}]; f[a_, s1_, x1_, x_] := a Exp[-((...

1

Try this: expr1 = Inactivate[D[2 y/x, y/x], D] /. y :> z*x Then Activate[expr1] (* 2 *) Have fun!

0

I played a little with the integration limits and the offset for the fit. I used the definitions below with integration limits {0.1,200}, fitting offset 10. Note that I am using much smaller precision and accuracy goals. Experiments with that code might produce better results. Clear[G]; G[d_?NumericQ, l_?NumericQ, p_?NumericQ] := NIntegrate[(y^(-3) - ...

1

If we set $Assumptions = x0 < 0;,we can prove this result by Mathematica !$Assumptions = x0 < 0; f[x_] = x^3; tan[x_, p_] = f[p] + f'[p] (x - p); p1[x0_] = x /. Solve[tan[x0, x] == f[x0] && x != x0, x] // First p2[x0_] = x /. Solve[tan[x, x0] == f[x] && x != x0, x] // First reg1 = ImplicitRegion[{x0 <= x <= p1[x0], tan[x, p1[...

3

One more way is as follows. ComplexRegionPlot[ Pi/4 < Arg [z] <= 5 Pi/4 && 1 <= Abs [z] < 2, {z, -2 - 0*I, 2 + 2*I},AspectRatio->Automatic]

4

With correct Mathematica syntax and increased range of x,y RegionPlot solves your problem: RegionPlot[ 3 Pi/4 < Arg[x + I y] <= 5 Pi/4 && 1 <= Abs [x + I y] < 2 , {x, - 2,2}, {y, - 2, 2} ] addendum Thanks to the comment @Bili Debili: Arg returns angle in the range -Pi...Pi, that's why the condition 3 Pi/4 < Arg[x + I y] <= 5 Pi/4 ...

3

ParametricPlot[ ReIm[r*Exp[I*θ]], {θ, π/4, (5 π)/4}, {r, 1, 2}, MeshFunctions -> {#3 &, #4 &}, Mesh -> {{{π/4, {Thick, Blue, Dashed}}, {(5 π)/ 4, {Thick, Blue}}}, {{1, {Thick, Red}}, {2, {Thick, Red, Dashed}}}}, BoundaryStyle -> None, PlotStyle -> Yellow]

0

You can also verify the result without draw the pictures. f[x_] = x^3; Do[tan[x_, p_] = f[p] + f'[p] (x - p); x0 = RandomReal[{-4, 0}] // Rationalize; x2 = x /. Solve[f[x0] + f'[x0] (x - x0) == f[x] && x != x0, x] // First; x1 = p /. Solve[tan[x0, p] == f[x0] && p != x0, p] // First; reg1 = ImplicitRegion[{x0 <= x <= x1, tan[x, ...

1

For 0<c<1/8 the integral is (numerically verified) (-((8192*c^3*(-1 + Sqrt[1 - 8*c] + 4*c*(-5 - 2*Sqrt[1 - 8*c] + 2*c)))/((-1 + Sqrt[1 - 8*c] - 4*c)^5*Sqrt[1 + Sqrt[1 - 8*c] + 4*c]*Sqrt[1 + Sqrt[1 - 8*c] - 4*c*(1 + c*(3 + Sqrt[1 - 8*c] + 4*c))])))* EllipticE[(2*Sqrt[1 - 8*c])/(1 + Sqrt[1 - 8*c] - 4*c*(1 + 2*c))] + (((4*(1 - Sqrt[1 - 8*c] - 4*c*(1 + 2*c)...

5

The tricky part is the sum: Sum[Log[ϵ + (1-ϵ)*((n*mPayload*(1-ϵ) + (n-k)*mp)/(n*mPayload*(1-ϵ) + (n-k+1)*mp))], {k, 1, n}] (* Log[Gamma[-(((mp - mPayload n) (-1 + ϵ))/mp)]/Gamma[1 - n + (mPayload n (-1 + ϵ))/mp - ϵ]] - Log[Gamma[(mPayload n (-1 + ϵ))/mp]/Gamma[-((n (mp + mPayload - mPayload ϵ))/mp)]] *) This difference of two logarithms can be ...

0

I found it: It is the same case as with the integration of compiled functions. One has to use a wrapper: foo[x_?NumericQ] := p[x]; The function foo can now be differentiated using ND.

1

You can integrate by parts twice and get an expression for the antiderivative: parts[u_, v_, {x_, n_}] := Sum[(-1)^m D[u, {x, m}] Nest[Integrate[#, x] &, v, m + 1], {m, 0, n - 1}] + (-1)^n Integrate[ D[u, {x, n}] Nest[Integrate[#, x] &, v, n], x]; parts[x^2, y1''[x], {x, 2}] int1[x_] = % /. First[sol]; int1 - int1[0.001] (* 2 ...

Top 50 recent answers are included