# Tag Info

23

The plan is first get the "external" contour and then use Green's theorem to find its area. r[t_] := {-9 Sin[2 t] - 5 Sin[3 t], 9 Cos[2 t] - 5 Cos[3 t], 0} (*find the intersections*) tr = Quiet@ToRules@Reduce[{r@t1 == r@t2, 0 < t1 < t2 < 2 Pi}, {t1, t2}]; pt = {t1, t2} /. {tr} // Flatten; pts = SortBy[pt, N@# &]; pps = Partition[pts, 2]; Now ...

19

Summary: Setting GenerateConditions -> False turns off safety checks. In my opinion, when the user does that and the result is erroneous, I would not call that a bug. Now WRI could decide to improve Mathematica in this case, but it might not be such a simple matter. On the other hand, it is entirely up to the user to decide whether or not he or she is ...

17

SphericalPlot3D is having problems where the radius goes to infinity. You can use RegionFunction to restrict the plotting region to a range where the function is well-behaved: SphericalPlot3D[ Sqrt[1/(Sin[θ]^2 Cos[2 ϕ] - Cos[θ]^2)], {θ, 0, π}, {ϕ, 0, 2 π}, MaxRecursion -> 4, PlotRange -> {-3, 3}, RegionFunction -> Function[{x, y, z, θ, ϕ, ...

16

$Version (* "8.0 for Microsoft Windows (64-bit) (October 7, 2011)" *) Direct attack fails: Timing[Convolve[Sinc[x], Exp[-x^2], x, y]] (* Out[218]= {59.296, Convolve[Sinc[x], E^-x^2, x, y]} *) or, equivalently, Timing[Integrate[Sinc[x] Exp[-(x - y)^2], {x, -∞, ∞}] ]$\left\{49.92,\int_{-\infty }^{\infty } e^{-(x-y)^2} \text{Sinc}[x] \, dx\right\}$... 16 The "canonical" way in Mathematica is f[x_, y_] := x^2 + y^2 g[x_, y_] := x^4 + 4 x y + 2 y^4 - 8 Maximize[{f[x, y], g[x, y] == 0}, {x, y}] If you want to make explicit usage of the Lagrange multiplier: ss = N@Solve[Grad[f[x, y] + λ g[x, y], {x, y}] == 0 && g[x, y] == 0, {x, y, λ}, Reals] gives the {x, y} coordinates of the maxs and mins. ... 14 Introduction as of 31 January 2015 This is a very interesting problem which on trying to solve it with Mathematica requires skilled handling in an interactive way ("man-machine-interaction"). What do you do when Mathematica refuses to solve an integral? How to help solving integrals which Mathematica declares (erroneously) as divergent. I have done an ... 14 Update a working approach using Fourier transforms The following method is based on the idea that the Laplace operator is just a multiplication with$-k^2$in Fourier space. Therefore, I first find the Fourier transform of the$1/r$potential and then do an inverse Fourier transform of that result, multiplied by$-k^2$(which I call laplacianFT in the ... 13 As you notniced, the result is not correct. This is not uncommon with definite integrals, as the system needs to detect any special behaviour inbetween the integration bounds, which is hard. For this reason it's a good idea to verify such integrals numerically. We can get the correct result like this (please evaluate in a fresh kernel without ... 13 [This is not a full response, but too much detail for a comment.] The general rule is that any integral that can behave differently on a measure zero set in the space of real values of parameters is a candidate for giving a result that will not be what you want. There are other caveats as well, for example in dealing with multiple integrals. And sometimes ... 13 Yes, this is a bug in the more general function RegionMeasure. I knew there were some edge cases in the handling of inexact numberics, but I was unaware of such a simple example. I will forward this bug internally. Workarounds include using the parametric (2-argument) form of ArcLength, and using DiscretizeRegion to preprocess regions before sending them ... 13 There is a specific function called CountRoots, which does exactly what you want: CountRoots[poly, x] (* 12 *) You can then compare this number to the number of roots given by the length of the coefficient list: Exponent[poly, x] (* 36 *) 13 Just wanted to update everyone that things are much simpler, - there is built in support for this: MeshFunctions -> {"ArcLength"} So for our case: Show[{ ParametricPlot3D[KnotData[{3, 1}, "SpaceCurve"][t], {t, 0, 2 Pi}, (* the trick *) Mesh -> 15, MeshFunctions -> {"ArcLength"}, (* styles *) MeshStyle -> Directive[Red, ... 13 Update: a way of joining (taking the union of) the first rectangle and first circle, so that the edge form only marks the exterior of the joined form p = 100; curve = Line@ Table[{.53, 0} + {.07, Sqrt[0.5]} {Sin[-2 Pi k /p], Cos[-2 Pi k /p]}, {k, -1 + p/2}]; poly = Polygon[Join[t1 = Table[{.47, 0} + {.07, Sqrt[0.5]} {Sin[-2 Pi k /p], ... 12$Version (* Out[228]= "8.0 for Microsoft Windows (64-bit) (October 7, 2011)" *) The result provided by Mathematica is correct: Integrate[a/(a^2 + Sin[t]^2), {t, 0, 2 π}] (* Out[213]= (2 π)/(Sqrt[1 + 1/a^2] a) *) Now the same procedure as "always" which "explains" the zero result. The indefinite integral is Integrate[a/(a^2 + Sin[t]^2), t] (* ...

11

$$\mathbb{P} \big(\sum_{i=1}^{m} (A_i + S_i) \le L < \sum_{i=1}^{m+1} (A_i + S_i) \big) = \frac{\mu ^m (2 \lambda +\mu )}{2^{m+1} (\lambda +\mu )^{m+1}}$$ Observing: d1 = TransformedDistribution[ a + s, {a \[Distributed] ExponentialDistribution[λ], s \[Distributed] ExponentialDistribution[μ]}] (* ...

11

Chances are it will behave similarly to Integrate[Exp[-x t^3 ], {t, 0, π/2}]. I show a numerical verification below. We'll use the dominant series term, which we can compute, for the above integral, and compare to numerical integrations of the one we cannot compute analytically. ee1[x_] := Exp[-x*t^3] ee2[x_] := Exp[-x*t^3*Cos[t]]; Let's get the candidate ...

11

I've got a theory.... Let's try to get the antiderivative: Integrate[(1 + 16*Tan[2*x - y]^2)/(1 + 4*Tan[2*x - y]^2), x, Assumptions -> y \[Element] Reals] (*returns 5 x - (5 y)/2 - ArcTan[2 Tan[2 x - y]] *) Seems legit. You can test with D[5 x - (5 y)/2 - ArcTan[2 Tan[2 x - y]], x] and plot or rearrange. This antiderivative is technically correct, ...

11

Just purely for fun: to illustrate approximation of volume integral. Manipulate[ Module[{dp, dpl, dpu, r = Range[0, 1, 1/n], f, cc, lb, la, ub, ua, ll, ul}, dp = Range[0, 1, 1/#] & /@ Range[2, 100]; dpl = Total[Pi Most[#]/(Length@# - 1)] & /@ dp; dpu = Total[Pi Rest[#]/(Length@# - 1)] & /@ dp; f = Sqrt /@ r; cc = Partition[{#, 0, ...

11

One quick way for rational functions is to leverage built-in control system functions: TransferFunctionPoles[TransferFunctionModel[1/(1 + s^2), s]][[1, 1]] {-I, I}

10

I have no idea why Mathematica complains about the convergence. When you expand the integrand, then integrate each of the terms and add, you will find the exact result: E^(-((201 x1^2)/101)) x1^2 (15 - 20 x1^2 + 4 x1^4) (5913508078417951503 + 1124782662003060300000 x1^2 - 2983951574394000000000 x1^4 + 2591462040000000000000 x1^6 - 797900000000000000000 ...

10

Here is a visualization using MeshFunctions (as mentioned by Daniel in the comment): f = x y z; g = x^2 + 10 y^2 + z^2; gp = With[{r = 3}, RegionPlot3D[g < 5, {x, -r, r}, {y, -r, r}, {z, -r, r}, PlotStyle -> Orange, PlotPoints -> 40, Mesh -> None, ViewPoint -> Front, PlotTheme -> "Classic"] ]; With[{r = 3}, Manipulate[ ...

10

It has to do with the default behavior of GenerateConditions in multivariate integrals. Setting it explicitly to True will help in this case. Some explanation may be found here or here. The gist is that, for multiple integration, automatically checking conditions and issuing provisos for all but the final integration is typically both too costly (in speed) ...

10

For what it's worth, you can monitor the progress of Integrate and NIntegrate. I'm not sure how helpful the tools below are, but I feel it is probably worth mentioning them. Integrate InternalIntegratedebugSwitch If you set InternalIntegratedebugSwitch to the magic number 10, it will print its (major) steps in searching for the answer. For instance: ...

10

This should be a comment, but it's too long... This isn't really a Mathematica solution, but here's some insight into the integral. I assume all parameters are positive. Call your integral $I$ and let $s = b^2/(2c)$. Substituting $t = \sqrt{x}$ transforms $I$ into the Laplace transform  I = \frac{1}{2} \mathcal{L}_t\left( J_0(n \sqrt{t} \,) \theta(a^2 - ...

10

However, is there something you can put in the code to set the speed? Animate takes an option AnimationRate I still haven't been able to draw the bottom part of the first animation on https://courses.engr.illinois.edu/tam212/avt.xhtml#avt and note how it continues to move to the right (without everything jumping around). I think you can get as ...

10

The problem here is the Sin[n] which has no limit since it is an oscillating function, but it is always bounded by $\pm 1$: if you change you code with the following: Limit[(n - Sqrt[1 + 10 n + n^2])^2, n -> Infinity] with 1 in place of Sin (or -1 if you want), you get the result: (*25*)

10

On a different occasion (Dirichlet coefficients as limits: wrong) I have shown that the sometimes limited capabilities of the function Limit[] can be improved by using an intermediate Series[]. Following this idea we can write for the limit in question Limit[Expand[ Normal[Series[(n - n Sqrt[1 + x])^2, {x, 0, 2}]] /. x -> 10/ n + Sin[n]/n^2], n ...

9

Besides trivial observations that one cannot evaluate numerically integrals involving symbolic constants there are more interesting aspects of the problem at hand. First one should realize that a standard numeric approach is not appropriate for this kind of problems, since the integrand involves singular points (zero in the denominator) thus it is not ...

9

Here's how to make Mathematica integrate this, but a lot is done by hand. Cos[β] Exp[I z Cos[β - α]] == Cos[β] Cos[z Cos[α - β]] + I Cos[β] Sin[z Cos[α - β]] The integral of the first summand is zero. Make the substitution γ == α - β and expand the Cos[β] to Cos[α] Cos[γ] + Sin[α] Sin[γ]. Also note that the function is Pi-periodic. The integral ...

9

It seems, that findExponentialGeneratingFunction would be equivalent to FindGeneratingFunction of the sequence {a1/0!,a2/1!,a3/2!,...}. Therefore findExponentialGeneratingFunction = FindGeneratingFunction[#1/(Factorial /@ Range[0, Length@#1 - 1]), #2] & findExponentialGeneratingFunction[{1,1,1,1,1,1,1,1},x] (* Exp[x] *)

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