# Tag Info

1

Here is a way to plot the sampling points of the integration process. (If this is what it is asked.) First using the functions Sow and Reap and the option EvaluationMonitor we gather the sampling points: Block[{x = 0.2, t = 0.2, m = 0.2}, res = Reap[ NIntegrate[ f[y] (2. Sin[(x^2 + y^2)/(8. (t - r)) + \[Pi]/ 4] Cos[(x y)/(4. (t - ...

3

You can see that: Fold[Sqrt[#1 + #2] &, 0, Reverse@Range[7]] is the Kasner number - - OEIS link with many references DiscretePlot[Fold[Sqrt[#1 + #2] &, 0, Reverse@Range[i]], {i, 50}]

3

protected = Unprotect[Dt] {"Dt"} Dt[f[___], g[___]] := R1 Dt[f[a, b], g[]] R1 Clear your definition Clear@Dt Dt[f[], g[]] 0 Remember If your definition is no longer needed Clear it - otherwise you might get unwanted results in other areas. And restore protection: Protect[Evaluate[protected]] {"Dt"}

1

f = x^3 + y x z + z^3; grad = Plus @@ Map[D[f, #] &, {x, y, z}] 3 x^2 + x y + x z + y z + 3 z^2 Or grad = Plus @@ Grad[f, {x, y, z}]; Same result vals = Function[{x, y, z}, Evaluate@grad] @@@ {{2, 2, 2}, {3, 1, 7}, {1, 6, 3}, {4, 5, 1}, {0,4, 9}} {36, 205, 57, 80, 279} Mod[vals, 4] {0, 1, 1, 0, 3} Put together: Function[{x, ...

0

Its useful to minimize the number of parameters in your expression as much as possible. Fundamentally you have things that look like this: Integrate[ t Log[a + b t], {t, 0, 1}] With just two parameters this quickly returns a conditional expression: ConditionalExpression[((2 a - b) b + 2 a^2 Log[a] + 2 (-a^2 + b^2) Log[a + b])/( 4 b^2), ...

2

I'm not sure, what exactly goes wrong on your machine, but here it works. Maybe one note, if your integral behaves well, then you could calculate the indefinite integral, which is faster by ages because Mathematica does not check conditions etc. expr = dy ((b x y + c x)/(e y + f) + g x) - dx (a/e x Log[e y + f]) /. {x -> px + t dx, y -> py + t ...

1

The fact that Mma returns the input means that it is not able to fulfill the task. In other words there is no analytical expression for your integral, but you might agree to get it numerically. Try this: lst = Table[{T, NIntegrate[(9.07 - 5.60 Cos[.87 t] - 2.71 I Sin[.87 t] + 2.14 Cos[1.74 t] + 2.70 I Sin[1.74 t])^0.5, {t, 0, T}]}, {T, 0, ...

1

Since the power is not an integer, for Expand[(x+y)^0.2] you can use Binomial series. r = Sum[Binomial[.2, k] x^k y^(.2 - k), {k, 0, Infinity}] (* 1. y^0.2 HypergeometricPFQ[{-0.2}, {}, -((1. x)/y)] *) To verify r /. {x -> 10, y -> 99} (* 2.55556 *) (x + y)^(0.2) /. {x -> 10, y -> 99} (*2.55556*) Here are few terms in the ...

0

Problem was fixed in version 10.

2

The integral probably can be evaluated analytically, but the answer will not be simple. Let $u$ be the integrand function. u = Exp[-((x - \[Mu])^2/(4 s^2)) - (I k (y - x)^2)/(2 R)] ((y - x)/(R^2 + (y - x)^2)) After the substitution v = u /. {y - x -> z, x -> y - z} /. y - \[Mu] -> \[Nu] we obtain $\frac{z}{R^2+z^2} \exp\left[-\frac{(\nu ... 0 Maybe this give you some idea: Plus[Sequence @@ Table[Gamma[k/2]/k!, {k, 1, 1000}]] // N so = Plus[ Sequence @@ Table[Gamma[k - 1/2]/(2 (2 k - 1)!), {k, 1, 1000}]] // N se = Plus[Sequence @@ Table[Gamma[k]/(2 k)!, {k, 1, 1000}]] // N se + so I know this is not an answer but I am under 50 reputation and can not post comment. 3 Just regularize: Sum[Gamma[k/2]/(2 k!), {k, 1, ∞}, Regularization -> "Abel"] 1/4 (2 π Erfi[1/2] + HypergeometricPFQ[{1, 1}, {3/2, 2}, 1/4]) 0 Although this question has been answered by a bug statement, I found it interesting to study some aspects in more detail. The study could not explain the internal cause of the bug. But, besides taking into account that the OP provides two different integrals the study shows that applying a simple unified regularization procedure leads to correct results. ... 5 Just going on the figures I see in the linked page, it seems if you are dealing with two continuous distribution functions, defined for all real numbers, then the overlap is just the minimum of the two at all points. If they are allowed to go negative, then a different definition is needed I think. We'll look at the overlap between two Gaussians ... 0 It's actually rather easy to specify the evenness or oddness of an integer in an assumption: Assuming[Mod[k, 2] == 0, Integrate[Sin[x]^k, {x, 0, 2 π}]] (* (2 Sqrt[π] Gamma[(1 + k)/2])/Gamma[1 + k/2] *) Assuming[Mod[k, 2] == 1, Integrate[Sin[x]^k, {x, 0, 2 π}]] (* 0 *) 1 Assuming[{n/2 ∈ Integers}, Integrate[Sin[x]^n, {x, 0, 2 π}]] // TraditionalForm Assuming[{n >= 0 && n/2 ∈ Integers}, Integrate[Sin[x]^n, {x, 0, 2 π}]] // TraditionalForm Assuming[{n >= 0 && n/2 ∈ Integers}, Integrate[Sin[x]^n, {x, 0, 2 π}] // FullSimplify] // TraditionalForm 0 I tried: Integrate[Sin[x]^(2 n), {x, 0, 2 Pi}, Assumptions -> {n >= 0 && n \[Element] Integers}] //TraditionalForm $$\frac{\sqrt{\pi } \left((-1)^{2 n}+1\right) \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}$$ 1 In the call to Limit, there doesn't seem to be any way to restrict the function to integers. When I try Limit[Ceiling[n] Sin[2 Pi E Factorial[Ceiling[n]]], n -> ∞] as @Sora suggested, it just hangs (and I'm too impatient to wait more than 10 minutes). Since the factorial is only defined for nonnegative integers, the limit is indeed$2\pi$as ... 1 I know the answer is a bit late, but - from a mathematician's point of view, I'm not that much of an expert when it comes to Mathematica - why don't you just replace every occurence of$n$by$\lceil n\rceil$? If$(a_n)_{n\in\mathbb N}$is a convergent real sequence, then$f:\mathbb R_{>0}\to\mathbb R,\ f(x):=a_{\lceil x\rceil}$will be a convergent real ... 5 With$Version (* 10.3.0 for Microsoft Windows (64-bit) (October 9, 2015) *) Mathemeatica returns a limit that depends on r, as it should. Plot[Limit[E^-(r^2)^(2*n)*r*2*Pi, n -> Infinity], {r, -2, 2}, AxesLabel -> {r, Lim}] Addendum Although Limit returns unevaluated for arbitrary r, it can be made to produce useful results by making ...

2

The problem boils down to the fact that Derivative[1][f[##] &] 0 & Which is, in my opinion unexpected. More in: Derivative of a pure function with SlotSequence The fix is to inject a sequence of Slots (#1, #2...) of length equal to the number of arguments our function accepts: series[n__] := With[{ l = Length[{n}], slots = ...

13

I believe there are at least three cases treated separately by Derivative. 1) A function defined by a Symbol. This follows the the rule cited in the documentation. g[x___] := f[x]; Derivative[1][g][x] // Trace (* { { g' , { g[#1] <-- Here the rule is being applied , f[#1] } , f'[#1] & } , (f'[#1] &)[x] , f'[x] } ...

8

Consider Derivative[1][f[##] &] // FullForm Function[0] Function[0][x] 0 So the snippet you quote from the docs might not be entirely accurate; might not be considering such an edge case as ##. I believe that Derivative is looking for head Slot when detects a pure function goes on to rewrite it, so it ignores head SlotSequence -- i.e., ...

1

b = 2*23.5; nu = 0.3; Ee = 10^5; h = 4.5; z = x + y*I; G = Ee/(2*(1 + nu)); Zline = (Ee*2*h)/(2*Pi*(1 - nu^2))*ArcSinh[Sqrt[z/b]]; Z = (Ee*2*h)/(2*Pi*(1 - nu^2))*1/Sqrt[z (z + b)]; u = ComplexExpand[((1 - 2 nu)*Re[Zline] - y*Im[Z])/(2 G), TargetFunctions -> {Re, Im}]; v = ComplexExpand[(2*(1 - nu)*Im[Zline] - y*Re[Z])/(2 G), TargetFunctions -> ...

2

With v[x, t] the dependent variable, and x the spatial independent variable ranging from x1 to x2, the requested boundary conditions are, v[x1, t] == 0 (D[v[x, t], x] /. x -> x2) == 0 In answer to the OP's comment below, these and any other boundary and initial conditions are included in the first argument of NDSolveValue, as illustrated by several of ...

1

Following suggestion above, and dropping 0.0027, as well introducing some additional parameters we get Intf[(phi_)?NumericQ, (a_)?NumericQ, (b_)?NumericQ] := NIntegrate[ Sin[phi*Pi]/(a*Exp[(-b)*w] + Exp[239*w] - Cos[phi*Pi]), {w, 0, 0.1}] Manipulate[ListLinePlot[Table[{t, Intf[t, e, b]}, {t, 0.01, 1, 0.05}]], {e, 0.5, 1.5}, {b, 100, 400}] ...

3

This integral is unknown to Mathematica and indeed it appears that there is no simple solution at all. You can try rooting around the DLMF, and using the relations of $\operatorname{erfi}$ to the incomplete gamma and Kummer $M$ functions to look for similar relations in the standard books of integrals (such as Gradshteyn & Ryzhik, Prudnikov et al., ...

2

I am not quite sure what you want but this may help. First I generate some data data = Table[{x, x + Sin[2 \[Pi] x]}, {x, 0.01, 4.82, 0.1}]; This plots as ListPlot[data, Frame -> True, PlotTheme -> "Scientific"] Now we make an interpolation function and then integrate f = Interpolation[data]; g = Integrate[f[x], x] Interestingly the x from ...

5

This can be done using SetDelayed and either of the following two definitions: myIntegrate[a_, b_, fun_] := Integrate[fun[1 - t], {t, a, b}] myIntegrate2[a_, b_, fun_] := Integrate[fun[t], {t, b - 1, a - 1}] The latter tip was provided in a comment by J.M. You need to make sure that the third argument is a function. Below are some examples: ...

4

The answer is: res[j_, j_, n_] := (3 j - 2)^2 /(9 n^6) /; j <= n res[j_, i_, n_] := (2 j - 1)^2 /(4 n^6) /; j <= i <= n Testing against your function: rr[j1_, i1_, n1_] := Block[{j = j1, i = i1, n = n1}, Integrate[f[x, i]*f[y, i]*f[x, j]*f[y, j], {x, 0, 1}, {y, 0, 1}]] list = Sort /@ RandomInteger[{1, 10^2}, {100, 3}]; ...

0

FullSimplify@Assuming[n > 0, Integrate[ UnitBox[2 (x - y)] UnitBox[n y], {y, -\[Infinity], +\[Infinity]}]] \$\begin{cases} \frac{1}{2} & (n>0\land 4 n x+2>n\land 4 n x+n\leq 2)\lor (n=2\land x=0) \\ \frac{1}{n} & n>2\land 4 n x+n\geq 2\land 4 n x+2\leq n \\ \frac{1}{2}-x & n=2\land 0<x<\frac{1}{2} \\ x+\frac{1}{2} ...

1

When you specify that this is a temporal problem and use FEM a spatial discretization method I get to a solution: eqn2 = \[Rho]*c*D[T[t, z], t] - k*Laplacian[T[t, z], {z}]; GD = DirichletCondition[T[t, z] == 100, z == .00048]; GN = NeumannValue[28/k, z == 0]; soln = NDSolveValue[{eqn2 == GN, GD, T[0, z] == Ti[z]}, T, {t, 0, .01}, {z, 0, .00048}, ...

8

As pointed out by J. M.♦, Simon Woods's approach in #48486 could be used. sharpregplot[ region_, {x_, x0_, x1_}, {y_, y0_, y1_}, {z_, z0_, z1_}, opts : OptionsPattern[] ] := Module[ {reg, preds}, reg = LogicalExpand[region && x0 <= x <= x1 && y0 <= y <= y1 && z0 <= z <= z1]; preds = Union@Cases[reg, ...

6

Here's one approach that uses MeshFunctions to highlight the parts of the bounding surfaces that belong to the region. So many different approaches are possible.... opts = Options[ParametricPlot3D]; SetOptions[ParametricPlot3D, {Mesh -> {{0}, 15, 15}, MeshStyle -> Opacity[0.], (* ignored -- bug? *) MeshShading -> {{{Automatic, ...

15

A simple alternative is to use Plot3D with both RegionFunction and Filling. Plot3D[y, {x, 0, 1}, {y, 0, 1}, RegionFunction -> Function[{x, y, z}, x^2 + y^2 <= 1 && x >= 0 && y >= 0 && z >= 0], Filling -> 0, FillingStyle -> Opacity[.75], PlotStyle -> Opacity[.5], AxesLabel -> (Style[#, 14, Bold] ...

1

As noted by J.M., the series can be summed explicitly. g = Sum[Exp[β n x], {n, 0, Infinity}] (* -(1/(-1 + E^(x β))) *) h = Sum[Exp[β n x] (1 + x)/(n + x), {n, 0, Infinity}] (* HurwitzLerchPhi[E^(x β), 1, x] + x HurwitzLerchPhi[E^(x β), 1, x] *) Repeating these summations with GenerateConditions -> True indicates that E^Re[x β] < 1 is needed for ...

1

I have rewritten your definition of the region as region2[a_?NumericQ]:=ImplicitRegion[2 x^3 (y-z)^2 (y+z)+2 x y (y-z)^2 z (y+z)+x^4 (3 y^2+2 y z+3 z^2)+y^2 z^2 (3 y^2+2 y z+3 z^2)+x^2 (3 y^4-2 y^3 z-18 y^2 z^2-2 y z^3+3 z^4)==3 a^2 (x+y)^2 (x+z)^2 (y+z)^2&&x>0&&y>0&&z>0&&z<=x&&z<=y,{x,y,z}] That ...

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