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9

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[ ...


8

Since Reduce doesn't seem to like the inequality, I tried FullSimplify with Assumptions instead. This works in three steps: differenceByTerm = SeriesCoefficient[(1 + x)^n - (1 + n x), {x, 0, m}] $$ \cases{ 0 & m=0 \\ \binom{n}{m} & m>1 \\ 0 & \text{True} \\ }$$ FullSimplify[ differenceByTerm >= 0, Assumptions -> n ...


6

The antiderivative is correct, in the sense its derivative gives back the original integrand Clear[x] integrand = 4279/Sqrt[6817/10000 + 3183/10000*(1 + x)^3]; mmaResult = Integrate[integrand, x]; integrandBack = Simplify[D[mmaResult, x]]; Plot[{integrandBack, integrand}, {x, 0, 20}, PlotTheme -> "Detailed"] But the definite integral does not give ...


5

You can gain insight by evaluating and then simplifying the indefinite integrals. Integrate[(1 + b x + c y)/(1 + e x + f y + I η), y, x, Assumptions -> {x ∈ Reals, y ∈ Reals, b ∈ Reals, c ∈ Reals, e ∈ Reals,f ∈ Reals, η ∈ Reals, η != 0}]; ans = Collect[%, {ArcTan[η/(1 + e x + f y)], Log[1 + e^2 x^2 + 2 f y + f^2 y^2 + 2 e (x + f x y) + η^2]}, ...


5

One way to approach this is to look for the source of the problem. In this case, the outer integral (on y) is irrelevant to the problem, because even the simpler integral int = Integrate[(1 + b x)/(1 + e x + I η), {x, -1/2, 1/2}, Assumptions -> {b ∈ Reals, c ∈ Reals, e ∈ Reals, η ∈ Reals, η != 0}] gives a conditional expression. But now the answer ...


4

Here is another way: multi[f_, x_, a_, b_, n_] := Inactive@Integrate @@ {f @@ Array[x[#] &, {n}]}~Join~Array[{x[#], a[#], b[#]} &, {n}] I used Inactive because I don't know if the purpose is purely typesetting, or if you want it to try to evaluate (if so remove it, or use Activate later). multi[g, y, ymin, ymax, 10] (* Inactive[Integrate][ ...


4

With using number of assumptions, and breaking thing step by step: (I do things step by step, just to see where the problem is when it shows up, much easier to debug this way) integrand = 1/(r g) Exp[-p (1 + r a) - b q ((1 + r a)/(1 + g x))] Exp[-a/r] Exp[-b] Exp[-x/g]; z0 = Assuming[Re[(1 + q + a q r + g x)/(1 + g x)] > 0, ...


4

Let us define the equation: Clear[eq]; eq[m_, f_] := 1/(x - 1) - (m + 1)/(x^(m + 1) - 1) == f; where x stays for Exp[f/(k t)]. If one applies the function Solve to it, Mma clearly answers that it cannot provide exact solution of this equation. It should not be expected, therefore, that one can find any other analytical solution. Numerically, it is ...


3

Because NDSolve cannot accommodate the x=0 boundary condition, it is necessary to perform this computation by discretizing the PDE in x. The resulting do-it-yourself procedure is discussed in Introduction to Method of Lines. For illustrative purposes, assume that x is divided into five equal segments. n = 5; h = 1/n; with a variable defined at each node, ...


2

You can split up the interval of integration: Integrate[DiracDelta[1 - x] DiracDelta[x] f[x], {x, 0, 1/2, 1}] (* 0 *) Still, it seems a bit inconvenient.


2

Using Sequence and Transpose to splice in the integration parameters: mi[f_, x_, a_, b_]:=Integrate[f @@ x, Sequence @@ Transpose@{x, a, b}]; Usage (exemplary): f[v_,w_,y_]:=v^2+w^2+y^2 x={x1,x2,x3}; a={a1,a2,a3}; b={b1,b2,b3}; mi[f,x,a,b] (* -(1/3)(a1-b1)(a2-b2)(a3-b3)(a1^2+a2^2+a3^2+a1 b1+b1^2+a2 b2+b2^2+a3 b3+b3^2) *)


2

f[x_] := 3 x^4 + 8 x^3 - 24 x^2 - 48x + 19 s = 4; (*slope*) xs = x /. NSolve[f'[x] == s, x](*here are the x values where f'[x]=4*) lines = s (x - xs) + f[xs] (*here are the tangents*) Plot[{lines, f[x]}, {x, Min[xs], Max[xs]}]


2

Evaluating the indefinite integral works better. sol = Integrate[1/(w^4 + 2 (2 v^2 - 1) w^2 w1^2 + w1^4), w1, Assumptions -> {w > 0, 1 > v > 0}] (* (-(ArcTan[w1/(Sqrt[-1 + 2*v^2 + 2*v*Sqrt[-1 + v^2]]*w)]/ Sqrt[-1 + 2*v^2 + 2*v*Sqrt[-1 + v^2]]) - ArcTanh[w1/(Sqrt[1 - 2*v^2 + 2*v*Sqrt[-1 + v^2]]*w)]/ Sqrt[1 - 2*v^2 + 2*v*Sqrt[-1 + ...


2

gradient[g_, vars_] := Table[D[g@@vars, vars[[j]]], {j, 1, Length[vars]}] system1[lstConst_, vars_] := Join[ Join@@ Table[gradient[lstConst[[j]], vars], {j, 1, Length[lstConst]}], Table[lstConst[[j]]@@vars,{j,1,Length[lstConst]}]]; system2[f_, lstConst_, vars_, lambda_] := Join[ gradient[f, vars] - Sum[ lambda[[j]]*gradient[lstConst[[j]], vars], ...


2

Your limit $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}-n-2}{2^{n}}=2$$ may be decomposed into the sum: $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}-n-2}{2^{n}}=\lim _{n\rightarrow \infty }\dfrac {2^{n+1}}{2^{n}}-\dfrac {n}{2^{n}}-\dfrac {2}{2^{n}}=2-0-0=2$$ And each limit: $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}}{2^{n}}=2$$ $$\lim _{n\rightarrow ...


2

Here's an inductive proof: Defining the function f[n_] := (1 + x)^n - (1 + n x) we have to prove that f[n] > 0 for n > 1 and x > -1. Now we observe that for f[n] we have the identity Simplify[f[n + 1] == (1 + x) f[n] + n x^2] (* True *) It is obvious "by eye" that f[n] > 0 because there are only positive quantities involved on the right ...


1

Something like this: f[x_, m_] := 1/(x - 1) - (1 + m)/(x^(1 + m) - 1) g[y_, m_] := x /. NSolve[f[x, m] == y, x]


1

I'm not sure if this will be complete enough for you, but here's a snippet of code that worked for me: vars = Table[x[i], {i, 1, 7}] Do[x[i, 0] = RandomInteger[{0, 5}], {i, 1, 7}] Do[x[i, 1] = RandomInteger[{6, 10}], {i, 1, 7}] Integrate[Plus @@ vars, Sequence @@ Table[{x[i], x[i, 0], x[i, 1]}, {i, 1, 7}]] In this case I create a list of variables x[i] ...


1

If you use: x0 = (p^2 + k^2 + m^2 - (p0 - k0)^2)/(2 p k); f[x_] := (p^2 + k^2 - 2 p k x)/(x - x0); res=Integrate[f[x], {x, -1, 1}, PrincipalValue->True] you will presented with a result within a few minutes' time, however, there are lots of conditions spilled out to be fulfilled, which I left out here for clarity: -4 k p - (k0^2 - m^2 - 2 k0 p0 + ...


1

Indeed, the integral should give zero even with finite bounds. This workaround seems to give the desired result: Limit[ Integrate[ DiracDelta[1 - x] DiracDelta[x] f[x], {x, ϵ, 1}], ϵ -> 0] (* ==> 0 *) But it works only because it effectively cuts off the lower bound and thus the lower delta function.


1

With v10.0.2 it appears to work only for infinite bounds or using NIntegrate $Version "10.0 for Mac OS X x86 (64-bit) (December 4, 2014)" Integrate[DiracDelta[1 - x] DiracDelta[x] f[x], {x, -Infinity, Infinity}] 0 Integrate[DiracDelta[x, 1 - x] f[x], {x, -Infinity, Infinity}] 0 NIntegrate[DiracDelta[1 - x] DiracDelta[x] f[x], {x, 0, ...


1

Your integral is too complicated for closed form. After fixing your syntax error, and integrating w.r.t. x only, you can see the result contains complex value and very complicated trig functions. Mathematica can't do the integration w.r.t. y at this stage f[x_] := 1/(a - c*x^2 + x*b I); g[y_] := 1/(a - c*y^2 + y*b I); h[z_] := 1/(a - c*z^2 + z*b I); j[x_, ...


1

Perhaps you will find a step-by-step solution helps you to understand how to approach a problem of this type. The function which defines the curve. f[x_] := 3 x^4 + 8 x^3 - 24 x^2 - 48 x + 19 The function giving the slope of the curve. slope[x_] = D[f[x], x] -48 - 48 x + 24 x^2 + 12 x^3 Finding the x values for which the slope is 4. xPts = x /. ...


1

It might be Errr... a bug or some singular points(not on the real axis) might be involved and Integrate chose a strange Integrate route. define f[x_] := Integrate[4279/Sqrt[0.6817 + 0.3183*(1 + x)^3], x] then fdat = Transpose@{Range[0, 20, 0.1], f[x] /. x -> Range[0, 20, 0.1]} plot the f generated by Integrate: ListLinePlot[{{#[[1]], ...


1

but why For the why part, you need little bit of math. You can always start by looking at series expansion of all terms expr = (Exp[x] - Exp[Sin[x]])/(x - Sin[x]); Limit[expr, x -> 0] (*1*) Now expand all in taylor (used 4 terms to make it easy to see) expr = (Normal@Series[Exp[x], {x, 0, 4}] - Normal@Series[Exp[Sin[x]], {x, 0, 4}])/(x - ...


1

First of all correcting the sign in the exponential we get f2 = Integrate[ Exp[r^2/(2 M^2)] (b^2 - 3 r^2)^2 (Abs[b^2 + r^2] - (b^2 - r^2))/( r (r^2 - b^2)), {r, 0, I \[Infinity]}, Assumptions -> {M > 0, b > 0] (* Out[278]= 18 b^2 M^2 + 4 b^4 E^(b^2/(2 M^2)) ExpIntegralEi[-(b^2/(2 M^2))] *) Taking the pricipal value has no influence as ...


1

Outline As you didn't provide boundary and initial conditions and the function pa'[t] this solution must be generic. Your equation for p[u,t] is linear (I guess pa'(t) means D[p[u,t],u]/.u->a) and can therefore be solved by standard mathematical methods once you provide the boundary and initial conditions. Physically it describes diffusion in a cylinder. A ...



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