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25

Fixed (see below) Here's an approach: r1 = Exp[-x^3 - y] - 1 == z; r2 = y == z; We create ImplicitRegions: reg1 = ImplicitRegion[r1, {x, y, z}]; reg2 = ImplicitRegion[r2, {x, y, z}]; The intersection of these regions is the line you seek: reg = RegionIntersection[reg1, reg2]; And here is the length (note the inclusion of the range of values in ...


18

Maximize[{y, x^2 + y^2 == (2 x^2 + 2 y^2 - x)^2}, {x, y}] {(3 Sqrt[3])/8, {x -> 3/8, y -> (3 Sqrt[3])/8}}


17

Since another answer provides misleading results I'm motivated to explain what's wrong there and suggest a correct solution. One can guess that the error there has been produced by DiscretizeRegion or by an improper parametrization of the curve with RegionIntersection (more likely the former reason). Let's start from the begining defining the following ...


16

After a lengthy study (I'm using version 8) I conclude that there is a bug in Mathematica in the Integrate function when applied to a Sqrt integrand. Ok. let's go (some patience is required because of the long text) Let us define the functions corresponding to your integrals. Remark: because of the relation $1 + cos(2x) = 2 cos^2(x)$ the two forms of ...


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, -∞, ∞}] ] $\left\{49.92,\int_{-\infty }^{\infty } e^{-(x-y)^2} \text{Sinc}[x] \, dx\right\}$ ...


15

Having experienced similar problematic issues with Mathematica I instantly thought that expanding the fraction in the integrand i.e. applying Appart could resolve the problem, and indeed it does: Integrate[ Apart[(1 - x)(1 + 2x)^6/Sqrt[1 - x^2]], {x, -1, 1}]/Pi 15 These arguments apply to this case as well Bug in mathematica analytic integration? i.e. ...


13

This is not an answer (yet). Rather it explores the question in more depth. n = 8; parameters = ConstantArray[{0, 1}, n]; variables = Symbol /@ CharacterRange["a", FromCharacterCode[ToCharacterCode["a"] + n - 1]]; The following takes a long time to evaluate, but the results it produces reveal give us a better view of the problem with Probability. ...


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

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 ...


12

In this second answer I give the cause for the mismatch in the integrals, show how to remove it, and make a suggestion to improve the function Integrate[]. Simplified restatement of the problem In order to focus on the core of the problem we consider the simpler integral $\int_0^1 \sqrt{\cos (2 π k r)+1} \, dr$. It has the square root and the cosine ...


12

[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 ...


12

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 ...


11

From Weierstrass Approximation Theorem we know there is such a polynomial, moreover there are infinitely many polynomials satisfying given criterion. Therefore we would like to find those ones of the minimal order. Since we are supposed to exploit series approximations we define a polynomial of n - th order approximating 6 ArcTan[x] for x such that ...


11

You could use Lagrange multipliers to maximize $f(x,y)=y$ subject to the constraint that $$g(x,y) = x^2 + y^2 - (2 x^2 + 2 y^2 - x)^2 = 0.$$ f[x_, y_] = y; g[x_, y_] = x^2 + y^2 - (2 x^2 + 2 y^2 - x)^2; eqs = {D[f[x, y], x] == lambda*D[g[x, y], x], D[f[x, y], y] == lambda*D[g[x, y], y], g[x, y] == 0}; Solve[eqs, {x, y, lambda}] // InputForm (* Out: { ...


11

In Version 9 currently, we can do (using the Undocumented form of Integrate): Integrate[Boole[z >= 0] z, {x, y, z} ∈ Sphere[{0, 0, 0}, 4]] 64 Pi Note: This is undocumented behaviour and functionality may change or behave differently in newer versions of Mathematica so use with caution e.g. as noted by Szabolcs in the comment, Sphere in V10 ...


11

I suspect you are correct in your assessment. Since there are approximate numbers in the input, Maximize punts to NMaximize, which uses penalty methods to enforce some constraints (not sure why it needs them here for linear constraints; I need to check into that). You can get better behavior by forcing real values. NMaximize[{Re[(h*10)/(300*(100 - (l^.5 + ...


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

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: Length@CoefficientList[poly, x] - 1 (* 36 *)


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, ...


10

Taylor polynomials of order n aren't necessarily the nth degree polynomials that optimally approximate a function on a given interval. We can use linear least squares to find the optimal polynomials for a fixed n. inn[f_, g_, x_] := Integrate[f g, {x, -1/Sqrt[3], 1/Sqrt[3]}] LeastSquarePolynomial[f_, x_, n_] := With[{pows = x^Range[0, n]}, pows . ...


10

Here is the derivation promised earlier. I have chosen to create a new answer in order not to mix things up and because it shows some handling which I like to call "man-machine" interaction with Mathematica. And, sorry for the "Greeks", but I find it very cumbersome to reedit them :-( $Version "8.0 for Microsoft Windows (64-bit) (October 7, 2011)" ...


10

The limit is definitely computed correctly. Keep in mind that Limit assumes the variable (n, in this case) is continuous. Thus, this is a specific example of the general fact that $f(x)\sin(g(x))$ oscillates back and forth over the whole real line, whenever $f(x)$ and $g(x)$ both increase to $\infty$ with $x$. A plot verifies this is correct. ...


10

Thanks to Artes for helping me sort out bugs in this answer. Solve[E^(-x^3 - y) - 1 - y == 0, y] {{y -> -1 + ProductLog[E^(1 - x^3)]}} To account for that fact that $-4 < y = z < 4$ we need to find the range of allowed $x$ values. FindRoot[-1 + ProductLog[E^(1 - x^3)] == 4, {x, 1}] {x -> -1.77681} FindRoot[-1 + ProductLog[E^(1 - ...


10

Brute force approach: g[y0_] := x /. FindRoot[ Exp[-(x^3 + y)] - 1 == y /. y -> y0 , {x, -4, 4}] line = Table[{g[y], y, y}, {y, -1, 4, .0001}]; Graphics3D[Line@line] Total[Norm@(Subtract @@ #) & /@ Partition[line, 2, 1]] 10.9513


10

Using a symbolic functionality like Maximize with an expression involving approximate numbers is not in general a good idea, even though Maximize calls automatically NMaximize in such cases. However if we rewrite the expression to an exact form, then Maximize will run very long time returning no symbolic results. The problem one encounters here is most ...


10

Solve also works Solve[y == 30 x^2 (1 - x)^2 && 0 < x < 1, x, Reals] {{x -> ConditionalExpression[Root[-y + 30 #1^2 - 60 #1^3 + 30 #1^4 &, 2], 0 < y < 15/8]}, {x -> ConditionalExpression[Root[-y + 30 #1^2 - 60 #1^3 + 30 #1^4 &, 3], 0 < y < 15/8]}} Use ToRadicals to get it in a nice looking form.


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

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 ...


9

Let's define the function f suitably: f[x_, a_, b_] := Log[x, 1 + (x^a - 1)*(x^b - 1)/(x - 1)] We would have defined f with appropriate conditions (I recommend to examine this post: Placement of Condition /; expressions), e.g. f[x_, a_, b_] /; 0 < a < 1 && 0 < b < 1 && x > 0 := ... however since we are to deal with ...


9

In Mathematica 10, this computation may be made as follows: Clear @ r volSphere9[r_] = RegionMeasure[Ball[ConstantArray[0, 10], r]] (π^5 r^10)/120 volSphere9[1000.] 2.55016*10^30



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