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0

Edit Because of the question's title, I assumed initially that you're interested in the general case of time-dependent driving. This complication is briefly addressed in the first part of my answer. However, the actual system of differential equations in the question has purely time-independent coefficients and is therefore easier to solve in principle. ...


1

purify[f_, x_] := Function @@ {f /. x -> #} fun = 30*x^2 (1 - x)^2; inv = InverseFunction[purify[fun, x]][x] // Quiet LogPlot[{fun, inv}, {x, 0, 1}, PlotTheme -> "Detailed"]


8

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.


6

Reduce appears to provide useful information: Reduce[{(30*x^2 (1 - x)^2) == y, 0 < x < 1}, x, Reals] (0 < y < 15/8 && (x == Root[-y + 30 #1^2 - 60 #1^3 + 30 #1^4 &, 2] || x == Root[-y + 30 #1^2 - 60 #1^3 + 30 #1^4 &, 3])) || (y == 15/8 && x == 1/2) ToRadicals can be used to put this into a more familiar form. ...


4

Your second approach is nearly correct. Modify it like so. f = 30*#^2 (1 - #)^2 &; g = InverseFunction[f] 1/30 (15 - Sqrt[15] Sqrt[15 - 2 Sqrt[30] Sqrt[#1]]) & Plot[f[g[x]], {x, 0, 1}]


1

ClearAll[f, r, obj, x]; f[r_, x_] := Exp[-(x + r*Cos[x])^2] obj[r_?NumericQ] := NIntegrate[f[r, x], {x, -1, 1}] NMaximize[obj[r], r] (* {1.49365,{r->-1.25605*10^-8}} *)


1

Here is a naive start. It will probably work on most calculus course material, but Solve is not guaranteed to invert every possible substitution. (For instance, it does not check the domain of integration in substitution of trigonometric functions, so it should not be hard to come up with an example where it does not work.) I have always wished there were ...


1

Mathematica V10 Integrate[1/4*(1 - t^2)/(1 - t^2 + t^2/a^2), {t, -1, 1}, Assumptions -> a > 1]


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In Mathematica v9: mmaF[a_] := Evaluate@Integrate[1/4*(1 - t^2)/(1 - t^2 + t^2/a^2), {t, -1, 1}, PrincipalValue -> True] yourF[a_] := a/(2 (a^2 - 1)^1.5)*(a*(a^2 - 1)^0.5 - ArcCosh[a]) Quiet @ Chop@Table[mmaF[x] - yourF[x], {x, 1, 3, .1}] (* {Indeterminate, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} *)


4

UPDATE: version 10.0.1 on a Mac produces the same behavior. An extended comment and observation. $Version "10.0 for Mac OS X x86 (64-bit) (June 29, 2014)" The double integral evaluates only with integration over c as the outer integral int1 = Integrate[1/(E^((a - c)^2 + (b - c)^2)), {c, 0, 1}, {b, -Infinity, Infinity}] Investigating the ...


1

For example, ReleaseHold[x^2 HoldForm[D[x,u]]/.x->ArcCos[u]] gives $$-\frac{ArcCos[u]^2}{\sqrt{1-u^2}}.$$


1

One way is to use the Mathematica substitution operator: $f(x)/.\{x\rightarrow u\}$ returns $f(u)$. Suppose you want to substitute $\cos x = u$, you can write out your integrand on one line then put $/.\{x\rightarrow \textrm{ArcCos}[u]\}$ after it. Bear in mind you will also need to fix the differential factor: $\sin x dx \rightarrow du$


4

try FindMaximum. FindMaximum[{(h*10)/(300*(100 - (l^.5 + d^.4 + H^.6))), (l + d + H + h) == 669, l > 0, d > 0, H > 0, h > 0}, {h, l, d, H}] (*{0.23866, {h -> 612.159, l -> 12.8158, d -> 5.77578, H -> 38.2499}}*)


10

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 + ...


9

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 ...


1

It might be interesting for you to compare @Jens' answer with FunctionDomain (new in V10): compare[fun_] := { fun[x], FunctionDomain[fun[x], x, Reals], FunctionDomain[fun[x], x, Complexes], analyticityCondition[fun, x], singularCondition[fun, x]} TableForm[ compare /@ {Sin, Tan, f, g, h}, TableHeadings -> {None, {"Function", "RealDomain", ...


5

I don't think there's a general function built in that can deal with all possible cases. But Reduce is quite powerful. Here is a function that seems to work for the last two examples given: singularCondition[func_, variable_] := Reduce[1/func[variable] == 0 || 1/func'[variable] == 0, variable, Reals] singularCondition[h, x] (* ==> x == -b *) ...


2

First I assume that the OP wants to compute the Cauchy principal value of the integral, since the integrand has a simple pole in the interior of the interval of integration. Dealing with such singularities is described in the tutorial NIntegrate Integration Strategies. I let the derivative be computed in the process of defining f. The symbols a and b are ...


1

I would think (having become cautious) that the integral is divergent since there is no prescription of how to circumvent the pole in the integrand at x = x0 = 1/Sqrt[1-b^2] which is > 1 for 0 < b < 1. As it is the integrand becomes 1/(x-x0) times a finite factor so that the integral is logarithmically divergent. Regards, Wolfgang


2

I can't find a way to integrate it symbolically, but here is a way to approximate it: data = Flatten[Table[{a, b, Im@NIntegrate[ Exp[-a x]/(Sqrt[x^2 - 1] (1 + x Sqrt[1 - b^2]) (x - (1/Sqrt[1 - b^2]))), {x, 1, Infinity}]}, {a, 1, 20, .5}, {b, 2, 20, .5}], 1]; model = k2 ...


1

One can also use FindInstance or with V10 NumberLinePlot: c = 1.1111; y[x_] = x - c Sin[x]; sol = FindInstance[y[x] == 0 && -10 < x < 10, x, Reals, 15] // Values // Flatten; {-0.786647, 0, 0.786647} p = Point @ Transpose[{sol, Table[0, {Length @ sol}]}]; Plot[y[x], {x, -1, 1}, Epilog -> {PointSize[0.02], Red, p}] ...


4

NSolve and Reduce can solve this equation by restricting search. As can be seen by inspection: x=0 is a solution and there are two other solutions symmetric about the origin (a plot reveals and noting if r is a root then so is -r: $c \sin(-r)- (-r)= -(c\sin(r)-r)=0$. Quiet@Reduce[y[x] == 0 && Abs@x < 1, x] yields: x == -0.786647 || x == 0 || ...


2

As it can be seen in Mathematica's help, NSolve deals primarily with linear and polynomial equations. A more general function dealing with numerical methods is FindRoot (starting points are based on the plot you've attached to your post): FindRoot[y[x], {x, 1}] (*{x -> 0.7866465}*) FindRoot[y[x], {x,-1}] (*{x -> -0.7866465}*) For x=0 ...


1

First: There might be a typo in your expression as b2 and c2 do not appear. This is the solution for the expression like it is written in your question: When you have more information on parameters, you can pass it to the Integrate function in the Assumptions option: Integrate[..., Assumptions -> b1 > 0 && b2 > 0 && c1^2 >= 0 ...


3

One can decrease the difficulty of the problem by reducing the Dyson series to a matrix ODE. Let's start from the definition $$ U(x,x_0) = 1 + \int_{x_0}^{x}{dy_1V(y_1)}+\int_{x_0}^x{dy_1\int_{x_0}^{y_1}{dy_2V(y_1)V(y_2)}}+\ldots $$ and take the derivative with respect to $x$ $$ \frac{\partial}{\partial x}U(x,x_0) = ...


4

A more geometrical approach based on CP3D surface-surface intersections boundary style... inter = ContourPlot3D[{h == 0, g == 0}, {x, -4, 4}, {y, -4, 4}, {t, -4, 4}, Mesh -> None, ContourStyle -> Directive[Orange, Opacity[0.3], Specularity[White, 30]], BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Blue}] reg = ...


0

Maybe trivial, but just for the sake of completeness: the inequality holds not only for x>=1 but already for x>=x0, where x0 = x /. FindRoot[x Sin[\[Pi]/x] == \[Pi] Cos[\[Pi]/x], {x, 0.7}, WorkingPrecision -> 100] (* Out[122]= 0.6991556596428411966369319152386816172807303061005296218612315287681429432614\ 994992474339172310490725 *) Remark: ...


3

Why not define the expressions in such a way that their dependence on the independent variables is explicit: ClearAll[f, g, x, y] g[x_, y_] = x f[y] + y (* ==> y + x f[y] *) D[g[x, y], y, x] (* ==> Derivative[1][f][y] *) D[g[x, y], x, y] (* ==> Derivative[1][f][y] *) The mixed derivatives now agree.


4

Rough approximation via plotting V9: One can estimate the length of the polygonal path in the ContourPlot from the graphics. It's a little easier to process if I adapt Daniel Lichtblau's (undocumented?) use of BoundaryStyle in his answer to Plotting implicitly-defined space curves. If the points in the plot lie exactly on the intersection, the length ...


4

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


16

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 ...


9

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 - ...


20

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 ...


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. ...


5

With V10 one can see that the special case p = -1 is explicitly excluded: FunctionDomain[Integrate[x^p, x], p, Reals] Reason: The general formula Integrate[x^p, x] would result in a division by zero error with p = -1: Limit[Integrate[x^p, x], p -> -1] Integrating over a certain interval one gets the expected results: Integrate[x^#, x] ...


2

The reason is because Mathematica does not know anything about f[x]. What if f[x] was 1? then the integrand will be zero. One way is to make your own rule for this case. ClearAll[x, y, h, f]; rul0[x_] := (x /. Integrate[1 - f[y_], {y_, a_, b_}] :> ((b - a) - Integrate[f[y], {y, a, b}])) Simplify[Integrate[1 - f[x], {x, 0, h}], TransformationFunctions ...


0

if you are looking at the final result then this could help. int = Integrate[ 1/Sqrt[(y1)^2 + (y2)^2 + (y3)^2]*1/(1 + (y3)^2)* Exp[-2*((y1)^2 + (y2)^2)/(1 + (y3)^2)], {y1, -Infinity, Infinity}, {y2, -Infinity, Infinity}]; NIntegrate[int, {y3, -10, 10}] (*10.2436*)


4

There exist few typos in your kernel definition. This is how your kernel looks assuming a=2 (we denote it as A while defining the kernel as Kpart in the following). Please utilize the code from here to solve your problem. Below I changed the constants and functional arguments of FredholmKind2 to fit your particular problem. n = 20;(*number of ...


3

Your equation has a sign error for the inverse Fourier transform. A simpler prescription is I InverseFourierTransform[FourierTransform[f[t], t, w]/w, w, x] Although there is a lrge range of function for which this works, it will of course fail for all those functions for which the Fourier transform can't be calculated. Anyway, it does work for ...


1

As I said in a comment I object to the downvaluing of my solution by somebody anonymus without giving the reasong for it. Now you can convince yourself here that my solution of the real problem is correct. And, consequently, that the second integral of "identy" (and sequel) is not. Here we go: I gather from your first lines that you wanted to calculate ...


5

In Version 10 you can now create ImplicitRegions and Integrate over them: region = ImplicitRegion[And @@ {0 <= x, x <= 1 - y, -1 <= y, x + 2 y <= 2}, {x, y}] Now Integrate[x + y, {x, y} ∈ region] 2/3


0

I think what you mean in your second integral is the integral between y^2/2 and \[Infinity]: Integrate[(E^-t ArcCos[y/(Sqrt[2] Sqrt[t])])/\[Pi], {t, 1/2 y^2, \[Infinity]}] It is returned unevaluated by Mathematica: (* $\int_{\frac{y^2}{2}}^{\infty } \frac{e^{-t} \text{ArcCos}\left[\frac{y}{\sqrt{2} \sqrt{t}}\right]}{\pi } \, dt$*) Therefore, let's ...



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