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0

The following also works. f[m_ /; m >= 4] := Catch[ Do[ a = 2^(m*(1 - c)) - 2.718/((m^.5)*2*3.14*c^(c*m + .5)*(1 - c)^((1 - c)*m + .5)); If[a <= 0, Throw[m/(m (1 - c) + 1)]], {c, 0.006, 1, 0.001}]; 2.] f[m_ /; m < 4] = 2. FindMinimum[f[m], {m, 4, 2500}, AccuracyGoal -> 5, PrecisionGoal -> 5] {1.11235, {m ...


3

Perhaps better: NMinimize[{m/(m (1 - c) + 1), 2^(m*(1 - c)) - 2.718/((m^.5)*2*3.14*c^(c*m + .5)*(1 - c)^((1 - c)*m + .5)) <= 0 && 0.006 <= c <= 1 && 4 <= m <= 2500}, {m, c}, MaxIterations -> 300] (* {1.11184, {m -> 4., c -> 0.350592}} *)


0

Here is your function: f[m_] := Catch[ Do[a = 2^(m*(1 - c)) - 2.718/((m^.5)*2*3.14*c^(c*m + .5)*(1 - c)^((1 - c)*m + .5)); If[a <= 0, Throw[m/(m (1 - c) + 1)]], {c, 0.006, 1, 0.001}]] and here is a plot of this over 4 to 2500 Plot[f[m], {m, 4, 2500}, PlotRange -> All] Pretty clearly the min is at f[4].


6

The expression in question is (we have replaced sigma by s) g = Integrate[ Exp[-(x - y)^2/(2 y s^2)] f[y] 1/Sqrt[2 \[Pi] y s^2], {y, 0, \[Infinity]}] $\int_0^{\infty } \frac{e^{-\frac{(x-y)^2}{2 s^2 y}} f[y]}{\sqrt{2 \pi } \sqrt{s^2 y}} \, dy$ First of all we notice that for y>0 "Limit[Exp[-(x - y)^2/(2 y s^2)] 1/Sqrt[2 \[Pi] y s^2], s -> 0] = ...


0

So what you want to do is evaluate the integral and assign the result to a variable, like this: Rev = 2 a + (b + a*x)^2 integrand = Integrate[Rev, {x, 0, 1}] Then you take that variable integrand, which is written in terms of a and b and perform a substitution like integrand /. {b -> 2.3, a -> .01} and it will return a numerical value. But now ...


1

My earlier answer resulting from misreading the question. Taken at face value, you are asking for sum=Sum[3 - 4/(1 - (-3)^(n + 1)), {n, 1436, 2015}] which produces a rationale number with an enormous number of digits. It is, however, almost precisely equal to 3*(2015 - 1436 + 1), or 1740, as one would expect. N[sum,1000] (* ...


1

Did you try using assumptions? Using assumptions with version 10.0.0, I get the same results as with version 10.0.2 $Version "10.0 for Mac OS X x86 (64-bit) (June 29, 2014)" Clear[f] f[n_Integer] = Integrate[HermiteH[n, x]*Exp[-x^2], {x, 0, Infinity}, Assumptions -> {Element[n, Integers]}] (2^(-1 + n)*Sqrt[Pi])/Gamma[1 - n/2] ...


0

Integrating each term separately obtains the desired result: Integrate[#, {x, 0, \[Infinity]}] & /@ Expand[HermiteH[50, x] Exp[-x^2]] results in 0. If each term of the integral is convergent, the whole integral must be convergent, so this must be a bug.


2

There are a very large number of syntax errors. The most serious relate to use of protected symbols: D, I. This is a correction that should work. Please adjust plot range to your needs: sirds[α_, β_, δ_, μ_] := {S[t], SS[t], i[t], R[t], d[t]} /. First@NDSolve[{S'[t] == -α*S[t]*i[t] - δ*SS[t], i'[t] == α*S[t]*i[t] - β*i[t] - μ*i[t], R'[t] ...


3

As a general rule, avoid using l' and m', which Mathematica interprets as derivatives. Mathematica has difficulty performing the integration for symbolic constants l, m, ll and mm, but can verify orthonormality for a finite range of those values: Table[Assuming[{l, ll, m, mm} \[Element] Integers, Integrate[ Conjugate[ ...


0

This does not answer the question why but I post for interest. The region of integration: ir = ImplicitRegion[0 < x < y + 35 && -1 < y < 3, {x, y}]; RegionPlot[ir] The integral must be < Area[ir]f[38,0]:369981. (not a helpful bound). $0<x<38 &&-1<y<3$ would be a closer. SetAttributes[dis, HoldFirst]; dis[u_] ...


2

The output of the Mathematica code is ConditionalExpression[ 1/2 (2 a - a Sqrt[1 - a^2] + Sqrt[3 + 2 a - a^2] - a Sqrt[3 + 2 a - a^2] - 2 b + b Sqrt[1 - b^2] - Sqrt[3 + 2 b - b^2]+ b Sqrt[3 + 2 b - b^2] + 4 ArcSin[(1 - a)/2] *Routine clean-up*-ArcSin[a] - 4 ArcSin[(1 - b)/2] + ArcSin[b]), -1 < b < 1 && ( -1 < a < b || b ...


0

If f is dependent on a variable k, the above definition leads to a conflict. It is better to rename k in a less common name such as var. multiTaylor[f_, {vars_?VectorQ, pt_?VectorQ, n_Integer?NonNegative}] := Sum[Nest[(vars - pt).# &, (D[f, {vars, var}] /. Thread[vars -> pt]),var]/var!, {var, 0, n}, Method ...


1

In V10,D can symbolically differentiate NIntegrate if it is Inactive. In the case that the integral can only be evaluated with NIntegrate the following defines an arbitrary Derivative of the OP's function numericalModelInternalEnergy. ClearAll[numericalModelInternalEnergy, numericalModelHeatCapacity]; Block[{NIntegrate, Alpha, ground, T, Td}, integrand = ...


2

First I'll provide a symbolic workaround, and then I'll explain why your attempt doesn't work. The integral can be symbolically evaluated, like this: ModelInternalEnergy[Td_, T_] := Evaluate[Simplify[ Alpha*T^4* Integrate[x^3/(Exp[x] - 1), {x, 0, Td/T}, Assumptions -> Td/T > 0] + ground]] which for reference gives ground - 1/15 ...


2

I apologize if I have misunderstood the aim here. $P(\Pi X_i>2.5)$ where $X$ is 10 iid CauchyDistribution[1,1]. It seems the easiest way to estimate this is by simulation, e.g: pdt[n_] := Module[{p = ProductDistribution @@ Table[CauchyDistribution[1, 1], {n}], rv}, rv = RandomVariate[p, 100000]; Length[Pick[rv, Times @@ # > 2.5 & /@ ...


6

Unevaluated[Limit[(c^(1 - g) - 1)/(1 - g), g -> d]] /. d -> 1 Hold[Limit[(c^(1 - g) - 1)/(1 - g), g -> d]] /. d -> 1 // ReleaseHold With[{d = 1}, Limit[(c^(1 - g) - 1)/(1 - g), g -> d]] Limit[(c^(1 - g) - 1)/(1 - g), g -> #] &@1 Block[{Limit}, SetAttributes[Limit, HoldAll]; Limit[(c^(1 - g) - 1)/(1 - g), g -> d] /. d -> 1] ...


5

Assuming you want to plot all three roots of $y^3=x$ with real and imaginary axes, and with the the complex phase and magnitude indicated by the line colour and thickness: With[{p = ParametricPlot3D[Evaluate@ Table[With[{r = Root[#1^3 - x &, i]}, {x, Re[r], Im[r]}], {i, 3}], {x, -3, 3}, AxesLabel -> {"x", "Re[x^(1/3)]", "Im[x^(1/3)]"}, ...


4

Since you have a reputation over 1000, I presume you're familiar with basic plotting and seek stylistic suggestions. I prefer to plot the real and imaginary parts separately: Plot[{Re[x^(1/3)], Im[x^(1/3)]}, {x, -3, 3}, PlotLegends -> {"real", "imaginary"}] You could, of course, also include Abs[x^(1/3)].



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