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15

Let me show how to roll your own numerical solution to a non-linear integral equation using a collocation method. It's fun! This will involve two approximations. First, we will approximate the function B[x] by its values at n particular points in the range {x, 0, 1}. The integral over x will be replaced by a weighted sum over n, i.e., a quadrature rule. ...


11

I am not interested in the closed form solution of M(T) - and that's a great pity, since your equation can be solved exactly by using Laplace transform - for the numbers you have, and in the general case you can go a long way using the same approach. Analytical solution Applying Laplace transform to both sides of your equation, $\int_{0}^{\infty}e^{-p ...


10

If what you really want is to find all natural numbers $a$ and $b$ on the surface for a given $n$, a search among the divisors of $2n$ will do it efficiently even for sizable values of $a$ and $b$. To do this, we find all divisors $k$ of $2n$ (Divisors), then all divisors of the quotient $2n/k$ (Divisors again), forming triples $(u,v,w)$ for which $uvw=2n$ ...


10

Mathematica is an incredible tool for checking conjectures and making sketches. I'm going to demonstrate it below. Let's start with checking that in case when $R_0$ (I replaced it with $R$) is a polynomial, the solution of this Volterra equation reduces to linear ODE. Lets take some derivatives of the equation: ClearAll[P, R, s, t]; eqn = P[t] == R[t] + ...


8

Something similar showed up on StackExchange a while back (last August). http://stackoverflow.com/questions/6974929/how-can-i-reference-a-specific-point-of-my-function-inside-ndsolve Here is a variant from one response that I cobbled together to illustrate your case. I chose the kernel function (your R_0) to be e^(-x) + x sin(x) - x cos(x^2). I attempt to ...


8

There are many sources that give practical advice on how to program such an integral equation. So I went a different route and asked myself how to get a solution by using the most literal application of the defining equation. By that I mean no explicit discretization, keeping the integral. That can be done using the assumption that the fixed-point theorem ...


7

Using your definitions (using the placeholder pattern f1[y_] instead of the absolute pattern f1[y] is usually a good idea if you want to use it as a function that works with numerical values, too. Also using := (SetDelayed) instead of = (Set) inserts the left hand side value y into the definition every time you use it, which is closer to the behavior you ...


6

This might help you get started. I think this is a variation on a method called Frobenius' method; the idea is to use the fact that the equation is linear to expand the solution over a set of basis functions (here B-Splines) and find the corresponding coefficients. It should provide you with an approximate solution (which you can improve upon while adding ...


6

The integral is conditionally convergent. You can progress using substitution: $u=2^{\frac{r}{b}}\iff r= b\log_2 u $ Hence,$\frac{dr}{du}=\frac{b}{u\ln 2}$ You can do these substitutions in Mathematica: f[r_, b_, la_, k_] := 2^(r/b) Exp[k (2^(r/b) - 1)/la]/(b la) exp = f[x, a1, a2, a3] /. {2^(x/a1) -> u}; ex = D[a1 Log[2, u], u]; ans = Integrate[a1 ...


6

Your s lives on the ray [1,infinity). Integrate can kinda figure out s is real. Tell it about the other variables and then we'll proceed from there. i1 = Integrate[(-(Cos[b*s] - (2*Sin[b*s]*(a*d - b*c))/(a^2 + b^2))/ E^(a*s))^2, {s, 1, \[Infinity]}, Assumptions -> {Element[{b, c, d}, Reals], a > 0}] (* Out[152]= 1/4 (1/a + 1/(2 a - 2 I b) + ...


5

I found a good solution. The new version of Mathematica is using the symbolic processing, so we can just turn it off. b[i_] := NIntegrate[(Sin[x] - 1) x^i, {x,-5,5}, Method->{Automatic,"SymbolicProcessing"->0}] source: Techniques for Accelerating NIntegrate Evaluations http://support.wolfram.com/kb/3442


5

Since OP seems to be interested in numerical calculations, this might be more suitable: f0[x_]:= 1/Sqrt[2*Pi*1^2]*E^(-(x + 1)^2/(2*1^2)) q[1, A_, B_, f_] := f q[n_ /; n > 1, A_, B_, f_][x_] := Module[{ω},tempIntegrate[Evaluate[q[n - 1, A, B, f]][ω] f[x - ω], {ω, A, B}]] p[n_, A_, B_, f_] := tempIntegrate[q[n, A, B, f][x], {x, -∞, B}] + ...


5

If you want to solve the Fredholm equation of the second kind which is an integral equation of the form $$f(x) - \lambda\int_{a}^{b} K(x,y)f(y)dy = g(x), \quad \forall x \in [a, b]$$ you can use the following code. We follow the simplistic source. One can make it much better by considering further implementation aspects as well as injecting more state of ...


5

Not knowing what the actual differential equation is, I can only point out one obvious problem: you should define Clear[r,f,y]; M1[r_] = Integrate[(1 + f*Exp[-(r/y)^2])*(r)^2, r] instead of using := NIntegrate because you seem to be looking for an indefinite integral there. For numerical integration you have to specify integration limits. The Clear is ...


4

For the scalar case: Set up a function, h, depending on EFermi, and use FindRoot to solve h[EFermi]==c Clear[g, f, h]; g[x_] = Exp[-x^2/5000]; f[x_, EFermi_] = (1 + Exp[(x - EFermi)/25])^-1; h[EFermi_?NumericQ] := NIntegrate[g[x] f[x, EFermi], {x, -200, 200}] c = 100; sol = FindRoot[h[e] == c, {e, 10}] h[e] /. sol (* {e -> 55.2154} *) (* 100. *) ...


4

I made this for you. It should get you started. Here you can type a function to integrate, for example: func[x_] := 1/Exp[x] This is your integrator. It takes a function f, a lower and upper bound a,b and a step size l: riemint[f_, a_, b_, l_] := Total[ Table[ f[i]*l, {i, a, b - l, l}]] You can try it out like this: riemint[ func, 1, 5, .0001] ...


3

It is not possible. Your integral can be written as (with c = r+b) Integrate[1/Sqrt[c + x], {x, 0, n}, Assumptions -> n > 0] (* 2 (-Sqrt[c] + Sqrt[n + c]) *) where $n \rightarrow \infty$ But as we know Sqrt grows unboundedly, so the integral will diverge for all input. Limit[2 (-Sqrt[c] + Sqrt[n + c]), n -> Infinity] (* Infinity *)


3

It seems to me that for your first stated problem there is a much better method than Solve or Reduce: {m, k, p} = {16, 3, 6}; IntegerPartitions[m, {k}, Range@p] {{6, 6, 4}, {6, 5, 5}} If you want all permutations just use Permutations: Join @@ Permutations /@ % {{6, 6, 4}, {6, 4, 6}, {4, 6, 6}, {6, 5, 5}, {5, 6, 5}, {5, 5, 6}} For your second ...


3

I was not going to post this because in a previous edit (now gone) you showed a very difficult problem. Anyway, the current question could be answered as follows: You want to find f[x] satisfying f[x] == Integrate[f[x] g[x], x] for a known g[x]. Differentiating: f'[x] == g[x] f[x] So for example g[x_] := Sin@x^2; fs = DSolve[f'[x] == g[x] ...


2

I think what you're trying to do is to solve $f(z)=h(z)\int _z^\infty dz'\,\left(1-g(z')\right)\rho(z')$ for $g$, with $f$, $h$ and $\rho$ known functions. This is an integral equation. However, dividing both sides by $h(z)$ (I'm assuming $h\neq0$) and differentiating with respect to $z$ we get ...


2

I replaced your syntax so that the := are all in one piece and I changed both Integrate to NIntegrate and I got a result. The usual caveat of it being better to not use variables starting with a capital letter so as not to accidentally confuse a built-in function applies. Clear[K, H, f0, f1] f0[x_] := 1/(E^(x^2/2)*Sqrt[2*Pi]) f1[x_] := (Sqrt[2]*(Pi/2 + ...


2

Try to read Documentation on functions Solve and Reduce and tutorial Solving Equations. Look through this forum, I bet this is a duplicate question. There is also a guide: Diophantine Equations. {x1, x2, x3} /. Solve[x1 + x2 + x3 == 16 && 1 <= x1 <= 6 && 1 <= x2 <= 6 && 1 <= x3 <= 6, {x1, x2, x3}, Integers] ...


2

Perhaps this will help:- eqn = a Integrate[Exp[-s^2]/(s - c x)^2, {s, -Infinity, Infinity}, Assumptions -> Im[c x] != 0] + b Integrate[Exp[-s^2]/(s - d x)^2, {s, -Infinity, Infinity}, Assumptions -> Im[d x] != 0]; a = 1.5; b = 2; c = 0.8; d = 1; Plot[eqn, {x, -1000, 1000}]


2

Your data/function generates complex value. So I plotted the Re part only. It was hard to read your code, so check that I translated it ok to Mathematica. f[t_?NumericQ] := Quiet@NIntegrate[ Exp[-I*x*t]/((4*Pi) ((x - 1)^2 + 25) (Exp[-100 x] + 1)) , {x, -Infinity,Infinity}]; data = Table[{t, f[t]}, {t, 0, 100, .5}]; ListLinePlot[Re@data, Frame -> ...


2

There are infinitely many solutions satisfying the system of the equations. If we set x[v, z] to be constant and add appropriate assumptions the both integrals vanish. Moreover we could find a few consecutive values of the integrals for e.g. x[v, z] == z^k and k ∈ {0, 1, 2, ...} putting it in a Table. Nevertheless we can do much more proving an appropriate ...


2

In this answer, I handle coefficients of the form $$ \frac{p(n)}{q(n)}, $$ where $p(n)$ and $q(n)$ are polynomials in the index symbol $n$, $\deg(p(n)) < \deg(q(n))$, and all of the roots of $q(n)$ are negative integers. If $\deg(p(n)) \geq \deg(q(n))$, we may have to differentiate a power series, and as I mentioned in my comment, Mathematica 9 doesn't ...


1

Using ErsekRootSearch package Needs["Ersek`RootSearch`"] f[x_] := x^4 - 3 x^3 - Cos[3 x] - 3; soln = x /. RootSearch[f[x] == 0, {x, -10, 10}] (* {-0.831448, 3.06992} *) Total[soln] (* 2.23848 *) You need to edit the RootSearch.m first and change $MinPrecision=-Infinity to $MinPrecision=Infinity (there are about 5-6 places) this is because the package ...


1

First, let's plot both graphs to get an idea of their points of intersection Plot[{x^4 - 3 x^3, Cos[3 x] + 3}, {x, -3, 4}, Filling -> {1 -> {2}}] We see that the intersections are around -1 and 3. Then using FindRoot we get: {r, s} = x /. FindRoot[x^4 - 3 x^3 == Cos[3 x] + 3, {x, #}] & /@ {-1, 3} {-0.831447617, 3.06992283} Total[{r, ...


1

It helps to tell Integrate a little about your parameters: int = Exp[-1/2 y1^2 y - 2 (y2^2 (z - b - a) + y1^2 (-z + b + x + a))] y1^2 Boole[r]; Integrate[int, {z, a, 1}, {x, a, z}, {y, 0, a}, Assumptions -> 0 < a < b < 1/2 && y1 > 0 && y2 > 0] (* -(1/(6 y1^2 y2^2)) E^(-((5 a y1^2)/2) - 2 b y1^2 - 2 (a + b) y1^2 - ...


1

Thanks to the comment of @b.gatessucks I have solved the issue by splitting the integral into 3 parts where I shift one part by $\pi$ like this: $\int_{\frac{\pi }{2}}^{\frac{3 \pi }{2}} \cos (\theta ) \cos \left(\tan ^{-1}\left(\frac{a^2 \tan (\phi )}{b^2}\right)+\pi \right) \, d\phi +\int_0^{\frac{\pi }{2}} \cos (\theta ) \cos \left(\tan ...



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