Tag Info

New answers tagged

1

you might handle this with a TransformationFunction: Simplify[ Integrate[ f[x] , {x, 0, 1} ] + Integrate[ f[x] , {x, 1, t } ] ] t[s_Plus?( Length[#] == 2 && Head[#[[1]]] == Integrate && Head[#[[2]]] == Integrate && #[[1, 1]] == #[[2, 1]] && #[[1, 2, 3]] == #[[2, 2, 2]] &)] := ...


7

Here is a trick that allows you to get exactly what you're looking for: FourierTransform[ InverseFourierTransform[ x/y DiracDelta[x - y], x, k], k, x] DiracDelta[x - y] What I did here is to apply the Fourier transform and its inverse, which is of course the identity and therefore is equivalent to the original expression. But in doing so, ...


1

DiracDelta must be inside an integral to have much meaning. From its documentation: "DiracDelta can be used in integrals, integral transforms, and differential equations. " Assuming[Element[y, Reals], Integrate[x/y DiracDelta[x - y], {x, -Infinity, Infinity}]] 1 Assuming[Element[x, Reals], Integrate[x/y DiracDelta[x - y], {y, -Infinity, Infinity}]] ...


3

It's not the exact output you requested but in case you are not aware of the second parameter of Rationalize: Rationalize[N[4/3 + I Sqrt[2]/3], 1*^-6] 4/3 + (272 I)/577 If your hybrid output really is desired then perhaps building on m_goldberg's deleted answer: # + Defer[#2 I] & @@ Rationalize /@ {Re@#, Im@#} & @ N[4/3 + I Sqrt[2]/3] 4/3 ...


0

Maybe you are looking for the function PowerExpand? PowerExpand[((3 + Sqrt[6]) nn^4 Sqrt[g])/(2 Sqrt[nn^7 P*T])] (* ((3+Sqrt[6]) Sqrt[g] Sqrt[nn])/(2 Sqrt[P] Sqrt[T]) *)


0

It works for me if I turn the simplifications into a list: {nn>0,P>0,T>0}.


1

This is your code for InverseLaplaceTransform. ig = (3.2526911934581187`*^7 s)/((424000.` + 923.` s) (142122.30337568675` + s^2)) I have tried using ComplexExpand . InverseLaplaceTransform[ig, s, t] // Simplify // ComplexExpand // FullSimplify or InverseLaplaceTransform[ig, s, t] // Simplify // ComplexExpand // Simplify // Chop -45.8409 ...


6

Referring to your own answer there is a much simpler form that you may use, recalling: Collect[expr, var, h] applies h to the expression that forms the coefficient of each term obtained. sol = x[t] /. s[[1]]; Collect[sol, _C, Simplify] (E^(I t ωd) f0)/(m ω0^2 + I m β ωd - m ωd^2) + E^(-(1/2) t (β + Sqrt[β^2 - 4 ω0^2])) C[1] + E^(1/2 t (-β + ...


3

One way to tackle the problem is to recognize that the solution is one giant sum with three terms. You can then convert this sum to a list, simplify the terms individually, and sum all elements of the list back together. sol = x[t] /. s[[1]]; Total[ FullSimplify[ Apply[ List , sol ] ] ] But this "hack" is somewhat unelegant, because it might fail if ...


2

It is too difficult for Mathematica to combine two sums. Even in the following simple example 2 Sum[a[n], {n, 1, q}] - Sum[2 a[n], {n, 1, q}] (* 2 Sum[a[n], {n, 1, q}] - Sum[2 a[n], {n, 1, q}] *) There is only one exception: if your expressions are exactly the same (not necessarily Sum) they will subtracted to 0 Sum[a[n], {n, 1, q}] - Sum[a[n], {n, 1, ...


-2

I can't get it to work with "q", but I can get it to work with the dummy variables m & n. Sum[f[n], {n, 1, 10}] - Sum[f[m], {m, 1, 10}] // FullSimplify 0 As you can see, I replaced your subscript with functional notation.


7

You can use TrigReduce to do what you want: TrigReduce[-2 Cos[v] g[u] Sin[v] Derivative[1][g][u]] -g[u] Sin[2 v] Derivative[1][g][u]


1

The problem is that you wrote aq where you meant a*q. Although, implicit multiplication with a space a q can be handy, I come more and more to the conclusion that it introduces far more mistakes than a explicit * would allow for. Anyway, with the corrected code you get f[n_] := Expand[(1 - a*q^{2 n})* Sum[FunctionExpand[QPochhammer[a q, q, n + j - ...


1

Just an outline really, but too long for a comment. The idea is along the following lines. u = {ux[x, y, z], uy[x, y, z], uz[x, y, z]}; cc = Curl[Curl[u, {x, y, z}, "Cartesian"], {x, y, z}, "Cartesian"]; dv = Div[u, {x, y, z}, "Cartesian"]; Take another round of derivatives to obtain more simplifying equations. derivs2 = Map[D[dv, #] &, {x, y, z}]; ...


10

Try TrigExpand TrigExpand[Cos[2 ArcCos[A]]] -1 + 2 A^2


1

This is the most efficient I can think of: With[{vars = (δA | δB | δC)}, expr /. {x: vars * y: vars -> 0, vars^n_ /; n>1 -> 0}]


7

First, you really don't want to modify Times, as this will have all sorts of unforeseen side-effects. Your best route will be to use ** (NonCommutativeMultiply), for which you'll have to write rules that enforce the behaviour you want. Here's how you might go about that: genericRules = { (* Move numeric factors outside of NonCommutativeMultiply. *) ...


5

expr = 1/(-4 - 2 x + 2 x^2 + 3 x^3 + x^4); expr2 = Factor[expr, GaussianIntegers -> True] // Apart expr == expr2 // Simplify True If the denominator cannot be factored with integers or Gaussian integers expr = 1/(x^2 + x + 1); expr2 = Numerator[expr]/ (Times @@ (x - (x /. Solve[Denominator[expr] == 0, x]))) // Apart expr == ...


1

eps = 0.001; repl = (z_Symbol == val_) :> (val - eps/2 <= z <= val + eps/2); r1 = Line[{{0, 0}, {1, 0}}]; r2 = Line[{{1, 0}, {2, 0}}]; r3 = ImplicitRegion[RegionMember[ RegionUnion[r1, r2], {x, y}] // Simplify, {x, y}] /. repl ImplicitRegion[(x | y) \[Element] Reals && -0.0005 <= y <= 0.0005 && 0 <= x ...


0

(version 10) I made function lastUnion , firstUnion and RectangleRegionSimplify instead of RegionSimplify. I think you might want to make RectangleRegionSimplify. lastUnion[coo_] := Flatten[Replace[SplitBy [ coo, Last], {f_, ___, l_} -> {f, l}, 1], 1] firstUnion[coo_] := Flatten[Replace[SplitBy [ coo, First], {f_, ___, l_} -> {f, l}, 1], 1] ...



Top 50 recent answers are included