New answers tagged simplifying-expressions
3
Although it's not quite so obvious, Apart will always choose a variable when one isn't explicitly given.
Apart[(a - b + c - d)/(a - b), c]
$$\frac{c}{a-b}+\frac{a-b-d}{a-b}$$
vs
Apart[(a - b + c - d)/(a - b)]
$$\frac{a-b+c}{a-b}-\frac{d}{a-b}$$
and will choose in alphabetical inverse order which is the variable. (the last one in the alphabet. )
...
0
Yet another possibility:
ComplexExpand[Sqrt[2 + I], TargetFunctions -> {Re, Im}] // FunctionExpand // Simplify
(5^(1/4) ((2 + I) + Sqrt[5]))/Sqrt[10 + 4 Sqrt[5]]
answered Sep 15 at 14:56
0
The identity works in a more broad case of $a\in\mathbb{C}$. Verification
Simplify[a^(b c) - (a^b)^c, Assumptions -> b ∈ Integers && c ∈ Integers]
(* 0 *)
4
For questions like these, FindInstance is your friend:
FindInstance[a^(b c) - (a^b)^c != 0, {a, b, c}]
(* {{a -> 99/5 + (12 I)/5, b -> -(28/5) + (79 I)/5, c -> 61/10 + (143 I)/10}} *)
So at least for complex numbers, there are issues. We can try to ask for solutions with real values as well:
FindInstance[{
a^(b c) - (a^b)^c != 0,
(a | b | c) ∈...
0
You could introduce the perturbative part as δ[a] and define an extra rule how multiplication works with those (δ can be entered quickly via EscdeltaEsc or via \[Delta]):
δ /: Times[___, _δ, _δ, ___] = 0;
Now every time at least two δ[_] symbols are multiplied they will be simplified to zero automatically. For example in
(* Input *)
(a + δ[a]) (b + δ[b])
%...
3
I don't think that what you're seeing is a bug. The problem is that Sum implicitly assumes that the summation bounds are ordered and differ by an integer amount (mentioned here). The following example illustrates how this might cause issues:
(* evaluate sum symbolically *)
Sum[i, {i, a, b}]
(* -(1/2) (-1 + a - b) (a + b) *)
(* insert problematic values ...
2
I can't reproduce the first part you say. On V 12 on windows 10, I get 2.25 in both cases (it has to be, as simplifying 2.25 to 1.5 would be a really bad bug)
But what I see, is that when changing the definition from delayed := to =, then now Mathematica gives 1.5
And this I think must be a bug.
Code
ClearAll[n, x, j, i, f, g];
f[n_, x_] :=
Sum[Sum[(-...
5
One way might be
expr = Sqrt[-1]/a^(3/2);
Assuming[a < 0, Simplify[ComplexExpand[expr]]]
PowerExpand[%]
Btw, you will not get the same exact expressions in both cases you showed.
This is because $\sqrt{a^3} = a^{3/2}$ only for $a\geq 0$ and you said that $a<0$.
ClearAll[a]
Reduce[ Sqrt[a^3] == a^(3/2) && Element[a, Reals]]
8
Many times Mathematica gives enormous results to simple problems
If Simplify still does not help reduce the antiderivative to what you like, you could always try Rubi
<< Rubi`
integrand = 1/Sqrt[1 + Sin[x]];
sol = Int[integrand, x]
D[sol, x] // Simplify
There is a page here which compares different integrators with the size of antiderivatives ...
0
Setting the first statement equal to lhs and the second to rhs, we can ask for:
FindInstance[lhs =!= rhs, {d, x, h}]
{{d -> -(12/5) - I/2, x -> -(9/5) + (3 I)/5, h -> -(24/5) - (21 I)/5}}
So they are not equal.
0
Rewriting problems can often be directly solved with replacements. Since there are a relatively small number of replacement rules involved, that would be my go to. I use ReplaceRepeated (//.) in dist below to implement these replacements, as there may be several layers of replacements to unravel as the distribution continues:
dist[expr_] :=
expr //. {Dot[...
1
{1, I}.(Sqrt[2 + I] // ReIm // ComplexExpand // FunctionExpand //
FullSimplify)
(* I Sqrt[1/2 (-2 + Sqrt[5])] + Sqrt[1/2 (2 + Sqrt[5])] *)
or
% // Simplify
(* (I Sqrt[-2 + Sqrt[5]] + Sqrt[2 + Sqrt[5]])/Sqrt[2] *)
Verifying that these are equivalent to the original form
Sqrt[2 + I] === (% // FullSimplify) === (%% // FullSimplify)
(* True *)
0
Try
Sqrt[2 + I] // ComplexExpand // FunctionExpand
It may not create a result simplified in the exact form you want, but it will be closer.
1
One approach is to define your integral as a function, which you can then call with your desired x[s]:
myHeavy[x_] := Integrate[HeavisideTheta[s - t] x, {s, 0, 1}, Assumptions -> {t > 0}];
Now, when you want to call this with a function x[s] = s^2, you evaluate
myHeavy[s^2]
To evaluate with Sin[s]
myHeavy[Sin[s]]
These seem to return plausible ...
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