New answers tagged simplifying-expressions
6
You can use the substitution rules based on the Eq.8.126 from Table Of Integrals, Series And Products by Gradshteyn and Ryzhik. They are called quadratic arithmetic-geometric mean (AGM) transformations for the complete elliptic integrals. I retype them below in original notations
$$K\!\left(\frac{2\sqrt{k}}{1+k}\right)=(1+k)K(k),$$
and
$$E\!\left(\frac{2\...
1
$Version
"12.1.1 for Mac OS X x86 (64-bit) (June 19, 2020)"
Clear["Global`*"]
expr = Sqrt[(B Bc)/(A^2 - B Bc)] Sqrt[
B Bc + A (-A + Sqrt[A^2 - B Bc])] Sqrt[-B Bc + A (A + Sqrt[A^2 - B Bc])];
Require all of the arguments of Sqrt to be positive
cons = FullSimplify[And @@ Cases[expr, Sqrt[z_] :> (z > 0), Infinity]]
(* B Bc + A ...
1
Here is why:
y=Sqrt[a (a+Sqrt[a^2-b])-b] Sqrt[a (-a+Sqrt[a^2-b])+b] Sqrt[b/(a^2-b)]
FullSimplify[y,Assumptions->a>0&&b>0&&a^2-b>0]
(*b*)
Plot[{y/.{b->-1},y/.{b->1}},{a,-2,2},
PlotTheme->{"BoldColor","Frame"},
PlotLegends->{-1,1},
PlotStyle->{Thickness[0.03],Directive[Thickness[0.01]]}]
3
The problem is outlined in the documentation of ToRadicals with a minimal example similar to
Root[#^3 - a &, 1]
ToRadicals[%]
{%%, %} /. {a -> 1}
The problem is that Root refers to the n-th root in its second argument. For roots involving parameters ToRadicals does not guarantee/preserve this ordering. That being said for a given set/ranges of ...
0
ComplexPlot3D[
Sqrt[Exp[-z/(4*Sqrt[2])]] Sqrt[Exp[z/(4*Sqrt[2])]], {z, -2 \[Pi] -
2 \[Pi] I, 2 \[Pi] + 2 \[Pi] I}, PlotLegends -> Automatic]
Mathematica chooses the branch cut for ๐งโ to lie along the negative real axis and the first value is {0,0} and this is 1.
This asks whether to trust PowerExpand or ComlexExpand.
Refine[Re[
ComplexExpand[
...
4
You can either write
ComplexExpand /@ sol2
or
FullSimplify[ComplexExpand /@ sol2, Trig -> False]
whichever you find most appropriate for your purpose.
1
As an example, say you want to solve a differential equation knowing your variable has some sort of gauge freedom $x=ax+b$, where $a,b$ are real values.
In[]:= sol=DSolve[y[x] y''[x] == 2 y'[x]^2, y[x], x]
In[]:= f[x_] = y[x] /. soltest[[1]]
Out[]:= C[2]/(x + C[1])
Now my question is, is there anyway to use Simplify[] (or another function) with an ...
2
There seems to be a weakness in series expansions of QPochhammer in Mathematica:
Series[QPochhammer[q, q, 2], {q, 0, 1}] //TeXForm
$1+q
\left(\text{QPochhammer}^{(0,1,0)}(0,0,2)+\text{QPochhammer}^{(1,0,0)}(0,0,2)\right)+O
\left(q^2\right)$
You can fix this by using FunctionExpand (essentially Vaclav Kotesovec's comment to Bill Watt's answer):
Series[...
0
I'm not sure how you get all the terms, but if you want to look at a finite number you can do something like this:
First get a series in q from
fn = q^(1/2 n (n + 1))/QPochhammer[q, q, n]
Series[fn, {q, 0, 100}] // Normal
Then apply the first few n
Plus @@ (Table[%, {n, 0, 100}] // Expand)
That should give you several coefficients to play with. If you ...
4
If you replace a few of the complex terms with constants the simplification is done within 15 seconds.
reps = {E^(-j k (d - t)) -> c[1], E^(-h1 j t) -> 1/c[2],
E^(h1 j t) -> c[2], E^(-h2 j t) -> 1/c[3], E^(h2 j t) -> c[3],
E^(-h3 j t) -> 1/c[4], E^(h3 j t) -> c[4], E^(-h4 j t) -> 1/c[5],
E^(h4 j t) -> c[5], E^(-j k t) ->...
1
Only provide another way similar to Eliminate. Not so beautiful.
Here we add extra condition.
a โ Reals, b โ Reals, x โ Reals, y โ Reals, m โ Reals, n โ Reals, a != b, a != m, b != n, m != b, n != a
Clear["`*"];
sys = {a*Tan[x] == b*Tan[y], a*Sin[x]^2 + b*Cos[x]^2 == m,
b*Sin[y]^2 + a*Cos[y]^2 == n, a โ Reals,
b โ Reals, x โ Reals, y โ ...
2
Weierstrass substitution makes Eliminate working:
Solutions of sys are 2Pi-periodic
sys={a*Tan[x] == b*Tan[y], a*Sin[x]^2 + b*Cos[x]^2 == m, b*Sin[y]^2 +a*Cos[y]^2 == n}
Weierstrass-substitution constraints the solution {x,y} to -Pi<x,y<Pi
sysu = TrigExpand[sys /. {x -> 2 ArcTan[ux], y -> 2 ArcTan[uy]}] // Simplify
cond=Eliminate[sysu, {ux, uy}]...
1
Because the keyword simplification is used:
Solve[a^(2 k) == 0 && k > 0 && Element[k, Integers], a, Complexes]
{{a -> ConditionalExpression[0, Element[k,Integers] && k >= 1]}}
or enter
a=ConditionalExpression[0, Element[k,Integers] && k >= 1]
The potentiation is not needed until explicitly required.
Refine and ...
2
I assume, according to your information, that all variables, except "a" appear only with exponent 1.
Setting every term in the sequence to 0, you want to express all the terms as functions of a. In this case, the first two term in the sequence do not contribute anything and can be neglected. You can then get expressions for the rest of variables in ...
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