7
Mathematica
Limit[((h + x)^(1/4) - ( x)^(1/4))/((h + x)^(1/2) - ( x)^(1/2)),h -> 0]
(*1/(2 x^(1/4))*)
evaluates the limit without problems
6
Actually the answer depends on parameters. The best what I get just playing with difference order and rational numbers is the following code
imax = 500;
t0 = 10/11;
(*E grid*)
DE = 1/20;
Emin = 1/100;
grid = Table[Emin + DE*k, {k, 0, imax}];
fd = NDSolve`FiniteDifferenceDerivative[Derivative[1], grid,
"DifferenceOrder" -> 8]; m = fd["...
4
What you want is:
for i in list1:
for j in list2:
for k in list3:
Times[i,j,k]
Which means, you need to construct a Cartesian product
F1 ,search it:
So we need Outer
According to F1
Outer[Times, {1, 2, 3, 4}, {a, b, c}]
this gives the Outer product of vector
Which is what you need.
Outer[Times, list1, list2, list3]
3
I found the answer in some documentation. The function to be used is called
GeneratingFunction.
Consider for example the following function - a term of a Master equation - :
$aP(a)$.
We can transform such function by making use of probability generating functions with the following syntax:
GeneratingFunction[a*P[a], a, z]
which yields the following answer:
...
3
Responding to the only question-like feature of the edited Question,
Assuming[{x > 0},
Limit[ /* more stuff here */ ]
]
2
Well, TensorProduct finishes the one-step job without the help of any other functions:
TensorProduct[list1, list2, list3]
2
Try to use Check. For example:
Check[{tmp1=SomeCode,1},{tmp1,0},MessageName];
1
Wolfram Inc. is still a properitary company. So the first established rules govern all. A guide is the price for the product. That is a crude simplificator. So some prodcuts include others at all or partly. These facts make it more difficult to decide. Comparing is really hard and depends on the intentions.
Desktop for example includes Wolfram Pro to some ...
1
Try the general solution:
dsol = DSolve[{k Sin[θ[t]] == θ''[t]}, θ, t]
The resulting equations for the initial condition cannot be solved symbolically.
ics = {θ[0] == θ0, θ'[0] == w0} /. dsol;
Solve[#, {C[1], C[2]}] & /@ ics
Solve::nsmet: This system cannot be solved with the methods available to Solve.
Solve::nsmet: This system cannot be solved with ...
1
You can simplify the solution and give r > 0 as an assumption:
solR = Simplify[Solve[rel3 /. solH /. sols, R, PositiveReals][[1]], Assumptions -> {r > 0}]
I doubt there is a better way to access the solution than [[1]] as Solve always returns a list, of which you are interested in the first element.
Only top voted, non community-wiki answers of a minimum length are eligible
