New answers tagged syntax
1
You can use NSolve instead of Solve followed by N:
NSolve[{x + 1/x == y - 1/y, x^2 + 1/x^2 == y^2 - 1/y^2}]
{{x -> -0.353553 - 0.935414 I, y -> 0.707107}, {x -> -0.353553 + 0.935414 I,
y -> 0.707107}, {x -> 0.353553 - 0.935414 I,
y -> -0.707107}, {x -> 0.353553 + 0.935414 I, y -> -0.707107}}
Or, you can use FullSimplify instead of Simplify:
...
1
Thank you Xminer for a hint. The solution is this:
{{PDF[ DiscreteMarkovProcess[1, P][Infinity], 1],PDF[ DiscreteMarkovProcess[1,P][Infinity], 2], ...}} // MatrixForm
where P is Markov transition matrix and the second PDF argument is the value of Boole.
1
A (much slower) procedural method.
First we find the number of digits of 123...(n-1)n.
numlength[n_] := With[
{pow = Floor@N@Log10[n]},
Range[pow].(9*10^Range[0, pow - 1]) + (pow + 1) (n - 10^pow + 1)
]
Here pow = Floor@N@Log10[n] gives the order of magnitude (OoM) of n, Range[pow] and (pow + 1) gives the number of digits in each OoM and the last OoM, ...
7
The number you are constructing is related to the base 10 Champernowne constant, except that the Champernowne is a real number that starts with 0.123... while your "number" is an integer with an infinite number of digits.
In Mathematica, the Champernowne constant in base 10 is given by ChampernowneNumber[]. To get the 10,000 digit, just use RealDigits:
...
1
If your ultimate goal is implementation of Romberg Integration Method, here is straightforward way to do it usingTable
ClearAll["Global`*"]
f[x_] := 1/x
romberg[a_, b_, n_] := Module[{}, h[j_] := (b - a)/2^j;
R[0, 0] := h[1] (f[a] + f[b]);
R[j_, 0] := 1/2 R[j - 1, 0] + h[j] \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\),
SuperscriptBox[\(2\), \(j - ...
2
Update: an alternative approach using FoldList:
nla = FoldList[Function[{x, y}, ((4 y - x)/3)] @@@ Partition[#, 2, 1] &, T,
Range[Length[T] - 1]];
TeXForm @ Grid[Flatten /@ PadLeft[nla]]
$\begin{array}{ccccc}
0.75 & 0.708333 & 0.697024 & 0.694122 & 0.693391 \\
0 & 0.694444 & 0.693254 & 0.693155 & 0.693148 \\
0 &...
11
This is a precedence issue.
"fa" -> (# + 1/2) & is interpreted as ( "fa" -> (# + 1/2) ) &.
Write "fa" -> ( (# + 1/2) & ) instead.
3
Maybe you can use the fact that TraditionalForm by default interprets parentheses as function calls:
fromCString[s_String] := ReplaceAll[
ToExpression[s, TraditionalForm, Defer],
{
sqrt -> Sqrt,
sqr -> square
}
]
Then:
fromCString["sqrt(sqr(x)+y*z)"]
Sqrt[square[x] + y z]
6
There is a handy package, Notation, built-in to Mathematica. With it, we can do the following:
Needs["Notation`"]
Notation[ParsedBoxWrapper[RowBox[{"sqrt", RowBox[{"(", "x_", ")"}]}]]
⟹
ParsedBoxWrapper[SqrtBox["x_"]]]
Notation[ParsedBoxWrapper[RowBox[{"sqr", RowBox[{"(", "x_", ")"}]}]]
⟹
ParsedBoxWrapper[...
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