This is my Mathematica cheat sheet. I keep in it useful things I learn about Mathematica and keep forgetting, and some things I see on the net. If something here is from the net, I try to make sure I put a reference or credit to where I saw it, else it will be something I wrote myself, in that case, all blames to me.
This was written using Latex and converted to HTML using tex4ht and to pdf using pdflatex.
Few are here
Here is a link to information about old Mathematica signal processing package. Currently as of Mathematica version 9, it contains number of build-in DSP functions.
Say you want to read source code of LaplaceTransform. First load it with some call then apply ?? on it
LaplaceTransform[x,x,t]; ClearAttributes[LaplaceTransform,ReadProtected] ??LaplaceTransform
I made this simple diagram to help me understand pure functions.
I made this diagram to show the installation tree structure.
Small note here
Append to the Path the folder name where the package is located in. In this example, assuming there is a package control.m located in folder C:\data then type the following to load the package
AppendTo[$Path, "C:\\data"] << control.m
use Context.
| Context[Integrate] | "System`" |
Use Contexts["packageName*"]
to do
?Apart
| To insert I |
esc ii esc |
| To enter \(\pi \) |
esc p esc |
| To enter E | esc ee esc |
| ctrl-6 wil make exponent |
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| ctrl^ will make superscript |
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Trace[Integrate[x, {x, 1, 2}], TraceInternal -> True]
ParametricPlot[{Sin[u], Cos[u]}, {u, 0, 2 Pi}, AspectRatio -> 1]
Suppose we are given \(z=x e^{-y}, x=\cosh (t), y=\cos (s)\) and need to find \(\frac {dz}{ds}\)
x[t_] := Cosh[t] y[s_] := Cos[s] z[x_, y_] := x[t] Exp[-y[s]] D[z[x, y], s]
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(Cosh[t]*Sin[s])/E^Cos[s]
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Another example: \(u=x^2 y^3 z, x=\sin (s+t), y=\cos (s+t), z=e^{s t}\) and we need to find \(\frac {du}{ds}\) and \(\frac {du}{dt}\)
rootsPlot[poly_, z_] := ListPlot[{Re[z], Im[z]} /. NSolve[poly == 0, z], PlotStyle -> Directive[PointSize[0.015]]]; rootsPlot[z^7 - 1, z]
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Where did I get this from?
"Mathematica can handle partial differential equations via the DSolveIntegrals package. These arise in chemical contexts in the 1D wave equation, 3D wave equation, 3D diffusion equation, Time-dependent and Time independent Schrödinger equation. Hermite showed that the quintic equation could be solved by elliptic functions"
For example, to integrate this below, for n positive integer we do
Integrate[Sin[n x]^2, {x, 0, Pi}]
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Pi/2 - Sin[2*n*Pi]/(4*n) |
Assuming[Element[n, Integers] && n > 0, Integrate[Sin[n x]^2, {x, 0, Pi}]]
| Pi/2 |
eq = f == x^3 + 6/x^3; Reduce[{eq, x^3 == z}]
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(x == z^(1/3) || x == (-(-1)^(1/3))*z^(1/3) || x == (-1)^(2/3)*z^(1/3)) && z != 0 && f == (6 + z^2)/z
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f[z_] := z^2 ParametricPlot[Through[{Re, Im}[f[x + I y]]], {x, -2, 2}, {y, -1, 1}, PlotStyle -> Red]
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one way
Clear[y, x] DSolve[{y'[x]^2 == 1 - y[x]^2, y[0] == 0}, y[x], x]
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{{y[x] -> -Sin[x]}, {y[x] -> Sin[x]}}
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| y[x] /. % |
{-Sin[x],Sin[x]} |
In[60]:= s = HoldForm[1 + 2] Out[60]= HoldForm[1 + 2] In[61]:= ReleaseHold[s] Out[61]= 3
One way is to use Show
By Bill Rowe http://forums.wolfram.com/mathgroup/archive/2004/Apr/msg00357.html
In[1]:= Timing[For[sum = 0; n = 1, n < 100001, n++, sum += n]] Out[1]= {1.2999999999999998*Second, Null} In[2]:= Timing[Plus @@ Range[10000]] Out[2]= {0.009999999999999787*Second, 50005000}
Use notation package
See using_zero_index_in_Mathematica
In[62]:= eq = x^2 + Sin[4*a] == 3 - Derivative[1][y][t] Out[62]= x^2 + Sin[4*a] == 3 - Derivative[1][y][t] In[63]:= lhs = eq /. (lhs_) == (rhs_) -> lhs Out[63]= x^2 + Sin[4*a] In[64]:= rhs = eq /. (lhs_) == (rhs_) -> rhs Out[64]= 3 - Derivative[1][y][t]
One way is to use Vitaliy Kaurov ManToGif. Another way is to run the manipulate and do screen capture using program such as LICEcap
see ?ListConvolve
see ?Piecewise
from the net. Using Times font family is the idea.
data = Table[{x, Random[Real, {0, x}]}, {x, 0, 10}]; ListPlot[data, Frame -> True, PlotStyle -> {FontFamily -> "Times"}, PlotLabel -> "64Cycles in FIFO", FrameLabel -> {"S/N dB", "RMS error mm"}]
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Rememebr: Position and Cases return result that can be used by Extract directly. But can’t be used by Part directly.
a = Table[RandomInteger[100], {4}, {4}]
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Out[69]= {{55, 63, 78, 45}, {13, 45, 67, 1}, {94, 32, 48, 90}, {31, 75, 43, 60}}
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a[[All,1]]
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Out[70]= {55, 13, 94, 31}
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Find rows which has elements in first column less than 3 in the following
a = {{1, 2, 3}, {4, 5, 8}, {7, 8, 9}}
Reference: how-to-extract-rows-from-matrix-based-on-value-in-first-entry
The solution using pattern below (by WRech) is interesting since the same pattern can be used by Cases and Position.
solution by me
pos = Position[a[[All,1]], _?(#1 <= 4 & )] Out[73]= {{1}, {2}} Extract[a, pos] Out[74]= {{1, 2, 3}, {4, 5, 8}}
by Simon
pos = Position[a, _List?(First[#1] <= 4 & ), {1}] Out[75]= {{1}, {2}} Extract[a, pos] Out[76]= {{1, 2, 3}, {4, 5, 8}}
By Asim
Pick[a, a[[All,1]], _?(#1 <= 4 & )] Out[77]= {{1, 2, 3}, {4, 5, 8}}
By WReach
Cases[a,{n_,__}/;n<=4,{}] Out[78]= {{1,2,3},{4,5,8}}
By WReach
pos=Position[a,{n_,__}/;n<=4,{}] Extract[a,pos] Out[79]= {{1},{2}} Out[80]= {{1,2,3},{4,5,8}}
Random values on the diagonal
DiagonalMatrix[Table[Random[], {3}]]
Ones on the diagonal
DiagonalMatrix[Table[1, {3}]]
one way, using SparseArray
Normal[SparseArray[{{i_, i_} :> 2*i, {i_, j_} :> i + j /; i < j}, {5, 5}]] Out[81]= {{2, 3, 4, 5, 6}, {0, 4, 5, 6, 7}, {0, 0, 6, 7, 8}, {0, 0, 0, 8, 9}, {0, 0, 0, 0, 10}}
Or using Table. But notice that in SparseArray, the ’zeros’ are already the default case, so using SparseArray is simpler.
Table[If[i == j, 2*i, If[i < j, i + j, 0]], {i, 5}, {j, 5}] Out[82]= ... same as above
see ?Tr[a]
a = {{1, 2, 3}, {4, 5, 8}, {7, 8, 9}} Tr[a, Times] Out[84]= 45
by Jon MacLoone
DiagonalQ[m_List] /; ArrayDepth[m] === 2 && Equal @@ Dimensions[m] := And @@ Flatten[MapIndexed[#1 === 0 || Equal @@ #2 & , m, {2}]]; DiagonalQ[m_] := Return[False]; a = {{1, 2}, {2, 4}} b = {{1, 0}, {0, 2}} DiagonalQ[a] Out[89]= False DiagonalQ[b] Out[90]= True
By Paul Abbott
Clear[DiagonalQ]; DiagonalQ[(m_)?MatrixQ] /; SameQ @@ Dimensions[m] := m === DiagonalMatrix[Tr[m, List]] DiagonalQ[a] Out[93]= False DiagonalQ[b] Out[94]= True
Find location of zeros in this matrix
a = {{1, 2, 3}, {4, 0, 8}, {7, 8, 0}}
one way
Position[a, 0] Out[96]= {{2, 2}, {3, 3}}
Another way
Position[a, _?(#1 == 0 & )] Out[97]= {{2, 2}, {3, 3}}
find all elements between 4 and 8
a = {{1, 2, 3}, {4, 0, 8}, {7, 8, 0}} Position[a, _?(#1 >= 4 && #1 <= 8 & )] Out[99]= {{2, 1}, {2, 3}, {3, 1}, {3, 2}} Extract[a, %] Out[100]= {4, 8, 7, 8}
Using Part to inser 99 in position (1,1)
a = {{1, 2, 3}, {4, 0, 8}, {7, 8, 0}} a[[1,1]] = 99; a Out[103]= {{99, 2, 3}, {4, 0, 8}, {7, 8, 0}}
a = {{1, 2, 3}, {4, 0, 8}, {7, 8, 0}}
To insert this row in the second row in matrix above
row = {97, 98, 99}; newa = Insert[a, row, {2}] Out[106]= {{1, 2, 3}, {97, 98, 99}, {4, 0, 8}, {7, 8, 0}}
or just use ’2’, it will also work
newa = Insert[a, row, 2] Out[107]= {{1, 2, 3}, {97, 98, 99}, {4, 0, 8}, {7, 8, 0}}
a = {{1, 2, 3}, {4, 0, 8}, {7, 8, 0}}
To insert this column in the second column position in above matrix
column = {97, 98, 99};
one way
newa = Transpose[Insert[Transpose[a], column, 2]] Out[110]= {{1, 97, 2, 3}, {4, 98, 0, 8}, {7, 99, 8, 0}}
another way
Normal[SparseArray[{{i_, j_} :> column[[i]] /; j == 2, {i_, j_} :> a[[i,j]] /; j == 1, {i_, j_} :> a[[i,j - 1]] /; j > 1}, {3, 4}]] Out[111]= {{1, 97, 2, 3}, {4, 98, 0, 8}, {7, 99, 8, 0}}
Another way by Leonid Shifrin how-to-insert-a-column-into-a-matrix-the-correct-mathematica-way
MapThread[Insert, {a, column, Table[2, {Length[column]}]}] Out[112]= {{1, 97, 2, 3}, {4, 98, 0, 8}, {7, 99, 8, 0}}
Another by Leonid Shifrin
ArrayFlatten[{{a[[All,1 ;; 1]], Transpose[{column}], a[[All,2 ;; All]]}}] Out[113]= {{1, 97, 2, 3}, {4, 98, 0, 8}, {7, 99, 8, 0}}
Given
a = {{1, 2, 3}, {4, 0, 8}, {7, 8, 0}}
and we want to make matrix { {a,a},{a,a} }
b = ArrayFlatten[ {{a, a}, {a, a}}] Out[118] {{1, 2, 3, 1, 2, 3}, {4, 0, 8, 4, 0, 8}, {7, 8, 0, 7, 8, 0}, {1, 2, 3, 1, 2, 3}, {4, 0, 8, 4, 0, 8}, {7, 8, 0, 7, 8, 0}}]
Given
a = {{1, 2, 3}, {4, 0, 8}, {7, 8, 0}}
and we want to apply the this function to it
f[x_] := x + 2*Sin[x]
Then using Map
r = Map[f[#1] & , a, {2}] Out[123]= {{1 + 2*Sin[1], 2 + 2*Sin[2], 3 + 2*Sin[3]}, {4 + 2*Sin[4], 0, 8 + 2*Sin[8]}, {7 + 2*Sin[7], 8 + 2*Sin[8], 0}}
Remove["Global`*"] Refine[Sin[x]^2 + Cos[x]^2 == q, q == 1] Out[125]= True
One way
$PrePrint = If[MatrixQ[#], MatrixForm[#], #] &; m = {{1, 2}, {3, 4}}
Another otpion is to use TraditionalForm. Can change default form from the menu, so this way no need to change $PrePrint
Use Boxed -> False
Plot3D[2 x + 7 y, {x, -4, 4}, {y, -4, 4}, Boxed -> False, AxesEdge -> {{-1, -1}, {-1, -1}, {-1, -1}}, AxesLabel -> {x, y, z}]
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One way to use Transpose
x = Table[i, {i, 0, 2 Pi, Pi/10}]; y = Sin[x]; data = Transpose[{x, y}]; ListPlot[data, Joined -> True]
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v = {1, 2, 3}; ListCorrelate[v, v, {-1, 1}, 0] Out[139]= {3, 8, 14, 8, 3}
In Matlab it is
v=[1,2,3] xcorr(v,v) ans = 3 8 14 8 3
From the net, lost reference
See also how-to-draw-a-spring
SetDirectory[$BaseDirectory]; FileNames["*"]
Possible locations for init.m files include the following:
I have my init.m in the following folder
C:\Documents and Settings\All Users\Application Data\Mathematica\Kernel\init.m
maxy = Pi; Row[{VerticalSlider[Dynamic[maxy], {-2 Pi, 2 Pi}, Appearance -> "LeftArrow"], Dynamic@Plot[Sin[x], {x, -Pi, Pi}, PlotRange -> {{-Pi, Pi}, {-maxy, maxy}}, ImageSize -> 200]}]
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some notes below
Precision means the variability between estimates Accuracy means the amount of deviation between the estimate and the "true value"
The condition number is the ratio of the output error to the input error. if the condition number is about 10k, then one loses about k digits of accuracy.
The main sources of inaccuracy (= error) is truncation error and round-off error.
From the above dart diagram, then we can say this: a value is accurate if it is near the bull-eye. But if is away from the bull-eye, but it is always away from the bull-eye and in the same place, then it is precise. So something can be precise but not accurate. So precise has to do with repeated values. i.e. one can’t say a value is precise, but must talk about an experiment being precise, it is produced same result each time (or very close results each time).
So, it is best of course to be both accurate and precise. So what does it mean to be accurate but not precise? using the above dart diagram, it means values generated from the experiment are always close to the pull eye, but not in the same locations.
From http://forums.wolfram.com/mathgroup/archive/2010/Jan/msg00917.html
The definition of precision in Mathematica is this. Suppose x is a number known up to an error of epsilon, that is it can be viewed as lying in the interval (x-epsilon/2,x+epsilon/2). Then its precision is -Log[10,epsilon/x]. Its accuracy is -Log[10,epsilon]. The two are related by the equation: Precision[x] - Accuracy[x] == RealExponent[x] The interpretation in terms of digits is only approximate. Both accuracy and precision can be negative - this depends on the scale of the number i.e. RealExponent. A number will have negative accuracy if its absolute error is large. It is easy to produce such numbers by cancellation With[{x = N[10^100, 50] - N[10^100, 50]}, Accuracy[x]] -50.301 On the other hand, since $MinPrecision 0 You won't normally in Mathematica see numbers with negative Precision. Precision is the main concept, Accuracy is only used because Precision is singular at 0 (remember - its relative error). It's all perfectly documented so this tired scape goat is not available this time.
In math italicize single Roman letters that are variables or functions (example, x,y,f(x),t 2. Exception to above is capital letters for points like P and Q in geometry. 3. Do not italicize Greek letters (example, alpha, gamma, beta, etc..), and units like sec or rad, or punctuation like ( ). 4. Styling the control labels is optional. 5. Do not use strings with <> for such formatting. Use Row[{ }] 6. put () around units in plot labels. As (sec) or (hz) 7. do not italicize function names longer than one letter. So Style["exp",Italic] should just be "exp" 8. The t in delta(t) should be italic--but not the delta, Greek letter are not Italian letters is how I remember that. 9. Log should be log. 10. Is j^2= -1? Better say so in the caption for non EE.
see also http://demonstrations.wolfram.com/guidelines.html
I made small copy here so I do not have to keep looking for this all the time.
To put label on plot, example
To typeset math for display on the demo use this type of coding
Text@Style[TraditionalForm[HoldForm[Sin[x] + Cos[y]], 12]]
To add invisible space use ESC is ESC
By Bob Hanlon from math group:
Clear[x, $PrePrint] expr = {E^x, x, x^2, Log[x]}; Position[expr, _?( !PolynomialQ[#1, x] & ), 1] Out[146]= {{1}, {4}}
on windows, V 8, example data is located in
C:\Program Files\Wolfram Research\Mathematica\8.0\Documentation\English\System\ExampleData
and it can be read like this
str = OpenRead["ExampleData/USConstitution.txt"] Out[147]= InputStream[ExampleData/USConstitution.txt, 127]
Use SphericalRegion->True
ListPlot3D[Table[RandomReal[], {5}, {5}], AxesLabel -> {"x", "y", "z"}, ImageSize -> {200, 200}, ImagePadding -> 20, SphericalRegion -> True];
This question was posted on the net. Given
b:=Table[{x,y},{x,1,6},{y,1,6}]
select from it elements \(x,y\) which satsify \(x+y>9\)
some answers
(me) Select[Flatten[b, 1], #1[[1]] + #1[[2]] > 9 & ] Out[152]= {{4, 6}, {5, 5}, {5, 6}, {6, 4}, {6, 5}, {6, 6}}
Adriano Pascoletti answer
Cases[b, {x_Integer, y_Integer} /; x + y > 9, {2}]
Bill Row answer
Cases[Flatten[b, 1], _?(Total[#1] > 9 & )]
Murray Eisenberg answer
Select[Flatten[b, 1], First[#1] + Last[#1] > 9 & ]
Given a matrix, say which has Indeterminate and we want to change all these entries in the matrix by zero.
mat = {{-1., -1., Indeterminate, -1., -1.}, {-1., -1., Indeterminate, -1., -1.}, {Indeterminate, Indeterminate, Indeterminate, Indeterminate, Indeterminate}, {-1., -1., Indeterminate, -1., -1.}, {-1., -1., Indeterminate, -1., -1.}} p = Position[mat, Indeterminate] mat = ReplacePart[mat, p -> 0] Out[169]= {{-1., -1., 0, -1., -1.}, {-1., -1., 0, -1., -1.}, {0, 0, 0, 0, 0}, {-1., -1., 0, -1., -1.}, {-1., -1., 0, -1., -1.}}
another example, given a matrix of values, replace those values which are less than \(0.5\) by NULL
n = 5; a = Table[RandomReal[], {n}, {n}]; p = Position[a, x_ /; x < 0.5]; a = ReplacePart[a, p -> Null] Out[173]= {{Null, Null, Null, 0.6781657418995635, 0.7290662037036753}, {Null, 0.7084980071179792, Null, Null, 0.5811489862295911}, {Null, Null, 0.8467863882617719, Null, 0.8891915946646993}, {0.8173279058333203, 0.7272894246356278, Null, Null, 0.8665880423275274}, {Null, Null, 0.662026816962838, 0.5982839657423036, 0.6603967280952212}}
see full-documentation-for-appearanceelements
list = { "AutorunPlayButton", "BookmarksButton", "BookmarksPlayButton" "ContentResizeArea", "DirectionButton", "FasterSlowerButtons", "HideControlsButton", "InteractiveTradingChartMenu", "InteractiveTradingChartSnapshotButton", "InteractiveTradingChartResetButton", "InputField", "InlineInputField", "ManipulatePlayButton", "ManipulateMenu", "PlayPauseButton", "ProgressSlider", "ResetButton", "SnapshotButton", "StepLeftButton", "StepRightButton", "UpdateButton", None};
one way
p = Plot[Sin[x], {x, 0, Pi}, MaxRecursion -> 0, PlotPoints -> 10]; data = Cases[Normal[p], x_Line :> First[x], Infinity]; Show[p, ListPlot[data, PlotStyle -> Red]]
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p = ContourPlot3D[x^2 + y^3 - z^3 == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, PlotPoints -> Automatic] data = (InputForm@p)[[1, 1, 1]]
and
data = Reap[DensityPlot[Sin[x*y], {x, 0, 2 Pi}, {y, 0, 2 Pi}, EvaluationMonitor :> Sow[{x, y, Sin[x*y]}]]][[2, 1]]; ListPlot3D[data]
Useful notes taken from different places from Mathematica documentation. And some added by me.
(added 7/5/14) There is a race condition between when updating a variable in the second argument of dynamics, which is also updated in the Manipulate expression. The problem is easy to explain
Manipulate[.... n=n+1;...., Manipulator[Dynamic[f,{f=#;n=0}&....]
Now, what happens, sometimes, is that when moving the slider, and setting \(n=0\), that this update to \(n\) can be lost, since it depends on the timing of when this happens. The Manipulate expression could be in the middle on updating \(n\) itself. This is classical lost update problem in multi-threading. The way to avoid this, is to follow this rule of thumb: When using second argument of dynamics in a Manipulate control, do not update a shared variable which can also be updated inside the Manipulate expression, as the update can be lost. The write to the shared variable/updating should only happen either inside the Manipulate expression or in the second argument of dynamics. But not in both places
Module variables should *never* appear inside Dynamics or Manipulates internal to that Module. To be clear with some examples (all using Dynamic, but they could equally well use Manipulate, which is implemented using Dynamic)... (* OK *) Dynamic[Module[{a}, a]] (* OK *) Module[{a}, (* expression involving a*); Dynamic[(* expression *not* involving a *)] (* BAD *) Module[{a}, Dynamic[a]]
By John Fultz on math group, jan 24/2012
"Generally, you should construct controls so that they’re not inside Dynamics that will trigger while you’re interacting with those controls, since this can create instability"
By John Fultz on math group, feb 3/2012
"CDF files which you expect to deploy cannot rely on Shift+Enter evaluations to prime the pump. You need to make sure that all of the code dependencies are in the dynamic evaluations somewhere. Some possible ways of doing this, all of which have been discussed at various points on MathGroup, include:
* Using the Initialization option of Dynamic, DynamicModule, or Manipulate
* Using the SaveDefinitions option of Manipulate
* Adding code to the NotebookDynamicExpression option of the notebook (if it’s initialization code, then wrapped in Refresh[#,None]& to prevent it from evaluating more than once per session).
* Not such a good idea for function definitions, but if you simply have code that needs to run before Dynamic code runs, nesting a DynamicWrapper around it might be appropriate, too."
Notes from WRI tech support
This is the support explanation of why this error came showed up: The issue is specifically with the section: Evaluate@env[{{age, 100, "age"}, 10, 200, 1}] Manipulate doesn't really evaluate until it gets to the Initialization option, but it will check its input for correct form. Mathematica reads the main body of the Manipulate before running the Initialization option. This is can be verified by using a Print statement: Initialization -> (Print["Test"]; makeCustomEnvironmentAlt = Function[Null, Function[code, With @@ Hold[{##}, code], HoldAll], HoldAll]; env = makeCustomEnvironmentAlt[$age = 1, $salary = 2]; Print["Test"]) Test does not print. Getting around this will be probably not be clean. .... Having the code for the controller for age depend on evaluation of some function which must be initialized does not appear to be possible with simply Manipulate.
see how-to-define-constants-for-use-with-with-in-one-place-and-then-apply-them-lat
Some useful posts and links on dynamics why-wont-this-work-dynamic-in-a-select
When DynamicModule is first evaluated, initial assignments for local variables are made during the evaluation. Any setting for the Initialization option is evaluated only when the output of DynamicModule is displayed.
see how-to-make-dynamicmodule-work-without-an-extra-enter
Here is a trick to allow one to control one slider based on another
Manipulate[{a, b}, Grid[{ {"a", Manipulator[Dynamic[a, {(a = #) &, (a = #; If[b > a, b = a]) &}], {1, 10, 1}], Dynamic[a]}, {"b", Manipulator[Dynamic[b, {(b = #) &, (b = #; If[b > a, b = a]) &}], {1, 10, 1}], Dynamic[b]}} ], {{a, 5}, None}, {{b, 3}, None} ]
There is no build-in struct or record in Mathematica. But this is what I do. Since in M a matrix can include non-numeric data, I use a list for a record, and then use a matrix to make an array of records (or array of structs). I just need to make a constant to indicate the field name in the record, to make it easier to reference. Here is an example
id = 1; (*first field name*) pop = 2; (*second field name*) name = 3; (*third field name*) (*now build the array of record, each row is a record*) m = {{1, 3000, "London"}, {1, 6000, "New York"}, {3, 7300, "Moscow"}}; (*now can iterate over the records*) Do[ Print@m[[i, id]]; m[[i, pop]] += 1, {i, Length[m]} ]
Ok, not very fancy, but easy to setup and data works with all M other functions, since it is just a list of lists.
Some more links on the subject
Remove[a, b, c, d, e, f] Apply[#1 + 3*#2 & , {{a, b}, {c, d}, {e, f}}, 1] Out[183]= {a + 3*b, c + 3*d, e + 3*f}
Or
Apply[#1 + 3*#2 & , {{a, b}, {c, d}, {e, f}}, {1}] Out[184]= {a + 3*b, c + 3*d, e + 3*f}
ListAnimate[Flatten[Reap[Do[Sow@Plot[Sin[x - c], {x, 0, 4 Pi}, Ticks -> None], {c, 0, 2 Pi - Pi/10, Pi/10}]]]]
Thanks to Alexey Popkov for this
SetOptions[EvaluationNotebook[], AutoStyleOptions -> {"CommentStyle" -> {FontWeight -> Plain, FontColor -> GrayLevel[0.6], ShowAutoStyles -> False, ShowSyntaxStyles -> False, AutoNumberFormatting -> False, FontFamily -> "Consolas"}}]
This came about when I was trying to convert 1/(1-x^2/2) to normal form, i.e. tell Mathematica to change the above to 1+x^2/2 But doing Simplify[1/(1-x^2/2)] or Expand does not work. So the only solution I found is to use Series command, as follows
Normal[Series[1/r, {x, 0, 2}]] Out[189]= 1 + x^2/2
Use these in parameter "declaration" of functions to make more robust. From the help
Clear[s]; f1 = 2/(s + 3); f2 = 7/(s^2 + 2.5*s + 7); Simplify[Together[f1 + f2]] Out[193]= (35. + 12.*s + 2.*s^2)/(21. + 14.5*s + 5.5*s^2 + s^3)
Options[myFun] = {form -> "linear"}; myFun[x_, OptionsPattern[]] := Module[{}, Print["x=", x, "form=", OptionValue[form]]; ] myFun[3, form -> "quadratic"]
This below is also a useful post by David Park on the net on options usage in packages
by Andrzej Kozlowski on math group, July 2010:
Suppose in the expression 2/3 I + x/y I you wish to replace all fractions (that is 2/3 and x/y) by r and I by d. Without worrying about evaluation you can do this as follows:
Unevaluated[Unevaluated[(2/3)*I + (x/y)*I] /. HoldPattern[(x_)/(y_)] -> r] /. HoldPattern[I] -> d Out[200]= -d + d*(1 - x^2/2)
If you allow the expression to evaluate the patterns will no longer match. For example, with only one Unevaluated you will get
Unevaluated[(2/3)*I + (x/y)*I] /. HoldPattern[(x_)/(y_)] -> r /. HoldPattern[I] -> d Out[201]= -I + I*(1 - x^2/2)
question: I want to replace y for x everywhere except in Exp[x].
Answer by Bob Hanlon on the net. messageID=7120881&tstart=0
Remove["Global`*"] expr = a*x + b*x^2 - c*Exp[x]; expr /. {Exp[x] -> z, x -> y} /. z -> Exp[x] Out[211]= (-c)*E^x + a*y + b*y^2
Thanks to Mike for these commands, see http: //stackoverflow.com/questions/8583521/why-do-i-get-security-warning-message-this-file-contains-potentially-unsafe-dyn
CurrentValue[$FrontEnd, {"NotebookSecurityOptions", "TrustedPath"}] Out[212]= {FrontEnd`FileName[{$InstallationDirectory}], FrontEnd`FileName[{$BaseDirectory}], FrontEnd`FileName[{$UserBaseDirectory}]} CurrentValue[$FrontEnd, {"NotebookSecurityOptions", "UntrustedPath"}] Out[213]= {FrontEnd`FileName[{FrontEnd`$DesktopDirectory}], FrontEnd`FileName[{FrontEnd`$DownloadsDirectory}], FrontEnd`FileName[{FrontEnd`$LocalApplicationDataDirectory}], FrontEnd`FileName[{FrontEnd`$RemoteApplicationDataDirectory}], FrontEnd`FileName[{FrontEnd`$ProgramFilesDirectory}], FrontEnd`FileName[{FrontEnd`$ProgramFilesX86Directory}], FrontEnd`FileName[{$TemporaryPrefix}]}
Now to find if your current notebook is on the trusted path type NotebookDirectory[] and see if the output shows up in the trusted path of not. To add a folder to trusted path go to "Preferences > Advanced > Open Options Inspector". Then under Global Preferences search for trusted
Block and Module have values, the last expression evaluated is their value, we can see this by making a Grid (or just printing). But module leaked symbols have $ signs
Remove["Global`*"] Grid[{{Module[{x}, x]}}, Frame -> All]
Modules and Blocks both execute if they are in the path of code, without calling them. Block:
Remove["Global`*"] x = 4; Block[{}, If[x == 4, x = 3]]; x Out[217]= 3
Module:
Remove["Global`*"] x = 4; Module[{}, If[x == 4, x = 3]]; x Out[221]= 3
These are the steps I use to make TOC which is just HTML links to internal tags in the notebook, where these cell tags are sections. This way, when I exprt to HTML, I end up with TOC which is hyperlinks to internal locations within the web page.
Now, using the mouse again, select the text you just pasted, and do right-click and select MAKE hyperlink. This will bring up a menu like this
lst = {{x -> 4, y -> 7}, {x -> 2, y -> 5}, {x -> -1, y -> 10}}
one way
({#1[[1,2]], #1[[2,2]]} & ) /@ lst Out[223]= {{4, 7}, {2, 5}, {-1, 10}}
another way
Cases[lst, {_ -> n_, _ -> m_} :> {n, m}] Out[224]= {{4, 7}, {2, 5}, {-1, 10}}
One way us to use Item
mat = Table[Random[], {3}, {3}]; Framed[ Item[Grid[mat, Frame -> All, Alignment -> Center], Alignment -> {Center, Top}], ImageSize -> {300, 200}]
|
|
One way us to use Item
Grid[{{"row1,row1"}, {"row2"}, {"row3"}}, Frame -> All] Grid[{{"row1,row1"}, {Item["row2", Alignment -> Left]}, {"row3"}}, Frame -> All]
Use NumberForm
NumberForm[1./10^6, ExponentFunction -> (Null & )] Out[227] 0.000001
Sometimes I get the case that the notebook retain old definitions and symbols even after I deleted them from the notebook. This happened when I was using a Demonstration stylesheet and had an separate initilization cell, and had added SaveDefinitions->True in the Manipulate cell.
To make sure the notebook clears any old symbols, enter this command in the notebook once
SetOptions[EvaluationNotebook[], PrivateNotebookOptions -> {"FileContents" -> {"NotebookData"}, "FileOutlineCache" -> False}]
In addition, I change the preferences like this:
The call for each is as follows
A list has a Head at its zero index position
lst = {1, 2, 3}; Head[lst] Out[242]= List lst[[0]] Out[243]= List
By changing the head we use Apply. For example, to add the numbers of the above lst, we need to change the Head from List to Plus. There is a command in Mathematica to change the Head, called Apply
Plus @@ lst Out[244]= 6
We could have used the zero index trick, but it is better to use Apply:
lst[[0]] = Plus Out[245]= Plus lst Out[246]= 6
If we have a list of lists, like this
lst = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}} Out[247]= {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
And we wanted to change the head of each list in it, for example, we want the product of each list shown, then we need to change the head of each list to Times. To do that, the follwing short but might be strange looking command
Apply[Times, lst, {1}] Out[248]= {6, 120, 504}
Another way to do the above is to Map the Apply function
(Times @@ #1 & ) /@ lst Out[249]= {6, 120, 504}
or, little shorter version of the above:
(Times @@ #1 & ) /@ lst Out[250]= {6, 120, 504}
Mathematica default display of polynomial is reverse the traditional form:
poly = x^3 + a*x + b*x^2 + c + d Out[251]= 27 + 3*a + 9*b + c + d
use Traditional Form with ParameterVariables to make it appear as in text books
TraditionalForm[poly, ParameterVariables :> {a, b, c, d}]
See this article by David Wagner http://www.mathematica-journal.com/issue/v6i2/columns/wagner/wagner62.pdf
And why-are-some-function-names-red
Some note here
Sort[] sorts numbers from small to large by default. By supplying a function, one can change this as needed
lst = {1, 2, 5, 3, 7}; Sort[lst] Out[255]= {1, 2, 3, 5, 7} Sort[lst, #1 < #2 & ] Out[256]= {1, 2, 3, 5, 7} Sort[lst, #1 > #2 & ] Out[257]= {7, 5, 3, 2, 1}
To find say all names in NDSolve and options used by that name if any do (this example is for NDSolve)
getList[name_String] := Module[{options, idx}, options = Names[name <> "`*"]; options = ToExpression /@ options; options = {#, Options[#]} & /@ options; idx = Range[Length[options]]; options = {#[[1]], TableForm[#[[2]]]} & /@ options; options = Insert[options[[#]], #, 1] & /@ idx; options = Insert[options, {"#", "Option", "Options to this option"}, 1]; Grid[options, Frame -> All, Alignment -> Left, FrameStyle -> Directive[Thickness[.005], Gray]] ];
then call it with
getList["NDSolve"]
It will produce large table. Here is part of it
getList["FindMinimum"]
if one types in 1,2,3,4 is there is a way to select these and have {} automatically put around them to make a list {1,2,3,4} using escape key shortcut?
Answer by Chris Degnen who wrote this
FrontEndExecute[FrontEnd`AddMenuCommands["DuplicatePreviousOutput", {Delimiter, MenuItem["Make List", FrontEnd`KernelExecute[nb = SelectedNotebook[]; sel = NotebookRead[nb]; NotebookWrite[nb, Cell[BoxData[RowBox[{"{", sel, "}"}]]]]], MenuKey["u", Modifiers -> {"Control"}], MenuEvaluator -> Automatic]}]]
put it in the init file to load at start-up. See wrap-text-selection-in-brackets-in-mathematica
I needed to do this when I was parsing some output. The problem is like this: given a string say "foo[] boo[] more goo[] more" and wanted to look for pattern like this "__["
In other words, a letter or more that end with "[", but needed to find the first one. Hence in the above, I wanted to find "foo".
2 ways to do this:
s = "foo[] boo[] more goo[] more"; StringCases[s, RegularExpression["^\\w*\\["]] Out[265]= {foo[}
and
StringCases[s, Shortest[StartOfString~~__~~"["], Overlaps -> False] Out[266]= {foo[}
Manipulate[x, {x, {True, False}}, Grid[{ {Dynamic@If[x, Control[{y, {True, False}}], Control[{z, 0, 1, .1}] ] } } ] ]
Takes function and applies it to each element in a list
f /@ {a, b, c} Out[267]= {f[a], f[b], f[c]} (1 + g[#1] & ) /@ {a, b, c} Out[268]= {1 + g[a], 1 + g[b], 1 + g[c]} f /@ {a, b, c} Out[269]= {f[a], f[b], f[c]}
Use when function needs to be called with arguments taken from more than one list, else use Map if argument come from one list
Thread[f[{a, b, c}]] Out[270]= {f[a], f[b], f[c]} f /@ {a, b, c} Out[271]= {f[a], f[b], f[c]} Thread[f[{a, b, c}, {1, 2, 3}]] Out[272]= {f[a, 1], f[b, 2], f[c, 3]}
MapThread[f, {{a, b, c}, {1, 2, 3}}] Out[273]= {f[a, 1], f[b, 2], f[c, 3]}
In this case gives the same answer as using Thread
Thread[f[{a1, a2, a3}, {b1, b2, b3}]] Out[274]= {f[a1, b1], f[a2, b2], f[a3, b3]}
This is only when the lists are one level. For 2 levels we have to use MapThread. This shows the difference
MapThread[f, {{{a, b}, {c, d}}, {{1, 2}, {3, 4}}}] Out[275]= {f[{a, b}, {1, 2}], f[{c, d}, {3, 4}]} Thread[f[{{{a, b}, {c, d}}, {{1, 2}, {3, 4}}}]] Out[276]= {f[{{a, b}, {c, d}}], f[{{1, 2}, {3, 4}}]}
see tutorial/PatternsOverview this below from tutorial/PuttingConstraintsOnPatterns
See also what-is-the-recommended-way-to-check-that-a-list-is-a-list-of-numbers-in-argumen
foo[(x_)?(Element[#1, Integers] & )] := x foo[x_Integer] := x foo[x_Integer] := x
foo[(x_)?(IntegerQ[#1] && #1 > 0 & )] := x foo[x_Integer /; x > 0] := x foo[(x_Integer)?Positive] := x foo[x_Integer /; x > 0] := x
foo[(x_)?(IntegerQ[#1] && #1 < 0 & )] := x foo[x_Integer /; x < 0] := x foo[(x_Integer)?Negative] := x foo[x_Integer /; x < 0] := x
foo[(x_)?(IntegerQ[#1] && #1 >= 0 & )] := x foo[x_Integer /; x >= 0] := x foo[(x_Integer)?NonNegative] := x foo[x_Integer /; x >= 0] := x
foo[(x_)?(IntegerQ[#1] && #1 <= 0 & )] := x foo[x_Integer /; x <= 0] := x foo[(x_Integer)?NonPositive] := x foo[x_Integer /; x <= 0] := x
foo[x_Integer /; x > 3 && x < 7] := x
foo[x_?(Element[#, Reals] &)] := x foo[x_Real] := x
foo[x_Real /; x > $MachineEpsilon] := x foo[x_Real /; x > $MachineEpsilon] := x foo[x_Real /; Positive[x]] := x foo[x_ (Element[#, Reals] && Positive[#] &)] := x
foo[x_Real /; x < $MachineEpsilon] := x foo[x_Real /; x < $MachineEpsilon] := x foo[x_Real /; Negative[x]] := x foo[x_?(Element[#, Reals] && Negative[#] &)] := x
foo[x_Real /; x >= $MachineEpsilon] := x foo[x_Real /; x >= $MachineEpsilon] := x foo[x_Real /; Positive[x] || x == 0] := x foo[x_ (Element[#, Reals] && (Positive[#] || # == 0) &)] := x
foo[x_Real /; x <= $MachineEpsilon] := x foo[x_Real /; x <= $MachineEpsilon] := x foo[x_Real /; Negative[x] || x == 0] := x foo[x_ (Element[#, Reals] && (Negative[#] || # == 0) &)] := x
foo[x_ (Element[#, Reals] && ((# - 3) > $MachineEpsilon && (7 - #) > $MachineEpsilon) &)] := x foo[x : _Real /; (x - 3) > $MachineEpsilon && (7 - x) > $MachineEpsilon] := x
foo[x_?(Element[#, Booleans] &)] := x
foo[x_?(Element[#, Reals] &)] := x foo[x_?(NumericQ[#] &)] := x foo[x : _?NumericQ] := x
foo[x_?(NumberQ[#] &)] := x
foo[x_Complex] := x foo[x_?(Not@FreeQ[#, _Complex] &)] := x
foo[x_List] := x
foo[x_?(VectorQ[#] &)] := x
foo[x_?(VectorQ[#, NumericQ] &)] := x
foo[x_?(VectorQ[#, NumericQ] &)] := x foo[x : {_?NumericQ ..}] := x foo[x : {__?NumericQ }] := x foo[x_?(VectorQ[#, IntegerQ] &)] := x
foo[x_?(MatrixQ[#, NumericQ] &)] := x foo[x : {{_?NumericQ ..}}] := x foo[x : {{__?NumericQ }}] := x
foo[x_?(MatrixQ[#, NumericQ] && FreeQ[#, _Complex] &)] := x
foo[x_?(MatrixQ[#, StringQ] &)] := x
use MatchQ
MatchQ[1/3, _Rational] Out[277]= True MatchQ[3, _Integer] Out[278]= True Or for the above can do IntegerQ[3] Out[279]= True
Grid[{ {Item[a, Alignment -> Center], b}, {SpanFromAbove, c}}, Frame -> All] Grid[{ {Item[a, Alignment -> Center], Item[Column[{b, c}]]}}, Frame -> All] Grid[{ {Item[a, Alignment -> Center], Item[Column[{b, c}, Frame -> All]]}}, Frame -> All] Grid[{ {a, Item[b, Alignment -> Center]}, {c, SpanFromAbove}}, Frame -> All] Grid[{ {Item[a, Alignment -> Center], Item[b, Alignment -> Center], c}, {SpanFromAbove, SpanFromAbove, d}, {SpanFromAbove, e, f}}, Frame -> All]
From help
See select-and-blank
test = {{"String1", "a"}, {"String2", "b"}, {"String3", "a"}, {"String4", "a"}}; Cases[test, {_String, "a"}] Out[281]= {{String1, a}, {String3, a}, {String4, a}} Select[test, MatchQ[#1, {_String, "a"}] & ] Out[282]= {{String1, a}, {String3, a}, {String4, a}}
See given-a-symbolic-expression-how-to-find-if-starts-with-a-minus-or-not
Clear[x] p = (_.)*_?Negative; MatchQ[-3*x^2, p] Out[285]= True MatchQ[3*x^2, p] Out[286]= False expr = -3*x^2; (expr /. Thread[Variables[expr] -> 1]) < 0 Out[288]= True expr = 3*x^2; (expr /. Thread[Variables[expr] -> 1]) < 0 Out[290]= False
See convert_manipulate_to_dynamicModule
Suppose we have \(u''(t)+u'(t)+u(t)=3 \cos (2 t)\) and we wanted to find a particula solution by replacing \(u\) in the differential equation by some guess for a particular solution. Then do
ode = Derivative[2][u][t] + Derivative[1][u][t] + u[t] == 3*Cos[2*t]; ode /. u -> (c1*Cos[#1] + c2*Sin[#1] & ) Out[2]= c2*Cos[t] - c1*Sin[t] == 3*Cos[2*t]
Watch out for adding the extra third argument to trigger as show below (which is 1 now). This seems to cause a problem. Was using it in Manipulate and when I added, sometimes the trigger stops firing on its own. When I remove it, it never does stop.
Trigger[Dynamic[t0, {t0 = #} &], {0, 10000, 1}, ....] ToString[#] & /@ a| is same as ToString /@ a
see how-to-select-and-delete-all-output-cells
To make a matrix, which contains on its diagonal matrices, Matlab uses the command blkdiag. In Mathematica use the following
a = {{1, 2, 3}, {4, 5, 6}} b = {{7, 8}, {9, 10}} SparseArray[Band[{1, 1}] -> {a, b}]
Below is from "accuracy and stability of numerical algorithms", by Highma, page 36
ClearAll["Global`*"] lst = {{a, b}, {c, d}}; MapThread[f, lst] Out[14]= {f[a, c], f[b, d]} Thread[f[Sequence @@ lst]] Out[15]= {f[a, c], f[b, d]}
ClearAll[x, y, z, g, foo]; p1 = Conjugate[x_]*Conjugate[y_] :> Conjugate[x*y]; p2 = (x_)*Conjugate[x_] :> Abs[x]^2; p3 = Abs[(x_)*(y_)]^(n_.) :> Abs[x]^n*Abs[y]^n; p4 = (x_)*Conjugate[y_] + (y_)*Conjugate[x_] :> 2*(Re[x]*Re[y] + Im[x]*Im[y]); p5 = (x_) + Conjugate[x_] :> 2*Re[x]; allRules = {p1, p2, p3, p4, p5};
test it
expr = {{foo = x*Conjugate[y] + y*Conjugate[x]; foo, foo //. allRules}, {foo = x*Conjugate[y] + y*Conjugate[x] + z*Conjugate[g] + g*Conjugate[z]; foo, foo //. allRules}, {foo = x*Conjugate[x]; foo, foo //. allRules}, {foo = x*y*Conjugate[x*y]; foo, foo //. allRules}, {foo = x*y*z*Conjugate[x*y*z]; foo, foo //. allRules}, {foo = x + Conjugate[x]; foo, foo //. allRules}, {foo = x*y + Conjugate[x*y], foo; foo //. allRules}, {foo = x*y*z + Conjugate[x*y*z]; foo, foo //. allRules}, {foo = x*y + Conjugate[x]*Conjugate[y]; foo, foo //. allRules}, {foo = x*y*z*g + Conjugate[x]*Conjugate[y]*Conjugate[z]*Conjugate[g]; foo, foo //. allRules}}; Grid[expr, Frame -> All, Spacings -> {0.5, 1}, Alignment -> Left]
See mathematica/guide/ListingOfNamedCharacters.html
From extract-values-for-viewmatrix-from-a-graphics3d/3538
by Yu-Sung Chang (Wolfram Research)
eq = E^(0.002/t) + E^(0.032/t) == 2*E^(0.03/t) Thread[Log[eq], Equal] Out[26]= Log[E^(0.002/t) + E^(0.032/t)] == Log[2*E^(0.03/t)]
I wrote these for an answer here http://community.wolfram.com/groups/-/m/t/153862?p_p_auth=GLyok3xN
The manipulate expression is anything between the start of Manipulate and the first ","
Manipulate[Plot[Sin[x (1 + a x)], {x, 0, 6}], {a, 0, 2}]
It has the form
Manipulate[expression, controlVariables, Initialization :> ()]
foo[] := Plot[Sin[x (1 + a x)], {x, 0, 6}] Manipulate[Evaluate@foo[], {a, 0, 2}]
move foo[] in the above example to inside Manipulate, in the initialization section, add literal symbol a so Manipulate will track it
Manipulate[a; foo[], {a, 0, 2}, Initialization :> (foo[] := Plot[Sin[x (1 + a x)], {x, 0, 6}]) ]
Notice the trick above. a was added so that it appears in the Manipulate expression. Otherwise, it will not be tracked. You’ll get an initial plot, but nothing will happen when moving the slider
move foo[] to global context, but have to tell Manipulate that LocalizeVariable is false
foo[] := Plot[Sin[x (1 + a x)], {x, 0, 6}]; Manipulate[a; foo[], {a, 0, 2}, LocalizeVariables -> False]
But notice, we still need to put a somewhere in the expression for it to tracked. Not enough just to say LocalizeVariables -> False
It is not enough to use TrackedSymbols :>a, if the symbol itself do not show up in the expression. Hence this does not work
foo[] := Plot[Sin[x (1 + a x)], {x, 0, 6}]; Manipulate[foo[], {a, 0, 2}, LocalizeVariables -> False, TrackedSymbols :> a]
Notice there is no a in the expression. Manipulate will not track a even though we told it to !
Same as case 4, even if we put the function inside the Initialization section, it will still not track a . One will get an initial plot, but that is all.
Manipulate[foo[], {a, 0, 2}, TrackedSymbols :> a, Initialization :> (foo[] := Plot[Sin[x (1 + a x)], {x, 0, 6}];)]
Putting the function definition of foo[] itself inside Manipulate expression, now Manipulate sees a there and will automatically track it. No need to do anything more:
Manipulate[Module[{}, foo[] := Plot[Sin[x (1 + a x)], {x, 0, 6}]; foo[]], {a, 0, 2}]
Or simply
Manipulate[Plot[Sin[x (1 + a x)], {x, 0, 6}], {a, 0, 2}]
This is the method I use myself. Put all the functions inside the initialization section, but pass the dependent variables by argument call.
Manipulate[foo[a], {a, 0, 2}, Initialization :> ( foo[a_] := Module[{x}, Plot[Sin[x (1 + a x)], {x, 0, 6}]] )]
I like this, becuase it achieves both the goal of having all the slider symbols inside the Manipulate expression, hence Manipulate will track them, and at the same time, it avoid the function be in global context, and it is the modular way, since one can see which parameter each function depends on by looking at the signature of the function.
Similar to case 7, but the function itself is now in the global context. This is a fine solution as well, if this function needs to be called from somewhere else as well other than from the Manipulate. Otherwise, it is best not to have in the global context and use case 7.
foo[a_] := Module[{x}, Plot[Sin[x (1 + a x)], {x, 0, 6}]]; Manipulate[foo[a], {a, 0, 2}]
Watch out when defining a function such as this:
f[x_] := Integrate[x - t, {t, 0, x}]
The problem is this:
In[45]:= f[t] Out[45]= 0
This is becuase the replacement of "x" by "t" changed the integrand to zero.
The correct way is to always use Module symbols for everything inside the function, like this
f[x_] := Module[{t}, Integrate[x - t, {t, 0, x}]]
Now it gives the correct answer regardless of the symbol used as argument
In[46]:= f[t] Out[46]= t^2/2 In[47]:= f[x] Out[47]= x^2/2
StringCases[#, ___ ~~ "Distribution" ~~ ___ :> #] & /@ Names["System`*"]; DeleteCases[%, {}]
Gives
{{"ArcSinDistribution"}, {"BarabasiAlbertGraphDistribution"}, {"BatesDistribution"}, {"BeckmannDistribution"}, \ {"BenfordDistribution"}, {"BeniniDistribution"}, {"BenktanderGibratDistribution"}, {"BenktanderWeibullDistribution"}, \ {"BernoulliDistribution"}, {"BernoulliGraphDistribution"}, {"BetaBinomialDistribution"}, {"BetaDistribution"}, \ {"BetaNegativeBinomialDistribution"}, {"BetaPrimeDistribution"}, {"BinomialDistribution"}, {"BinormalDistribution"}, \ {"BirnbaumSaundersDistribution"}, {"BorelTannerDistribution"}, {"CauchyDistribution"}, {"CensoredDistribution"}, \ {"ChiDistribution"}, {"ChiSquareDistribution"}, {"CompoundPoissonDistribution"}, {"CopulaDistribution"}, {"CoxianDistribution"}, \ {"DagumDistribution"}, {"DataDistribution"}, {"DavisDistribution"}, {"DegreeGraphDistribution"}, {"DirichletDistribution"}, \ {"DiscreteUniformDistribution"}, {"DistributionChart"}, {"DistributionDomain"}, {"DistributionFitTest"}, \ {"DistributionParameterAssumptions"}, {"DistributionParameterQ"}, {"EmpiricalDistribution"}, {"ErlangDistribution"}, \ {"EstimatedDistribution"}, {"ExpGammaDistribution"}, {"ExponentialDistribution"}, {"ExponentialPowerDistribution"}, \ {"ExtremeValueDistribution"}, {"FailureDistribution"}, {"FindDistributionParameters"}, {"FirstPassageTimeDistribution"}, \ {"FisherHypergeometricDistribution"}, {"FisherZDistribution"}, {"FRatioDistribution"}, {"FrechetDistribution"}, \ {"GammaDistribution"}, {"GeometricDistribution"}, {"GompertzMakehamDistribution"}, {"GraphPropertyDistribution"}, \ {"GumbelDistribution"}, {"HalfNormalDistribution"}, {"HistogramDistribution"}, {"HotellingTSquareDistribution"}, \ {"HoytDistribution"}, {"HyperbolicDistribution"}, {"HyperexponentialDistribution"}, {"HypergeometricDistribution"}, \ {"HypoexponentialDistribution"}, {"InverseChiSquareDistribution"}, {"InverseGammaDistribution"}, {"InverseGaussianDistribution"}, \ {"JohnsonDistribution"}, {"KDistribution"}, {"KernelMixtureDistribution"}, {"KumaraswamyDistribution"}, {"LandauDistribution"}, \ {"LaplaceDistribution"}, {"LevyDistribution"}, {"LindleyDistribution"}, {"LogGammaDistribution"}, {"LogisticDistribution"}, \ {"LogLogisticDistribution"}, {"LogMultinormalDistribution"}, {"LogNormalDistribution"}, {"LogSeriesDistribution"}, \ {"MarginalDistribution"}, {"MaxStableDistribution"}, {"MaxwellDistribution"}, {"MeixnerDistribution"}, {"MinStableDistribution"}, \ {"MixtureDistribution"}, {"MoyalDistribution"}, {"MultinomialDistribution"}, {"MultinormalDistribution"}, \ {"MultivariateHypergeometricDistribution"}, {"MultivariatePoissonDistribution"}, {"MultivariateTDistribution"}, \ {"NakagamiDistribution"}, {"NegativeBinomialDistribution"}, {"NegativeMultinomialDistribution"}, {"NoncentralBetaDistribution"}, \ {"NoncentralChiSquareDistribution"}, {"NoncentralFRatioDistribution"}, {"NoncentralStudentTDistribution"}, {"NormalDistribution"}, \ {"OrderDistribution"}, {"ParameterMixtureDistribution"}, {"ParetoDistribution"}, {"PascalDistribution"}, {"PearsonDistribution"}, \ {"PERTDistribution"}, {"PoissonConsulDistribution"}, {"PoissonDistribution"}, {"PolyaAeppliDistribution"}, {"PowerDistribution"}, \ {"PriceGraphDistribution"}, {"ProbabilityDistribution"}, {"ProductDistribution"}, {"RayleighDistribution"}, \ {"ReliabilityDistribution"}, {"RiceDistribution"}, {"SechDistribution"}, {"SinghMaddalaDistribution"}, {"SkellamDistribution"}, \ {"SkewNormalDistribution"}, {"SliceDistribution"}, {"SmoothKernelDistribution"}, {"SpatialGraphDistribution"}, \ {"SplicedDistribution"}, {"StableDistribution"}, {"StandbyDistribution"}, {"StationaryDistribution"}, {"StudentTDistribution"}, \ {"SurvivalDistribution"}, {"SuzukiDistribution"}, {"TransformedDistribution"}, {"TriangularDistribution"}, \ {"TruncatedDistribution"}, {"TsallisQExponentialDistribution"}, {"TsallisQGaussianDistribution"}, {"TukeyLambdaDistribution"}, \ {"UniformDistribution"}, {"UniformGraphDistribution"}, {"UniformSumDistribution"}, {"VarianceGammaDistribution"}, \ {"VoigtDistribution"}, {"VonMisesDistribution"}, {"WakebyDistribution"}, {"WalleniusHypergeometricDistribution"}, \ {"WaringYuleDistribution"}, {"WattsStrogatzGraphDistribution"}, {"WeibullDistribution"}, {"WignerSemicircleDistribution"}, \ {"ZipfDistribution"}}
see how-to-return-the-value-of-automatic-when-it-is-used-in-a-mathematica-function
Thanks to Bob Hanlon for this method:
p1 = Plot[Sin[x], {x, 0, Pi}, PlotPoints -> Automatic] Cases[p1, Line[pts_] :> Length[pts], Infinity] (*259*)
Another method due to Simon Woods
Trace[ Plot[Sin[t], {t, 0, 2 Pi}], HoldPattern[PlotPoints | MaxRecursion -> _], TraceInternal -> True] // Flatten // Union {MaxRecursion -> 6, MaxRecursion -> Automatic, PlotPoints -> 50, PlotPoints -> Automatic}
SetDirectory[NotebookDirectory[]]; list = {{3, 4, 5}, {4, 5, 6}}; Export["data.txt", list]
To read it later, say after closing and restarting Mathematica again to continue working on the data
SetDirectory[NotebookDirectory[]]; list = ToExpression@Import["data.txt", "List"] {{3, 4, 5}, {4, 5, 6}}
For an ode, that has just ordinary point at \(x=x_0\), we can find power series solution near \(x_0\) as follows. Assume the ode is
\[ u''(t)+\frac {1}{10} u(t)^3 u'(t)^2+4 u(t)=0 \] with intitial conditions \(u(0)=1,u'(0)=1\), then
findSeriesSolution[t_, nTerms_] := Module[{pt = 0, u, ode, s0, s1, ic, eq, sol}, ic = {u[0] -> 1, u'[0] -> 1}; ode = u''[t] + 4 u[t] + 1/10 u[t]^3 u'[t]^2; s0 = Series[ode, {t, pt, nTerms}]; s0 = s0 /. ic; roots = Solve@LogicalExpand[s0 == 0]; s1 = Series[u[t], {t, pt, nTerms + 2}]; sol = Normal[s1] /. ic /. roots[[1]] ]
and now call it with
seriesSol = findSeriesSolution[t, 6]
It returns \[ \frac {445409479 t^8}{840000000}+\frac {8878343 t^7}{10500000}-\frac {277427 t^6}{600000}-\frac {12569 t^5}{50000}+\frac {1607 t^4}{2000}-\frac {29 t^3}{50}-\frac {41 t^2}{20}+t+1 \]
make sure to copy the notebook first, just in case.
Module[{nb}, nb = EvaluationNotebook[]; NotebookFind[EvaluationNotebook[], "Input", All, CellStyle]; NotebookDelete[nb]]
To understand what it does, these three do the same thing
Map[If[# == 1, Unevaluated@Sequence[], #] &, {1, 2, 3}]; If[# == 1, Unevaluated@Sequence[], #] & /@ {1, 2, 3}; If[# == 1, Sequence @@ {}, #] & /@ {1, 2, 3}; If[# == 1, ## &[], #] & /@ {1, 2, 3};
All above give {2,3}. So the effect of ## &[] is to remove the entry completely (so we do not end up with a Null or empty slot in there).
Given lists a={1,2,3}, b={4,5,6} and we want to do operation from slot 1 from a with slot 1 from b, and so on. For this, we can use ‘MapThread‘. Suppose we want to add each corresponding slot, then
a = {1, 2, 3} b = {4, 5, 6} MapThread[(#1 + #2) &, {a, b}] (* {5, 7, 9} *)
Of course, in this simple example, doing a+b will work, but this is just an example.
see https://mathematica.stackexchange.com/questions/5212/automating-esc-esc-formatting
Thanks to Rolf Mertig reference for help on this.
Make a file foo.m such as
AppendTo[$Echo, "stdout"] SetOptions[ $Output, FormatType -> OutputForm ]; Integrate[Sin[x],x]
Now type, from DOS window
"C:\Program Files\Wolfram Research\Mathematica\10.1\math.exe" < foo.m
This will send input and output to screen. To send output to file do
"C:\Program Files\Wolfram Research\Mathematica\10.1\math.exe" < foo.m > log.txt
You can modify the PATH on windows to add the above to environment variable so not to have to type the long command each time
Type this in the chrome window
chrome://flags/#enable-npapi
and click enable to enable NPAPI. Chrome now disables NPAPI and so CDF no longer runs inside browser. After doing the above, restart the browser again. Now it should run
>sudo bash Mathematica_10.1.0_LINUX.sh [sudo] password for me: Mathematica 10.1.0 for LINUX Installer Archive Verifying archive integrity. Extracting installer. ............. Wolfram Mathematica 10.1 Installer Copyright (c) 1988-2015 Wolfram Research, Inc. All rights reserved. WARNING: Wolfram Mathematica is protected by copyright law and international treaties. Unauthorized reproduction or distribution may result in severe civil and criminalpenalties and will be prosecuted to the maximum extent possible under law. Enter the installation directory, or press ENTER to select /usr/local/Wolfram/Mathematica/10.1: Now installing... [****************** Type the directory path in which the Wolfram Mathematica script(s) will be created, or press ENTER to select /usr/local/bin: > Installation complete. >which Mathematica /usr/local/bin/Mathematica >export DISPLAY=:0 >Mathematica
One method is to just type WolframAlpha["command here"] and then click on the show step by step on top right corner of the result that displays on the notebook, assuming Wolfram Alpha gives an answer.
If the above does not work, try
WolframAlpha["Integrate[x Sin[x],{x,0,Pi}]", {{"Input", 2}, "Content"}, PodStates -> {"Input__Step-by-step solution"}]
Another example
WolframAlpha["solve y''=-y", IncludePods -> "DifferentialEquationSolution", AppearanceElements -> {"Pods"}, TimeConstraint -> {20, Automatic, Automatic, Automatic}, PodStates -> {"DifferentialEquationSolution__Step-by-step solution"}]
or for text output
WolframAlpha["solve y''=-y", {{"DifferentialEquationSolution", 2}, "Plaintext"}, PodStates -> {"DifferentialEquationSolution__Step-by-step solution"}]
Another example
WolframAlpha["q*c'[x]==k*c[x], c[0]==c0", PodStates -> {"Step-by-step solution"}]
see http://mathematica.stackexchange.com/questions/15480/how-do-i-designate-arguments-in-a-nested-map
From about url by Halirutan:
Map[Function[p2, Map[Function[p1, f[p1, p2]], list1]], list2]
How to make TeXForm handle higher derivatives better.
See http://mathematica.stackexchange.com/questions/134936/texform-handling-of-derivative-higher-than-two
From above URL by Carl Woll
Derivative /: MakeBoxes[Derivative[n_Integer?Positive][h_],TraditionalForm] := SuperscriptBox[MakeBoxes[h,TraditionalForm], StringJoin@ConstantArray["\[Prime]",n] ] TeXForm[y'''''[t] + 2 x'''[t] - y'[t] == x[t]]
Given an ode, such as \(x(1-x) y''(x)+(c-(a+b+1)x)y'(x)-a b y(x)=0\) and we want to classify the singular points (finite and infinite).
We write it as \(y''(x)+p(x) y'(x)+q(x)=0\) and then follow standard procedures. In this example, we want to classify points \(0,1,\infty \).
This small function I wrote for a HW will do this. Pass it \(p(x),q(x)\) and list of the points to classify.
checkForSingularity[p_, q_, at_List, x_] := Module[{p0, q0, t, r, r0}, r = First@Last@Reap@Do[ If[at[[n]] === Infinity, p0 = p /. x -> (1/t); p0 = (2 t - p0)/t^2; q0 = (q /. x -> (1/t))/t^4; r0 = checkForSingularity[p0, q0, {0}, t]; Sow[Flatten[{Infinity, Rest[Flatten@r0]}]]; , r0 = {at[[n]], Limit[(x - at[[n]]) p, x -> at[[n]]], Limit[(x - at[[n]])^2 q, x -> at[[n]]]}; Sow@r0 ] , {n, 1, Length@at} ]; r ]
To use it on the above example
ClearAll[c, a, b, x]; m = checkForSingularity[(c - (a + b + 1) x)/(x (1 - x)), (-a b)/(x (1 - x)), {0, 1, Infinity}, x]; Grid[Join[{{"point", "limit x p(x)", "limit x^2 q(x)"}}, m], Frame -> All]
Gives
{{"point", "limit x p(x)", "limit x^2 q(x)"}, {0, c, 0}, {1, 1 + a + b - c, 0}, {\[Infinity], 1 - a - b, a b} }
Since all points have finite limits, then all \(0,1,\infty \) are removable singularities.
I did not test this function too much, but it works for the HW I did :)
Given ode \(y^{'''}=x y(x)\) we want to replace \(y(x)\) by \(e^{s(x)}\)
ClearAll[y,x,s] eq = y'''[x] == x y[x] eq = eq /. y -> Function[{x}, Exp[s[x]]] Simplify@Thread[eq/Exp[s[x]], Equal]
Gives \(x=s^{'''}(x)+s'(x)^3+3 s'(x) s''(x)\)
Needs["GeneralUtilities`"]; PrintDefinitions@Charting`FindTicks
Thanks to https://mathematica.stackexchange.com/questions/132568/extract-ticks-from-plot
The above works if the function is not read protected.
See also this link
Given y'[x]+3 Sin[x]-3-Cos[x]-2 y[x]==0 how to rewrite it to be y'[x]==3+Cos[x]-3 Sin[x]+2 y[x] i.e. put y'[x] on one side, and the rest on the other side?
ClearAll[f,x,y] expr=y'[x]+3 Sin[x]-3-Cos[x]-2 y[x]==0; lhs=expr/.lhs_==rhs_:>lhs; rhs=expr/.lhs_==rhs_:>rhs; rhs=-(lhs/.(y'[x]+any_.):>any)+rhs; expr=y'[x]==rhs
This makes it easier to make it in form \(y'(x)=f(x,y)\).
sometime code is posed at Mathematica stacexchange which is hard to read. To convert it to clear 2D math code, copy into a mathematica cell (in an open notebook) on the computer, then do
CTRL-SHIFT-N
Use this
Block[{DSolve`print=Print}, DSolve[{ode,ic},y[t],t] ]
see https://mathematica.stackexchange.com/questions/120364/why-cant-dsolve-find-a-solution-for-this-ode/120650#120650 by Michael E2.
ClearAll[withTimedIntegrate]; SetAttributes[withTimedIntegrate, HoldFirst]; withTimedIntegrate[code_, tc_] := Module[{$in}, Internal`InheritedBlock[{Integrate}, Unprotect[Integrate]; i : Integrate[___] /; ! TrueQ[$in] := Block[{$in = True}, TimeConstrained[i, tc, Inactivate[i, Integrate]] ]; Protect[Integrate]; code ] ]; withTimedIntegrate[{dsol} = DSolve[ode == 0, y, x], 1]; // AbsoluteTiming dsol
Given \[ a \sqrt {z}+\sin \left (c \sqrt {r+4}+20\right )+\frac {e^{-c} \sqrt {9 e^{2 c} x^2+8}-x+99}{4 x}+4 \sqrt {h} e^y-99 \] Replace only the subexpression of the form anything*Exp[anyting]*Sqrt[anything] by the square of the pattern found if any.
expr /. patternName : (Exp[any3_]*any_.)*Power[Any_, Rational[1 | -1, 2]] -> patternName^2
This gives \[ a \sqrt {z}+\sin \left (c \sqrt {r+4}+20\right )+\frac {e^{-2 c} \left (9 e^{2 c} x^2+8\right )-x+99}{4 x}+16 h e^{2 y}-99 \]
The code patternName : above gives a name for the pattern found, this way we can use the name to easily do any transformation needed. In this example, it was just squaring it. But it can be anything else we want.
Optional arguments are very important and useful. It is easy to make these in Mathematica. lets say we want to make function foo that takes person name and age but the sex is provided as optional with default value as say male. To do this, two things are needed to be defined.
The function foo itself, but with additional argument called opt : OptionsPattern[{foo}] and we need to define an Options[foo] =... which sets up the default values. Here is an example
Clear["Global`*"]; (*this sets up the DEFAULT values for each optional argument *) Options[processPerson] = {"sex" -> "male"}; (*this is the definition of the function itself*) processPerson[name_String, age_?NumericQ, opt : OptionsPattern[{processPerson}]] /; age > 0 := Module[{sexOfPerson}, (* the following line reads the optional value sex->"" if provided *) (* else Mathematica will automatically return the default value set up*) sexOfPerson = OptionValue["sex"]; Print["Name = ", name]; Print["age = ", age]; Print["sex = ", sexOfPerson] ]
Now we are ready to call the function.
processPerson["me",10]
Gives
Name = me age = 10 sex = male
We see in the above, that variable sex was automatically set to the default value thanks to the definitions Options[foo]=...
Now we call it with explicitly passing in the optional argument
processPerson["me",10,"sex"->"female"]
Gives
Name = me age = me sex = female
In the above, the option argument sex->"..." must always be the last one passed in. You can not do this for example
processPerson["joe doe","sex"->"female",10]
Here is another example
Clear["Global`*"]; Options[dsolve] = {"ic" -> {}, "hint" -> "None"}; dsolve[ode_, y_[x_], x_, OptionsPattern[]] := Module[{ic = OptionValue["ic"], hint = OptionValue["hint"]}, Print["ic=", ic, " hint=", hint] ]; dsolve[y''[x] + y[x] == 1, y[x], x, "ic" -> {x[0] == 1}] dsolve[y''[x] + y[x] == 1, y[x], x] dsolve[y''[x] + y[x] == 1, y[x], x, "hint" -> "linear", "ic" -> {x[10] == 0}]
Which prints
ic={x[0]==1} hint="None" ic={} hint="None" ic={x[10]==0} hint="linear"
To find all options supported by function do as an example Options[LinearSolve] and that will print {Method -> Automatic, Modulus -> 0, ZeroTest -> Automatic}
It is better to use string for the name of the hint as in the above, so not to clash with the same symbol being already defined.
The above showed how to use optional value. This below shows how to check and what to do if an optional value was not what is expected. This is done by explicitly checking, after reading the optional value, that its value is one that is expected. Like this
Clear["Global`*"]; Options[processPerson]={"sex"->"male"}; processPerson::msg="`1`"; processPerson[name_String,age_?NumericQ,opt:OptionsPattern[{processPerson}]]/;age>0:=Module[{sex}, sex = OptionValue["sex"]; If[Not[MemberQ[{"male","female"},sex]], Message[processPerson::msg,"Invalid value for sex found. "<>sex<>" but expected one of {\"male\", \"female\"}"]; Abort[] ]; Print["Name = ",name]; Print["age = ",age]; Print["sex = ",sexOfPerson] ]
Now we are ready to call the function. Lets call it with bad value for the optional parameter
processPerson["me",10,"sex"->"car"] processPerson::msg: Invalid value for sex found. car but expected one of {"male", "female"} $Aborted
Copy the Mathematica code from the notebook, paste it into http://steampiano.net/msc/ and click convert. Then copy the output and paste that into the post at stackexchange.
Clear["Global`*"]; expr = y[z] + x + Pi + m/Sin[a] + b[0]; DeleteDuplicates@Cases[expr, any_ /; Head[any] === Symbol && Not[MemberQ[Attributes[any], Constant]], Infinity]
Gives {x,n,a,x}. Note that the above does not detected indexed variables, such as \(b[0]\). The Attributes check is so not to include \(\pi \) since this is also a symbol. But we do not want to be there.
Use
Clear["Global`*"]; ode=y''[x]+3*y'[x]+3==0; Internal`ProcessEquations`DifferentialOrder[ode,{x},{y}] (* {{2}} *)
So it is order 2. Notice the input must be lists [ode,{x},{y}], it will hang if you use [ode,x,y]
Degree of an ode is the degree of the highest derivative in the oder. This needs a helper function
(*This function thanks to Carl Woll, see https://mathematica.stackexchange.com/questions/151850/using-cases-and-when-to-make-input-a-list-or-not*) Clear["Global`*"]; getPatterns[expr_, pat_] := Last@Reap[expr /. a : pat :> Sow[a], _, Sequence @@ #2 &];
And now do
getDegreeOfOde[ode_,y_,x_]:=Module[{maxDer,p,d,der}, der = getPatterns[ode,Power[Derivative[_.][y][x],_.]] ; p = Flatten[Internal`ProcessEquations`DifferentialOrder[#,{x},{y}]&/@der]; maxDer = Extract[der,Position[p,Max[p]]]; Abs[First[maxDer/.(Derivative[_.][y][x])^(d_.):>d]] ]
Examples of usage
ode=y'''[x]+y''[x]^4+3*y'[x]^8+3*y[x]^8==0; getDegreeOfOde[ode,y,x] (*1*) ode=y''[x]^4+3*y'[x]^8+3*y[x]^8==0; getDegreeOfOde[ode,y,x] (*4*) ode=y''[x]+3*y'[x]^8+3*y[x]^8==0; getDegreeOfOde[ode,y,x] (*1*)