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4

a1 = NDSolve[{y1''[x] - 0.1 y1[x]^0.25 == 0, y1'[0] == 0, y1[1] == 1}, y1, {x, 0, 1}] Plot[y1[x] /. a1, {x, 0, 1}, GridLines -> Automatic] Mathematica 10.4.1 on Windows 10 (64 bit)


2

Adding links to comment by MarcoB Note the Attributes of RegionPlot Attributes[RegionPlot] (* {HoldAll, Protected, ReadProtected} *) Since RegionPlot and other plot functions have attribute HoldAll you need to use Evaluate a = {x^2 < y^3 + 1, y^2 < x^3 + 1}; RegionPlot[Evaluate[a], {x, -2, 5}, {y, -2, 5}]


2

A PDF is not a distribution. To convert a PDF to its associated distribution use ProbabilityDistribution Clear[λ, pb, Ps, μ] f[h_] := 1/Gamma[m]*(m/P)^m*h^(m - 1)*Exp[-(m*h)/P]; m = 1; P = 1; α = 4; δ = 2/α; int = Assuming[{A > 0, r > 0, λ > 0}, Integrate[(1 - Exp[-x*h*r^(-1/δ)])*pb*λ*π, {r, A, Infinity}, GenerateConditions -> False]] ...


2

This is an extended comment. f[h] as defined while clearly a probability density function is not of the type of function that Expectation expects. Consider the difference between the two functions f and g below: f[h_] := 1/Gamma[m]*(m/P)^m*h^(m - 1)*Exp[-(m*h)/P] (* Head *) Head[f] (* Symbol *) (* PDF *) f[h] (* (E^(-((h m)/P)) h^(-1 + m) ...


2

I think that value injection using With will work for you. Let's first define the sample g value you want: g = 1 + x^2 Now we define myfunc injecting the current definition of g inside its definition: Clear[myfunc] With[{g = g}, myfunc[f_] := Simplify[f/g]] Now let's change the value of g and check whether the definition of myfunc is affected: g = ...


1

Maybe something like this? g = 1 + x^2 Unprotect[saveg] saveg = g; myfunc[f_] := Simplify[(f/saveg)] Protect[saveg]


1

See ReplaceAll and What are the most common pitfalls awaiting new users?. Say you want to find a root, you would code something like rule = FindRoot[x^3 - Tanh[x], {x, 1.0, 1.2}] and get as result: {x -> 0.893395} Next step ist (as proposed by @Alexei Boulbitch) expr/.rules applies a rule or list of rules in an attempt to transform each ...


7

To be very explicit and related to the situation in your question, consider this simple function f and the different fs that all might be ways to compute the derivative at a first glance. ClearAll[f, fs, fs2, fs3, fs4]; f[x_] := x^2; fs[x_] := D[f[x], x]; fs2[x_] := f'[x]; fs3[x_] = D[f[x], x]; fs4[x_] := Block[{t}, D[f[t], t] /. t -> x]; Let's check ...


2

Braces [], brackets {}, and parenthesis () all have different meanings in mathematica. Sorry for the post. I will read the documentation and tutorials more completely.


0

I hope someone else explains this better! It seems that expressions don't get automatically evaluated inside the Plot function. Try including an Evaluate function f[x_, N_] := FourierSinSeries[x - 1, x, N]; Plot[{Evaluate@f[x, 1], Evaluate@f[x, 5]}, {x, 0, 3}]


2

Wrap your Monitor call in a Function object endowed with a holding attribute: Table[Pause[n/10], {n, 10}] // Function[{input}, Monitor[input, n], HoldAll] This will work as though you had wrapped Monitor around your Table.


2

You can use Unevaluated Unevaluated@Table[Pause[n/10], {n, 5}] // Monitor[#, n] & The neater way will be to use the infix version as suggested by wxffles in the comments, but your question specifically asked for postfix so... For completeness, here it is Table[Pause[n/10], {n, 5}] ~Monitor~ n


3

I think you are looking for something like this: ordered = {#[[1]], StringRiffle@sort[StringSplit@#[[2]], weights]} & /@ needsreorder Export["ordered50parts.csv", ordered] (*{{"PARTNUM", "DESC"}, {"DUK175189", "Blower Hose 14"}, {"ROU4060436", "Halogen Lamp 100 120 Volt Watt"}} *)


6

It seems that it will align the column to the first character of the string you put there. So it's an undocumented extension to: "c" - align on the character "c" Column[{ "OffOutputasodas", "OutputOffo", "adssoadasOffOutput" }, "Output" ] Column[{ "OffOutputasodas", "oOutputOffo", "adssoadasOffOutput" }, "output" ]


1

A set solution. cond = {{x}, {y}, {z}, {m}} ∈ Interval[{2, 4}] && {x, y, z, m} ∈ Integers Set x, y, z, and m to be Integers on the Interval 2 to 4. We can use cond to be certain it is defined as expected. Solve[x + y + z + m == 8 && cond, {x, y, z, m}] (* {{x -> 2, y -> 2, z -> 2, m -> 2}} *) This gives the expected result. ...


0

And @@ Thread@Table[Or @@ Thread@Table[j == i, {i, {2, 3, 4}}], {j, {x, y, z, m}}]


11

My offering: And @@ Or @@@ Outer[Equal, {x, y, z, m}, {2, 3, 4}]


3

In version 10, you can use this simpler syntax without Thread: Solve[x+2y+z+m==14&&2<=var<=3,var,Integers] {{x->2,y->3,z->3,m->3},{x->3,y->3,z->2,m->3},{x->3,y->3,z->3,m->2}}


3

This is one aspect of a classic problem: Plot is HoldAll. So, moreProp is not being evaluated. I think the most effective way to work around this is to use With (cf. this question) to inject the values into Plot, e.g. newPlot[func_, range_, prop___] := With[{mp = moreProp}, Plot[func, range, prop, mp]] Then, it works: Alternatively, you can use ...


5

You could do : Fpmat = Switch[Sign[Im[{EEV1,EEV2,EEV3,EEV4}]], {1, 1, _, _}, Transpose[{EVEC1, EVEC2}], {1, -1, 1, _}, Transpose[{EVEC1, EVEC3}], {1, -1, -1, 1}, Transpose[{EVEC1, EVEC4}], {-1, 1, 1, _}, Transpose[{EVEC2, EVEC3}], {-1, 1, -1, 1}, Transpose[{EVEC2, EVEC4}], {-1, -1, 1, 1}, Transpose[{EVEC3, ...


2

How about defining: f[{ 1, 1, _, _}] := Transpose[{EVEC1, EVEC2}] f[{-1, 1, 1, _}] := Transpose[{EVEC2, EVEC3}] f[{ 1, -1, 1, _}] := Transpose[{EVEC1, EVEC3}] f[{-1, -1, 1, 1}] := Transpose[{EVEC3, EVEC4}]; f[{ 1, -1, -1, 1}] := Transpose[{EVEC1, EVEC4}]; f[{-1, 1, -1, 1}] := Transpose[{EVEC2, EVEC4}]; f[_] := Print["choose another value of ...


3

The number of different actions that can result from the various conditions is much smaller than the number of conditions. What you have to do is collect all the conditions leading to identical actions, and then combine these conditions with Or (written usually as ||). That's all there is to it. I'll leave the typing to you.


4

This is a method that is similar to Oleksandr R.'s answer: The following tells Mathematica to render Primed symbols in superscripted box form: MakeBoxes[Primed[x_], StandardForm] := SuperscriptBox[ToBoxes[x], "\[Prime]"] And this tells Mathematica to parse a superscripted box structure internally as Primed, and not as Derivative: ...



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