Tag Info

Hot answers tagged

6

Something like heldint = HoldForm[Integrate[r^3, {t, 0, 2 Pi}, {r, 0, 2}, {z, r^2, 4}]]; int = ReleaseHold[heldint]; Row[{heldint, " \[LongEqual] ", int, " \[TildeTilde] ", N[int]}] // TraditionalForm


5

TraditionalForm looks nice but it also incurs additional complexity in handling expressions. If you attempt to evaluate a TraditionalForm cell you will be asked if you wish to interpret it as input, and the equivalence is not always complete. Fortunately Mathematica is quite cabable of displaying formatted integrals in StandardForm. If you merely wrap ...


4

Mathematica 10 introduced a new set of functions that can help with this: Inactive[Integrate][r^3, {t, 0, 2 Pi}, {r, 0, 2}, {z, r^2, 4}] Or, obtaining same output: Inactivate[Integrate[r^3, {t, 0, 2 Pi}, {r, 0, 2}, {z, r^2, 4}]] And then: Activate[%] (* 32 Pi/3 *)


4

dt = RandomInteger[10, 10]; The following reproduces the issue (Version 9.0.1.0 Windows 8 64bit) with large enough font size for the frame labels: lp1 = ListPlot[dt, ImageSize -> 400, Frame -> {True, True, False, False}, FrameLabel -> {None, Style["Probability of extinction", 16], None, None}] You can wrap the labels with Framed or Pane ...


4

As Yves said replace Print with List output. Here is an example using Sow and Reap along with Block and Mathematica 10 notation for Composition. (It would be better to avoid Print from the beginning but I am trying to make this an easy substitution for you.) myfun[x_, y_: 0] := Block[{ans, myplot, summary, Print = Sow@*Row@*List}, Print["The first ...


2

My favorite way to check that I've input correctly is to apply Hold and TraditionalForm TraditionalForm[Hold[Integrate[r^3, {t, 0, 2Pi}, {r, 0, 2}, {z, r^2, 4}]]] yields: A second option is to use the "Basic Math Input" Palette which you can find in the Palette toolbar, and input the integral from the get-go in pretty-print.


2

Interpreting the word "difference" in your question as "different" (Not sure, please check): Select[lst, ( 9 == Length[Union @@ #] &)] (* {{{1, 2, 3}, {-1, -2, -4}, {5, 6, 10}}} *) Select[lst, ( And @@ (Unequal @@@ Transpose[#]) &)] (* {{{1, 2, 3}, {-1, -2, -4}, {5, 6, 10}}, {{-2, 5, -3}, {1, 10, 7}, {5, 6, 11}}} *)


2

Normal[Series[ K[(Subscript[x, 0] - Subscript[x, 0] 0) t + Subscript[x, 0] 0, (Subscript[x, 1] - Subscript[x, 1] 0) t + Subscript[x, 1] 0, (Subscript[x, 2] - Subscript[x, 2] 0) t + Subscript[x, 2] 0, (Subscript[x, 3] - Subscript[x, 3] 0) t + Subscript[x, 3] 0], {t, 0, 2}]] /. t -> 1 % /. K_[0, 0, 0, 0] -> K


2

It seems there's a problem when mathematica renders the output image in some cases: Thats label for 18-22 font sizes, evidently it doesn't smoothly work for 19, 20, 21. But since you increase the quality - everthing becomes perfect: Rasterize[ListPlot[{{1, 2}}, FrameLabel -> {"", "Probability of extinction!"}, Frame -> True, FrameStyle -> ...


1

wiki answer. All the above solutions are fine, but I think Maple does this much simpler and I thought to show how Maple does it, may be future versions of M can adopt this. (in some way, can't use the same Upper/Lower trick ofcourse, since M commands already start with UpperCase) In Maple, there is what is called the Inert form. Command that starts with ...


1

χ /. First[NSolve[SetV == #/(Cpf (1 - χ)), χ]] & /@ demand or equivalently χ /. First@Solve[SetV == dem/(Cpf (1 -χ)), χ] (-dem + Cpf SetV)/(Cpf SetV) and now (-# + Cpf SetV)/(Cpf SetV)& /@ demand {(-1.92*10^6 + Cpf SetV)/(Cpf SetV), (-2.07*10^6 + Cpf SetV)/( Cpf SetV), (-2.37*10^6 + Cpf SetV)/( Cpf SetV), (-2.72*10^6 + Cpf SetV)/( ...


1

Simple replace should do the trick χ /. {{χ -> (-1.92`*^6 + Cpf SetV)/(Cpf SetV)}} Yields {(-1.92*10^6 + Cpf SetV)/(Cpf SetV)}



Only top voted, non community-wiki answers of a minimum length are eligible