Skip to main content

plotting - Why does evaluating an integral within a Plot expression take so long?



I want to evaluate


Plot[Integrate[Sin[((1 + 1/2) x)] (1/(2 Sin[(x/2)]) - 1/x), {x, 0, t}], {t, 0, 5}]

but it takes ages to load on Corei3 computer with Mathematica 9.


I'm still a novice when it comes to using Mathematica. Is there a way to get this expression to be evaluated faster? (I'm not used to waiting more than a minute while computing, that's why it feels odd to me).



Answer



Oska nailed it in the comments before me


The trick here is to evaluate the integral only once. In your code it is being evaluated once for every point. You can evaluate the following



Integrate[Sin[((1 + 1/2) x)] (1/(2 Sin[(x/2)]) - 1/x), {x, 0, t}]

The outcome of this is


ConditionalExpression[1/2 (t + 2 Sin[t] - 2 SinIntegral[(3 t)/2]), 
 t \[Element] Reals]

You know that t will always be a real number (at least in your plot), so you can just copy paste the first argument of this ConditionalExpression and make a function out of it.


func[t_] := 1/2 (t + 2 Sin[t] - 2 SinIntegral[(3 t)/2])

We can then do



Plot[func[t], {t, 0, 5}]

which gives


enter image description here


To define the function "automatically", you could use the code in Oska's answer. I slightly prefer the following


func[t_] := Evaluate@
  Integrate[Sin[((1 + 1/2) x)] (1/(2 Sin[(x/2)]) - 1/x), {x, 0, t}, 
   Assumptions -> t \[Element] Reals]

Which gets rid of the ConditionalExpression and uses SetDelayed rather than Set. The code only works if t does not have value, so watch out.



Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more