Skip to main content

Plotting discrete data but not using discreteplot function


For a given function:


Plot[Sqrt[Abs[x]], {x, -Pi, Pi}]

I have the code to draw the function (with its Abs remove), partial sums and cesaro means as:


f[x_] := Sqrt[x]
s[k_, x_] := \frac{2\sqrt{\pi}}{3}+(-Sqrt[2] FresnelS[Sqrt[2] Sqrt[n]] + 2 Sqrt[n] Sin[n \[Pi]])/(n^(3/2) Sqrt[\[Pi]]) Cos[n x], {n, 1, k}]
partialsums[x_] = Table[s[n, x], {n, {4}}];
c[n_, x_] := (1/n) Sum[s[m, x], {m, 0, n - 1}]

Plot[Evaluate[{f[x], partialsums[x], c[4, x]}], {x, -Pi, Pi},
PlotLegends -> {"f(x)=x", "Fourier, 4 terms", "Cesaro, 4 terms"},
PlotStyle -> {{Blue}, {Dashed, Thickness[0.006]}, {Red, Thickness[0.006]}}]

This code fails on my computer and hence I resolve to manual computation.


Updates: It turn out that I can easily solve this issue by removing the k with any number rather than letting it to be indefinite. Although I am not certain the graph is right for k=4 as both graphs(Partial and Cesaro) coincides with each other.



Answer



Maybe


plt = Plot[f[k], {k, 0, 50}, Frame -> True, PlotStyle -> Red,ImageSize -> 300];
dplt = DiscretePlot[cesaro[k], {k, 0, 50}, Frame -> True, PlotRange -> PlotRange[plt],

PlotStyle -> Directive[{Blue, Dashed}], Joined -> True, ImageSize -> 300];
Row[{plt, dplt, Show[plt, dplt]}]

enter image description here


Update: or, perhaps, this:?


 dplt2 = DiscretePlot[cesaro[k], {k, 0, 50}, Frame -> True, Filling -> None, 
PlotRange -> PlotRange[plt], PlotStyle -> Blue, Joined -> True, ImageSize -> 300];
Row[{plt, dplt2, Show[plt, dplt2]}]

enter image description here



or, using Interpolation on cesaro[k] and


 intFCsr = Interpolation[Table[{k, cesaro[k]}, {k, 0, 50}]];
Plot[{f[k], intFCsr[k]}, {k, 0, 50}, Frame -> True,PlotStyle -> {Red, Blue}]

enter image description here


Update 2:


 intFCsr = Interpolation[Table[{k, cesaro[k]}, {k, 0, 50}]];
intFPrtlSms = Interpolation[Table[{k, part[k]}, {k, 0, 50}]];
Plot[{f[k], intFCsr[k], intFPrtlSms[k]/15}, {k, 0, 50}, ImagePadding -> 45,
Frame -> True, PlotStyle -> {Red, Blue, Black}, ImageSize -> 500,

FrameLabel -> {{Style["f, cesaro", 12], Style["partial sum", 12]},
{Style["k", 12], Style["plot label", 14]}},
FrameTicks -> {{Join[{#, #, {.01, 0}} & /@ Range[0, 4.],
{#, " ", {.0075, 0}} & /@ Range[0.2, 4., .2]],
Join[{#, 15 #, {.01, 0}} & /@ Range[0, 4.],
{#, " ", {.0075, 0}} & /@ Range[0.2, 4., .2]]}, {Automatic, Automatic}}]

enter image description here


Comments

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...

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...

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-...