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I am trying to plot the function $\phi(x)=\sum_{n=1}^{250} \psi(x)\psi(a/3)^*$, where $\psi(x)=\sqrt(2/a)\sin(n \pi x/a)$. For some reason my plot is showing up choppy and is missing a lot of points. I have tried setting PlotPoints to 1 million, but it still isn't smooth, this is my code:

ClearAll
a = 1;
ψ[x_, n_] = Sqrt[2/a] Sin[n Pi x/a];
ϕ[x_, n_] = 
  Sum[ψ[x, i]*Conjugate[ψ[a/3, i]], {i, 0, n}];
Plot[{ϕ[x, 250]}, {x, 0, 1}, PlotRange -> Full]
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    – Michael E2
    Feb 9 at 21:21
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the problem vanishes if you define the functions with := like

ClearAll
a = 1;
\[Psi][x_, n_] := Sqrt[2/a] Sin[n Pi x/a];
\[Phi][x_, n_] := 
Sum[\[Psi][x, i]*Conjugate[\[Psi][a/3, i]], {i, 0, n}];
Plot[{\[Phi][x, 250]}, {x, 0, 1}, PlotRange -> Full]
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  • $\begingroup$ Thank you! I am not sure why I didn't try this before. $\endgroup$ Feb 9 at 20:19
  • 2
    $\begingroup$ Also note, using an evaluation of the summation before plotting Plot[{Evaluate[\[Phi][x, 250]]},... increases the plot-speed significantly. $\endgroup$ Feb 9 at 22:43
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ClearAll["Global`*"]

a = 1;

ψ[x_, n_] = Sqrt[2/a] Sin[n Pi x/a];

Use ComplexExpand to simplify

ϕ[x_, n_] = 
 Sum[ψ[x, i]*Conjugate[ψ[a/3, i]], {i, 0, n}] // ComplexExpand // 
  FullSimplify

(* (-((Sqrt[3] Cos[(n π)/3] + Sin[(n π)/3]) Sin[n π x]) + 
   2 Sin[(n π)/3] Sin[(1 + n) π x])/(-1 + 2 Cos[π x]) *)

Plot[ϕ[x, 250], {x, 0, 1}, PlotRange -> Full]

enter image description here

EDIT: More generally, for real a

Clear[a]

ϕ2[x_, n_, a_ : 1] = 
 Sum[ComplexExpand[ψ[x, i]*Conjugate[ψ[a/3, i]]], 
  {i, 0, n}] // FullSimplify

(* (Cos[1/3 π (a + a n - 3 n x)] - Cos[1/3 π (a + a n + 3 n x)] - 
   Cos[1/3 π (a n - 3 (1 + n) x)] + 
   Cos[(a n π)/3 + (1 + n) π x])/(2 (Cos[(a π)/3] - Cos[π x])) *)

For a == 1 this reduces to the previous result

ϕ2[x, n] == ϕ[x, n] // Simplify

(* True *)
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The problem seems to be from round-off error in the imaginary parts in the symbolic formula returned by Sum[] when evaluated at floating-point x. An alternative is to use arbitrary-precision numbers:

Plot[{ϕ[x, 250]}, {x, 0, 1},
 PlotRange -> Full, WorkingPrecision -> 16]

enter image description here

Another way to simplify the sum:

Assuming[{n, i} ∈ Integers && n >= i >= 0 && 0 < x < 1,
 ϕ[x_, n_] = 
  Sum[ψ[x, i]*Conjugate[ψ[a/3, i]] // Simplify, {i, 0, n}] //
     ExpToTrig // FullSimplify
 ]
(*
1/2 (-Sec[1/6 π (2 + 3 x)] Sin[1/6 (1 + 2 n) π (-1 + 3 x)] - 
   Csc[1/6 (π + 3 π x)] Sin[1/6 (1 + 2 n) π (1 + 3 x)])
*)
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Other way:

For your graphics labels, use:

<< MaTeX`

Definition of the $\phi(x,n)$ function:

a = 1;
p[f_] := Plot[f, {x, 0, 1}, PlotRange -> {{-0.08, 1.08}, {-60, 280}}, PlotStyle -> {Blue, Thickness[0.0025]}, AxesStyle -> {{Black, Arrowheads[{0, 0.03}]}, {Black, Arrowheads[{0, 0.03}]}}, LabelStyle -> Directive[Black, Bold, Tiny], AxesLabel -> {MaTeX["x", Magnification -> 1], MaTeX["\\phi\\left(x,250\\right)", Magnification -> 1]}, AspectRatio -> 0.85]
ψ[x_, n_] := Sqrt[2/a] Sin[n Pi x/a];
φ[x_, n_] := FullSimplify[Refine[ComplexExpand[Sum[ψ[x, i]*Conjugate[ψ[a/3, i]], {i, 0, n}]], Assumptions -> {x > 0, n > 0}]]

The function $\phi(x,n)$ is given by $$ \phi(x,n) = \displaystyle\frac{2 \sin \left(\frac{\pi n}{3}\right) \sin (\pi (n+1) x)-\sin (\pi n x) \left(\sin \left(\frac{\pi n}{3}\right)+\sqrt{3} \cos \left(\frac{\pi n}{3}\right)\right)}{2 \cos (\pi x)-1} $$

φ[x, n]
(* (-(Sqrt[3] Cos[(n π)/3] + Sin[(n π)/3]) Sin[n π x] + 
2 Sin[(n π)/3] Sin[(1 + n) π x])/(-1 + 2 Cos[π x]) *)

We take $n = 250$:

φ2[x_] := (-(Sqrt[3] Cos[(n π)/3] + Sin[(n π)/3]) Sin[n π x] + 2 Sin[(n π)/3] Sin[(1 + n) π x])/(-1 + 2 Cos[π x]) /. n -> 250
φ2[x]
(*(Sqrt[3] Sin[250 π x] - Sqrt[3] Sin[251 π x])/(-1 + 2 Cos[π x])*)

Finally, we plot the function $\phi(x,250)$:

p[φ2[x]]

enter image description here

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