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I am using Mathematica 10. I have the following code yielding a ragged plot

    market [r_, a_] = (1 + 2*r + a^2)/(1 - 2*I*(Sqrt[a^2 - r^2])  - a^2);


stock11[l_, x_, r_, a_] = (
   Abs[market[r, a]]^2* Exp[- 5.59 * 10^9 * x])/
   2 (Cosh[(1.1174 * 10^10)/2 x] - Exp[- l* x] Cos[5.302*10^9 x]);

stock22[l_, x_] = 
  Exp[- 5.59 * 10^9* x]/
   2 (Cosh[(1.1174 * 10^10)/2 x] + Exp[- l*x] Cos[5.302*10^9 x]);

VarianceS[l_, x_, r_, a_] = 
  1 - 4*stock11[l, x, r, a] + 8*stock11[l, x, r, a]*stock22[l, x]  + 
   Abs[market[r, a]]^2* Exp[-2*5.59 * 10^9*x] (Exp[-2*l*x] - 1) ; 

Λ[l_, x_, r_, a_] = 
 1 - 2 stock11[l, x, r, 
    a]; (* The two time correlation Subscript[C, 01] = Subscript[c, \
12] = Λ*)

VarianceT[l_, x_, r_, a_] = 
 2 Λ[l, x, r, a] - Λ[l, 2*x, r, a];

differenceV[l_, x_, r_, a_] = 
  VarianceS[l, x, r, a] - VarianceT[l, x, r, a];

Plot[differenceV[0, 1.7889*10^-10*x, 0.001596, 2.228*10^-3], {x, 0, 
  20}]

Mathematica graphics

Is there any way of smoothing this raggedness/peaks?

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5
  • $\begingroup$ You use F. How is it defined? $\endgroup$
    – Coolwater
    Jan 23, 2018 at 17:40
  • $\begingroup$ Perhaps the undefined symbol noticed by @Coolwater can fix it, but as written I can't find a plot scale on which your function is not ragged: list = Table[ differenceV[0, 1.7889*10^-10*x, 0.001596, 2.228*10^-3], {x, 0, 20, 0.001}]; ListLinePlot@list yields this: i.sstatic.net/XTI4W.png $\endgroup$
    – Jason B.
    Jan 23, 2018 at 17:42
  • $\begingroup$ @Coolwater, sorry. F[r,a] was a typo. It is actually market[r,a]. $\endgroup$
    – H. Kenan
    Jan 23, 2018 at 17:45
  • $\begingroup$ I expect the difference of two smooth functions to be a smooth curve. Could it be possible to obtain a smooth curve? $\endgroup$
    – H. Kenan
    Jan 23, 2018 at 17:49
  • 2
    $\begingroup$ I added a plot. It rather makes the problem obvious to those who understand numerics (it's machine-precision rounding error of the difference of two things that are approximately equal). If a question is about how a plot looks, probably the plot should be included anyway. $\endgroup$
    – Michael E2
    Jan 24, 2018 at 13:06

1 Answer 1

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After replacing the inexact numbers by exact numbers, FullSimplify turns the function you plot into 0

a1 = 559/100;
a2 = 11174/10000;
a3 = 5302/1000;

market[r_, a_] = (1 + 2*r + a^2)/(1 - 2*I*(Sqrt[a^2 - r^2]) - a^2);
stock11[l_, x_, r_, a_] = (Abs[market[r, a]]^2*Exp[-a1*10^9*x])/2 (Cosh[(a2*10^10)/2 x] -
                           Exp[-l*x] Cos[a3*10^9 x]);
stock22[l_, x_] = Exp[-a1*10^9*x]/2 (Cosh[(a2*10^10)/2 x] + Exp[-l*x] Cos[a3*10^9 x]);

VarianceS[l_, x_, r_, a_] = 1 - 4*stock11[l, x, r, a] + 8*stock11[l, x, r, a]*
  stock22[l, x] + Abs[market[r, a]]^2*Exp[-2*a1*10^9*x] (Exp[-2*l*x] - 1);
Λ[l_, x_, r_, a_] = 1 - 2 stock11[l, x, r, a];
VarianceT[l_, x_, r_, a_] =  2 Λ[l, x, r, a] - Λ[l, 2*x, r, a];

differenceV[l_, x_, r_, a_] = VarianceS[l, x, r, a] - VarianceT[l, x, r, a];

differenceV[0, 17889/10000*10^-10*x, 1596/1000000, 2228/1000*10^-3] // FullSimplify

0

So to get a non-ragged plot use Plot[0, {x, 0, 20}]

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  • $\begingroup$ its actually zero even if you leave the a's and other constants symbolic. $\endgroup$
    – george2079
    Jan 23, 2018 at 18:05

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