I want to plot a function with very small values, close to zero, but even though the limit is -3/2 as x->0, I can't get any graphs. So I'm wondering if I could really get any graphs with any modifications in the Plot command options.

Limit[15 Integrate[Cos[t^2] - 1, {t, 0, x^ 3}]/x^15, x -> 0, Direction -> "FromAbove"]
(* Ans: -3/2*)
Plot[{15 NIntegrate[(Cos[t^2] - 1), {t, 0, x^3}, MinRecursion -> 4]/x^15, -3/2}, {x, -0.001, 0}]

enter image description here

Edit: The function and series representation is

$\dfrac{15\int_{0}^{x^3}(\cos (t^2)-1)\, dt }{x^{15}}=-\dfrac{3}{2}+\dfrac{5 x^{12}}{72}+O\left(x^{16}\right)$

  • $\begingroup$ Integrate your function exactly instead of approximately and then use more precision for the plot z=15*Integrate[(Cos[t^2]-1), {t,0,x^3}]/x^15; Plot[z, {x,-10^-5,10^-5}, WorkingPrecision->128, PlotRange->{-1.6,.2}] Then reduce the precision to 64 and see that Plot makes a mistake. Who knows what the plot REALLY REALLY looks like. $\endgroup$ – Bill Nov 8 '18 at 17:09
  • $\begingroup$ "...Who knows what the plot REALLY REALLY looks like...": So you think there is no way to get the graph? $\endgroup$ – wmora2 Nov 8 '18 at 20:13
  • 1
    $\begingroup$ Evaluate the integral symbolically, 15 Integrate[Cos[t^2] - 1, {t, 0, x^3}, Assumptions -> x < 0]/x^15, and then use ListPlot[Table[{x, %}, {x, -1.`100, -0.000000001`100, .0001`100}], Joined -> True]. $\endgroup$ – AccidentalFourierTransform Nov 8 '18 at 23:29
  • $\begingroup$ It doesn't work, no graphic $\endgroup$ – wmora2 Nov 9 '18 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.