Re: Mathematica Weirdness
- To: mathgroup at smc.vnet.net
- Subject: [mg116700] Re: Mathematica Weirdness
- From: Gary Wardall <gwardall at gmail.com>
- Date: Thu, 24 Feb 2011 06:23:35 -0500 (EST)
- References: <ik2nab$9kk$1@smc.vnet.net>
On Feb 23, 4:25=C2 am, Steve Heston <shes... at rhsmith.umd.edu> wrote: > My question is why I get a negative integral of a positive > function? > > Integrate[1000000*Exp[x^2-12*x]*x^14,{x,0,1}]//N > Integrate[1000000*Exp[x^2-12*x]*x^14,{x,0.,1}]//N > NIntegrate[1000000*Exp[x^2-12*x]*x^14,{x,0,1}] > > The first line gives a negative answer, while the second two lines give > identical positive answers. =C2 Something is strange here. > > Steve > > Steven L. Heston > Associate Professor > Finance Department > Robert H. Smith School of Business > 4447 Van Munching Hall Van Munching Hall > University of Maryland > College Park, MD 20742-1815 > 301-405-9686 TEL > 301-405-0359 FAX > shes... at rhsmith.umd.eduhttp://www.rhsmith.umd.edu Steve, There is a problem, a very interesting problem. Define the function: f[x_] := 1000000*Exp[x^2 - 12*x]*x^14 Then the first and second derivatives are: f1[x_] := =E2=80=A8 14000000 E^(-12 x + x^2) x^13 + 1000000 E^(-12 x + x^2) x^14 (-12 + 2 x) f2[x_] := =E2=80=A8 182000000 E^(-12 x + x^2) x^12 + 28000000 E^(-12 x + x^2) x^13 (-12 + 2 x) + =E2=80=A8 1000000 x^14 (2 E^(-12 x + x^2) + E^(-12 x + x^2) (-12 + 2 x)^2) F[x_]:=Integrate[1000000*Exp[t^2 - 12*t]*t^14, {t, 0, x}] Which is also: 15625/2 (-174657787110 + 2065912371945 DawsonF[6] + =E2=80=A8 E^((-12 + x) x) (174657787110 + =E2=80=A8 x (29981073375 + =E2=80=A8 2 x (2614326570 + =E2=80=A8 x (463615155 + =E2=80=A8 2 x (41841090+ x (7695585 + =E2=80=A8 2 x (722070 + x (138465 + =E2=80=A8 2 x (13590 + x (2735 + 2 x (282 + x (59 + 2 x (6 + x)))))))))))) - =E2=80=A8 2065912371945 DawsonF[6 - x])) (Check these results in Mathematica.) Since the first derivative is positive and continuous on [0,1] then f[x] is a rising function on [0,1]. (The first derivative has zeros at x=0, 3 - Sqrt[2], and 3 + Sqrt[2]. At x=0 the zero has a multiplicity 13. The other roots are not on [0,1].) Since f[x] is rising on [0,1] and f[0]=0 then the integral function F[x] must be positive on [0,1]. Make plots of the functions: Plot[f[x], {x, 0, 1}] Plot[f1[x], {x, 0, 1}] Plot[f2[x], {x, 0, 1}] These functions and their graphs are not that "weird". Plot also F[x]. Plot[F[x], {x, 0, 1}] This graph is "weird" and not a correct representation of the function. Note also by the Fundamental Theorem of Calculus that the first derivative of the F[x] must be the function f[x]. Use the D operation on F[x]. You should get the the function f[x]. But it's not quite right. (This can be seen graphically and numerically quite easily.) So either Mathematica 8 is not symbolically integrating or not symbolically differentiating correctly or both. If Mathematica is not symbolically integrating correctly it would explain why Integrate[1000000*Exp[x^2 - 12*x]*x^14, {x, 0, 1}] // N Integrate[1000000*Exp[x^2 - 12*x]*x^14, {x, 0., 1}] // N are not giving good approximations. The result obtained from: NIntegrate[1000000*Exp[x^2 - 12*x]*x^14, {x, 0, 1}] is a good approximation. This I verified independently using the Trapezoid and Simpson's 1/3 rules. Another interesting item: Suppose you define the function g[x] as the function you get from doing: D[F[x],x] This should be the same as f[x] the original function. Make a plot of g[x]/f[x] on say, [-3,3]. If g[x] and f[x] were the same them the plot: Plot[g[x]/f[x], {x, -3, 3}] should be the line y=1, except at zero of course. The plot is not the line y=1, but you will see the line in it. Make the plot: Plot[g[x]/f[x], {x, 10, 20}, PlotRange -> {-2, 2}] The line y=1 appears in this plot too. This would give credence to the conclusion that Mathematica 8 did correctly do D[F[x],x] after all, well correctly away from the origin. Perhaps, just perhaps, the first derivative having a multiple zero at zero with a multiplicity of 13 has something to do with it, perhaps not. Well, a most interesting problem. More could be said, I'm sure. Gary Wardall Retired Instructor University of Wis.-Green Bay