Re: if using Mathematica to solve an algebraic problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg109035] Re: if using Mathematica to solve an algebraic problem*From*: Murray Eisenberg <murray at math.umass.edu>*Date*: Sun, 11 Apr 2010 04:31:25 -0400 (EDT)

One thing you "missed" is that the integral is not "elementary" in the technical sense of the term, or what many folks would call "closed-form". And that's one of the things we often discuss in a calculus class -- that this or that function, although itself elementary, does not have an elementary antiderivative (even though the Fundamental Theorem of Calculus guarantees that it does have an antiderivative, since it's continuous). With paper-and-pencil, one apply this or that method in an attempt to get a closed-form antiderivative, only to fail. But having Erf pop-up in Mathematica as the answer always gives students a surprise: "What is Erf?" And it can serve to motivate the whole discussion of elementary functions and a (usually, necessarily, only brief) discussion of the existence of the Risch algorithm for deciding whether a function has an elementary anti-derivative. On 4/10/2010 3:21 PM, David Park wrote: > Well, I'm not very hot at integration problems but I started out by trying > some of the standard techniques (using Mathematica), such as substitution of > variable, or integration by parts, but that didn't seem to get anywhere. > > So then I decided to look at the answer. > > f0[x_] = Integrate[Exp[-x^2], x] > 1/2 Sqrt[\[Pi]] Erf[x] > > That doesn't seem to be any of the standard functions so I looked up Erf in > Help, and also in Gradshteyn& Ryzbik, and in Wikipedia: > > http://en.wikipedia.org/wiki/Error_function > > and in MathWorld: > > http://mathworld.wolfram.com/Erf.html > > and the result is that Erf is DEFINED to be the integral. The real question > is how to evaluate it. One way to evaluate a non-elementary integral is to > expand the integrand in a series: > > Series[Exp[-x^2], {x, 0, 8}] > 1-x^2+x^4/2-x^6/6+x^8/24+O[x]^9 > > and then integrate term by term. So I did that, experimenting with how many > terms to take and settled on: > > Series[Exp[-x^2], {x, 0, 400}] // Normal; > f1[x_] = Integrate[#, x]& /@ %; > > The two results appear to lie on top of each other for -5< x< 5. > > Plot[{f0[x], f1[x]}, {x, -5, 5}] > > Plotting the difference also shows the finite series works fairly well in > this domain. > > Plot[f1[x] - f0[x], {x, -5, 5}, > PlotRange -> All] > > If we look at the series for the two expressions we see they are the same. > (Because I suppose that was the basic definition of Erf.) > > Take[f1[x], 10] > > Series[f0[x], {x, 0, 20}] // Normal > > Having obtained the series solution, I tried taking the first half dozen > terms and then using FindGeneratingFunction. > > Take[f1[x], 6] > List @@ % /. x -> 1 > Riffle[%, 0, {1, 12, 2}] > FindGeneratingFunction[%, x] > > which gives the integral 1/2 Sqrt[\[Pi]] Erf[x], bypassing the Integrate > command. But again it is just because that series is the definition of the > function and Mathematica had that definition stored away somewhere. > > If there is much more to learn I suppose it is in special methods of summing > the series in various domains. We could do more in getting the analytic > expressions for the series coefficients and perhaps working with them. > > So I got some experience in using term by term integration and saw how > special functions are sometimes defined. I usually don't do much of that or > know a lot about it. How much would I have done without Mathematica? > Probably none at all. Am I going to calculate a 400 term function by hand > and plot it? But WITH Mathematica it is difficult to resist a Murray > challenge. > > Did I find out what was instructive, or completely miss it? I'm not really > certain. > > > David Park > djmpark at comcast.net > http://home.comcast.net/~djmpark/ > > > > From: Murray Eisenberg [mailto:murray at math.umass.edu] > > > An instructive example when one is teaching techniques of integration: > > Integrate[ Exp[-x^2],x ] > > Why this is instructive is left as an exercise to the reader. > > > On 4/9/2010 3:34 AM, Bill Rowe wrote: >> On 4/8/10 at 8:03 AM, dave at removedbailey.co.uk (David Bailey) wrote: >> >>> I also think that if Mathematica had been available to me back then, >>> I would have felt it was a bit like an addictive substance - very >>> interesting in small doses, but also terribly dangerous. There would >>> have always been the possibility of becoming skilled in answering >>> questions through Mathematica, rather than actually learning the >>> subject! >> >> While I think I understand the concern you express here, I still >> wonder. If one becomes very skilled at using Mathematica to >> solve problems correctly wouldn't there have to be some >> corresponding gain in understanding of how the same problems >> would be solved without Mathematica? >> >> The point I am trying to get at is areas Mathematica fails or >> shows limitations invariably require understanding of details of >> the problem and computer arithmetic. It seems becoming highly >> skilled at getting correct results from Mathematica requires >> mastery of these details to a large degree. And it also seems >> understanding those details are exactly what is required to >> solve the problem without Mathematica. > > -- Murray Eisenberg murray at math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305

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**Re: if using Mathematica to solve an algebraic problem**

**Re: if using Mathematica to solve an algebraic problem**