21,209 reputation
245103
bio website sites.google.com/a/unca.edu/…
location Asheville, NC
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visits member for 2 years, 9 months
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I've been a professor of mathematics at The University of North Carolina - Asheville since 1997. I've been using Mathematica since I started graduate school in mathematics at Ohio State in 1989. At that time, we used version 1.1 (as I recall) to teach calculus in our Calculus and Mathematica classes. I've used it pretty much continuously in my teaching and research since then.

In addition to my posts on SE, you can find some of my papers, teaching notebooks and other Mathematica based oddities strewn throughout my website.

In recent years, I've also worked as a part-time consultant to Wolfram Research focusing on development of mathematical content for WolframAlpha.


2d
comment The Riemann Zeta Function From Parametric to 3D
Also, if you'd like the question to be reopened, I think you need to further clarify how the Riemann zeta function and the $i(n^{s}+x^{2s})/(4x^{s})$ term fit in.
2d
comment Accessing Reduce from DSolve
Seems so simple - thanks!
2d
comment The Riemann Zeta Function From Parametric to 3D
I've edited your question so that the Mathematica code works and typeset your mathematics using LaTeX, rather than a code block. I'm not certain that I've retained your intended meaning, however. I also must agree with Artes and others that your question is unclear. If you genuinely found his links unuseful, perhaps that's just because he had difficulty discerning your meaning.
Oct
16
comment Getting errors when I try to evaluate code taken from Trott's Mathematica Guidebook for Graphics
Have a look at mathematica.stackexchange.com/questions/20912/…
Oct
8
comment Contour integration *directly*
@Dror So WolframAlpha["integrate 1/z along a closed path about the origin"] would be what you're looking for (if it worked), right?
Oct
8
comment Contour integration *directly*
@Dror I think I understand the question quite well, namely "How can I evaluate this complex integral without knowing what a complex integral is"? You are right the only possible answer in any possible system is - "you can't". Well, perhaps WolframAlpha could as the intention there is to generally guess the intention of the user. Assuming one knows at least the basics, then this answer shows quite clearly how to approach the problem. Not that I expect an accept as the question deserved to be closed.
Oct
8
comment Contour integration *directly*
@george2079 You can certainly do things like Integrate[x^2*y^2, Element[{x, y}, DiscretizeRegion[Circle[]]]]. Note that the geometric region functionality works in $\mathbb R^2$, though, not in $\mathbb C$ so I suppose the OP would raise the same objections as a translation step would be necessary. Furthermore, the result is a numerical estimate, as that's the emphasis of the region based functionality.
Oct
8
comment Contour integration *directly*
@Dror I guess I don't understand your objection to the responses. If you desire a specific output then, surely, you need appropriate input. In this particular example, you've got to describe the contour (or family of contours) along which you want to integrate somehow - right? A parametrization seems to be the way to do it here. Of course, if you want a more general approach then you can use the residue theorem and Mathematica can certainly compute residues.
Oct
8
comment Contour integration *directly*
I agree with Artes that this is a duplicate. I also think that Nasser's answer is exactly what you want. He shows how to compute a contour integral given a particular parametrized path. The idea is easy to generalize by simply changing the parametrization and integrand.
Oct
5
comment Graphics3D not rendering Ball as a filled object
@m_goldberg Show[{DiscretizeRegion[Ball[],{{0,1},{0,1},{0,1}}],Graphics3D[Sphere[]]}] works fine for me. I really don't understand your object to the answer at all. Regardless, the assertion that it's not relevant seems patently false.
Oct
4
comment Graphics3D not rendering Ball as a filled object
@m_goldberg Well, you asked if it was a bug. I replied "this is a minor bug" with explanation as to why I consider it to be minor and suggested a workaround.
Oct
3
comment StreamPlot for Bifurcation Diagram
@Stephen I'm glad you found the code useful! It sounds like you might need to incorporate a SaveDefinitions option into your code, if you want the dynamic content to work when you open the notebook. It's hard to say for sure, though, without seeing your code.
Sep
27
comment Enumerating all connected graphs with 9 vertices
Should be faster if you use FindIsomorphism[g,g,All] on a Graph object. On my iPhone right now so I can't try it just now.
Sep
27
comment Fitting ellipse to 5 given points on the plane
@Jens My assumption is that the OP simply wants to add more points that are know to be on the same ellipse, rather than find a best fit.
Sep
27
comment Fitting ellipse to 5 given points on the plane
@user153012 If you'd like to plot more points in a list, say morePoints, simply add another Point primitive containing your list to the Epilog.
Sep
27
comment Enumerating all connected graphs with 9 vertices
@EdenHarder I believe the Graph6 representation of a graph depends on the ordering of the vertices and, as such, is not unique so there's no reason to expect "HhcWJEA" to be on that list. Presumably, one of the automorphisms of your graph leads to your g6 code.
Sep
27
comment Plotting Semi-hollow spheres
In light of the post of @belisarius, I wonder if the answers here contributed significantly to your cover image? If so, attribution (which I don't see, but might have missed in the article) would be appropriate.
Sep
27
comment Enumerating all connected graphs with 9 vertices
getConnG[7] in about 6 minutes.
Sep
27
comment Bifurcation diagrams for multiple equation systems
@belisarius I saw that question but was not quite able to figure a good way to approach it. At least, not given the kind of time I've got these days!
Sep
27
comment Enumerating all connected graphs with 9 vertices
I modified your code slightly so that getConnG[6] runs in 5 seconds as opposed to 30 seconds on my machine. The major point is that DeleteDuplicates has quadratic time complexity when using a user defined comparison function. Thus, breaking the list of graphs in smaller parts using the total number of edges helps a lot. Of course, DeleteDuplicates can run like $n\log(n)$, when the data is linearly ordered, but that's not generally possible with a user defined comparison function.