2,864 reputation
1611
bio website stephenluttrell.com
location West Malvern, United Kingdom
age
visits member for 2 years, 4 months
seen 16 mins ago

I have been a Mathematica user since 1981, using version 0 to do QCD calculations for my PhD, and all subsequent versions to do research into various sorts of adaptive information processing. The use of Mathematica has been the key to almost every new result that I have discovered, because I can do "experiments" at the speed at which I think (more or less), so I don't get bored or lose my train of thought. Mathematica and I are a well-matched symbiotic pair!


10h
comment Strange integration
BTW, I like your use of the 1/(1-z) == Sum[z^k, {k, 0, \[Infinity]}] trick.
10h
comment Strange integration
Based on a ListPlot of the terms in my Sum I had assumed that there must be subtle tricks going on behind the scenes in V10 in order for it to get its result to come out - in fact, I was rather surprised that it worked at all.
16h
answered Strange integration
Oct
15
comment Integrate yields complex value, while after variable transformation the result is real. Bug?
I tried to drill down to the origin of this bug. Define the integrand f[z_] := 1/Sqrt[7/10 + 3/10 (1 + z)^3], then evaluate the indefinite integral int[z_] = Integrate[f[z], z]. Close inspection reveals that int[z] has a discontinuity that shouldn't be there - e.g. compare int[1.6527048] with int[1.6527049]. If you evaluate ContourPlot[Arg[int[x + I y]], {x, -3, 3}, {y, -3, 3}] it shows the context for this discontinuity. It appears that the wrong result is due to Mathematica jumping between different branches of Arg, or something like that ...
Oct
14
answered Maximizing a trigonometric function
Oct
13
comment Segmentation of a microscopy image with uneven illumination
You could try Import["http://i.stack.imgur.com/5ni1u.jpg"] // EdgeDetect // DeleteSmallComponents // FillingTransform. The cell at the top has an "invisible" left boundary, so it doesn't get filled in.
Oct
2
comment NIntegrate giving message NIntegrate::slwcon:
If you change the range of integration to {x, -r, 2r}, where r is a random real, then the problem persists, and if you include the option Exclusions -> 0 then the problem goes away. Interestingly, z = 0 is the point in the complex plane z = x + I y where Arg[f[z]] is singular, and it lies exactly 1/3 of the way along the {x, -r, 2r} range of integration. Just wondering ...
Sep
30
comment How to get NMinimize to work
In NMaximize and NMinimize use Method -> "SimulatedAnnealing", MaxIterations -> 500.
Sep
24
awarded  Autobiographer
Sep
21
answered Bug in ClusteringComponents yielding different results on practically same input?
Sep
21
comment Bug in ClusteringComponents yielding different results on practically same input?
The k-means algorithm can get trapped in a local minimum during optimisation, so try using the "RandomSeed" option of ClusteringComponents to vary its initial guess. I found that "RandomSeed" -> 12345 makes the results the same in both cases.
Sep
18
revised How can I create a set (list) provide a kind of rule/algorithm?
added 2 characters in body
Sep
18
answered How can I create a set (list) provide a kind of rule/algorithm?
Sep
16
awarded  Nice Answer
Sep
10
comment Hardware required to perform large symbolic calculations
@mitochondrial You say "Standard licence allows Mathematica use four core only" but it says on here that the standard licence allows you to run 8 "computation kernels". A 4-core i7 CPU hyperthreads 8 computation kernels (modulo the collisions that occur in hyper threading) but a 4-core i5 CPU does NOT hyperthread, so 4-core i7 (rather than 4-core i5) seems to be well-matched to the Mathematica "standard licence". I agree with other comments here that maxing out the RAM size is the top priority.
Sep
9
comment How to calculate the analytical result of this singular integral?
I noticed that the substitution $x = \cosh(t)$, $dx = \sinh(t) dt$, $\cosh(t)^2 - \sinh(t)^2 = 1$ transforms the integrand into a simpler form. However, this transformation doesn't seem to get you any closer to a closed-form solution to the integral, and you also need to know which way to go around the pole on the real line.
Sep
5
comment symbolic integration error
This also works: displace the endpoints of the integration, evaluate the integration, then move the endpoints back to their original locations: Integrate[1/\[Pi] ((1 - x) (1 + 2 x)^6)/Sqrt[1 - x^2], {x, -1 + a, 1 + a}] /. a -> 0.
Sep
4
revised Rewrite an expression as a sum of $SU(2)$ characters?
deleted 3 characters in body
Sep
4
comment Rewrite an expression as a sum of $SU(2)$ characters?
Yes. The second example is the expression at the top of the question. I added it to the end of my answer after its typos were corrected, but it doesn't really demonstrate anything more than what is in the body of my answer.
Sep
4
answered Rewrite an expression as a sum of $SU(2)$ characters?