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17h
comment Can ParallelDo speed up a counting problem?
This might be too far from what you want, but maybe evs = Table[ParallelSubmit[Boole[RandomReal[] > 0.5]], {10000}]; and then Total@WaitAll[evs]
1d
comment Making single letters non-italic in Graphics/Text/Style objects
check out SingleLetterItalics
2d
comment Automatically highlight certain cells
@Kuba I am quite sure it won't, (for example if the cells you want to highlight in this manner are not connected) but this should show how one could use Item and the options it takes to style as wanted. Personally I wouldn't use a frame and rely on background.
Apr
14
comment How to produce functions
@george2079 I agree, I was trying to coax more information if possible.
Apr
14
comment How to produce functions
Is this closer? Array[Integrate[Symbol["f"<>ToString[#]][x],{x,Symbol["x"<>ToString[2#-1]],Symb‌​ol["x"<>ToString[2#]]}]&,4]
Apr
14
comment Machine precision infinity
@nikie that makes sense
Apr
14
comment Machine precision infinity
something like $MaxMachineNumber?
Apr
11
comment Setting an initial value for ListAnimate?
why not frames = Table[Plot[Sin[p x],{x,-1,1}],{p,1,4}]; Manipulate[Part[frames,p],{{p,3},1,4,1}]? If you are going to precompute the frames anyway...
Apr
8
comment Symbolically Expanding One-Variable q-hypergeometric Series
I was about to say the same thing Series[Sum[(-1)^n q^(6 n^2)/( QPochhammer[-q,q,3n]QPochhammer[q^3,q^3,n]),{n,0,20}],{q,0,13}]
Apr
7
comment Smoothing ListContourPlot contours
NonlinearModelFit might also be appropriate as well. In either case, I suspect they don't want to "smooth" as much as fit. The picture they show doesn't look to terrible to model.
Apr
7
comment PlotLegends and ImageSize
Yes, sorry I see I was using a slightly different plot. I will correct this.
Apr
7
comment PlotLegends and ImageSize
Ah very true indeed.
Apr
7
comment PlotLegends and ImageSize
Not a complete answer, but using Export["Neg_long_OU_FBM.pdf", %, ImageSize -> 9 cm]and not setting the ImageSize in the Plot gets you much closer.
Apr
1
comment Why IntegerPart[x/(x/2)]=1?
Then I guess it comes down to what you want IntegerPart[1.9999999999999998] to return?
Apr
1
comment Why IntegerPart[x/(x/2)]=1?
Look at the FullForm of the result of x/(x/2).
Mar
24
comment ToExpression an array
You will likely need to find a function to split strings.
Mar
18
comment ReadList with header and foot lines
You can also Skip rather than Read the data you are going to ignore.
Mar
11
comment Combine Plots into a single non-rasterized Graphic with no spacing
changed to make result a Graphics
Mar
11
comment Combine Plots into a single non-rasterized Graphic with no spacing
Yikes you are right...
Mar
3
comment Does RandomInteger[{0,n}] disfavor n?
As @Rojo says it looks like a Histogram binning thing: Table[{k,ListPlot[Tally[RandomInteger[k,20000]],Filling-> Axis,PlotRange->{0,Automatic}]},{k,10,100,10}]