## Rasterization problem in version 11.3.0 that was not in 8.0.1

I am running MacOS 10.11.5 and I have recently upgraded from Mathematica version 8.0.1 to version 11.3.0. I created the animation for this answer on Mathematics using the Mathematica code

f[x_] := ParametricPlot3D[{(t (1 - t))^(3/2) Cos[2 Pi s], t , (t (1 - t))^(3/2) Sin[2 Pi s]}, {t, 0, 1}, {s, 0, 1}, Axes -> False, Boxed -> False, Mesh -> {15, {1/1200, 1/12, 2/12, 3/12, 4/12, 5/12, 6/12, 7/12, 8/12, 9/12, 10/12, 11/12}}, ViewVector -> {{10 Cos[x], 10 Sin[x] + 1/2, 0}, {0, 1/2, 0}}, ViewVertical -> {0, 0, 1}, ViewAngle -> 1/36, ImageSize -> {400, 100}, PlotTheme -> {"Classic"}, MeshStyle -> Directive[Thickness[1/144]]]

The PlotTheme -> {"Classic"} is ignored in 8.0.1, but is needed in 11.3.0 to get the same lighting and color scheme.

Under 8.0.1, I exported the GIF animation with

Export["potatoes8.gif", Table[f[Pi k/32], {k, 0, 31}], "GraphicsList", "DisplayDurations" -> .05]

and under 11.3.0, I exported the GIF animation with

Export["potatoes11.gif", Table[f[Pi k/32], {k, 0, 31}], "GraphicsList", "DisplayDurations" -> .05, AnimationRepetitions -> Infinity]

The output from 11.3.0 is much worse than that from 8.0.1. The MeshStyle option in the code above was added to try to make the mesh thicknesses similar, as was the PlotStyle option for the lighting.

I have tried different Thicknesses for the MeshStyle, but those that give unbroken mesh lines, give lines that are way too thick.

An interesting thing is that under 8.0.1, in the Notebook, the rendered image and the rasterized image look identical:

However, under 11.3.0, in the Notebook, the rendered image is at a higher resolution than under 8.0.1 (this is a good thing), but the rasterized image looks worse than under 8.0.1 (and at the same resolution):

It would be great to get the higher resolution images from 11.3.0 in the animation, but I would be happy if the animation under 11.3.0 were at least as good as that under 8.0.1.

Does anyone have any suggestions of what might be done to improve the rasterization quality under 11.3.0?

## How to evaulate salesmen objectively? [migrated]

Each salesman will sell product to three kinds of person,each kind of customers has a different response rate,I will mark these customers level A,level B and level C.

According data here,how to evaluate if a saleman is better by their response rate.Any idea is highly appreciated.

+----------+------------------+----------------+------------------+----------------+------------------+----------------+ | sales_no | levelA_customers | levelA_success | levelB_customers | levelB_success | levelC_customers | levelC_success | +----------+------------------+----------------+------------------+----------------+------------------+----------------+ | no1      |              158 |              5 |             1254 |             11 |             5423 |              8 | | no2      |              300 |              6 |             2236 |             15 |             3687 |             11 | | no3      |              198 |              4 |             1727 |             18 |             2567 |             14 | | no4      |              234 |              7 |              987 |             23 |               99 |              0 | | no5      |              388 |              2 |              876 |              4 |             1116 |             23 | +----------+------------------+----------------+------------------+----------------+------------------+----------------+ 

sales_no is id of saleman levelA_customers is how many level A customers have been selled levelA_success is how many level A customers have buy the product

## Changing ContourStyle based on an inequality

I would like to create a bifurcation diagram with ContourPlot. Suppose I want to plot the contour 1-2*a*Q+3*Q^3==0 with a the x-axis, Q the y-axis. I also want linestyle/thickness of the contour plot to change depending on the expression expr = 9*Q^2-2*a. If expr>0, I want it to be dashed; if expr<0, I want it to be a thick line. How to accomplish this?

Some previous posts about similar topic suggested putting an If statement inside the plot, but when I try something like

ContourPlot[If[expr>0, 1-2*a*Q+3*Q^3 == 0], {a,-2,2}, {Q,-2,2}] 

it gives error.

## Solving time dependent boundary conditions heat PDE

I am attempting to solve the heat equation $$\frac{\partial T}{\partial t}=\nabla^2T$$, where $$T=T(x,y,z,t)$$, subject to the following boundary conditions:

$$\frac{\partial T}{\partial x}|_{x=10}=\frac{\partial T}{\partial x}|_{x=-10}=0$$

$$\frac{\partial T}{\partial y}|_{y=10}=0$$

$$T(0,0,z_{crit}(t),t)=f(t)$$ where $$z_{crit}(t)=t$$ and $$f(t)=e^{-t}$$

$$\frac{\partial T}{\partial y}|_{y=0}=\frac{\partial T}{\partial z}|_{z=0}=0$$

But my attempted code has not worked properly and I encounter frequently many different errors.

Is there any way to fix this issue?

## Why can’t I plot this surface?

I can’t plot (x^z)-(y^z)=x*y. I’ve tried it with Geogebra and with

Plot3D[x^z – y^z == x*y, {x, -1, 1}, {y, -1, 1}]

in Mathematica, but in both images only appears the axes.

Is there any other Mathematica tool to do it?

I’m started to think that the function is “non-plottable” in some mathematically meaning. Is that possible? Well, it is plottable for any constant z I’ve tried (I mean setting a value of z and making it a curve), but its shape changes a lot even between very close values… Do “non-plottable” functions exist?

Any idea will help, thank you!

## Average value of function with 2 variables

How would I determine the average value of a function of two intervals, both of which have given bounds if the function f[x,y] (below) has bounds x,y=[0,5]. I assume by using the integration function, but I’ve only done that with a function of 1 variable.

Sqrt[1 + .25y^2 ((2.5 + 1.5 Cos[Pi/3 x] - 2.75)^2 + y^2(2.4 +    1.1 Sin[Pi/3 x] - 3.75)^2)] 

## KL divergence of two independent variables

I’m missing something obviously this is the way to start this:

$$\sum_{x\in X}^{}p(x)log(p(x)) – \sum_{x\in X}^{}p(x)log(q(x))$$

how to continue from here?

It looks intuitive that if the variables are independent then the sum of all the partial variables ($$p_{1},…,p_{i}$$) expression will be equal to the KL divergence. Is my intuition ok? How can it be shown in a mathematical way?

## Function of Maximum and Minimum Functions of Two Functions

I try to answer this question The Maximum and Minimum Functions of Two Functions

I wrote the following code

f[x_, y_] := 1 + 2*x + 3*y^3 g[x_, y_] := y + x^2 maxi[x_, y_] :=   Refine[{(f[x, y] + g[x, y])/2 + Abs[f[x, y] - g[x, y]]/2},    Assumptions -> {0 <= x <= 1, 0 <= y <= 1}] Plot3D[maxi[x, y], {x, 0, 1}, {y, 0, 1}] 

Is there any way to find function of maxi?

## 2D function on a 3D plot

I’m trying to put an ellipse given by:

{2.5 + 1.5 Cos[Pi/3 t], 2.4 + 1.1 Sin[Pi/3 t]}, {t, 0, 6}

onto a 3D plot of a large function composed of several different parts, which essentially displays a mountain range. How do I get the 2D ellipse to show up on the 3D plot? For reference, my 3D plot is given by:

Plot3D[m[x, y], {x, 0, 5}, {y, 0, 5}

I won’t put m[x,y] here because it will take up very much space, but it is essentially 4-5 peaks over the given interval.

## Possible bug in Sum?

Consider the following (minimal?) working example:

f[k_, lim_] := Module[{q, r},    {q, r} = QuotientRemainder[lim - k (k + 1)/2 + 1, k] ];  Sum[f[k, 10^12], {k, 1000000}]  (* {14142725474025, 249498313774} *)  Sum[f[k, 10^12], {k, 1000001}]  (* Set::shape: Lists {q$$28245507,r$$28245507} and        QuotientRemainder[1000000000001-1/2 k (1+k),k] are not the same shape. *)  (* {14142725974023, 249498313776} *)  f[1000001, 10^12]  (* {499998, 2} *)  Total[Table[f[k, 10^12], {k, 1000001}]]  (* {14142725974023, 249498313776} *) 

The function f returns a pair of integers through some arithmetic. If I sum up the first million, things work fine. If I sum up 1000001 of them, I get an error message. Even stranger, if I convert Sum to Total[Table, the error message goes away. Finally, simply evaluating f at the offending value (1000001) does not produce an error message.

Lastly, I just realized that I could simply remove the {q,r}= and the corresponding Module, leaving a one-line function for f. When I do this, the error vanishes.