## Plot of the derivative of a function

I have a function functionSL as a function of t (t<0) where I want to find the extremum of the function and also find at which t it occurs. I took the derivative of functionSL with t which I wrote it as the function functionSLD.

d = 3; torootL[a_?NumericQ, t_?NumericQ, zl_?NumericQ, zh_?NumericQ] := a - NIntegrate[(zl y^d)/Sqrt[(1 - (zl/zh)^(d + 1) y^(d + 1)) (1 + t^2 (1 - (zl/zh)^(d + 1))^-1 - y^(2 d))], {y, 0, 1}, PrecisionGoal -> 6, Method -> "GlobalAdaptive"] zs[a_?NumericQ, t_?NumericQ, zh_?NumericQ] := zl /. FindRoot[torootL[a, t, zl, zh], {zl, 0.5, 0, 1}] intSL[a_?NumericQ, t_?NumericQ, zh_?NumericQ] := NIntegrate[With[{b = zs[a, t, zh]/zh}, (((-1)/(d - 1)) (zs[a, t, zh]^(2 d) (1 + t^2 (1 - (zs[a, t, zh]/zh)^(d + 1))^-1))^-1 zs[a, t, zh]^(2 d)) x^d ((1 - (b x)^(d + 1))/(1 - (zs[a, t, zh]^(2 d) (1 + t^2 (1 - (zs[a, t, zh]/zh)^(d + 1))^-1))^-1 (zs[a, t, zh] x)^(2 d)))^(1/2) - ((b^(d + 1) (d + 1))/(2 (d - 1))) x ((1 - (zs[a, t, zh]^(2 d) (1 + t^2 (1 - (zs[a, t, zh]/zh)^(d + 1))^-1))^-1 (zs[a, t, zh] x)^(2 d))/(1 - (b x)^(d + 1)))^(1/2) + (b^(d + 1)x)/((1 - (b x)^(d + 1)) (1 - (zs[a, t, zh]^(2 d) (1 + t^2 (1 - (zs[a, t, zh]/zh)^(d + 1))^-1))^-1 (zs[a, t, zh] x)^(2 d)))^(1/2)], {x, 0, 1}, MinRecursion -> 20, MaxRecursion -> 20, AccuracyGoal -> 12, PrecisionGoal -> 10, Method -> {"GlobalAdaptive", "SingularityHandler" -> Automatic}] functionSL[a_?NumericQ, t_?NumericQ, zh_?NumericQ] := ((-((1 - (zs[a, t, zh]^(2 d) (1 + t^2 (1 - (zs[a, t, zh]/zh)^(d + 1))^-1))^-1 zs[a, t, zh]^(2 d)) (1 - (zs[a, t, zh]/zh)^(d + 1)))^(1/2)/(d - 1)) + intSL[a, t, zh] + 1)/(4 zs[a, t, zh]^(d - 1)) functionSLD[t_] := Evaluate[Derivative[0, 1, 0][functionSLL][0.01, t, 1]] 

I took some sample values of functionSLD for some t,

In[44]:= functionSLD[0]  Out[44]= -3.58024*10^-12  In[48]:= functionSLD[-10]  Out[48]= 0.15527  In[90]:= functionSLD[-15]  Out[90]= 0.0477369  In[91]:= functionSLD[-16]  Out[91]= 0.041289  In[93]:= functionSLD[-16.5]  Out[93]= 0.039934  In[59]:= functionSLD[-17] // Quiet  Out[59]= 0.0424448  In[60]:= functionSLLP[-17.5]  Power::infy: Infinite expression 1/0. encountered.  NIntegrate::izero: Integral and error estimates are 0 on all integration subregions. Try increasing the value of the MinRecursion option. If value of integral may be 0, specify a finite value for the AccuracyGoal option.  NIntegrate::izero: Integral and error estimates are 0 on all integration subregions. Try increasing the value of the MinRecursion option. If value of integral may be 0, specify a finite value for the AccuracyGoal option.  Power::infy: Infinite expression 1/0. encountered.  NIntegrate::izero: Integral and error estimates are 0 on all integration subregions. Try increasing the value of the MinRecursion option. If value of integral may be 0, specify a finite value for the AccuracyGoal option.  General::stop: Further output of NIntegrate::izero will be suppressed during this calculation.  Power::infy: Infinite expression 1/0. encountered.  General::stop: Further output of Power::infy will be suppressed during this calculation.  FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.  FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.  FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.  General::stop: Further output of FindRoot::lstol will be suppressed during this calculation.  FindRoot::jsing: Encountered a singular Jacobian at the point {zl} = {0.825567}. Try perturbing the initial point(s).  FindRoot::jsing: Encountered a singular Jacobian at the point {zl} = {0.825567}. Try perturbing the initial point(s).  FindRoot::jsing: Encountered a singular Jacobian at the point {zl} = {0.825567}. Try perturbing the initial point(s).  General::stop: Further output of FindRoot::jsing will be suppressed during this calculation.  In[61]:= functionSLD[-17.5] // Quiet  Out[61]= \$  Aborted 

I expect functionSL to have an extremum for at least two values (looking only at t<0) so that functionSLD has at least two roots, I think I got one root at t=0 which is clear in the plot of $$\frac{dS}{dt}$$ and it really confirms my expectations, the other is located somewhere else (it seems clear in the plot).

You can see that at t=0, functionsSLD = -3.58024*10^-12 which is essentially zero, as t goes to lower values functionSLD rises and then goes down again and it looks like it is going to be essentially zero again but as you can see at t=-17.5 I aborted the calculation (in the sample values of functionSLD) because it just takes so long and it seems like there is a problem.

In the end, what I want to see is a plot of functionSLD[t] vs. t ($$\frac{dS}{dt}$$).

I would also like somebody to check my NIntegrate Rules if there is something wrong with it, or can it be improved. I added a singularity handler because an error occurred which I pasted in the above sample values code.

## How to plot a set of complex numbers with given argument and absoulute value bounds

I want to plot the following complex numbers $$z \in \text{(complex numbers)}:\pi/4 < \arg (z) \leq 5 \pi/4,\ 1 \leq |z| < 2$$

I don’t know how to graph it so that it would look like 2D without any unecessary details. The closest I found how I want it to look is, when I looked at how graphing of parametric function looks. I tried to use contour plot to graph it, but I just can’t seem to do it…

ContourPlot[ Im[F[z[x, y]]], {3 pi/4 < arg (z) <= 5 pi/4, 1 <= abs (z) < 2}, {x, -.2, .2}, {y, -.2, .2}, PlotRange -> All, Contours -> Range[-5, 5, .5], ContourLabels -> True]

Does anybody know how to graph my set?

## 3D plot from an array

I have an array on the following form:

Array[1, {20, 20, 3}]; 

and i want to visualise the data in 3D plot, i tried many times but usually Mathematica reply that there is an error.

here is a part of the code

part of the data

{{{0, 0, {0.20667}}, {0, 1, {0.584291}}, {0, 2, {0.948858}}, {0,3, {0.922532}}}, {{1, 0, {0.20667}}, {1, 1, {0.188105}}, {1,2, {0.000314782}}, {1, 3, {0.163525}}}, {{2, 0, {0.20667}}, {2,1, {0.948938}}, {2, 2, {0.0784778}}, {2, 3, {0.00361825}}}, {{3,0, {0.20667}}, {3, 1, {0.772576}}, {3, 2, {0.466603}}, {3,3, {0.00774734}}}} 

## Plot markers For QuantilePlot

I want to change the plot markers of the data points in QuantilePlot. This seems pretty straightforward:

QuantilePlot[                 RandomVariate[NormalDistribution[0, 1], 32], NormalDistribution[mu, sigma],                 ReferenceLineStyle -> Directive[Red, Dashing[{Large}]], PlotMarkers -> {"\[EmptyCircle]"}             ]  

However we can see that we get a "Null" plot marker at either end of the reference line:

Is there a specific command required in QuantilePlot to customise the plot markers for the data points?

## How to plot the surface defined by x^3+4y^2=10z; use (-3,3) as the range for x & y

Plot the surface defined by x^3+4y^2=10z; use (-3,3) as the range for x & y

## How do i use Streamplot to plot a non homogenous differential equation

I have the following equations:

x’ = x-y y’ = x+y-2xy

I used the following code to do the streamplot: I apologize for only having an image, as I am using Mathematica through my school servers I cannot copy and paste the data.

But I do not get any results, so then I linearize the equations and plot them separately, but I cannot figure out how to combine the plots to show in one plot.:

Is there any way to use Streamplot with nonhomogenous DE’s or a better way to combine the equations?

## How to store outputs as coordinates and use ListPlot to plot them?

so I have a program that outputs a desired value for incrementing values of alpha. Right now, my program prints a list of these for 40 values of alpha in the form {alpha, newValue}.

d = 1000; For[k = 1, k <= 40, k++,   alpha = 0.1*k;   fd = 3*d^2/(Pi^2);    fs = fd*FareySequence[d];   count = 0;    For[i = 1, i < fd, i++,     For[j = (i - 1), j > 0, j--,       If[(fs[[i]] - fs[[j]]) < alpha, count++, Break[]]];];     newValue = count/d^2;   Print["{", alpha, ",", newValue, "}"];   ]; 

Here are the first few outputs:

{0.1,0} {0.2,0} {0.3,0} {0.4,1911/100000} {0.5,13593/250000} {0.6,90521/1000000} .. .. .. {3.9,1155653/1000000} {4.,74163/62500} 

I want to make a plot all of these outputs as coordinates. I want to plot these point on a graph with alpha on the x axis and newValue on the y axis. Any help would be appreciated.

## Logplot and linear plot in the same plot

I have the following code:

q := 1.6*10^-19; me := 9.1*10^-31; (* Free electron rest mass in kg *) h :=  6.63*10^-34;  (* Reduced Planck's constant in J.s *) kb := 1.38*10^-23;(* Boltzmann constant in J/K *) LogPlot[Abs[Jschottky[V, 77]], {V, -0.5, 0.5}, PlotRange -> All,   Frame -> True,   FrameLabel -> {"Voltage (V)",     "\!$$\*FractionBox[\(J$$, SuperscriptBox[$$T$$, $$3/2$$]]\)"},   BaseStyle -> {FontSize -> 15}, PlotStyle -> {Thick, Red} ,   AspectRatio -> GoldenRatio, ImageSize -> 400, FrameStyle -> Black,   FrameTicks -> {{{#, Superscript[10, Log10@#]} & /@ ({10^-21, 10^-11,         10^-1, 10^9, 10^19}), None}, {Automatic, None}}]  Plot[Abs[Jschottky[V, 77]], {V, -0.5, 0.5}, PlotRange -> All,   Frame -> True,   FrameLabel -> {"Voltage (V)",     "\!$$\*FractionBox[\(J$$, SuperscriptBox[$$T$$, $$3/2$$]]\)"},   BaseStyle -> {FontSize -> 15}, PlotStyle -> {Thick, Blue} ,   AspectRatio -> GoldenRatio, ImageSize -> 400, FrameStyle -> Black] 

I get the following results:

Now I want to plot them on the same plot with the logplot on the left y axis and the linear plot on the right yaxis. What should I do? Also any recommendations for a good grayscale plot of the same?

## Make a density plot from a file where the data makes a shape with a hole in it

I have a .dat file and I want to make preferably a density (or surface) plot with it.

ListDensityPlot[f, ColorFunction -> "Rainbow", Axes -> False,   Background -> Black,   PlotLegends ->    Placed[BarLegend[Automatic, LegendMargins -> {{0, 0}, {10, 6}},      LegendMarkerSize -> {15, 300}, LegendLabel -> "F"], After] ,   AspectRatio -> 0.5,  PlotRange -> All] 

The data has the shape of a not perfectly round donut seen from above, so there’s a hole inside and the density and surface plot commands apply the color of the ColorFunction as if it’s 0 to the part in the center that corresponds to the hole. Is there a way that the part in the middle doesn’t get colored? Or can I assign specifically black to the values that are "0" (but where in reality there’s not data) and still use the Color Function?

## Given a plot of a network Graph[] how can the {x,y} screen coordinates be output?

Is there a general way to get the screen or "world" coordinates for every vertex in the graphic output of something like this?

Table[Graph[Table[i \[UndirectedEdge] i + 1, {i, 20}],    GraphLayout -> l,    PlotLabel -> l],  {l, {"CircularEmbedding", "SpiralEmbedding"}}] 

Trying to build an algorithm to generate novel layouts and want to use a large amount of other algorithms for training data.