Skip to main content

plotting - LogLogPlot plugs in zero


Bug introduced in 10.0 and fixed in 10.0.2




In testing the answers to Plot result of non analytical-integral in V10.0.1, I came across this behavior of LogLogPlot. All three version give the correct plot, but the first two give error messages.


ClearAll[f1, f2, f3, x];
k = 1/x;

kfn[x_?NumericQ] := 1/x;
f1[x_?NumericQ] := NIntegrate[k*y, {y, 0, 1}]; (* kludgy *)
f2[x_?NumericQ] := NIntegrate[kfn[x]*y, {y, 0, 1}]; (* correct way, but gives errors *)
f3[x_?Positive] := NIntegrate[k*y, {y, 0, 1}]; (* also correct and no errors *)

LogLogPlot[f1[x], {x, 1, 10}]
LogLogPlot[f2[x], {x, 1, 10}]
LogLogPlot[f3[x], {x, 1, 10}]

The first two give the message (twice):




NIntegrate::inumr: The integrand y/x has evaluated to non-numerical values for all sampling points in the region with boundaries {{0,1}}. >>



The second gives an additional message (twice):



Power::infy: Infinite expression 1/0 encountered. >>



Why does this happen? Is this a bug?


Notes: (1) It does not happen on V9. I seem to recall this coming up before (years ago). If folks have earlier versions, please test the code above. (2) Reported to WRI. (3) WRI confirmed it in 10.0.1 and stated it was resolved in 10.0.2.



Answer




The issue in V10.0.1 is that LogLogPlot uses x = 0 as a test point, instead of a value inside the specified plot domain.


The issue is resolved in V10.0.2.


Confirmed by WRI.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?