Skip to main content

equation solving - Find zeros of a transcendental function


Mathematica 10 can not give me solution of this equation


NSolve[τ - ArcCos[1/(-11.` + 6.16949` E^(0.05` τ))]/Sqrt[
0.380626` + (
0.08062600000000003` -
0.04130349999999999` E^(0.05` τ))/(-1.` +
1.` E^(0.05` τ) - 0.25` E^(0.1` τ))] == 0, τ ]


That's running too long time. Is there a way to have an alternative issue.


Any help is welcome



Answer



fun = t - ArcCos[1/(-11. + 6.16949 E^(0.05 t))]/
Sqrt[0.3806 + (0.080626 - 0.0413035 E^(0.05 t))/
(-1. + 1. E^(0.05 t) - 0.25 E^(0.1 t))];

To get a "feeling" for the function we first plot it. We find a reasonable plot range with FunctionDomain.


FunctionDomain[fun, t]


13.3062 < t < 13.8629 || t > 13.8629 || t < 9.65863

Plot[fun, {t, -5, 14}]

enter image description here


There are three zeros which we find by mapping FindRoot over the range:


(sol = FindRoot[fun == 0, {t, #}] & /@ Range@12) // TableForm

enter image description here



We get rid of the "duplicates" (values with very small differences) and chop the imaginary parts with


uni = Union[Chop@sol[[All, 1, 2]], SameTest -> (Abs[#1 - #2] < 0.1 &)]


{4.51773, 8.36369, 11.5824}



The zero-points are


p = Point@Transpose[{uni, ConstantArray[0, Length@uni]}];

Let's plot them



Plot[fun, {t, -5, 14}, Epilog -> {PointSize@Large, Red, p}]

enter image description here


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[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

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...