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Re: Re: Complex Plot
*To*: mathgroup at smc.vnet.net
*Subject*: [mg88314] Re: [mg88288] Re: Complex Plot
*From*: Andrzej Kozlowski <akozlowski at gmail.com>
*Date*: Thu, 1 May 2008 03:22:34 -0400 (EDT)
*References*: <fv9a83$mjo$1@smc.vnet.net> <200804301103.HAA04939@smc.vnet.net>
On 30 Apr 2008, at 20:03, Jean-Marc Gulliet wrote:
> Chris Degnen wrote:
>
>> Can anyone tell me how I can re-express this
>> equation in terms of x for plotting?
>>
>> Sin[x^2*y]==Log[x/y]
>>
>> last sighted: http://tinyurl.com/3s4hfd
>
> Hi Chris,
>
> If you are interested in the graphical representation of this
> function,
> you may want to use *ContourPlot* and experiment with its various
> options. For instance,
>
> ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23},
> ImageSize -> 500]
>
> ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23},
> ImageSize -> 500, ColorFunction -> (White &), ContourStyle -> Black,
> MaxRecursion -> 0, PlotPoints -> 500]
>
> You can see the result at
>
> http://homepages.nyu.edu/~jmg336/mathematica/
> chrisdegnencomplexplot.pdf
>
> Note that the second plot may take several minutes to complete due to
> the huge number of sampling points, but it should be very close to
> what
> you have seen on the Internet.
>
> Hope this helps,
> -- Jean-Marc
>
I do not think this gives a realistic plot, due to numerous artifices
caused by ContourPlot. Here is one way to get a realistic graph
without using ContourPlot.
We want to plot the set of solutions to:
Sin[x^2*y] == Log[x/y]
We first use a substitution:
x == t y
The equation that we get in terms of y and t can now be completely
solved by Mathematica:
Reduce[Sin[t^2*y^3] ==
Log[t], y, Reals]
Element[C[1], Integers] &&
1/E <= t <= E &&
(y == Root[-#1^3 -
ArcSin[Log[t]]/t^2 +
(2*Pi*C[1])/t^2 +
Pi/t^2 & , 1] ||
y == Root[-#1^3 +
ArcSin[Log[t]]/t^2 +
(2*Pi*C[1])/t^2 & , 1])
We can use Plot to get a graph of this:
g = Plot[Evaluate[Flatten[
Table[{Root[
-#1^3 - ArcSin[
Log[t]]/t^2 +
(2*Pi*n)/t^2 +
Pi/t^2 & , 1],
Root[-#1^3 +
ArcSin[Log[t]]/
t^2 + (2*Pi*n)/
t^2 & , 1]},
{n, -100, 100}]]],
{t, 1/E, E}];
Now we can convert this graph into a graph of the relationship we want:
Show[g1 = g /. {t_?NumberQ, y_?NumberQ} -> {t y, y}, AspectRatio ->
Automatic,
PlotRange -> All]
This is rather different from the graph displayed at the link above
but quite similar to:
Show[ContourPlot[Sin[x^2*y] == Log[x/y], {x, -3, 3}, {y, -3, 3}],
Axes -> True]
Andrzej Kozlowski
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