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I'm trying to solve for the slope with this implicit differentiation problem at x=1, but NSolve is unable to solve it. How can I get around this issue?
eqn[x_, y_] := x*Sin[y] - y*Sin[x] == 2 (*note: bound is -5<=x<=5,-5<=y<=5*)
yPrime = Solve[D[eqn[x, y[x]], x], y'[x]] /. {y[x] -> y,
y'[x] -> y'} // Simplify
{{Derivative[1][y] -> (y Cos[x] - Sin[y])/(x Cos[y] - Sin[x])}}
NSolve[eqn[x, y] /. x -> 1, y] (*this doesn't work*)
NSolve is not really the right tool for the job.
From Wolfram Documentation
If your equations involve only linear functions or polynomials, then
you can use NSolve to get numerical approximations to all the
solutions. However, when your equations involve more complicated
functions, there is in general no systematic procedure for finding all
solutions, even numerically. In such cases, you can use FindRoot to
search for solutions. You have to give FindRoot a place to start its
search.
Since your function is not really polynomial the following works :
FindRoot[eqn/.x->1,{y,-2}]
{y -> -2.7881400978783035` }
which you can plug into yprime.
Picking the starting point is not evident, however making a quick ContourPlot always works:
ContourPlot[x*Sin[y] - y*Sin[x], {x, -5, 5}, {y, -5, 5}, Contours -> {2}]
It shows that your equation is not unique for all x.
I asked this question a little while back that did help in reaching a solution. I've arrived at a somewhat acceptable approach but still not fully where I want it. Suppose there are two functions f1[x] and g1[y] that I want to determine the value of x and y for the common tangent(s). I can at least determine x and y for one of the tangents for example with the following:
f1[x_]:=(5513.12-39931.8x+23307.5x^2+(-32426.6+75662.x-43235.4x^2)Log[(1.-1.33333x)/(1.-1.x)]+x(-10808.9+10808.9x)Log[x/(1.-1.x)])/(-1.+x)
g1[y_]:=(3632.71+3806.87y-51143.6y^2+y(-10808.9+10808.9y)Log[y/(1.-1.y)]+(-10808.9+32426.6y-21617.7y^2)Log[1.-(1.y)/(1.-1.y)])/(-1.+y)
Show[
Plot[f1[x],{x,0,.75},PlotRange->All],
Plot[g1[y],{y,0,.75},PlotRange->All]
]
Chop[FindRoot[
{
(f1[x]-g1[y])/(x-y)==D[f1[x],x]==D[g1[y],y]
},
{x,0.0000001},{y,.00000001}
]
[[All,2]]
]
However, you'll notice from the plot that there exists another common tangent at slightly larger values of x and y (say x ~ 4 and y ~ 5). Now, interestingly if I slightly change the above expressions for f1[x] and g1[y] to something like the following:
f2[x_]:=(7968.08-59377.8x+40298.7x^2+(-39909.6+93122.4x-53212.8x^2)Log[(1.-1.33333x)/(1.-1.x)]+x(-13303.2+13303.2x)Log[x/(1.-1.x)])/(-1.+x)
g2[y_]:=(5805.16-27866.2y-21643.y^2+y(-13303.2+13303.2y)Log[y/(1.-1.y)]+(-13303.2+39909.6y-26606.4y^2)Log[1.-(1.y)/(1.-1.y)])/(-1.+y)
Show[
Plot[f2[x],{x,0,.75},PlotRange->All],
Plot[g2[y],{y,0,.75},PlotRange->All]
]
Chop[FindRoot[
{
(f2[x]-g2[y])/(x-y)==D[f2[x],x]==D[g2[y],y]
},
{x,0.0000001},{y,.00000001}
]
[[All,2]]
]
And use the same method to determine the common tangent, Mathematica chooses to find the larger values of x and y for the positive sloping tangent.
Finally, my question: is it possible to have Mathematica find both the high and low x and y values for the common tangent and store these values in a similar way that allows me to make a list plot? The functions f and g above are all complex functions of another variable, z, and I am currently using something like the following to plot the tangent points (should be two x and two y) as a function of z.
ex[z_]:=Chop[FindRoot[
{
(f[x,z]-g[y,z])/(x-y)==D[f[x],x]==D[g[y],y]
},
{x,0.0000001},{y,.00000001}
]
[[All,2]]
]
ListLinePlot[
Table[{ex[z][[i]],z},{i,1,2},{z,1300,1800,10}]
]
To find estimates for {x, y} that would solve your equations, you could plot them in ContourPlot and look for intersection points. For example
f1[x_]:=(5513.12-39931.8 x+23307.5 x^2+(-32426.6+75662. x-
43235.4 x^2)Log[(1.-1.33333 x)/(1.-1.x)]+
x(-10808.9+10808.9 x) Log[x/(1.-1.x)])/(-1.+x)
g1[y_]:=(3632.71+3806.87 y-51143.6 y^2+y (-10808.9+10808.9y) Log[y/(1.-1.y)]+
(-10808.9+32426.6 y-21617.7 y^2) Log[1.-(1.y)/(1.-1.y)])/(-1.+y)
plot = ContourPlot[{f1'[x] == g1'[y], f1[x] + f1'[x] (y - x) == g1[y]},
{x, 0, 1}, {y, 0, 1}, PlotPoints -> 40]
As you can see there are 2 intersection points in the interval (0,1). You could then read off the points from the graph and use these as starting values for FindRoot:
seeds = {{.6,.4}, {.05, .1}};
sol = FindRoot[{f1'[x] == g1'[y], f1[x] + f1'[x] (y - x) == g1[y]},
{x, #1}, {y, #2}] & ### seeds
To get the pairs of points from sol you can use ReplaceAll:
points = {{x, f1[x]}, {y, g1[y]}} /. sol
(*
==> {{{0.572412, 19969.9}, {0.432651, 4206.74}},
{{0.00840489, -5747.15}, {0.105801, -7386.68}}}
*)
To show that these are the correct points:
Show[Plot[{f1[x], g1[x]}, {x, 0, 1}],
{ParametricPlot[#1 t + (1 - t) #2, {t, -5, 5}, PlotStyle -> {Gray, Dashed}],
Graphics[{PointSize[Medium], Point[{##}]}]} & ### points]
OK, so let's quickly rewrite what you've done so far:
Using your f1 and g1, we have the plot
plot = Plot[{f1[x], g1[x]}, {x, 0, .75}]
and the first shared tangent at
sol1 = Chop[FindRoot[{(f1[x] - g1[y])/(x - y) == D[f1[x], x] == D[g1[y], y]},
{x, 0.0000001}, {y, .00000001}]]
(* {x -> 0.00840489, y -> 0.105801} *)
Define the function
l1[t_] = (1 - t) {x, f1[x]} + t {y, g1[y]} /. sol1
then, you can plot the tangents using
Show[plot, Graphics[Point[{l1[0], l1[1]}]],
ParametricPlot[l1[t], {t, -1, 2}],
PlotRange -> {{-.2, .4}, {-10000, 10000}}]
I briefly note (for my own sake) that the equations you used
(e.g., to generate sol1 above)
come from requiring that the tangent line for f1 at x
tangentially hits g1 at some point y, i.e.,
LogicalExpand[{x, f[x]} + t {1, f'[x]} == {y, g[y]} && f'[x] == g'[y]]
To investigate where the shared tangents lie, you can use a Manipulate:
Manipulate[Show[plot, ParametricPlot[{x, f1[x]} + t {1, f1'[x]}, {t, -1, 1}]],
{x, 0, .75, Appearance -> "Labeled"}]
which produces something like
Using the eyeballed values for x and y, you can get the actual solutions using
sol = Chop[Table[
FindRoot[{(f1[x] - g1[y])/(x - y) == D[f1[x], x] == D[g1[y], y]},
{x, xy[[1]]}, {y, xy[[2]]}], {xy, {{0.001, 0.01}, {0.577, 0.4}}}]]
define the two tangent lines using
l[t_] = (1 - t) {x, f1[x]} + t {y, g1[y]} /. sol
then
Show[plot, Graphics[Point[Flatten[{l[0], l[1]}, 1]]],
ParametricPlot[l[t], {t, -1, 2}, PlotStyle -> Dotted]]
This process could be automated, but I'm not sure how to do it efficiently.
Because PolarPlot should type r=... type of command.
But y=x will cause r to disappear.
How to draw that line with PolarPlot?
First, consider Plot[] which "generates a plot of f as a function of x from xmin to xmax" (I'm quoting the Mathematica documentation). You can't use it to plot a vertical line satisfying the equation x = x0, because the latter is not a function of x: instead of being single-valued, it has infinitely many values at x0.
Similarly, PolarPlot[] cannot be used to draw a straight line that passes through the origin, because its equation is not a function of θ: it has infinitely many values at a particular θ (equal to Pi/4 in the case requested), but none at all elsewhere. (Well, one could also allow the complementary angle 3Pi/4 as well.)
So I maintain it can't be done using the tools specified, short of the cheat
PolarPlot[0, {\[Theta], 0, 1},
Epilog -> Line[{Scaled[{1, 1}], Scaled[{0, 0}]}]]
I suggest you use a new function for things like this.
PolarParametricPlot[
{rT : {_, _} ..} | rT : {_, _},
uv : {_, _, _} ..,
opts : OptionsPattern[]
] :=
ParametricPlot[
Evaluate[# {Cos##2, Sin##2} & ### {rT}],
uv,
opts
]
Usage:
PolarParametricPlot[{t, 45 Degree}, {t, -10, 10}]
Here is the general polar form for a line of the form y = m x + b:
In[155]:= r /.
Solve[Eliminate[{x == r Cos[t], y == r Sin[t], y == m x + b}, {x,
y}], r]
Out[155]= {-(b/(m Cos[t] - Sin[t]))}
The solution vanishes when the y-intercept b is zero. This makes sense, since such lines are drawn at a constant angle, which is problematic since PolarPlot works by varying the angle.
You could approximate such a line by using a very small value for b, but there are probably better approaches.
You could draw the line using ListPolatPlot:
ListPolarPlot[{{Pi/4, 5}, {5 Pi/4, 5}}, Joined -> True]
I have 2 curves illustrated with the following Mathematica code:
Show[Plot[PDF[NormalDistribution[0.044, 0.040], x], {x, 0, 0.5}, PlotStyle -> Red],
Plot[PDF[NormalDistribution[0.138, 0.097], x], {x, 0, 0.5}]]
I need to do 2 things:
Find the x and y coordinates where the two curves intersect and
Find the area under the red curve to the right of the x coordinate in the
above intersection.
I haven't done this kind of problem in Mathematica before and haven't found a way to do this in the documentation. Not certain what to search for.
Can find where they intersect with Solve (or could use FindRoot).
intersect =
x /. First[
Solve[PDF[NormalDistribution[0.044, 0.040], x] ==
PDF[NormalDistribution[0.138, 0.097], x] && 0 <= x <= 2, x]]
Out[4]= 0.0995521
Now take the CDF up to that point.
CDF[NormalDistribution[0.044, 0.040], intersect]
Out[5]= 0.917554
Was not sure if you wanted to begin at x=0 or -infinity; my version does the latter. If the former then just subtract off the CDF evaluated at x=0.
FindRoot usage would be
intersect =
x /. FindRoot[
PDF[NormalDistribution[0.044, 0.040], x] ==
PDF[NormalDistribution[0.138, 0.097], x], {x, 0, 2}]
Out[6]= 0.0995521
If you were working with something other than probability distributions you could integrate up to the intersection value. Using CDF is a useful shortcut since we had a PDF to integrate.
Daniel Lichtblau
Wolfram Research
I am doing a brute force search for "gradient extremals" on the following example function
fv[{x_, y_}] = ((y - (x/4)^2)^2 + 1/(4 (1 + (x - 1)^2)))/2;
This involves finding the following zeros
gecond = With[{g = D[fv[{x, y}], {{x, y}}], h = D[fv[{x, y}], {{x, y}, 2}]},
g.RotationMatrix[Pi/2].h.g == 0]
Which Reduce happily does for me:
geyvals = y /. Cases[List#ToRules#Reduce[gecond, {x, y}], {y -> _}];
geyvals is the three roots of a cubic polynomial, but the expression is a bit large to put here.
Now to my question: For different values of x, different numbers of these roots are real, and I would like to pick out the values of x where the solutions branch in order to piece together the gradient extremals along the valley floor (of fv). In the present case, since the polynomial is only cubic, I could probably do it by hand -- but I am looking for a simple way of having Mathematica do it for me?
Edit: To clarify: The gradient extremals stuff is just background -- and a simple way to set up a hard problem. I am not so interested in the specific solution to this problem as in a general hand-off way of spotting the branch points for polynomial roots. Have added an answer below with a working approach.
Edit 2: Since it seems that the actual problem is much more fun than root branching: rcollyer suggests using ContourPlot directly on gecond to get the gradient extremals. To make this complete we need to separate valleys and ridges, which is done by looking at the eigenvalue of the Hessian perpendicular to the gradient. Putting a check for "valleynes" in as a RegionFunction we are left with only the valley line:
valleycond = With[{
g = D[fv[{x, y}], {{x, y}}],
h = D[fv[{x, y}], {{x, y}, 2}]},
g.RotationMatrix[Pi/2].h.RotationMatrix[-Pi/2].g >= 0];
gbuf["gevalley"]=ContourPlot[gecond // Evaluate, {x, -2, 4}, {y, -.5, 1.2},
RegionFunction -> Function[{x, y}, Evaluate#valleycond],
PlotPoints -> 41];
Which gives just the valley floor line. Including some contours and the saddle point:
fvSaddlept = {x, y} /. First#Solve[Thread[D[fv[{x, y}], {{x, y}}] == {0, 0}]]
gbuf["contours"] = ContourPlot[fv[{x, y}],
{x, -2, 4}, {y, -.7, 1.5}, PlotRange -> {0, 1/2},
Contours -> fv#fvSaddlept (Range[6]/3 - .01),
PlotPoints -> 41, AspectRatio -> Automatic, ContourShading -> None];
gbuf["saddle"] = Graphics[{Red, Point[fvSaddlept]}];
Show[gbuf /# {"contours", "saddle", "gevalley"}]
We end up with a plot like this:
Not sure if this (belatedly) helps, but it seems you are interested in discriminant points, that is, where both polynomial and derivative (wrt y) vanish. You can solve this system for {x,y} and throw away complex solutions as below.
fv[{x_, y_}] = ((y - (x/4)^2)^2 + 1/(4 (1 + (x - 1)^2)))/2;
gecond = With[{g = D[fv[{x, y}], {{x, y}}],
h = D[fv[{x, y}], {{x, y}, 2}]}, g.RotationMatrix[Pi/2].h.g]
In[14]:= Cases[{x, y} /.
NSolve[{gecond, D[gecond, y]} == 0, {x, y}], {_Real, _Real}]
Out[14]= {{-0.0158768, -15.2464}, {1.05635, -0.963629}, {1.,
0.0625}, {1., 0.0625}}
If you only want to plot the result then use StreamPlot[] on the gradients:
grad = D[fv[{x, y}], {{x, y}}];
StreamPlot[grad, {x, -5, 5}, {y, -5, 5},
RegionFunction -> Function[{x, y}, fv[{x, y}] < 1],
StreamScale -> 1]
You may have to fiddle around with the plot's precision, StreamStyle, and the RegionFunction to get it perfect. Especially useful would be using the solution for the valley floor to seed StreamPoints programmatically.
Updated: see below.
I'd approach this first by visualizing the imaginary parts of the roots:
This tells you three things immediately: 1) the first root is always real, 2) the second two are the conjugate pairs, and 3) there is a small region near zero in which all three are real. Additionally, note that the exclusions only got rid of the singular point at x=0, and we can see why when we zoom in:
We can then use the EvalutionMonitor to generate the list of roots directly:
Map[Module[{f, fcn = #1},
f[x_] := Im[fcn];
Reap[Plot[f[x], {x, 0, 1.5},
Exclusions -> {True, f[x] == 1, f[x] == -1},
EvaluationMonitor :> Sow[{x, f[x]}][[2, 1]] //
SortBy[#, First] &];]
]&, geyvals]
(Note, the Part specification is a little odd, Reap returns a List of what is sown as the second item in a List, so this results in a nested list. Also, Plot doesn't sample the points in a straightforward manner, so SortBy is needed.) There may be a more elegant route to determine where the last two roots become complex, but since their imaginary parts are piecewise continuous, it just seemed easier to brute force it.
Edit: Since you've mentioned that you want an automatic method for generating where some of the roots become complex, I've been exploring what happens when you substitute in y -> p + I q. Now this assumes that x is real, but you've already done that in your solution. Specifically, I do the following
In[1] := poly = g.RotationMatrix[Pi/2].h.g /. {y -> p + I q} // ComplexExpand;
In[2] := {pr,pi} = poly /. Complex[a_, b_] :> a + z b & // CoefficientList[#, z] & //
Simplify[#, {x, p, q} \[Element] Reals]&;
where the second step allows me to isolate the real and imaginary parts of the equation and simplify them independent of each other. Doing this same thing with the generic 2D polynomial, f + d x + a x^2 + e y + 2 c x y + b y^2, but making both x and y complex; I noted that Im[poly] = Im[x] D[poly, Im[x]] + Im[y] D[poly,[y]], and this may hold for your equation, also. By making x real, the imaginary part of poly becomes q times some function of x, p, and q. So, setting q=0 always gives Im[poly] == 0. But, that does not tell us anything new. However, if we
In[3] := qvals = Cases[List#ToRules#RReduce[ pi == 0 && q != 0, {x,p,q}],
{q -> a_}:> a];
we get several formulas for q involving x and p. For some values of x and p, those formulas may be imaginary, and we can use Reduce to determine where Re[qvals] == 0. In other words, we want the "imaginary" part of y to be real and this can be accomplished by allowing q to be zero or purely imaginary. Plotting the region where Re[q]==0 and overlaying the gradient extremal lines via
With[{rngs = Sequence[{x,-2,2},{y,-10,10}]},
Show#{
RegionPlot[Evaluate[Thread[Re[qvals]==0]/.p-> y], rngs],
ContourPlot[g.RotationMatrix[Pi/2].h.g==0,rngs
ContourStyle -> {Darker#Red,Dashed}]}]
gives
which confirms the regions in the first two plots showing the 3 real roots.
Ended up trying myself since the goal really was to do it 'hands off'. I'll leave the question open for a good while to see if anybody finds a better way.
The code below uses bisection to bracket the points where CountRoots changes value. This works for my case (spotting the singularity at x=0 is pure luck):
In[214]:= findRootBranches[Function[x, Evaluate#geyvals[[1, 1]]], {-5, 5}]
Out[214]= {{{-5., -0.0158768}, 1}, {{-0.0158768, -5.96046*10^-9}, 3}, {{0., 0.}, 2}, {{5.96046*10^-9, 1.05635}, 3}, {{1.05635, 5.}, 1}}
Implementation:
Options[findRootBranches] = {
AccuracyGoal -> $MachinePrecision/2,
"SamplePoints" -> 100};
findRootBranches::usage =
"findRootBranches[f,{x0,x1}]: Find the the points in [x0,x1] \
where the number of real roots of a polynomial changes.
Returns list of {<interval>,<root count>} pairs.
f: Real -> Polynomial as pure function, e.g f=Function[x,#^2-x&]." ;
findRootBranches[f_, {xa_, xb_}, OptionsPattern[]] := Module[
{bisect, y, rootCount, acc = 10^-OptionValue[AccuracyGoal]},
rootCount[x_] := {x, CountRoots[f[x][y], y]};
(* Define a ecursive bisector w/ automatic subdivision *)
bisect[{{x1_, n1_}, {x2_, n2_}} /; Abs[x1 - x2] > acc] :=
Module[{x3, n3},
{x3, n3} = rootCount[(x1 + x2)/2];
Which[
n1 == n3, bisect[{{x3, n3}, {x2, n2}}],
n2 == n3, bisect[{{x1, n1}, {x3, n3}}],
True, {bisect[{{x1, n1}, {x3, n3}}],
bisect[{{x3, n3}, {x2, n2}}]}]];
(* Find initial brackets and bisect *)
Module[{xn, samplepoints, brackets},
samplepoints = N#With[{sp = OptionValue["SamplePoints"]},
If[NumberQ[sp], xa + (xb - xa) Range[0, sp]/sp, Union[{xa, xb}, sp]]];
(* Start by counting roots at initial sample points *)
xn = rootCount /# samplepoints;
(* Then, identify and refine the brackets *)
brackets = Flatten[bisect /#
Cases[Partition[xn, 2, 1], {{_, a_}, {_, b_}} /; a != b]];
(* Reinclude the endpoints and partition into same-rootcount segments: *)
With[{allpts = Join[{First#xn},
Flatten[brackets /. bisect -> List, 2], {Last#xn}]},
{#1, Last[#2]} & ### Transpose /# Partition[allpts, 2]
]]]