I had a math problem I solved like this:
In[1]:= Solve[2x(a-x)==0, x]
Out[1]= {{x->0}, {x->a}}
In[2]:= Integrate[2x(a-x), {x,0,a}]
Out[2]= (a^3)/3
In[3]:= Solve[(a^3)/3==a, a]
Out[3]= {{a->0}, {a->-Sqrt[3]}, {a->Sqrt[3]}}
My question is if I could rewrite this to compute it in one step, rather than having to manually input the result from the previous line. I could easily replace the integral used in step three with the Integrate command from step two. But what I can't figure out is how I would use the result from step 1 as the limits of integration in the integral.
You could combine step 1 and 2 by doing something like
Integrate[2 x (a - x), {x, ##}] & ## (x /. Solve[2 x (a - x) == 0, x]);
If you agree to delegate the choice of the (positive oriented) domain to Integrate, by means of using Clip or Boole:
In[77]:= Solve[
Integrate[
Clip[2 x (a - x), {0, Infinity}], {x, -Infinity, Infinity}] == a, a]
Out[77]= {{a -> 0}, {a -> Sqrt[3]}}
or
In[81]:= Solve[
Integrate[
2 x (a - x) Boole[2 x (a - x) > 0], {x, -Infinity, Infinity}] ==
a, a]
Out[81]= {{a -> 0}, {a -> Sqrt[3]}}
The reason only non-negative roots are found, is that Integrate will integrate from the smallest root to the largest root, i.e. from {x,0,a} for positive a and {x,a,0} for negative a.
I have a function f[x_,y_,z_]:=Limit[g[x+eps,y,z],eps->0]; and I plot f[x,y,z] in the next step. Earlier, I used to evaluate the limit and copy the expression in the definition of f. I tried to make it all in one step. However, the evaluation of the Limit is done only when I try to plot f. As a result, every time I change around the variables and replot, the limit is evaluated all over again (it takes about a min to evaluate, so it becomes annoying). I tried evaluating the limit first, and then doing f[x_,y_,z_]:=%. But that doesn't work either. How do I get the function to evaluate the limit upon declaration?
The function you need is logically called Evaluate and you can use it within the Plot command.
Here is a contrived example:
f[x_, y_, z_] := Limit[Multinomial[x, y, z], x -> 0]
Plot3D[ Evaluate[ f[x, y, z] ], {y, 1, 5}, {z, 1, 5}]
Addressing your follow-up question, perhaps all you seek is something like
ff = f[x, y, z]
Plot3D[ff, {y, 1, 5}, {z, 1, 5}]
or possibly merely
ClearAll[f, x, y, z]
f[x_, y_, z_] = Limit[Multinomial[x, y, z], x -> 0]
Plot3D[f[x, y, z], {y, 1, 5}, {z, 1, 5}]
It would be helpful if you would post a more complete version of your code.
An alternative to Mr Wizard's solution is that you can also put the Evaluate in the function's definition:
f[x_, y_, z_] := Evaluate[Limit[Multinomial[x, y, z], x->0]]
Plot3D[f[x, y, z], {y, 1, 5}, {z, 1, 5}]
You can compare the two versions with the one without an Evaluate by Timing the Plot.
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]
]]]
I'm currently doing some normalization along the lines of:
J = Integrate[Psi[x, 0]^2, {x, 0, a}]
sol = Solve[J == 1, A]
A /. sol
For this type of normalization, the negative square root is extraneous. The result of this calculation is:
In[49]:= J = Integrate[Psi[x, 0]^2, {x, 0, a}]
Out[49]= 2 A^2
In[68]:= sol = Solve[J == 1, A]
Out[68]= {{A -> -(1/Sqrt[2])}, {A -> 1/Sqrt[2]}}
Even if I try giving it an Assuming[...] or Simplify[...], it still gives me the same results:
In[69]:= sol = Assuming[A > 0, Solve[J == 1, A]]
Out[69]= {{A -> -(1/Sqrt[2])}, {A -> 1/Sqrt[2]}}
In[70]:= sol = FullSimplify[Solve[J == 1, A], A > 0]
Out[70]= {{A -> -(1/Sqrt[2])}, {A -> 1/Sqrt[2]}}
Can anyone tell me what I'm doing wrong here?
I'm running Mathematica 7 on Windows 7 64-bit.
ToRules does what the box says: converts equations (as in Reduce output) to rules. In your case:
In[1]:= ToRules[Reduce[{x^2==1,x>0},x]]
Out[1]= {x->1}
In[2]:= {ToRules[Reduce[{x^2==1},x]]}
Out[2]= {{x->-1},{x->1}}
For more complex cases, I have often found it useful to just check the value of the symbolic solutions after pluging in typical parameter values. This is not foolproof, of course, but if you know there is one and only one solution then it is a simple and efficient method:
Solve[x^2==someparameter,x]
Select[%,((x/.#)/.{someparameter-> 0.1})>0&]
Out[3]= {{x->-Sqrt[someparameter]},{x->Sqrt[someparameter]}}
Out[4]= {{x->Sqrt[someparameter]}}
Solve doesn't work like this. You might try Reduce, instead, e.g.
In[1]:= Reduce[{x^2 == 1, x > 0}, x]
Out[1]= x == 1
It's then a little tricky to transform this output to replacement rules, at least in the general case, because Reduce might use arbitrary many logical connectives. In this case, we could just hack:
In[2]:= Solve[Reduce[{x^2 == 1, x > 0}, x], x]
Out[2]= {{x->1}}