Related
I can plot the curve corresponding to an implicit equation:
ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1}]
But I cannot find a way to color the contour line depending on the location of the point. More precisely, I want to color the curve in 2 colors, depending on whether x² + y² < k or not.
I looked into ColorFunction but this is only for coloring the region between the contour lines.
And I was not able to get ContourStyle to accept a location-dependent expression.
you could use RegionFunction to split the plot in two:
Show[{
ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1},
RegionFunction -> Function[{x, y, z}, x^2 + y^2 < .5],
ContourStyle -> Red],
ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1},
RegionFunction -> Function[{x, y, z}, x^2 + y^2 >= .5],
ContourStyle -> Green]
}]
Maybe something like this
pl = ContourPlot[x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1}]
points = pl[[1, 1]];
colorf[{x_, y_}] := ColorData["Rainbow"][Rescale[x, {-1, 1}]]
pl /. {Line[a_] :> {Line[a, VertexColors -> colorf /# points[[a]]]}}
which produces
This does not provide a direct solution to your question but I believe it is of interest.
It is possible to color a line progressively from within ContourPlot using what I think is an undocumented format, namely a Function that surrounds the Line object. Internally this is similar to what Heike did, but her solution uses the vertex numbers to then find the matching coordinates allowing styling by spacial position, rather than position along the line.
ContourPlot[
x^2 + (2 y)^2 == 1, {x, -1, 1}, {y, -1, 1},
BaseStyle -> {12, Thickness[0.01]},
ContourStyle ->
(Line[#, VertexColors -> ColorData["DeepSeaColors"] /# Rescale##] & ## # &)
]
For some of the less adept, less information is more. Time was wasted browsing for a way to set the color of contour lines until I chanced onto Roelig's edited answer. I just needed ContourStyle[].
Show[{ContourPlot[
x^2 + 2 x y Tan[2 # ] - y^2 == 1, {x, -3, 3}, {y, -3.2, 3.2},
ContourStyle -> Green] & /# Range[-Pi/4, Pi/4, .1]},
Background -> Black]
When I try to calculate the following integral in Mathematica 8, I get this strange result:
In[1]:= Integrate[y/((1 + x^2 + y^2)^(3/2)), {y, 0, 1}]
Maple 14 can solve this one easily:
Why is Mathematica giving me a different result?
Try this
r = Integrate[y/((1 + x^2 + y^2)^(3/2)), {y, 0, 1}]
r = Assuming[Element[x, Reals], Simplify[r]];
Together[r]
which gives
(-Sqrt[1+x^2]+Sqrt[2+x^2])/(Sqrt[1+x^2] Sqrt[2+x^2])
Which is the same as Maple's :
Following code returns different values for NExpectation and Expectation.
If I try the same for NormalDistribution[] I get convergence erors for NExpectation (but the final result is still 0 for all of them).
What is causing the problem?
U[x_] := If[x >= 0, Sqrt[x], -Sqrt[-x]]
N[Expectation[U[x], x \[Distributed] NormalDistribution[1, 1]]]
NExpectation[U[x], x \[Distributed] NormalDistribution[1, 1]]
Output:
-0.104154
0.796449
I think it might actually be an Integrate bug.
Let's define your
U[x_] := If[x >= 0, Sqrt[x], -Sqrt[-x]]
and the equivalent
V[x_] := Piecewise[{{Sqrt[x], x >= 0}, {-Sqrt[-x], x < 0}}]
which are equivalent over the reals
FullSimplify[U[x] - V[x], x \[Element] Reals] (* Returns 0 *)
For both U and V, the analytic Expectation command uses the Method option "Integrate" this can be seen by running
Table[Expectation[U[x], x \[Distributed] NormalDistribution[1, 1],
Method -> m], {m, {"Integrate", "Moment", "Sum", "Quantile"}}]
Thus, what it's really doing is the integral
Integrate[U[x] PDF[NormalDistribution[1, 1], x], {x, -Infinity, Infinity}]
which returns
(Sqrt[Pi] (BesselI[-(1/4), 1/4] - 3 BesselI[1/4, 1/4] +
BesselI[3/4, 1/4] - BesselI[5/4, 1/4]))/(4 Sqrt[2] E^(1/4))
The integral for V
Integrate[V[x] PDF[NormalDistribution[1, 1], x], {x, -Infinity, Infinity}]
gives the same answer but multiplied by a factor of 1 + I. This is clearly a bug.
The numerical integral using U or V returns the expected value of 0.796449:
NIntegrate[U[x] PDF[NormalDistribution[1, 1], x], {x, -Infinity, Infinity}]
This is presumably the correct solution.
Edit: The reason that kguler's answer returns the same value for all versions is because the u[x_?NumericQ] definition prevents the analytic integrals from being performed so Expectation is unevaluated and reverts to using NExpectation when asked for its numerical value..
Edit 2:
Breaking down the problem a little bit more, you find
In[1]:= N#Integrate[E^(-(1/2) (-1 + x)^2) Sqrt[x] , {x, 0, Infinity}]
NIntegrate[E^(-(1/2) (-1 + x)^2) Sqrt[x] , {x, 0, Infinity}]
Out[1]= 0. - 0.261075 I
Out[2]= 2.25748
In[3]:= N#Integrate[Sqrt[-x] E^(-(1/2) (-1 + x)^2) , {x, -Infinity, 0}]
NIntegrate[Sqrt[-x] E^(-(1/2) (-1 + x)^2) , {x, -Infinity, 0}]
Out[3]= 0.261075
Out[4]= 0.261075
Over both the ranges, the integrand is real, non-oscillatory with an exponential decay. There should not be any need for imaginary/complex results.
Finally note that the above results hold for Mathematica version 8.0.3.
In version 7, the integrals return 1F1 hypergeometric functions and the analytic result matches the numeric result. So this bug (which is also currently present in Wolfram|Alpha) is a regression.
If you change the argument of your function u to avoid evaluation for non-numeric values all three methods gives the same result:
u[x_?NumericQ] := If[x >= 0, Sqrt[x], -Sqrt[-x]] ;
Expectation[u[x], x \[Distributed] NormalDistribution[1, 1]] // N;
N[Expectation[u[x], x \[Distributed] NormalDistribution[1, 1]]] ;
NExpectation[u[x], x \[Distributed] NormalDistribution[1, 1]];
{% === %% === %%%, %}
with the result
{True, 0.796449}
I am attempting to solve the linear biharmonic equation in mathematica using DSolve. I think this issue is not just limited to the biharmonic equation but MATHEMATICA just spits out the equation when I attempt to solve it.
I've tried solving other partial differential equations and there was no trouble.
The biharmonic equation is just:
Laplacian^2[f]=0
Here is my equation:
DSolve[
D[f[x, y], {x, 4}] + 2 D[D[f[x, y], {x, 2}, {y, 2}]] +
D[f[x, y], {y, 4}] == 0,
f,
{x, y}]
The solution is spit out as
DSolve[(f^(0,4))[x,y]+2 (f^(2,2))[x,y]+(f^(4,0))[x,y]==0,f,{x,y}]
That is obviously not the solution. What gives? What am I missing? I've solved other PDEs without boundary conditions.
How about try it in polar coordinates? If f(r, \[Theta]) is symmetric with respect to azimuth \[Theta], the biharmonic equation reduces to something Mathematca can solve symbolically (c.f. http://mathworld.wolfram.com/BiharmonicEquation.html):
In[22]:= eq = D[r D[D[r D[f[r],r],r]/r,r],r]/r;
eq//FullSimplify//TraditionalForm
Out[23]//TraditionalForm= f^(4)(r) + (2 r^2 f^(3)(r) - r f''(r)
+ f'(r))/r^3
In[24]:= DSolve[eq==0,f,r]
Out[24]= {{f -> Function[{r},
1/2 r^2 C[2] - 1/4 r^2 C[3] + C[4] + C[1] Log[r]
+ 1/2 r^2 C[3] Log[r]
]}}
In[25]:= ReplaceAll[
1/2 r^2 C[2]-1/4 r^2 C[3]+C[4]+C[1] Log[r]+1/2 r^2 C[3] Log[r],
r->Sqrt[x^2+y^2]
]
Out[25]= 1/2 (x^2+y^2) C[2]-1/4 (x^2+y^2) C[3]+C[4]+C[1] Log[Sqrt[x^2+y^2]]+
1/2 (x^2+y^2) C[3] Log[Sqrt[x^2+y^2]]
DSolve[D[f[x, y], {x, 4}] + 2 D[f[x, y], {x, 2}, {y, 2}] +
D[f[x, y], {y, 4}] == 0, f, {x, y}]
This ought to be the actual syntax
What's the best way of getting Mathematica 7 or 8 to do the integral
NIntegrate[Exp[-x]/Sin[Pi x], {x, 0, 50}]
There are poles at every integer - and we want the Cauchy principle value.
The idea is to get a good approximation for the integral from 0 to infinity.
With Integrate there is the option PrincipleValue -> True.
With NIntegrate I can give it the option Exclusions -> (Sin[Pi x] == 0), or manually give it the poles by
NIntegrate[Exp[-x]/Sin[Pi x], Evaluate[{x, 0, Sequence##Range[50], 50}]]
The original command and the above two NIntegrate tricks give the result 60980 +/- 10. But they all spit out errors. What is the best way of getting a quick reliable result for this integral without Mathematica wanting to give errors?
Simon, is there reason to believe your integral is convergent ?
In[52]:= f[k_Integer, eps_Real] :=
NIntegrate[Exp[-x]/Sin[Pi x], {x, k + eps, k + 1 - eps}]
In[53]:= Sum[f[k, 1.0*10^-4], {k, 0, 50}]
Out[53]= 2.72613
In[54]:= Sum[f[k, 1.0*10^-5], {k, 0, 50}]
Out[54]= 3.45906
In[55]:= Sum[f[k, 1.0*10^-6], {k, 0, 50}]
Out[55]= 4.19199
It looks like the problem is at x==0. Splitting integrand k+eps to k+1-eps for integer values of k:
In[65]:= int =
Sum[(-1)^k Exp[-k ], {k, 0, Infinity}] Integrate[
Exp[-x]/Sin[Pi x], {x, eps, 1 - eps}, Assumptions -> 0 < eps < 1/2]
Out[65]= (1/((1 +
E) (I + \[Pi])))E (2 E^(-1 + eps - I eps \[Pi])
Hypergeometric2F1[1, (I + \[Pi])/(2 \[Pi]), 3/2 + I/(2 \[Pi]),
E^(-2 I eps \[Pi])] +
2 E^(I eps (I + \[Pi]))
Hypergeometric2F1[1, (I + \[Pi])/(2 \[Pi]), 3/2 + I/(2 \[Pi]),
E^(2 I eps \[Pi])])
In[73]:= N[int /. eps -> 10^-6, 20]
Out[73]= 4.1919897038160855098 + 0.*10^-20 I
In[74]:= N[int /. eps -> 10^-4, 20]
Out[74]= 2.7261330651934049862 + 0.*10^-20 I
In[75]:= N[int /. eps -> 10^-5, 20]
Out[75]= 3.4590554287709991277 + 0.*10^-20 I
As you see there is a logarithmic singularity.
In[79]:= ser =
Assuming[0 < eps < 1/32, FullSimplify[Series[int, {eps, 0, 1}]]]
Out[79]= SeriesData[eps, 0, {(I*(-1 + E)*Pi -
2*(1 + E)*HarmonicNumber[-(-I + Pi)/(2*Pi)] +
Log[1/(4*eps^2*Pi^2)] - 2*E*Log[2*eps*Pi])/(2*(1 + E)*Pi),
(-1 + E)/((1 + E)*Pi)}, 0, 2, 1]
In[80]:= Normal[
ser] /. {{eps -> 1.*^-6}, {eps -> 0.00001}, {eps -> 0.0001}}
Out[80]= {4.191989703816426 - 7.603403526913691*^-17*I,
3.459055428805136 -
7.603403526913691*^-17*I,
2.726133068607085 - 7.603403526913691*^-17*I}
EDIT
Out[79] of the code above gives the series expansion for eps->0, and if these two logarithmic terms get combined, we get
In[7]:= ser = SeriesData[eps, 0,
{(I*(-1 + E)*Pi - 2*(1 + E)*HarmonicNumber[-(-I + Pi)/(2*Pi)] +
Log[1/(4*eps^2*Pi^2)] - 2*E*Log[2*eps*Pi])/(2*(1 + E)*
Pi),
(-1 + E)/((1 + E)*Pi)}, 0, 2, 1];
In[8]:= Collect[Normal[PowerExpand //# (ser + O[eps])],
Log[eps], FullSimplify]
Out[8]= -(Log[eps]/\[Pi]) + (
I (-1 + E) \[Pi] -
2 (1 + E) (HarmonicNumber[-((-I + \[Pi])/(2 \[Pi]))] +
Log[2 \[Pi]]))/(2 (1 + E) \[Pi])
Clearly the -Log[eps]/Pi came from the pole at x==0. So if one subtracts this, just like principle value method does this for other poles you end up with a finitely value:
In[9]:= % /. Log[eps] -> 0
Out[9]= (I (-1 + E) \[Pi] -
2 (1 + E) (HarmonicNumber[-((-I + \[Pi])/(2 \[Pi]))] +
Log[2 \[Pi]]))/(2 (1 + E) \[Pi])
In[10]:= N[%, 20]
Out[10]= -0.20562403655659928968 + 0.*10^-21 I
Of course, this result is difficult to verify numerically, but you might know more that I do about your problem.
EDIT 2
This edit is to justify In[65] input that computes the original regularized integral. We are computing
Sum[ Integrate[ Exp[-x]/Sin[Pi*x], {x, k+eps, k+1-eps}], {k, 0, Infinity}] ==
Sum[ Integrate[ Exp[-x-k]/Sin[Pi*(k+x)], {x, eps, 1-eps}], {k, 0, Infinity}] ==
Sum[ (-1)^k*Exp[-k]*Integrate[ Exp[-x]/Sin[Pi*x], {x, eps, 1-eps}],
{k, 0, Infinity}] ==
Sum[ (-1)^k*Exp[-k], {k, 0, Infinity}] *
Integrate[ Exp[-x]/Sin[Pi*x], {x, eps, 1-eps}]
In the third line Sin[Pi*(k+x)] == (-1)^k*Sin[Pi*x] for integer k was used.
Simon, I haven't spent much time with your integral, but you should try looking at stationary phase approximation. What you have is a smooth function (exp), and a highly oscillatory function (sine). The work involved is now in brow-beating the 1/sin(x) into the form exp(if(x))
Alternatively, you could use the series expansion of the cosecant (not valid at poles):
In[1]:=Series[Csc[x], {x, 0, 5}]
(formatted) Out[1]=1/x + x/6 + 7/360 x^3 + 31/15120 x^5 +O[x]^6
Note that for all m>-1, you have the following:
In[2]:=Integrate[x^m Exp[-x], {x, 0, Infinity}, Assumptions -> m > -1]
Out[2]=Gamma[1+m]
However, summing the series with the coefficients of cosecant (from wikipedia), not including 1/x Exp[-x] case, which doesn't converge on [0,Infinity].
c[m_] := (-1)^(m + 1) 2 (2^(2 m - 1) - 1) BernoulliB[2 m]/Factorial[2 m];
Sum[c[m] Gamma[1 + 2 m - 1], {m, 1, Infinity}]
does not converge either...
So, I'm not sure that you can work out an approximation for the integral to infinity, but I if you're satisfied with a solution upto some large N, I hope these help.
I have to agree with Sasha, the integral does not appear to be convergent. However, if you exclude x == 0 and break the integral into pieces
Integrate[Exp[-x]/Sin[Pi x], {x, n + 1/2, n + 3/2}, PrincipalValue -> True]
where n >= 0 && Element[n, Integers], then it seems you may get an alternating series
I Sum[ (-1/E)^n, {n, 1, Infinity}] == - I / (1 + E )
Now, I only took it out to n == 4, but it looks reasonable. However, for the integral above with Assumptions -> Element[n, Integers] && n >= 0 Mathematica gives
If[ 2 n >= 1, - I / E, Integrate[ ... ] ]
which just doesn't conform to the individual cases. As an additional note, if the pole lies at the boundary of the integration region, i.e. your limits are {x, n, n + 1}, you only get DirectedInfinitys. A quick look at the plot implies that you with the limits {x, n, n + 1} you only have a strictly positive or negative integrand, so the infinite value may be due to the lack of compensation which {x, n + 1/2, n + 3/2} gives you. Checking with {x, n, n + 2}, however it only spits out the unevaluated integral.