What are the alternatives for drawing a simple curve for a function like
0, from -∞ to 0
x^2, from 0 to (3)^(1/3)
1, from (3)^(1/3) to ∞
look that my attempt is incomplet
curve(x^2, from=0, to=(3)^(1/3), xlab="x", ylab="y")
I hope I didn't write a duplicate question.
You need to specify a function that fits your problem:
> f <- function(x) ifelse(x < 0, 0, ifelse(x<3^(1/3), x^2, 1))
> curve(f, from = -5, to = 5)
Related
The following was my question given by my teacher,
Generate a sequence of N = 1000 independent observations of random variable with distribution: (c) Exponential with parameter λ = 1 , by
inversion method.
Present graphically obtained sequences(except for those generated in point e) i.e. e.g. (a) i. plot in the coordinates (no. obs.,
value of the obs) ii. plot in the coordinates (obs no n, obs. no n +
i) for i = 1, 2, 3. iii. plot so called covariance function for some
values. i.e. and averages:
I have written the following code,
(*****************************************************************)
(*Task 01(c) and 02(a)*)
(*****************************************************************)
n = 1000;
taskC = Table[-Log[RandomReal[]], {n}];
ListPlot[taskC, AxesLabel->{"No. obs", "value of the obs"}]
i = 1;
ListPlot[Table[
{taskC[[k]], taskC[[k+i]]},
{k, 1, n-i,1}],
AxesLabel->{"obs.no.n", "obs.no.n+1"}]
i++;
ListPlot[Table[
{taskC[[k]], taskC[[k+i]]},
{k, 1, n-i,1}],
AxesLabel-> {"obs.no.n", "obs.no.n+2"}]
i++;
ListPlot[Table[
{taskC[[k]], taskC[[k+i]]},
{k,1,n-i,1}],
AxesLabel->{"obs.no.n", "obs.no.n+3"}]
avg = (1/n)*Sum[taskC[[i]], {i,n}];
ListPlot[Table[1/(n-tau) * Sum[(taskC[[i]]-avg)*(taskC[[i+tau]] - avg), n], {tau, 1,100}],
Joined->True,
AxesLabel->"Covariance Function"]
He has commented,
The plots of co-variance functions should start from 0-shift. Note
that for larger than 0 shifts you are estimating co-variance between
independent observations which is zero, while for 0 shift you are
estimating variance of observation which is large. Thus the contrast
between these two cases is a clear indication that the observations
are uncorrelated.
What did I do wrong?
How can I correct my code?
Zero-shift means calculating the covariance for tau = 0, which is simply the variance.
Labeled[ListPlot[Table[{tau,
1/(n - tau)*Sum[(taskC[[i]] - avg)*(taskC[[i + tau]] - avg), {i, n - tau}]},
{tau, 0, 5}], Filling -> Axis, FillingStyle -> Thick, PlotRange -> All,
Frame -> True, PlotRangePadding -> 0.2, AspectRatio -> 1],
{"Covariance Function K(n)", "n"}, {{Top, Left}, Bottom}]
Variance[taskC]
0.93484
Covariance[taskC, taskC]
0.93484
(* n = 1 *)
Covariance[Most[taskC], Rest[taskC]]
0.00926913
The following code gives me the first k eigenvalues of a certain big matrix. Because of the symmetries of the matrix, the eigenvalues are in pairs, one positive and the other negative, with the same absolute value. This is indeed the case if I run the code with the exact matrices, without using the sparse version. However when I make them sparse, the resulting eigenvalues appear to lose the sign information, as now the pairs can be both negative, or both positive, depending on the number I put on "nspins" (which controls the size of the matrix). The variable "sparse" controls whether I use sparse matrices or not.
This issue gives me considerable trouble. Can anybody tell me why the sparse version of the computation gives wrong signs, and how to fix it?
sparse = 1; (*Parameter that controls whether I will use sparse \
matrices, 0 means not sparse, 1 means sparse*)
(*Base matrices of my big matrix*)
ox = N[{{0, 1}, {1, 0}}];
oz = N[{{1, 0}, {0, -1}}];
id = N[{{1, 0}, {0, 1}}];
(*Transformation into sparse whether desired*)
If[sparse == 1,
ox = SparseArray[ox];
oz = SparseArray[oz];
id = SparseArray[id];
]
(*Dimension of the big matrix, must be even*)
nspins = 8;
(*Number of eigenvalues computed*)
neigenv = 4;
(*Algorithm to create big matrices*)
Do[
Do[
If[j == i, mata = ox; matc = oz;, mata = id; matc = id;];
If[j == 1,
o[1, i] = mata;
o[3, i] = matc;
,
o[1, i] = KroneckerProduct[o[1, i], mata];
o[3, i] = KroneckerProduct[o[3, i], matc];
];
, {j, 1, nspins}];
, {i, 1, nspins}];
(*Sum of big matrices*)
ham = Sum[o[1, i].o[1, i + 1], {i, 1, nspins - 1}] +
o[1, nspins].o[1, 1] + 0.5*Sum[o[3, i], {i, 1, nspins}];
(*Print the desired eigenvalues*)
Do[Print [Eigenvalues[ham, k][[k]]], {k, 1, neigenv}];
I am trying to solve the captioned problem numerically using Mathematica, to no avail. Imagine a rod of length L. The speed of sound in the rod is c. A pressure impulse of gaussian shape whose width is comparable to L/c is applied at one end. I would like to solve for the particle displacement function u(t,x) inside the rod. The Mathematica codes are given are follows,
c = 1.0 (*speed of wave*)
L = 1.0 (*length of medium*)
Subscript[P, 0] = 0.0 (*pressure of reservoir at one end*)
Subscript[t, 0] = 5.0*c/L; (*mean time of pressure impulse*)
\[Delta]t = 2.0*c/L; (*Std of pressure impulse*)
K = 1.0; (* proportionality constant, stress-strain *)
Subscript[P, max ] = 1.0; (*max. magnitude of pressure impulse*)
Subscript[P, 1][t_] :=
Subscript[P, max ]
PDF[NormalDistribution[Subscript[t, 0], \[Delta]t], t];
PDE = D[func[t, x], t, t] == c^2 D[func[t, x], x, x]
BC1 = -K func[t, 0] == Subscript[P, 1][t]
BC2 = -K func[t, L] == Subscript[P, 0]
IC1 = func[0,
x] == (-Subscript[P, 1][0]/K) (x/L) + (-Subscript[P, 0]/K) (1 - x/L)
IC2 = Derivative[1, 0][func][0, x] == 0.0
sol = NDSolve[{PDE, BC1, BC2, IC1, IC2},
func, {t, 0, 2 Subscript[t, 0]}, {x, 0, L}]
The problem is that the program keeps running for minutes without giving any output. Given the simplicity of the problem (i.e. that an analytical solution exists), I think there should be a quicker way to arrive at a numerical solution. Would someone please give me some suggestions?
Following George's advice, the equation was solved.
BC1 and BC2 given in the question should be modified as follows
BC1 = -kk Derivative[0, 1][func][t, 0] == p1[t]
BC2 = -kk Derivative[0, 1][func][t, ll] == p0
Also t0 and [Delta]t has been modified,
t0 = 2.0*c/ll (*mean time of pressure impulse*)
\[Delta]t = 0.5*c/ll (*Std of pressure impulse*)
The problem can be solved to within the accuracy requirement for the time interval 0 < t < 2 t0. I solved the problem for a longer time interval 0 < t < 4 t0 in order to look for something interesting.
Here is a plot of 3D plot of pressure (versus x and t)
Here is a plot of the pressure at one end of the bar where impulse is applied. The pressure is a gaussian, as expected.
Here is a plot of the pressure in the middle of the bar. Note that although the applied pressure is a gaussian, and the pressure at the other end is held at P0=0, the pressure becomes negative for some time tc.
I'm working on a script in mathematica that will take simulate a string held at either end and plucked, by solving the wave equation via numerical methods. (http://en.wikipedia.org/wiki/Wave_equation#Investigation_by_numerical_methods)
n = 5; (*The number of discreet elements to be used*)
L = 1.0; (*The length of the string that is vibrating*)
a = 1.0/3.0; (*The distance from the left side that the string is \
plucked at*)
T = 1; (*The tension in the string*)
[Rho] = 1; (*The length density of the string*)
y0 = 0.1; (*The vertical distance of the string pluck*)
[CapitalDelta]x = L/n; (*The length of each discreet element*)
m = ([Rho]*L)/n;(*The mass of each individual node*)
c = Sqrt[T/[Rho]];(*The speed at which waves in the string propogate*)
I set all my variables
Y[t] = Array[f[t], {n - 1, 1}];
MatrixForm(*Creates a vector size n-1 by 1 of functions \
representing each node*)
I define my Vector of nodal position functions
K = MatrixForm[
SparseArray[{Band[{1, 1}] -> -2, Band[{2, 1}] -> 1,
Band[{1, 2}] -> 1}, {n - 1,
n - 1}]](*Creates a matrix size n by n governing the coupling \
between each node*)
I create the stiffness matrix relating all the nodal functions to one another
Y0 = MatrixForm[
Table[Piecewise[{{(((i*L)/n)*y0)/a,
0 < ((i*L)/n) < a}, {(-((i*L)/n)*y0)/(L - a) + (y0*L)/(L - a),
a < ((i*L)/n) < L}}], {i, 1, n - 1}]]
I define the initial positions of each node using a piecewise function
NDSolve[{Y''[t] == (c/[CapitalDelta]x)^2 Y[t].K, Y[0] == Y0,
Y'[0] == 0},
Y, {t, 0, 10}];(*Numerically solves the system of second order DE's*)
Finally, This should solve for the values of the individual nodes, but it returns an error:
"NDSolve::ndinnt : Initial condition [Y0 table] is not a number or a rectangular array"
So , it would seem that I don't have a firm grasp on how matrices work in mathematica. I would greatly appreciate it if anyone could help me get this last line of code to run properly.
Thank you,
Brad
I don't think you should use MatrixForm when defining the matrices. MatrixForm is used to format a list of list as a matrix, usually when you display it. Try removing it and see if it works.
I'm trying to use Mathematica's NDSolve[] to compute a geodesic along a sphere using the coupled ODE:
x" - (x" . x) x = 0
The problem is that I can only enter initial conditions for x(0) and x'(0) and the solver is happy with the solution where x" = 0. The problem is that my geodesic on the sphere has the initial condition that x"(0) = -x(0), which I have no idea how to tell mathematica. If I add this as a condition, it says I'm adding True to the list of conditions.
Here is my code:
s1 = NDSolve[{x1''[t] - (x1[t] * x1''[t] + x2[t] * x2''[t] + x3[t]*x3''[t]) * x1[t] == 0, x2''[t] - (x1[t] * x1''[t] + x2[t] * x2''[t] + x3[t]*x3''[t]) * x2[t] == 0, x3''[t] - (x1[t] * x1''[t] + x2[t] * x2''[t] + x3[t]*x3''[t]) * x3[t] == 0, x1[0] == 1, x2[0] == 0, x3[0] == 0, x1'[0] == 0, x2'[0] == 0, x3'[0] == 1} , { x1, x2, x3}, {t, -1, 1}][[1]]
I would like to modify this so that the initial acceleration is not zero but -x(0).
Thanks
Well, as the error message says -- NDSolve only accepts initial conditions for derivatives of orders strictly less than the maximal order appearing in the ODE.
I have a feeling this is more of a mathematics question. Mathematically, {x''[0]=-x0, x[0]==x0}, doesn't define a unique solution - you'd have to do something along the lines of {x0.x''[0]==-1, x[0]==x0, x'[0]-x0 x0.x'[0]==v0} for that to work out (NDSolve would still fail with the same error). You do realize you will just get a great circle on the unit sphere, right?
By the way, here is how I would have coded up your example:
x[t_] = Table[Subscript[x, j][t], {j, 3}];
s1 = NDSolve[Flatten[Thread /# #] &#{
x''[t] - (x''[t].x[t]) x[t] == {0, 0, 0},
x[0] == {1, 0, 0},
x'[0] == {0, 0, 1}
}, x[t], {t, -1, 1}]
I fixed this problem through a mathematical rearrangement rather than addressing my original issue:
Let V(t) be a vector field along x(t).
x . V = 0 implies d/dt (x . V) = (x' . V) + (x . V') = 0
So the equation D/dt V = V' - (x . V') x = V' + (x' . V) x holds
This means the geodesic equation becomes: x" + (x' . x') x = 0 and so it can be solved using the initial conditions I originally had.
Thanks a lot Janus for going through and pointing out the various problems I was having including horrible code layout, I learnt a lot through your re-writing as well.