I am using this octave code for solving differential equation.
# Define the right-hand side of the equation:
xvall= -11 ;#xvall
xvalu= 10 ;#xvalu
range=5000;
function ret=f(x,t);ret= t ;end;
# ywill be the values of the function at these moments of time.
t=linspace(xvall,xvalu,range);
y=lsode ('f', 2, linspace(xvall,xvalu,range));
y
plot(t,y);
i got the graph like this .
But when the same conditions are passed to wolfram alpha
I am getting the graph from 60 to 0 for y value
graph is
why is the graph behaving differently in two situations.
https://www.wolframalpha.com/input/?i=Runge-Kutta+method%2C+dy%2Fdx+%3D+x%2C+y%280%29+%3D+2%2C+from+-11+to+10%2C+h+%3D+0.25
To specify an initial value problem for an ordinary differential equation you need to define the initial condition. Here for Octave you have specified x(-11) = 2 since xvall = -11 and for Wolfram Alpha you have specified y(0) = 2. That is why you have two different solutions.
Octave
Octave's lsode (f,x_0,ts) solves the following initial value problem
dx/dt = t
x(t_0) = x_0
t in ts
Here ts is specified as a set of points in the interval [t_0,t_1]. You have specified t_0 = -11, t_1 = 10.
In closed form the solution to this problem is x = (t^2 - 117) / 2
Wolfram
For Wolfram you have used the semi-formal syntax:
Runge-Kutta method, dy/dx = x, y(0) = 2, from -11 to 10, h = 0.25
In closed form the solution to this problem would be y = (x^2 + 4) /2
The corresponding initial value problem is clearly different. Hence different results.
Related
The analysis with wavelets seems to be carried out as a discrete transform via matrix multiplication. So it is not surprising, I guess, that when plotting, for example, D4, the R package wmtsa returns the plot:
require(wmtsa)
filters <- wavDaubechies("d4")
plot(filters)
The question is how to go from this discretized plot to the plot in the Wikipedia entry:
Please note that I'm not interested in generating these curves precisely with wmtsa. Any other package will do - I don't have Matlab or Mathematica. But I wonder if the way to go is to start with translating this Mathematica chunk of code in this paper into R, rather than using built-in functions:
Wave1etTransform.m
c[k-1 := c[k] = Daubechies[4][[k+l]];
phi[l] = (l+Sqrt[3])/2 // N;
phi[2] = (l-Sqrt[3])/2 // N;
phi[xJ; xc=0 II x>=3] : = 0
phi[x-?NumberQ] := phi[x] =
N[Sqrt[2]] Sum[c[k] phi[2x-k],{k,0,3}];
In order to plot the wavelet and scaling function all you need are the four numbers shown in the first two plots. I'll focus on plotting the scaling function.
Integer shifts of the scaling function, 𝜑, form an orthonormal basis of the subspace V0 of the multiresolution analysis. We also have that V-1 ⊆ V0 and that 𝜑(x/2) ∈ V-1. Using this gives us the identity
𝜑(x/2) = ∑k ∈ ℤ hk𝜑(x-k)
Now we just need the values of hk. For the Daubechies wavelet these are the values show in the discrete plot you gave (and zero for every other value of k). For an exact value of the hk, first let 𝜇 = (1+sqrt(3))/2. Then we have that
h0 = 𝜇/4
h1 = (1+𝜇)/4
h2 = (2-𝜇)/4
h3 = (1-𝜇)/4
and hk = 0 otherwise.
Using these two things we are able to plot the function using what is known as the cascade algorithm. First notice that 𝜑(0) = 𝜑(0/2) = h0𝜑(0) + h1𝜑(0-1) + h2𝜑(0-2) + h3𝜑(0-3). The only way this equation can hold is if 𝜑(0) = 𝜑(-1) = 𝜑(-2) = 𝜑(-3) = 0. Extending this will show that for x ≦ 0 we have that 𝜑(x) = 0. Furthermore, a similar argument can show that 𝜑(x) = 0 for x ≥ 3.
Thus, we only need to worry about x = 1 and x = 2 to find non-zero values of 𝜑 for integer values of x. If we put x = 2 into the identity for 𝜑(x/2) we get that 𝜑(1) = h0𝜑(2) + h1𝜑(1). Putting x = 4 into the identity gives us that 𝜑(2) = h2𝜑(2) + h3𝜑(1).
We can rewrite the above two equations as a matrix multiplied by a vector equals a vector. In fact, it will be in the form v = Av (v is the same vector on both sides). This means that v is an eigenvector of the matrix A with eigenvalue 1. But v = (𝜑(1), 𝜑(2)) and so by finding this eigenvector using the standard methods we will be able to find the values of 𝜑(1) and 𝜑(2).
In fact, this gives us that 𝜑(1) = (1+sqrt(3))/2 and 𝜑(2) = (1-sqrt(3))/2 (this is where those values in the Mathematica code sample come from). Also note that we need to specifically chose the eigenvector of magnitude 2 for this algorithm to work so you must use those values for 𝜑(1) and 𝜑(2) even though you could rescale the eigenvector.
Now we can find the values of 𝜑(1/2), 𝜑(3/2), and 𝜑(5/2). For example, 𝜑(1/2) = h0𝜑(1) and 𝜑(3/2) = h1𝜑(2) + h2𝜑(1).
With these values, you can then find the values of 𝜑(1/4), 𝜑(3/4), and so on. Continuing this process will give you the value of 𝜑 for all dyadic rationals (rational numbers in the form k/2j.
The same process can be used to find the wavelet function. You only need to use the four different values shown in the first plot rather than the four shown in the second plot.
I recently implemented this Python. An R implementation will be fairly similar.
import numpy as np
import matplotlib.pyplot as plt
def cascade_algorithm(j: int):
mu = (1 + np.sqrt(3))/2
h_k = np.array([mu/4, (1+mu)/4, (2-mu)/4, (1-mu)/4])
# Array to store all the value of phi.
phi_vals = np.zeros((2, 3*2**j+1), dtype=np.float64)
for i in range(3*2**j+1):
phi_vals[0][i] = i/(2**j)
calced_vals = np.zeros((3*2**j+1), dtype=np.bool)
# Input values for 1 and 2.
phi_vals[1][1*2**j] = (1+np.sqrt(3))/2
phi_vals[1][2*2**j] = (1-np.sqrt(3))/2
# We now know the values for 0, 1, 2, and 3.
calced_vals[0] = True
calced_vals[1*2**j] = True
calced_vals[2*2**j] = True
calced_vals[3*2**j] = True
# Now calculate for all the dyadic rationals.
for k in range(1, j+1):
for l in range(1, 3*2**k):
x = l/(2**k)
if calced_vals[int(x*2**j)] != True:
calced_vals[int(x*2**j)] = True
two_x = 2*x
which_k = np.array([0, 1, 2, 3], dtype=np.int)
which_k = ((two_x - which_k > 0) & (two_x - which_k < 3))
phi = 0
for n, _ in enumerate(which_k):
if which_k[n] == True:
phi += h_k[n]*phi_vals[1][int((two_x-n)*2**j)]
phi_vals[1][int(x*2**j)] = 2*phi
return phi_vals
phi_vals = cascade_algorithm(10)
plt.plot(phi_vals[0], phi_vals[1])
plt.show()
If you just want to plot the graphs, then you can use the package "wavethresh" to plot for example the D4 with the following commands:
draw.default(filter.number=4, family="DaubExPhase", enhance=FALSE, main="D4 Mother", scaling.function = F) # mother wavelet
draw.default(filter.number=4, family="DaubExPhase", enhance=FALSE, main="D4 Father", scaling.function = T) # father wavelet
Notice that the mother wavelet and the father wavelets will be plotted depending on the variable "scaling.function". If true, then it plots the father wavelet (scaling), else it plots the mother wavelet.
If you want to generate it by yourself, without packages, I'd suggest you follow Daubechies-Lagarias algorithm, in this paper. It is not hard to implement.
I have two points which form one line: (1,4) and (3,6), and another two which form another line: (2,1) and (4,2). These lines are continuous and I can find their intersection points by finding the equation for each line, and then equating them to find the x value at the intersection point, and then the y value.
i.e. for the first line, the equation is y = x + 3, and the second is y = 0.5x. At the intersection the y values are the same so x + 3 = 0.5x. So x = -6. Subbing this back into either of the equations gives a y value of -3.
From those steps, I now know that the intersection point is (-6,-3). The problem is I need to do the same steps in Excel, preferably as one formula. Can anyone give me some advice on how I would start this?
Its long but here it is:
Define x1,y1 and x2,y2 for the 1st line and x3,y3 and x4,y4 for the second.
x = (x2y1-x1y2)(x4-x3)-(x4y3-x3y4)(x2-x1) / [ (x2-x1)(y4-y3) - (x4-x3)(y2-y1) ]
y = (x2y1-x1y2)(y4-y3)-(x4y3-x3y4)(y2-y1) / [ (x2-x1)(y4-y3) - (x4-x3)(y2-y1) ]
Note that the denominators are the same. They will be ZERO! when the system has no solution. So you may want to check that in another cell and conditionally compute the answer.
Essentially, this formula is derived by solving a system of equations for x and y by hand using generic points (x1,y1), (x2,y2), (x3,y3), and (x4,y4). Easier yet, is solving the system by hand using well developed linear algebra concepts.
Wikipedia outlines this procedure well: Line-line intersection.
Also, this website describes all the different formulas and lets you put in whatever data you have in any mixed format and provides many details of the solutions: Everything about 2 lines.
Here's a matrix based solution:
x - y = -3
0.5*x - y = 0
Written as a matrix equation (I apologize for the poor typesetting):
| 1.0 -1.0 |{ x } { -3 }
| 0.5 -1.0 |{ y } = { 0 }
You can invert this matrix or use LU decomposition to solve it to get the answer. That method will work for any number of cases where you have one equation for each unknown.
This is easy to do by hand:
Subtract the second equation from the first: 0.5*x = -3
Divide both sides by 0.5: x = -6
Substitute this result into the other equation: y = 0.5*x = -3
Hey so I'm reading this article by Chris Hecker where he has an image of a Parabola surrounded by the a vector field of it's derivative:
However he never mentions how exactly he got the vector field equation, and never even states it. He does say he overlayed the vector field of the slopes in Figure 1, by drawing the solution to the slope equation, dy/dx = 2x, as a short vector at each coordinate on the grid.
How do you create a vector field of the slopes of an equation in the vector field syntax of
V = xi + yj
The Figure title would be clearer if it read:
The curve y = x^2, and the vector field dy/dx = 2x for the general case y = x^2 + C
There are three equations at work in the graph above:
y = x^2 - The equation for the parabola drawn - This is the one long solid curve
y = x^2 + C -The equation for all parabolas that fit on the vector field - C is a constant. This is the equation for all parabolas that fit on that vector field
dy/dx = 2x The equation for the slope field. - This is the slope or derivative of the both the curve drawn and all the possible curves that can be drawn with y = x^2 + C for all constant Cs.
Note that C is a constant, since the derivative of y = x^2 + C with any C is 2x. So the vector field shows how to draw all the different parabolas with different Cs.
So there are two ways to calculate the vector field:
Iterate over your desired range of x and y and calculate the slope, dy/dx- 2x independent of y in this case - at each point. This is how the author did it.
Draw a bunch of parabolas by slowly varying C in y = x^2 + C over a desired range of - let's say - x calculating y.
For a differential equation dy/dx = f(x,y) (e.g., dy/dx = 2x in this case, with f(x,y) = 2x), the vector field (F) will be F = i + f(x,y)j (so in your case, F = i + 2x j )
I have a value, for example 2.8. I want to find 10 numbers which are on an exponential curve, which sum to this value.
That is, I want to end up with 10 numbers which sum to 2.8, and which, when plotted, look like the curve below (exponential decay). These 10 numbers should be equally spaced along the curve - that is, the 'x-step' between the values should be constant.
This value of 2.8 will be entered by the user, and therefore the way I calculate this needs to be some kind of algorithm that I can program (hence asking this on SO not Math.SE).
I have no idea where to start with this at all - any ideas?
You want to have 10 x values equally distributed, i.e. x_k = a + k * b. They shall fulfill sum(exp(-x_k)) = v with v being your target value (the 2.8). This means exp(-a) * sum(exp(-b)^k) = v.
Obviously, there is a solution for each choice of b if v is positive. Set b to an arbitrary value, and calculate a from it.
E.g. for v = 2.8 and b = 0.1, you get a = -log(v / sum(exp(-b)^k)) = -log(2.8/sum(0.90484^k)) = -log(2.8/6.6425) = -log(0.421526) = 0.86387.
So for this example, the x values would be 0.86387, 0.96387, ..., 1.76387 and the y values 0.421526, 0.381412, 0.345116, 0.312274, 0.282557, 0.255668, 0.231338, 0.209324, 0.189404, 0.171380.
Update:
As it has been clarified that the curve can be scaled arbitrarily and the xs are preferred to be 1, 2, 3 ... 9, this is much more simple.
Assuming the curve function is r*exp(-x), the 10 values would be r*exp(-1) ... r*exp(-9). Their sum is r*sum(exp(-x)) = r*0.58190489. So to reach a certain value (2.8) you just have to adjust the r accordingly:
r = 2.8/sum(exp(-x)) = 4.81178294
And you get the 10 values: 1.770156, 0.651204, 0.239565, 0.088131, 0.032422, 0.011927, 0.004388, 0.001614, 0.000594.
If I understand your question correctly then you want to find x which solves the equation
It can be solved as
(just sum numbers as geometric progression)
The equation under RootOf will always have 1 real square different from 1 for 2.8 or any other positive number. You can solve it using some root-finding algorithm (1 is always a root but it does not solve original task). For constant a you can choose any number you like.
After computing the x you can easily calculate 10 numbers as .
I'm going to generalize and assume you want N numbers summing to V.
Since your numbers are equally spaced on an exponential you can write your sum as
a + a*x + a*x^2 + ... + a*x^(N-1) = V
Where the first point has value a, and the second a*x etc.
You can take out a factor of a and get:
a ( 1 + x + x^2 + ... + x^(N-1) ) = V
If we're free to pick x then we can solve for a easily
a = V / ( 1 + x + x^2 + .. x^(N-1) )
= V*(x+1)/(x^N-1)
Substituting that back into
a, a*x, a*x^2, ..., a*x^(N-1)
gives the required sequence
At present I have a control to which I need to add the facility to apply various acuteness (or sensitivity). The problem is best illustrated as an image:
Graph http://img87.imageshack.us/img87/7886/control.png
As you can see, I have X and Y axess that both have arbitrary limits of 100 - that should suffice for this explanation. At present, my control is the red line (linear behaviour), but I would like to add the ability for the other 3 curves (or more) i.e. if a control is more sensitive then a setting will ignore the linear setting and go for one of the three lines. The starting point will always be 0, and the end point will always be 100.
I know that an exponential is too steep, but can't seem to figure a way forward. Any suggestions please?
The curves you have illustrated look a lot like gamma correction curves. The idea there is that the minimum and maximum of the range stays the same as the input, but the middle is bent like you have in your graphs (which I might note is not the circular arc which you would get from the cosine implementation).
Graphically, it looks like this:
(source: wikimedia.org)
So, with that as the inspiration, here's the math...
If your x values ranged from 0 to 1, the function is rather simple:
y = f(x, gamma) = x ^ gamma
Add an xmax value for scaling (i.e. x = 0 to 100), and the function becomes:
y = f(x, gamma) = ((x / xmax) ^ gamma) * xmax
or alternatively:
y = f(x, gamma) = (x ^ gamma) / (xmax ^ (gamma - 1))
You can take this a step further if you want to add a non-zero xmin.
When gamma is 1, the line is always perfectly linear (y = x). If x is less than 1, your curve bends upward. If x is greater than 1, your curve bends downward. The reciprocal value of gamma will convert the value back to the original (x = f(y, 1/g) = f(f(x, g), 1/g).
Just adjust the value of gamma according to your own taste and application needs. Since you're wanting to give the user multiple options for "sensitivity enhancement", you may want to give your users choices on a linear scale, say ranging from -4 (least sensitive) to 0 (no change) to 4 (most sensitive), and scale your internal gamma values with a power function. In other words, give the user choices of (-4, -3, -2, -1, 0, 1, 2, 3, 4), but translate that to gamma values of (5.06, 3.38, 2.25, 1.50, 1.00, 0.67, 0.44, 0.30, 0.20).
Coding that in C# might look something like this:
public class SensitivityAdjuster {
public SensitivityAdjuster() { }
public SensitivityAdjuster(int level) {
SetSensitivityLevel(level);
}
private double _Gamma = 1.0;
public void SetSensitivityLevel(int level) {
_Gamma = Math.Pow(1.5, level);
}
public double Adjust(double x) {
return (Math.Pow((x / 100), _Gamma) * 100);
}
}
To use it, create a new SensitivityAdjuster, set the sensitivity level according to user preferences (either using the constructor or the method, and -4 to 4 would probably be reasonable level values) and call Adjust(x) to get the adjusted output value. If you wanted a wider or narrower range of reasonable levels, you would reduce or increase that 1.5 value in the SetSensitivityLevels method. And of course the 100 represents your maximum x value.
I propose a simple formula, that (I believe) captures your requirement. In order to have a full "quarter circle", which is your extreme case, you would use (1-cos((x*pi)/(2*100)))*100.
What I suggest is that you take a weighted average between y=x and y=(1-cos((x*pi)/(2*100)))*100. For example, to have very close to linear (99% linear), take:
y = 0.99*x + 0.01*[(1-cos((x*pi)/(2*100)))*100]
Or more generally, say the level of linearity is L, and it's in the interval [0, 1], your formula will be:
y = L*x + (1-L)*[(1-cos((x*pi)/(2*100)))*100]
EDIT: I changed cos(x/100) to cos((x*pi)/(2*100)), because for the cos result to be in the range [1,0] X should be in the range of [0,pi/2] and not [0,1], sorry for the initial mistake.
You're probably looking for something like polynomial interpolation. A quadratic/cubic/quartic interpolation ought to give you the sorts of curves you show in the question. The differences between the three curves you show could probably be achieved just by adjusting the coefficients (which indirectly determine steepness).
The graph of y = x^p for x from 0 to 1 will do what you want as you vary p from 1 (which will give the red line) upwards. As p increases the curve will be 'pushed in' more and more. p doesn't have to be an integer.
(You'll have to scale to get 0 to 100 but I'm sure you can work that out)
I vote for Rax Olgud's general idea, with one modification:
y = alpha * x + (1-alpha)*(f(x/100)*100)
alt text http://www4c.wolframalpha.com/Calculate/MSP/MSP4501967d41e1aga1b3i00004bdeci2b6be2a59b?MSPStoreType=image/gif&s=6
where f(0) = 0, f(1) = 1, f(x) is superlinear, but I don't know where this "quarter circle" idea came from or why 1-cos(x) would be a good choice.
I'd suggest f(x) = xk where k = 2, 3, 4, 5, whatever gives you the desired degre of steepness for &alpha = 0. Pick a value for k as a fixed number, then vary α to choose your particular curve.
For problems like this, I will often get a few points from a curve and throw it through a curve fitting program. There are a bunch of them out there. Here's one with a 7-day free trial.
I've learned a lot by trying different models. Often you can get a pretty simple expression to come close to your curve.