Is there a way to specify the domain for a function in wolfram alpha? I have a function which I want to plot. I want to restrict the function f(x,y) = xy(3-x-y) to x>=0, y>=3 and y <=3-x
You can specify boundaries of the function parameters in Wolfram Alpha.
They are more of a hint to the plotting library how much to scale the plot than exact boundaries, so in your case y is not calculated only with a value of 3. Although the boundaries in your case do not make much sense, as noted by #Sjoerd, this answer could be useful for setting sensible boundaries.
To plot your function, write the following in Wolfram Alpha:
f(x,y) = xy(3-x-y) for x>=0, y=3..3-x
It works better with boundaries set around zero and univariate functions:
plot y^2 cos(x), x=-6..6, y=-2..2
plot f(x) = -8/x + sqrt(7/x^2+1/x+1) for x=1..50
Reference: Wolfram Alpha Blog
Your restrictions make no sense. For x>=0 it follows that y<=3-x equals y<=3, but you also have y>=3, so this means that you want a plot for y=3. So, what you want to plot is apparently f(x)= -3x^2.
Other than that, I don't think WA currently allows a RegionFunction as in Mathematica
Related
I'm trying to display 2D data with axis labels using both contour and pcolormesh. As has been noted on the matplotlib user list, these functions obey different conventions: pcolormesh expects the x and y values to specify the corners of the individual pixels, while contour expects the centers of the pixels.
What is the best way to make these behave consistently?
One option I've considered is to make a "centers-to-edges" function, assuming evenly spaced data:
def centers_to_edges(arr):
dx = arr[1]-arr[0]
newarr = np.linspace(arr.min()-dx/2,arr.max()+dx/2,arr.size+1)
return newarr
Another option is to use imshow with the extent keyword set.
The first approach doesn't play nicely with 2D axes (e.g., as created by meshgrid or indices) and the second discards the axis numbers entirely
Your data is a regular mesh? If it doesn't, you can use griddata() to obtain it. I think that if your data is too big, a sub-sampling or regularization always is possible. If the data is too big, maybe your output image always will be small compared with it and you can exploit this.
If you use imshow() with "extent" and "interpolation='nearest'", you will see that the data is cell-centered, and extent provided the lower edges of cells (corners). On the other hand, contour assumes that the data is cell-centered, and X,Y must be the center of cells. So, you need to be care about the input domain for contour. The trivial example is:
x = np.arange(-10,10,1)
X,Y = np.meshgrid(x,x)
P = X**2+Y**2
imshow(P,extent=[-10,10,-10,10],interpolation='nearest',origin='lower')
contour(X+0.5,Y+0.5,P,20,colors='k')
My tests told me that pcolormesh() is a very slow routine, and I always try to avoid it. griddata and imshow() always is a good choose for me.
I have sampled data and plot it with imshow():
I would like to interpolate just in horizontal axis so that I can easier distinguish samples and spot features.
Is it possible to make interpolation just in one direction with MPL?
Update:
SciPy has whole package with various interpolation methods.
I used simplest interp1d, as suggested by tcaswell:
def smooth_inter_fun(r):
s = interpolate.interp1d(arange(len(r)), r)
xnew = arange(0, len(r)-1, .1)
return s(xnew)
new_data = np.vstack([smooth_inter_fun(r) for r in data])
Linear and cubic results:
As expected :)
This tutorial covers a range of interpolation available in numpy/scipy. If you want to just one direction, I would work on each row independently and then re-assemble the results. You might also be interested is simply smoothing your data (exmple, Python Smooth Time Series Data, Using strides for an efficient moving average filter).
def smooth_inter_fun(r):
#what ever process you want to use
new_data = np.vstack([smooth_inter_fun(r) for r in data])
Basically, I have a set of up to 100 co-ordinates, along with the desired tangents to the curve at the first and last point.
I have looked into various methods of curve-fitting, by which I mean an algorithm with takes the inputted data points and tangents, and outputs the equation of the cure, such as the gaussian method and interpolation, but I really struggled understanding them.
I am not asking for code (If you choose to give it, thats acceptable though :) ), I am simply looking for help into this algorithm. It will eventually be converted to Objective-C for an iPhone app, if that changes anything..
EDIT:
I know the order of all of the points. They are not too close together, so passing through all points is necessary - aka interpolation (unless anyone can suggest something else). And as far as I know, an algebraic curve is what I'm looking for. This is all being done on a 2D plane by the way
I'd recommend to consider cubic splines. There is some explanation and code to calculate them in plain C in Numerical Recipes book (chapter 3.3)
Most interpolation methods originally work with functions: given a set of x and y values, they compute a function which computes a y value for every x value, meeting the specified constraints. As a function can only ever compute a single y value for every x value, such an curve cannot loop back on itself.
To turn this into a real 2D setup, you want two functions which compute x resp. y values based on some parameter that is conventionally called t. So the first step is computing t values for your input data. You can usually get a good approximation by summing over euclidean distances: think about a polyline connecting all your points with straight segments. Then the parameter would be the distance along this line for every input pair.
So now you have two interpolation problem: one to compute x from t and the other y from t. You can formulate this as a spline interpolation, e.g. using cubic splines. That gives you a large system of linear equations which you can solve iteratively up to the desired precision.
The result of a spline interpolation will be a piecewise description of a suitable curve. If you wanted a single equation, then a lagrange interpolation would fit that bill, but the result might have odd twists and turns for many sets of input data.
I want to generate a grid of plots, of several arrays, with positive and negative values, with log scale, sharing the same colorbar.
I've achieved the sharing part of the colorbar (using ImageGrid and common max and min values), and I know that I could get a logarithmic scale using LogNorm() on the imshow call in the case of only positive values. But given the presence of negative values, I would need a colorbar on symmetric logarithmic scale.
I have found what would be the solution on https://stackoverflow.com/a/7741317/1101750 , but running the sample code Yann provides gives me very different results, cleary wrong:
Reviewing the code, I'm not able to grasp what's going on.
In addition to that, I've discovered that on Matplotlib 1.2, scale.SymmetricalLogScale.SymmetricalLogTransform asks for a new argument not explained on the documentation (linscale, which looking at the code of other transforms I assume that leaving it as 1 is a safe value).
Is the easiest solution subclassing LogNorm?
I've used a pretty simple recipe in the past to do exactly this, without the need to do any subclassing. matplotlib.colors.SymLogNorm provides most of the functionality you need, except that I've found it necessary to generate the tick marks by hand. Note that this solution uses matplotlib 1.3.0, and I may be using features that weren't available with 1.2.
def imshow_symlog(my_matrix, vmin, vmax, logthresh=5):
img=imshow( my_matrix ,
vmin=float(vmin), vmax=float(vmax),
norm=matplotlib.colors.SymLogNorm(10**-logthresh) )
maxlog=int(np.ceil( np.log10(vmax) ))
minlog=int(np.ceil( np.log10(-vmin) ))
#generate logarithmic ticks
tick_locations=([-(10**x) for x in xrange(minlog,-logthresh-1,-1)]
+[0.0]
+[(10**x) for x in xrange(-logthresh,maxlog+1)] )
cb=colorbar(ticks=tick_locations)
return img,cb
Since 1.3 matplotlib has a SymLogNorm. http://matplotlib.org/api/colors_api.html#matplotlib.colors.SymLogNorm
edit
So based on the answers so far (thanks for taking your time) I'm getting the sense that I'm probably NOT looking for a Normal Distribution function. Perhaps I'll try to re-describe what I'm looking to do.
Lets say I have an object that returns a number of 0 to 10. And that number controls "speed". However instead of 10 being the top speed, I need 5 to be the top speed, and anything lower or higher would slow down accordingly. (with easing, thus the bell curve)
I hope that's clearer ;/
-original question
These are the times I wish I remembered something from math class.
I'm trying to figure out how to write a function in obj-C where I define the boundries, ex (0 - 10) and then if x = foo y = ? .... where x runs something like 0,1,2,3,4,5,6,7,8,9,10 and y runs 0,1,2,3,4,5,4,3,2,1,0 but only on a curve
Something like the attached image.
I tried googling for Normal Distribution but its way over my head. I was hoping to find some site that lists some useful algorithms like these but wasn't very successful.
So can anyone help me out here ? And if there is some good sites which shows useful mathematical functions, I'd love to check them out.
TIA!!!
-added
I'm not looking for a random number, I'm looking for.. ex: if x=0 y should be 0, if x=5 y should be 5, if x=10 y should be 0.... and all those other not so obvious in between numbers
alt text http://dizy.cc/slider.gif
Okay, your edit really clarifies things. You're not looking for anything to do with the normal distribution, just a nice smooth little ramp function. The one Paul provides will do nicely, but is tricky to modify for other values. It can be made a little more flexible (my code examples are in Python, which should be very easy to translate to any other language):
def quarticRamp(x, b=10, peak=5):
if not 0 <= x <= b:
raise ValueError #or return 0
return peak*x*x*(x-b)*(x-b)*16/(b*b*b*b)
Parameter b is the upper bound for the region you want to have a slope on (10, in your example), and peak is how high you want it to go (5, in the example).
Personally I like a quadratic spline approach, which is marginally cheaper computationally and has a different curve to it (this curve is really nice to use in a couple of special applications that don't happen to matter at all for you):
def quadraticSplineRamp(x, a=0, b=10, peak=5):
if not a <= x <= b:
raise ValueError #or return 0
if x > (b+a)/2:
x = a + b - x
z = 2*(x-a)/b
if z > 0.5:
return peak * (1 - 2*(z-1)*(z-1))
else:
return peak * (2*z*z)
This is similar to the other function, but takes a lower bound a (0 in your example). The logic is a little more complex because it's a somewhat-optimized implementation of a piecewise function.
The two curves have slightly different shapes; you probably don't care what the exact shape is, and so could pick either. There are an infinite number of ramp functions meeting your criteria; these are two simple ones, but they can get as baroque as you want.
The thing you want to plot is the probability density function (pdf) of the normal distribution. You can find it on the mighty Wikipedia.
Luckily, the pdf for a normal distribution is not difficult to implement - some of the other related functions are considerably worse because they require the error function.
To get a plot like you showed, you want a mean of 5 and a standard deviation of about 1.5. The median is obviously the centre, and figuring out an appropriate standard deviation given the left & right boundaries isn't particularly difficult.
A function to calculate the y value of the pdf given the x coordinate, standard deviation and mean might look something like:
double normal_pdf(double x, double mean, double std_dev) {
return( 1.0/(sqrt(2*PI)*std_dev) *
exp(-(x-mean)*(x-mean)/(2*std_dev*std_dev)) );
}
A normal distribution is never equal to 0.
Please make sure that what you want to plot is indeed a
normal distribution.
If you're only looking for this bell shape (with the tangent and everything)
you can use the following formula:
x^2*(x-10)^2 for x between 0 and 10
0 elsewhere
(Divide by 125 if you need to have your peek on 5.)
double bell(double x) {
if ((x < 10) && (x>0))
return x*x*(x-10.)*(x-10.)/125.;
else
return 0.;
}
Well, there's good old Wikipedia, of course. And Mathworld.
What you want is a random number generator for "generating normally distributed random deviates". Since Objective C can call regular C libraries, you either need a C-callable library like the GNU Scientific Library, or for this, you can write it yourself following the description here.
Try simulating rolls of dice by generating random numbers between 1 and 6. If you add up the rolls from 5 independent dice rolls, you'll get a surprisingly good approximation to the normal distribution. You can roll more dice if you'd like and you'll get a better approximation.
Here's an article that explains why this works. It's probably more mathematical detail than you want, but you could show it to someone to justify your approach.
If what you want is the value of the probability density function, p(x), of a normal (Gaussian) distribution of mean mu and standard deviation sigma at x, the formula is
p(x) = exp( ((x-mu)^2)/(2*sigma^2) ) / (sigma * 2 * sqrt(pi))
where pi is the area of a circle divided by the square of its radius (approximately 3.14159...). Using the C standard library math.h, this is:
#include <math>
double normal_pdf(double x, double mu, double sigma) {
double n = sigma * 2 * sqrt(M_PI); //normalization factor
p = exp( -pow(x-mu, 2) / (2 * pow(sigma, 2)) ); // unnormalized pdf
return p / n;
}
Of course, you can do the same in Objective-C.
For reference, see the Wikipedia or MathWorld articles.
It sounds like you want to write a function that yields a curve of a specific shape. Something like y = f(x), for x in [0:10]. You have a constraint on the max value of y, and a general idea of what you want the curve to look like (somewhat bell-shaped, y=0 at the edges of the x range, y=5 when x=5). So roughly, you would call your function iteratively with the x range, with a step that gives you enough points to make your curve look nice.
So you really don't need random numbers, and this has nothing to do with probability unless you want it to (as in, you want your curve to look like a the outline of a normal distribution or something along those lines).
If you have a clear idea of what function will yield your desired curve, the code is trivial - a function to compute f(x) and a for loop to call it the desired number of times for the desired values of x. Plot the x,y pairs and you're done. So that's your algorithm - call a function in a for loop.
The contents of the routine implementing the function will depend on the specifics of what you want the curve to look like. If you need help on functions that might return a curve resembling your sample, I would direct you to the reading material in the other answers. :) However, I suspect that this is actually an assignment of some sort, and that you have been given a function already. If you are actually doing this on your own to learn, then I again echo the other reading suggestions.
y=-1*abs(x-5)+5