quadratic

I am trying to do a simple quadratic function that would return number of roots and their values via an enum: enum QuadraticResult { None, OneRoot(f32), TwoRoots(f32, f32), } fn solveQuadratic(a: f32, b: f32, c: f32) -> QuadraticResult { let delta = b * b - 4.0 * a * c; match delta { < 0 => return QuadraticResult::None, > 0 => return QuadraticResult::TwoRoots(0.0, 1.0), _ => return QuadraticResult::OneRoot(0.0), } } This doesn't compile as it complains about '<' and '>'. Is there a way to achieve this with match or do I need to use if You can use a match guard , but that feels more verbose

I'm looking for a good easy to use Java based Quadratic Programming (QP) solver. Googling around I came across ojAlgo ( http://ojalgo.org ). However, I was wondering if there are any other/better alternatives. Have a look at Apache Commons Math . I haven't used ojalgo, and I really can't say I've used Commons Lang enough to be able to provide you with a lot of details, but it did do what I needed. Description from their website: Commons Math is a library of lightweight, self-contained mathematics and statistics components addressing the most common problems not available in the Java

The only Google search result I found is QuadProg++ but it can not solve the quadratic programming problem whose matrix is not applicable for Cholesky decomposition. So can anyone give me some suggestion on other library? Thanks. CGAL looks great for quadratic programming. There is even a manual . // by default, we have a nonnegative QP with Ax <= b Program qp (CGAL::SMALLER, true, 0, false, 0); // now set the non-default entries: const int X = 0; const int Y = 1; qp.set_a(X, 0, 1); qp.set_a(Y, 0, 1); qp.set_b(0, 7); // x + y <= 7 qp.set_a(X, 1, -1); qp.set_a(Y, 1, 2); qp.set_b(1, 4); // -x +

I'm very new to python programming and to this site. I'm currently working on a problem and can't seem to understand the error. import math # Problem number 5. A5 = 5 B5 = 0 C5 = 6.5 # Root1 x9 = (-B5 + math.sqrt(B5**2 - 4*A5*C5))/(2*A5) # Root2 x10 = (-B5 + math.sqrt(B5**2 - 4*A5*C5))/(2*A5) # Print solution print() print('Problem #5') print('Root 1: ',x9) print('Root 2: ',x10) I get this after i run it: x9 = (-B5 + math.sqrt(B5**2 - 4*A5*C5))/(2*A5) ValueError: math domain error I did the problem on paper and got an answer for both... If you got an answer, it must have been a complex number

I am having some issues calculating the nearest point on a quadratic curve to the mouse position. I have tried a handful of APIs, but have not had any luck finding a function for this that works. I have found an implementation that works for 5th degree cubic bezier curves, but I do not have the math skills to convert it to a quadratic curve. I have found some methods that will help me solve the problem if I have a t value, but I have no idea how to begin finding t. If someone could point me to an algorithm for finding t, or some example code for finding the nearest point on a quadratic curve

I currently have a class called Polynomial, The initialization looks like this: def __init__(self, *termpairs): self.termdict = dict(termpairs) I'm creating a polynomial by making the keys the exponents and the associated values are the coefficients. To create an instance of this class, you enter as follows: d1 = Polynomial((5,1), (3,-4), (2,10)) which makes a dictionary like so: {2: 10, 3: -4, 5: 1} Now, I want to create a subclass of the Polynomial class called Quadratic. I want to call the Polynomial class constructor in the Quadratic class constructor, however im not quite sure how to do

I have the following linear regression: import statsmodels.formula.api as sm model = sm.ols(formula = 'a ~ b + c', data = data).fit() I want to add a quadratic term for b in this model. Is there a simple way to do this with statsmodels.ols? Is there a better package I should be using to achieve this? Although the solution by Alexander is working, in some situations it is not very convenient. For example, each time you want to predict the outcome of the model for new values, you need to remember to pass both b**2 and b values which is cumbersome and should not be necessary. Although patsy does