derivative

问题 I have written a function that uses derivative product rule to find derivative of a term: def find_term_derivative(term): x , y = term new_term = (y*x, y-1) return new_term def find_derivative(function_terms): new_function = [] for term in function_terms: new_term = find_term_derivative(term) new_function.append(new_term) filtered_terms = filter(zero_filter, new_term) find_derivative[(4, 3), (-3, 1)] Ouputs [(12, 2), (-3, 0)] However I want to use the filter function to remove any terms which

问题 So I have a time series of MODIS NDVI values (vegetation values from 0-1 for the non-geographic geeks), and I'm trying to approximate the derivative by using a for loop. This is a sample of my data: > m2001 date value valnorm 1 1 0.4834 0.03460912 2 17 0.4844 0.03664495 3 33 0.5006 0.06962541 4 49 0.4796 0.02687296 5 65 0.5128 0.09446254 6 81 0.4915 0.05109935 7 97 0.4664 0.00000000 8 113 0.5345 0.13864007 9 129 0.8771 0.83611564 10 145 0.9529 0.99043160 11 161 0.9250 0.93363192 12 177 0.9450

问题 I'm trying reproduce results from a paper, in which they convolve the image with an horizontal partial derivative of a Gaussian kernel. I haven't found any way to achieve that with OpenCV. Is that possible ? Do I have to get Gaussian filter and then compute the partial derivatives by hand ? 回答1: OpenCV doesn't have built-in function to calculate Gaussian partial derivatives. But you may use cv::getGaussianKernel and cv::filter2D to do so. For example: cv::Mat kernel = cv::getGaussianKernel(3,

问题 I am having problems with the derivativeAt() method when using the BSpline path because it is not returning the correct values, however the derivativeAt() method works find for the CatmullRomSpline path so this is something specific to SPline. The wiki and API do not mention that the SPline needs to be treated differently from other paths...is there something that I’m missing with SPline, or is this a bug with LibGDX? I made a simple program to show the problem. It draws the control points

问题 I want to find the first derivative of exp(sin(x)) on the interval [0, 2/pi] using a discrete Fourier transform. The basic idea is to first evaluate the DFT of exp(sin(x)) on the given interval, giving you say v_k , followed by computing the inverse DFT of ikv_k giving you the desired answer. In reality, due to the implementations of Fourier transforms in programming languages, you might need to reorder the output somewhere and/or multiply by different factors here and there. I first did it

问题 I don't have a problem computing the derivative..it's just that I don't know to handle an entire polynomial in standard algebraic notation :p 回答1: For computer algebra in Python, sympy is the way to go. Computing the derivative of a polynomial in sympy is straightforward: >>> import sympy as sp >>> x = sp.symbols('x') >>> sp.diff(3*x**4 + 8*x**2 - 3*x + 1) 12*x**3 + 16*x - 3 回答2: The code is not concise, because I want to clearly show each step how it calculates. import re def FirstDerivative

问题 I am currently writing a calculator application. I am trying to write a derivative estimator into it. The formula below is a simple way to do it. Normally on paper you would use the smallest h possible to get the most accurate estimate. The problem is doubles can't handle adding really small numbers to comparatively huge numbers. Such as 4+1E-200 will just result in 4.0. Even if h was just 1E-16, 4+1E16 will in fact give you the right value but doing math it it is inaccurate because anything

问题 I am using density {stats} to construct a kernel "gaussian' density of a vector of variables. If I use the following example dataset: x <- rlogis(1475, location=0, scale=1) # x is a vector of values - taken from a rlogis just for the purpose of explanation d<- density(x=x, kernel="gaussian") Is there some way to get the first derivative of this density d at each of the n=1475 points 回答1: The curve of a density estimator is just the sum of all the kernels, in your case a gaussian (divided by

问题 I defined the derivative of a function in Mathematica without defining the function itself, i.e. I have a function definition that looks like this: y'[x_] := constant * f'[x]. I can't figure out how to clear it out. If I use Clear[y'] or `ClearAll[y'], I get an error message: ClearAll::ssym: y' is not a symbol or a string. Clear[y] and ClearAll[y] do nothing to remove the definition of y' . Any ideas on how I can remove the definition of y' ? 回答1: This should do what you want: y'[x_] =. See

问题 I would like to compute the derivative of a complex-valued function (Holomorphic function) numerically in MATLAB. I have computed the function in a grid on the complex plane, and I've tried to compute the derivative using the Cauchy–Riemann relations. Given: u = real(f), v = imag(f), x = real(point), y = imag(point) The derivative should be given by: f' = du/dx + i dv/dx = dv/dy - i du/dy where 'd' is the derivative operator. I've tried the following code: stepx = 0.01; stepy = 0.01; Nx = 2