numerical-stability | 易学教程

What are the constraints on the divisor argument of scipy.signal.deconvolve to ensure numerical stability?

阅读更多关于 What are the constraints on the divisor argument of scipy.signal.deconvolve to ensure numerical stability?

问题 Here is my problem: I am going to process data coming from a system for which I will have a good idea of the impulse response. Having used Python for some basic scripting before, I am getting to know the scipy.signal.convolve and scipy.signal.deconvolve functions. In order to get some confidence in my final solution, I would like to understand their requirements and limitations. I used the following test: 1. I built a basic signal made of two Gaussians. 2. I built a Gaussian impulse response.

numerically stable inverse of a 2x2 matrix

阅读更多关于 numerically stable inverse of a 2x2 matrix

问题 In a numerical solver I am working on in C, I need to invert a 2x2 matrix and it then gets multiplied on the right side by another matrix: C = B . inv(A) I have been using the following definition of an inverted 2x2 matrix: a = A[0][0]; b = A[0][1]; c = A[1][0]; d = A[1][1]; invA[0][0] = d/(a*d-b*c); invA[0][1] = -b/(a*d-b*c); invA[1][0] = -c/(a*d-b*c); invA[1][1] = a/(a*d-b*c); In the first few iterations of my solver this seems to give the correct answers, however, after a few steps things

Avoiding numerical overflow when calculating the value AND gradient of the Logistic loss function

阅读更多关于 Avoiding numerical overflow when calculating the value AND gradient of the Logistic loss function

问题 I am currently trying to implement a machine learning algorithm that involves the logistic loss function in MATLAB. Unfortunately, I am having some trouble due to numerical overflow. In general, for a given an input s , the value of the logistic function is: log(1 + exp(s)) and the slope of the logistic loss function is: exp(s)./(1 + exp(s)) = 1./(1 + exp(-s)) In my algorithm, the value of s = X*beta . Here X is a matrix with N data points and P features per data point (i.e. size(X)=[N,P] )

How can I avoid value errors when using numpy.random.multinomial?

阅读更多关于 How can I avoid value errors when using numpy.random.multinomial?

问题 When I use this random generator: numpy.random.multinomial, I keep getting: ValueError: sum(pvals[:-1]) > 1.0 I am always passing the output of this softmax function: def softmax(w, t = 1.0): e = numpy.exp(numpy.array(w) / t) dist = e / np.sum(e) return dist except now that I am getting this error, I also added this for the parameter ( pvals ): while numpy.sum(pvals) > 1: pvals /= (1+1e-5) but that didn't solve it. What is the right way to make sure I avoid this error? EDIT: here is function

Symmetrical Lerp & compiler optimizations

阅读更多关于 Symmetrical Lerp & compiler optimizations

问题 I had a function: float lerp(float alpha, float x0, float x1) { return (1.0f - alpha) * x0 + alpha * x1; } For those who haven't seen it, this is preferable to x0 + (x1-x0) * alpha because the latter doesn't guarantee that lerp(1.0f, x0, x1) == x1 . Now, I want my lerp function to have an additional property: I'd like lerp(alpha, x0, x1) == lerp(1-alpha, x1, x0) . (As for why: this is a toy example of a more complicated function.) The solution I came up with that seems to work is float lerp

Getting y from x co-ord for cubic bezier curve, fast Newton-Raphson method

阅读更多关于 Getting y from x co-ord for cubic bezier curve, fast Newton-Raphson method

问题 Given the points of a Bezier curve (P0, P1, P2, P3) in 2D, I would like to find the y co-ordinate for a given x co-ordinate. The problem is well defined because of the following restrictions: P0 = (0,0), P3 = (1,1) P1 = (t, 1-t) for t between 0, 1 P2 = 1 - P1 (x and y) I have the following function to calculate the answer, having put in all the restrictions above into the Bezier curve formula here CubicBezier.html. I am using Newton-Raphson to work out the parameter of the point I want, and I

Converting float to UInt32 - which expression is more precise

阅读更多关于 Converting float to UInt32 - which expression is more precise

问题 I have a number float x which should be in <0,1> range but it undergo several numerical operations - the result may be slightly outside <0,1>. I need to convert this result to uint y using entire range of UInt32 . Of course, I need to clamp x in the <0,1> range and scale it. But which order of operations is better? y = (uint)round(min(max(x, 0.0F), 1.0F) * UInt32.MaxValue) or y = (uint)round(min(max(x * UInt32.MaxValue, 0.0F), UInt32.MaxValue) In another words, it is better to scale first,

Interpreting error from computing Jordan form of 36-by-36 matrix

阅读更多关于 Interpreting error from computing Jordan form of 36-by-36 matrix

问题 I've been trying to compute the jordan normal form of a 36-by-36 matrix composed of only three distinct entries, 1 , 1/2 , and 0 . The matrix is a probability transition matrix so, given these entries, the matrix is obviously sparse. The issue I've been having is the following: whenever I try to compute [V, J] = jordan(A), or [V, J] = jordan(sym(A)), I get the following error message: Error using mupadmex Error in MuPAD command: Similarity matrix too large. Error in sym/mupadmexnout (line

What are the constraints on the divisor argument of scipy.signal.deconvolve to ensure numerical stability?

阅读更多关于 What are the constraints on the divisor argument of scipy.signal.deconvolve to ensure numerical stability?

Here is my problem: I am going to process data coming from a system for which I will have a good idea of the impulse response. Having used Python for some basic scripting before, I am getting to know the scipy.signal.convolve and scipy.signal.deconvolve functions. In order to get some confidence in my final solution, I would like to understand their requirements and limitations. I used the following test: 1. I built a basic signal made of two Gaussians. 2. I built a Gaussian impulse response. 3. I convolved my initial signal with this impulse response. 4. I deconvolved this convolved signal. 5

numerically stable inverse of a 2x2 matrix

阅读更多关于 numerically stable inverse of a 2x2 matrix

In a numerical solver I am working on in C, I need to invert a 2x2 matrix and it then gets multiplied on the right side by another matrix: C = B . inv(A) I have been using the following definition of an inverted 2x2 matrix: a = A[0][0]; b = A[0][1]; c = A[1][0]; d = A[1][1]; invA[0][0] = d/(a*d-b*c); invA[0][1] = -b/(a*d-b*c); invA[1][0] = -c/(a*d-b*c); invA[1][1] = a/(a*d-b*c); In the first few iterations of my solver this seems to give the correct answers, however, after a few steps things start to grow and eventually explode. Now, comparing to an implementation using SciPy, I found that the