4 petal rose formula that fits astroid (pedal curve) in desmos - math

https://www.desmos.com/calculator/g7m5l5zaxa
\left(\frac{6\sin\left(t\right)^{2}\cos\left(t\right)}{4\sin\left(t\right)^{2}+\cos^{2}\left(t\right)},\frac{3\sin\left(t\right)\cos^{2}\left(t\right)}{4\sin\left(t\right)^{2}+\cos\left(t\right)^{2}}\right)
I cannot fathom the quadrifolium formula (4 petal rose) that fits the astroid. Is it the pedal curve? I tried the pedal curve parametric equations but got a different result.
I tried the parametric equation from below link:
https://www.wikiwand.com/en/Pedal_curve

Related

Inferring/expressing the polynomial equation of a fitted smoothing spline?

If I smooth a data vector with a smoothing cubic spline my understanding is that each ‘segment’ between knots should be representable as a cubic polynomial.
Is it possible to infer the equation of each segment from the spline coefficients after e.g fitting by the smooth.spline function in R?
This is straightforward for an interpolating spine as the array of polynomial coefficients is generated explicitly. However, I’ve not been able to find an answer as to whether this is possible with a smoothing spline or regression spline?
The reason for wanting this is to in turn obtain an analytical expression for the derivative of a segment of a spline.

r fit logistic curve through three points

I have three points (1,4) (3,6) (2,5),the (2,5)is the midpoints .how to fit a particular kind of logistic curves like the following figure.

I fit a modified logistic with two parameters, "y = a / (1.0 + b*exp(-X))" with parameters a = 6.5094194977264266E+00 and b = 1.7053273551626427E+00 - see attached graph. I recommend not using this as it is a perfect fit with no errors, two parameters and two data points.

{Methcomp} – Deming / orthogonal regression – goodness of fit + confidence intervals

A question following this post. I have the following data:
x1, disease symptom
y1, another disease symptom
I fitted the x1/y1 data with a Deming regression with vr (or sdr) option set to 1. In other words, the regression is a Total Least Squares regression, i.e. orthogonal regression. See previous post for the graph.
x1=c(24.0,23.9,23.6,21.6,21.0,20.8,22.4,22.6,
21.6,21.2,19.0,19.4,21.1,21.5,21.5,20.1,20.1,
20.1,17.2,18.6,21.5,18.2,23.2,20.4,19.2,22.4,
18.8,17.9,19.1,17.9,19.6,18.1,17.6,17.4,17.5,
17.5,25.2,24.4,25.6,24.3,24.6,24.3,29.4,29.4,
29.1,28.5,27.2,27.9,31.5,31.5,31.5,27.8,31.2,
27.4,28.8,27.9,27.6,26.9,28.0,28.0,33.0,32.0,
34.2,34.0,32.6,30.8)
y1=c(100.0,95.5,93.5,100.0,98.5,99.5,34.8,
45.8,47.5,17.4,42.6,63.0,6.9,12.1,30.5,
10.5,14.3,41.1, 2.2,20.0,9.8,3.5,0.5,3.5,5.7,
3.1,19.2,6.4, 1.2, 4.5, 5.7, 3.1,19.2, 6.4,
1.2,4.5,81.5,70.5,91.5,75.0,59.5,73.3,66.5,
47.0,60.5,47.5,33.0,62.5,87.0,86.0,77.0,
86.0,83.0,78.5,83.0,83.5,73.0,69.5,82.5,78.5,
84.0,93.5,83.5,96.5,96.0,97.5)
x11()
plot(x1,y1,xlim=c(0,35),ylim=c(0,100))
library(MethComp)
dem_reg <- Deming(x1, y1)
abline(dem_reg[1:2], col = "green")
I would like to know how much x1 helps to predict y1:
normally, I’d go for a R-squared, but it does not seem to be relevant; although another mathematician told me he thinks a R-squared may be appropriate. And this page suggests to calculate a Pearson product-moment correlation coefficient, which is R I believe?
partially related, there is possibly a tolerance interval. I could calculated it with R ({tolerance} package or code shown in the post), but it is not exactly what I am searching for.
Does someone know how to calculate a goodness of fit for Deming regression, using R? I looked at MetchComp pdf but could not find it (perhaps missed it though).
EDIT: following Gaurav's answers about confidence interval: R code
Firstly: confidence intervals for parameters
library(mcr)
MCR_reg=mcreg(x1,y1,method.reg="Deming",error.ratio=1,method.ci="analytical")
getCoefficients(MCR_reg)
Secondly: confidence intervals for predicted values
# plot of data
x11()
plot(x1,y1,xlim=c(0,35),ylim=c(0,100))
# Deming regression using functions from {mcr}
library(mcr) MCR_reg=mcreg(x1,y1,method.reg="Deming",error.ratio=1,method.ci="analytical")
MCR_intercept=getCoefficients(MCR_reg)[1,1]
MCR_slope=getCoefficients(MCR_reg)[2,1]
# CI for predicted values
x_to_predict=seq(0,35)
predicted_values=MCResultAnalytical.calcResponse(MCR_reg,x_to_predict,alpha=0.05)
CI_low=predicted_values[,4]
CI_up=predicted_values[,5]
# plot regression line and CI for predicted values
abline(MCR_intercept,MCR_slope, col="red")
lines(x_to_predict,CI_low,col="royalblue",lty="dashed")
lines(x_to_predict,CI_up,col="royalblue",lty="dashed")
# comments
text(7.5,60, "Deming regression", col="red")
text(7.5,40, "Confidence Interval for", col="royalblue")
text(7.5,35, "Predicted values - 95%", col="royalblue")
EDIT 2
Topic moved to Cross Validated:
https://stats.stackexchange.com/questions/167907/deming-orthogonal-regression-measuring-goodness-of-fit

There are many proposed methods to calculate goodness of fit and tolerance intervals for Deming Regression but none of them widely accepted. The conventional methods we use for OLS regression may not make sense. This is an area of active research. I don't think there many R-packages which will help you compute that since not many mathematicians agree on any particular method. Most methods for calculating intervals are based on Resampling techniques.
However you can check out the 'mcr' package for intervals...
https://cran.r-project.org/web/packages/mcr/

How the length of averaged normal can be seen as a function of deviation of the angle?

Recently I read NVidia's Mipmapping_Normal_Maps
which says we can used the un-renormalized averaged normal to compute the standard deviation of the angle between averaged normal and sample normals.
By the first step, it assumes a Gaussian distribution of the angular deviation and give a figure (sorry but I cannot post an image as a new user, please refer to Figure_2 in that paper).
Then my question is, how the length of averaged normal is represented by a function of Standard Deviation of the angle(original function of Gaussian distribution, red curve in the figure)?

I believe the answer to your question is equation (1) in the paper. It shows how the averaged normal is equal to the reciprocal of 1 + sigma^2. Sigma is the standard deviation. Sometimes sigma^2 is called the variance.
At any rate, if you know the standard deviation, that's your value for sigma in the equations. Square it to get the variance, sigma^2.

Figure 2.5 in Elements of Statistical Learning

I met some difficulty in calculating the Bayes decision boundary of Figure 2.5. In the package ElemStatLearn, it already calcualted the probability at each point and used contour to draw the boundary. Can any one tell me how to calculate the probability? Thank you very much.
In traditional Bayes decision problem, the mixture distribution are usually normal distribution, but in this example, it uses two steps to generate the samples, so I have some difficulty in calculating the distribution.
Thank you very much.

Section 2.3.3 of ESL (accessible online) states how the data were generated. Each class is a mixture of 10 Gaussian distributions of equal covariance and each of the 10 means are drawn from another bivariate Gaussian, as specified in the text. To calculate the exact decision boundary of the simulation in Figure 2.5, you would need to know the particular 20 means (10 for each class) that were generated to produce the data but those values are not provided in the text.
However, you can generate a new pair of mixture models and calculate the probability for each of the two classes (BLUE & ORANGE) that you generate. Since each of the 10 distributions in a class are equally likely, the class-conditional probability p(x|BLUE) is just the average of the probabilities for each of the 10 distributions in the BLUE model.

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