I was seeking for solutions to inverse functions in R. The uniroot() indicates a possible to do this, however I don't know how.
My problem is quite simple, I guess. I have a nonlinear regression as trigonometric function:
function(d,x) d*sqrt(-0.9298239*(x-1)+0.0351246*sin(2.0737115*pi*x)+0.0037540*(1/tan(pi*x/2)))
I need to find the x value based on y value. How can I do that?
Edit: If I plot the curve using the expression (f(x)-y) I have something like the image below. Now, how could I extract the value that minimizes in zero?
minimization
Related
The autocorrelation command acf is slightly off the correct value. the command i am giving is acf(X,lag.max = 1) and the output is 0.881 while the same autocorrelation calculation done using the command cor(X[1:41],X[2:42]) gives the value 0.9452. I searched a lot to understand if there is any syntax I am missing or if the two computations have some fundamental difference but could not find suitable resources. Please help.
The equation for autocorrelation used by acf() and in all time series textbooks is
where
See https://otexts.com/fpp2/autocorrelation.html for example.
The equation for regular correlation is slightly different:
where
The simpler formula is better because it uses all available data in the denominator.
Note also that you might expect the top formula to include the multiplier $T/(T-k)$ given the different number of terms in the summations. But this is not included to ensure that the corresponding autocorrelation matrix is nonnegative definite.
I am trying to interpolate a series of data points using 2nd lagrange polinomial.
having
point1:(5;100)
point2: (9;17)
point3: (12;17)
and the formula
y=(x-x2)*(x-x3)/(x1-x2)*(x1-x3)*y1+
(x-x1)*(x-x3)/(x2-x1)*(x2-x3)*y2+
(x-x1)*(x-x2)/(x3-x1)*(x3-x2)*y3
It is obvious that a quadratic function might not fit the data.. It is just an example.
But i wonder why the value is surprisingly high for x=7.
If i am not wrong its y=1500.
Is the above formula correct?
answer:
In summary:
For the same x, you can't have two different y values; this violates the definition of a function.
you are missing brackets in your formula! Not (x-x2)*(x-x3)/(x1-x2)*(x1-x3), but ((x-x2)*(x-x3)) / ((x1-x2)*(x1-x3)).
back to 1>, note that the interpolation formula has x3-x2 in the denominator. If you have tied values, you will be dividing 0.
How can you make interpolation on such small data set? Yet you are asking for a quadratic interpolation!
follow-up:
1) fixed it. Accidentally i switched all the x and y values. So the points were in format (y,x).
Ah, haha, no wonder.
2) Thank you! The brackets improved the approximation. Regarding the missing brackets: I got the formula from the accepted answer here: Best way to find Quadratic Regression Curve in Java, but I don't understand this rule.
This is the famous, yet fundamental interpolation: Lagrange interpolation. You can verify its formula at Wolfram mathworld: Lagrange Interpolating Polynomial. I don't give you wikipedia link because this one looks more beautiful.
The link you found must contain a typo. Since you have suggest an edit to fix that, hopefully it will soon get approved.
3) It is a (significant larger (which answers your 4th question) time series. So it is impossible to have tied values.
Yes, time series won't have tied values.
formula should be correct.but when x=17,you have two different y value,it's might the cause of the trouble.you can try change anthor x.
I was using functional data analysis techniques to smooth a discrete dataset. I was using the fda package and the function smooth.basis. It gives me a fitted smooth object, yet now I need to get the inverse of it. The codes are something like the following:
basisobj<-create.bspline.basis(c(0,1),7)
fit=smooth.basis(argvals=Data$Time,y=Data$Score,fdParobj=basisobj)$fd
I wish to know what is the value of Data$Time when Data$Score is, say 20.
I did try the inverse.predict function but it could only give the result for lm function.
Thank you!
I am looking into time series data compression at the moment.
The idea is to fit a curve on a time series of n points so that the maximum deviation on any of the points is not greater than a given threshold. In other words, none of the values that the curve takes at the points where the time series is defined, should be "further away" than a certain threshold from the actual values.
Till now I have found out how to do nonlinear regression using the least squares estimation method in R (nls function) and other languages, but I haven't found any packages that implement nonlinear regression with the L-infinity norm.
I have found literature on the subject:
http://www.jstor.org/discover/10.2307/2006101?uid=3737864&uid=2&uid=4&sid=21100693651721
or
http://www.dtic.mil/dtic/tr/fulltext/u2/a080454.pdf
I could try to implement this in R for instance, but I first looking to see if this hasn't already been done and that I could maybe reuse it.
I have found a solution that I don't believe to be "very scientific": I use nonlinear least squares regression to find the starting values of the parameters which I subsequently use as starting points in the R "optim" function that minimizes the maximum deviation of the curve from the actual points.
Any help would be appreciated. The idea is to be able to find out if this type of curve-fitting is possible on a given time series sequence and to determine the parameters that allow it.
I hope there are other people that have already encountered this problem out there and that could help me ;)
Thank you.
I have data points that represent a logarithmic function.
Is there an approach where I can just estimate the function that describes this data using R?
Thanks.
I assume that you mean that you have vectors y and x and you try do fit a function y(x)=Alog(x).
First of all, fitting log is a bad idea, because it doesn't behave well. Luckily we have x(y)=exp(y/A), so we can fit an exponential function which is much more convenient. We can do it using nonlinear least squares:
nls(x~exp(y/A),start=list(A=1.),algorithm="port")
where start is an initial guess for A. This approach is a numerical optimization, so it may fail.
The more stable way is to transform it to a linear function, log(x(y))=y/A and fit a straight line using lm:
lm(log(x)~y)
If I understand right you want to estimate a function given some (x,y) values of it. If yes check the following links.
Read about this:
http://en.wikipedia.org/wiki/Spline_%28mathematics%29
http://en.wikipedia.org/wiki/Polynomial_interpolation
http://en.wikipedia.org/wiki/Newton_polynomial
http://en.wikipedia.org/wiki/Lagrange_polynomial
Googled it:
http://www.stat.wisc.edu/~xie/smooth_spline_tutorial.html
http://stat.ethz.ch/R-manual/R-patched/library/stats/html/smooth.spline.html
http://www.image.ucar.edu/GSP/Software/Fields/Help/splint.html
I never used R so I am not sure if that works or not, but if you have Matlab i can explain you more.