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.
Related
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 am looking to calculate the indefinite integral of an equation.
I have data from an accelerometer feed into R through a visual C program, and from there it was simple enough to come up with an equation to represent the acceleration curve. That is all well in good, however i need to calculate the impact velocity as well. From my understanding from the good ol' highschool days, the indefinite integral of my acceleration curve will yield the the equation for the velocity.
I know it is easy enough to perform numerical integration with the integrate() function, is there anything which is comparable for an indefinite integral?
library(Ryacas)
x <- Sym("x")
Integrate(sin(x), x)
gives
expression(-cos(x))
An alternative way:
yacas("Integrate(x)Sin(x)")
You can find the function reference here
If the NA's you mention are informative in the sense of indicating no acceleration input then they should be replace by zeros. Let's assume you have the data in acc.vec and the device recorded at a rate of rec_per_sec:
acc.vec[is.na(ac.vec)] <- 0
vel.vec <- cumsum(acc.vec)/recs_per_sec
I do not think constructing a best fit curve is going to improve your accuracy in this instance. To plot velocity versus time:
plot(1:length(acc.vec)/recs_per_sec, vel.vec,
xlab="Seconds", ylab="Integrated Acceleration = Velocity")
As Ben said, try the Ryacas package for calculating the antiderivative of a function. But you probably should ask yourself whether you really want to generate a continuous function which only approximates your data in the first place (fitting errors). I'd stick with numerical integration of your actual data. Keep in mind the uncertainty in each data point, of course.
Just a warning, I started using R a day ago...my apologies if anything seems idiotically simple.
Right now im trying to have R take in a .txt file with acelerometer data of an impact and calculate a Head injury criterion test for it. The HIC test requires that curve from the data be integrated on a certain interval.
The equation is at the link below...i tried to insert it here as an image but it would not let me. Apparently i need some reputation points before it'll let me do that.
a(t) is the aceleration curve.
So far i have not had an issue generating a suitable curve in R to match the data. The loess function worked quite well, and is exactly the kind of thing i was looking for...i just have no idea how to integrate it. As far as i can tell, loess is a non-parametric regression so there is no way to determine the equation of the curve iteslf. Is there a way to integrate it though?
If not, is there another way to acomplish this task using a different function?
Any help or insighful comments would be very much appreciated.
Thanks in advance,
Wes
One more question though James, how can i just get the number without the text and error when using the integrate() function?
You can use the predict function on your loess model to create a function to use with integrate.
# using the inbuilt dataset "pressure"
plot(pressure,type="l")
# create loess object and prediction function
l <- loess(pressure~temperature,pressure)
f <- function(x) predict(l,newdata=x)
# perform integration
integrate(f,0,360)
40176.5 with absolute error < 4.6
And to extract the value alone:
integrate(f,0,360)$value
[1] 40176.5
I'm creating a game where players can make an alloy. To make it less predictable and more interesting, I thought that the durability and hardness of an alloy should not be calculated by a simple formula, because it will be extremely easy to find extrema, where alloy have best statistics.
So the questions is, is there any formula for a function where extrema can be found only by investigating all points? Input values will be in percents: 0.0%-100.0%. I think it should look like this: half sound wave
A very simple way would be a couple of sin function, just vary the constants and the sign for each new player. Here is one example (sin(1.1*x) + sin(x) + sin(0.9 *x))^2
If you use this between 10pi and 20pi you have an by average increasing function with local minima.
Modulating a simple linear or exponential function with trigonometric functions whose frequency and amplitude are dependent on the input should get you what you want.
You don't need a formula, I think — throw a bunch of random values around your domain, and then interpolate (linear interpolation will do) between them. Then you can even change the "formula" completely each time the game is run, or once in a while, or change it slowly with time, etc, etc.
If you want something that is very hard to predict then I would suggest involving a random number generator with the same seed every time. You can use it as an envelope for whatever function you come up with (trig functions or what not) to make it more jagged.
An interesting formula to use would be that of gamma of the Black-Scholes options pricing model. It goes as follows:
You can easily replace the variables, here's a graph of how the function looks:
alt text http://www.sqbimmer.com/aalex/gamma.png
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