source("http://macosa.dima.unige.it/r.R") # If I have not already loaded the library ---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------S12Regression. Splines. Curve fitting.Linear, quadratic and cubicregression: the linear function with graph for a given point (p,q) and the generic linear, quadratic and cubic functions that "best" approximate some experimental data (x,y). Commands:regression(x,y,p,q),regression1(x,y),regression2(x,y),regression3(x,y)If the points are 2, 3 or 4 regression1, regression2 and regression3 find exactly the curve. We also see howsplines(approximating function graphs or curves) can be calculated.For a more complete method of studynglinear regressionsee here.If the "x" are too large [for example x=c(2007,2012,…)] an error message may appear. In this case you can modify x (x-2000) and find the regression rB (of x-2000), and then modify it [take r(x) = rB(x-2000)] x = c(-0.2, 1.3, 1.7, 3.2, 4.1, 4.6) y = c(9.1, 17.5, 20.7, 25.4, 35.7, 51.2) Plane(-1,6,-10,70); POINT(x,y,1)regression(x,y, 0,0); POINT(0,0,"blue")#9.829 * x f = function(x) 9.829 * x; graph(f,-1,6, "red")regression1(x,y)#7.628 * x + 7.911 g = function(x) 7.628 * x + 7.911; graph(g,-1,6, "blue")regression2(x,y)#1.37 * x^2 + 1.38 * x + 11.2 h = function(x) 1.37 * x^2 + 1.38 * x + 11.2; graph(h,-1,6, "violet")regression3(x,y)#1.21 * x^3 + -6.3 * x^2 + 11.9 * x + 11.6 k = function(x) 1.21 * x^3 + -6.3 * x^2 + 11.9 * x + 11.6; graph(k,-1,6, "green") POINT(0,0,"blue"); POINT(x,y,1)# On the right, the curve that goesexactlyfor (-1,-10),(0,-1),(1,-2),(2,5)Plane(-2,3,-20,20) POINT(-1,-10, 1); POINT(0,-1,"red"); POINT(1,-2,"green"); POINT(2,5,"blue") regression3( c(-1,0,1,2), c(-10,-1,-2,5) )#3 * x^3 + -5 * x^2 + 1 * x + -1 f = function(x) 3 * x^3 + -5 * x^2 + 1 * x + -1 graph(f,-2,3, 6) POINT(-1,-10, 1); POINT(0,-1,"red"); POINT(1,-2,"green"); POINT(2,5,"blue")# Thecubicsplinethat interpolates the points x,y (see):x = c(-0.2, 1.3, 1.7, 3.2, 4.1, 4.6) y = c(9.1, 17.5, 20.7, 25.4, 35.7, 51.2) Plane(-1,6,-10,70); POINT(x,y,1); L=spline(x,y,1000)# I calculated and put in L the 1000 points of the spline passing through the # points x,ypolyline(L$x,L$y,"red") # I traced it in red# The "extended" (blue) curve that passes by the same points, lowered by 10.POINT(x,y-10,1); L1=spline(x,y-10,1000, xmin=-1,xmax=5) polyline(L1$x,L1$y, "blue")# If I want to find a value of y I can do, for example: L=spline(x,y-10,1e6, xmin=-1,xmax=5);Yspline(0.5,L)#1.383594 POINT(0.5,1.383594, "red") #Yspline(0.5,L) [andXspline(0.5,L)] product the coordinate of the point withx=0.5 of # the spline L [I must use less the 1e8 points; in the example the number is 1e6] # If you have some points of a closed curve you can approximate it with the command #splineC(x,y,ColBorder,ColInside), orsplinCif you want a thin-stroke curve. If you # choose NULL as internal color the figure is transparent.BF=3; HF=3; PLANE(-2,8, -2,8) u=c( -1, 4, 7,6,3,0) v=c(-1.5,0,-1,6,7,5) splineC(u,v,"red",NULL); POINT(u,v,"seagreen") PLANE(-2,8, -2,8); splinC(u,v,"blue","yellow") # The points used to trace the curve are in splineCx(x,y) and splineCy(x,y) length(splineCx(u,v))#104 c( splineCx(u,v)[1],splineCy(u,v)[1] )#-1.0 -1.5 c( splineCx(u,v)[2],splineCy(u,v)[2] )#-0.8442124 -1.6377163 # If you want to have approximately theareaof the closed curve you can useareaPol:areaPol(splineCx(u,v),splineCy(u,v))#53.88609# If we want to approximate points with a curve y = A·exp(B·x) we can use regression1 # after transforming y-coordinates by the logarithmic function: see here # See here to see other methods ofcurve fitting(with the commandsF_RUN1,F_RUN2, #F_RUN3,F_RUN4). # # How to proceed for a regression of the second order when I have afixed point? # Without using specific methods, I can do the following: x = c(-0.2, 1.3, 1.7, 3.2, 4.1, 4.6) y = c(9.1, 17.5, 20.7, 25.4, 35.7, 51.2) Plane(-1,6,-10,70); POINT(x,y,"black") # Without a fixed point: regression2(x,y) # 1.37 * x^2 + 1.38 * x + 11.2 h = function(x) 1.37*x^2 + 1.38*x + 11.2; graph2(h,-1,6, "magenta") # If I have thefixed point (3,20)I can forcethe passage for alarge amount of points# (for example 1e6) that areequal to (3.20)! x1=c(rep(3,1e6),x); y1=c(rep(20,1e6),y) regression2(x1,y1) # 2.95 * x^2 + -6.54 * x + 13 h1 = function(x) 2.95 * x^2 + -6.54 * x + 13 graph2(h1,-2,7, "brown"); POINT(3,20, "red")##Another exampleof use of regression: # A car (initially) goes at a steady speed. The driver sees an obstacle and (after 1.5 # seconds) starts braking. Here are the photos taken every half second from the instant # in which he saw the obstacle. What was the initial speed of the car?# Times (in seconds) and spaces (in meters): s0 = c(1.5,2.0,2.5,3.0,3.5,4.0,4.5,5.0,5.5,6.0,6.5,7.0,7.5,8.0,8.5,9.0,9.5) m0 = c( 70, 92,114,133,151,168,183,198,209,221,230,238,244,250,253,255,257)# Since he has begun to brake:s = s0-1.5; m = m0-70; s; m # 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 # 0 22 44 63 81 98 113 128 139 151 160 168 174 180 183 185 187# The graph (time,space): Plane(-1/2,8.5, -1,190); POINT(s,m, "blue") abovex("sec"); abovey("m")# I want to impose that s(0)=0. I proceed as in the last example:s1 = c(rep(0,1e6), s); m1 = c(rep(0,1e6), m) regression2(s1,m1) # -2.89 * x^2 + 46.5 * x + -2.89e-07 # I neglect -2.89e-07 F = function(x) -2.89 * x^2 + 46.5 * x graph2(F,0,8, "brown") # See the chart above # I find the derivative (I can find it "by hand" but ...) deriv(F,"x") # 46.5 - 2.89 * (2 * x) # The initial speed is 46.5 m/s. 46.5/1000*60*60 # 167.4167 km/hOther examples of use