I found this equation in some textbook.
Then i asked myself if there is an easy way to solve this with R.
Of course, there is.
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
z = seq(-10, 10, by = 1) i<-10 y <- z[i] p = function( x, y ) { x^3 + y*x^2 - 2*y^2 - 10*y } x = seq(-5, 5, by = 0.001) plot(x, p(x,y), type = 'l', col = 'red', ylim = c(-150, 150)) # adding x-axis and y-axis abline(h = 0, v = 0, col = 'blue') |
|
1 2 3 4 5 6 7 |
# using polyroot() to find the solution(s) result <- polyroot( c( (- 2*y^2 - 10*y), 0, y, 1 ) ) i.result <- round( Im( result ), 2 ) which( i.result == 0 ) r.result <- Re(result)[which( i.result == 0 )] points(x = r.result, y = rep(0, length( r.result )), pch = 16, cex = 2, col = 'green') |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 |
z = seq(-100, 100, by = 0.1) #z <- -3 : 3 #z = seq(-10, 10, by = 0.001) #z = seq(-3, 50, by = 0.1) r_solution <- NULL for( i in 1:length(z) ){ y <- z[i] p = function( x, y ) { x^3 + y*x^2 - 2*y^2 - 10*y } result <- polyroot( c( (- 2*y^2 - 10*y), 0, y, 1 ) ) i.result <- round( Im( result ), 2 ) which( i.result == 0 ) r.result <- Re(result)[which( i.result == 0 )] r_solution <- data.frame( rbind( r_solution, cbind( rep( y, length(r.result) ), r.result ) ) ) } r_solution plot( r_solution[,2], r_solution[,1], type = "p", pch = 1, cex = 0.1, col = 'black' ) abline(h = 0, v = 0, col = 'blue') |
