How to get solution of following polynomial?

Hi, I want find all roots for the following polynomial :

a <- c(-0.07, 0.17); b <- c(1, -4); cc <- matrix(c(0.24, 0.00, -0.08,
-0.31), 2); d <- matrix(c(0, 0, -0.13, -0.37), 2); e <- matrix(c(0.2, 0,
-0.06, -0.34), 2)
A1 <- diag(2) + a %*% t(b) + cc; A2 <- -cc + d; A3 <- -d + e; A4 <- -e
fn <- function(z)
  {
   y <- diag(2) - A1*z - A2*z^2 - A3*z^3 - A4*z^4
   return(det(y))
  }; uniroot(fn, c(-10, 1))

Using uniroot function, I got only one solution of that. Is there any
function to get all four solutions? I looked at polyroot() function, but I
do not think it will work for my problem, because, my coef. are matrix, nor
number

On Sun, Jan 11, 2009 at 6:03 AM, RON70 <ron_michael70 at yahoo.com> wrote:

Hi, I want find all roots for the following polynomial :

a <- c(-0.07, 0.17); b <- c(1, -4); cc <- matrix(c(0.24, 0.00, -0.08,
-0.31), 2); d <- matrix(c(0, 0, -0.13, -0.37), 2); e <- matrix(c(0.2, 0,
-0.06, -0.34), 2)
A1 <- diag(2) + a %*% t(b) + cc; A2 <- -cc + d; A3 <- -d + e; A4 <- -e
fn <- function(z)
  {
   y <- diag(2) - A1*z - A2*z^2 - A3*z^3 - A4*z^4
   return(det(y))
  }; uniroot(fn, c(-10, 1))

Using uniroot function, I got only one solution of that. Is there any
function to get all four solutions? I looked at polyroot() function, but I
do not think it will work for my problem, because, my coef. are matrix, nor
number

Use curve to plot the curve of your function. Then, see where the
roots are, and use uniroot with a small interval around the roots to
determine their exact value.

Example:

f <- function(x) x^2-1
curve(f,-5,5)
uniroot(f,c(-2,-0.2))
uniroot(f,c(0.2,2))

Paul

1 - 1.18*x + 0.2777*x^2 - 0.2941*x^3 - 0.1004*x^4 + 0.3664*x^5 - 0.0636*x^6 + 0.062*x^7 - 0.068*x^8 

 Hi, I want find all roots for the following polynomial :
 
 a <- c(-0.07, 0.17); b <- c(1, -4); cc <- matrix(c(0.24, 0.00, -0.08,
 -0.31), 2); d <- matrix(c(0, 0, -0.13, -0.37), 2); e <- matrix(c(0.2, 
0,
 -0.06, -0.34), 2)
 A1 <- diag(2) + a %*% t(b) + cc; A2 <- -cc + d; A3 <- -d + e; A4 <- -e
 fn <- function(z)
    {
     y <- diag(2) - A1*z - A2*z^2 - A3*z^3 - A4*z^4
     return(det(y))
    }; uniroot(fn, c(-10, 1))
 
 Using uniroot function, I got only one solution of that. Is there any
 function to get all four solutions? I looked at polyroot() function, 
but I
 do not think it will work for my problem, because, my coef. are 
matrix, nor
 number
 
 Thanks
    
 
 
 -- 
 View this message in context: 
 Sent from the R help mailing list archive at Nabble.com.
 

Hi,

You can use the "polynomial" package to solve your problem.

The key step is to find the exact polynomial representation of fn(). 
Noting that it is a 8-th degree polynomial, we can get its exact form
using the poly.calc() function.  Once we have that, it is a simple matter
of finding the roots using the solve() function.

require(polynomial)

a <- c(-0.07, 0.17)
b <- c(1, -4)
cc <- matrix(c(0.24, 0.00, -0.08, -0.31), 2)
d <- matrix(c(0, 0, -0.13, -0.37), 2)
e <- matrix(c(0.2, 0, -0.06, -0.34), 2)
A1 <- diag(2) + a %*% t(b) + cc
A2 <- -cc + d
A3 <- -d + e
A4 <- -e

# I am vectorizing your function
fn <- function(z)
   {
    sapply(z, function(z) {
	y <- diag(2) - A1*z - A2*z^2 - A3*z^3 - A4*z^4
    	det(y)
	})
   }


x <- seq(-5, 5, length=9) # note we need 9 points to exactly determine a
8-th degree polynomial
y <- fn(x)

p <- poly.calc(x, y)  # uses Lagrange interpolation to determine
polynomial form
p

1 - 1.18*x + 0.2777*x^2 - 0.2941*x^3 - 0.1004*x^4 + 0.3664*x^5 -
0.0636*x^6 + 0.062*x^7 - 0.068*x^8 

# plot showing that p is the exact polynomial representation of fn(z)
pfunc <- as.function(p)
x1 <-seq(-5, 5, length=100)
plot(x1, fn(x1),type="l")
lines(x1, pfunc(x1), col=2, lty=2)   

solve(p)  # gives you the roots (some are, of course, complex)


Hope this helps,
Ravi.

Hi Ravi, Thanks for this reply. However I could not understand meaning of
"vectorizing the function". Can you please be little bit elaborate on that?
Secondly the package "polynomial" is not available in CRAN it seems. What is
the alternate package?

Thanks,

Ravi Varadhan wrote:

Hi,

You can use the "polynomial" package to solve your problem.

The key step is to find the exact polynomial representation of fn(). 
Noting that it is a 8-th degree polynomial, we can get its exact form
using the poly.calc() function.  Once we have that, it is a simple matter
of finding the roots using the solve() function.

require(polynomial)

a <- c(-0.07, 0.17)
b <- c(1, -4)
cc <- matrix(c(0.24, 0.00, -0.08, -0.31), 2)
d <- matrix(c(0, 0, -0.13, -0.37), 2)
e <- matrix(c(0.2, 0, -0.06, -0.34), 2)
A1 <- diag(2) + a %*% t(b) + cc
A2 <- -cc + d
A3 <- -d + e
A4 <- -e

# I am vectorizing your function
fn <- function(z)
   {
    sapply(z, function(z) {
	y <- diag(2) - A1*z - A2*z^2 - A3*z^3 - A4*z^4
    	det(y)
	})
   }


x <- seq(-5, 5, length=9) # note we need 9 points to exactly determine a
8-th degree polynomial
y <- fn(x)

p <- poly.calc(x, y)  # uses Lagrange interpolation to determine
polynomial form
p

1 - 1.18*x + 0.2777*x^2 - 0.2941*x^3 - 0.1004*x^4 + 0.3664*x^5 -
0.0636*x^6 + 0.062*x^7 - 0.068*x^8 

# plot showing that p is the exact polynomial representation of fn(z)
pfunc <- as.function(p)
x1 <-seq(-5, 5, length=100)
plot(x1, fn(x1),type="l")
lines(x1, pfunc(x1), col=2, lty=2)   

solve(p)  # gives you the roots (some are, of course, complex)


Hope this helps,
Ravi.

Hi Ravi, Thanks for this reply. However I could not understand meaning 
of "vectorizing the function". Can you please be little bit elaborate on

Secondly the package "polynomial" is not available in CRAN it seems. 
What is the alternate package?

Thanks,

Ravi Varadhan wrote:

Hi,

You can use the "polynomial" package to solve your problem.

The key step is to find the exact polynomial representation of fn(). 
Noting that it is a 8-th degree polynomial, we can get its exact 
form using the poly.calc() function.  Once we have that, it is a 
simple matter of finding the roots using the solve() function.

require(polynomial)

a <- c(-0.07, 0.17)
b <- c(1, -4)
cc <- matrix(c(0.24, 0.00, -0.08, -0.31), 2) d <- matrix(c(0, 0, 
-0.13, -0.37), 2) e <- matrix(c(0.2, 0, -0.06, -0.34), 2)
A1 <- diag(2) + a %*% t(b) + cc
A2 <- -cc + d
A3 <- -d + e
A4 <- -e

# I am vectorizing your function
fn <- function(z)
   {
    sapply(z, function(z) {
	y <- diag(2) - A1*z - A2*z^2 - A3*z^3 - A4*z^4
    	det(y)
	})
   }


x <- seq(-5, 5, length=9) # note we need 9 points to exactly 
determine a 8-th degree polynomial y <- fn(x)

p <- poly.calc(x, y)  # uses Lagrange interpolation to determine 
polynomial form p

1 - 1.18*x + 0.2777*x^2 - 0.2941*x^3 - 0.1004*x^4 + 0.3664*x^5 -
0.0636*x^6 + 0.062*x^7 - 0.068*x^8

# plot showing that p is the exact polynomial representation of 
fn(z) pfunc <- as.function(p)
x1 <-seq(-5, 5, length=100)
plot(x1, fn(x1),type="l")
lines(x1, pfunc(x1), col=2, lty=2)   

solve(p)  # gives you the roots (some are, of course, complex)


Hope this helps,
Ravi.