R-为功能参数的笛卡尔积求根

时间:2018-10-06 16:49:27

标签: r algorithm mathematical-optimization

给出x的函数f(x,c,d),它也取决于某些参数c和d。我想找到参数的某些值c_1,...,c_n和d_1,...,d_m的笛卡尔积的零,即x_ij,使得i的f(x_ij,c_i,d_j)= 0 = 1,...,n和j = 1,...,m。尽管不是那么关键,但我正在将Newton-Raphson算法应用于求根:

newton.raphson <- function(f, a, b, tol = 1e-5, n = 1000){
                  require(numDeriv) # Package for computing f'(x)
                  x0 <- a # Set start value to supplied lower bound
                  k <- n # Initialize for iteration results
                  # Check the upper and lower bounds to see if approximations result in 0
                  fa <- f(a)
                  if (fa == 0.0){
                      return(a)
                   }
                  fb <- f(b)
                  if (fb == 0.0) {
                     return(b)
                  }
                  for (i in 1:n) {
                    dx <- genD(func = f, x = x0)$D[1] # First-order derivative f'(x0)
                    x1 <- x0 - (f(x0) / dx) # Calculate next value x1
                    k[i] <- x1 # Store x1
                    # Once the difference between x0 and x1 becomes sufficiently small, output the results.
                    if (abs(x1 - x0) < tol) {
                          root.approx <- tail(k, n=1)
                          res <- list('root approximation' = root.approx, 'iterations' = k)
                          return(res)
                     }
                     # If Newton-Raphson has not yet reached convergence set x1 as x0 and continue
                     x0 <- x1
                   }
                     print('Too many iterations in method')
                   }

我感兴趣的实际功能更加复杂,但是以下示例说明了我的问题。

  test.function <- function(x=1,c=1,d=1){
                   return(c*d-x)
                   }

然后对于任何给定的c_i和d_j,我都可以轻松地通过以下方式计算零

  newton.raphson(function(x) test.function(x,c=c_i,d=d_j),0,1)[1]

这显然是乘积c_i * d_j。 现在,我试图定义一个函数,该函数为两个给定向量(c_1,...,c_n)和(d_1,...,d_m)查找所有组合的零。为此,我尝试定义

zeroes <- function(ci=1,dj=1){
           x<-newton.raphson(function(x) test.function(x,c=ci,d=dj),0,1)[1]
           return(as.numeric(x))
          }

然后使用外部功能,例如

 outer(c(1,2),c(1,2,3),FUN=zeroes)

不幸的是,这没有用。我收到错误消息

  Error during wrapup: dims [product 6] do not match the length of object [1]

也许对于我的问题还有更好的解决方案。我很高兴有任何投入。

0 个答案:

没有答案