在 R 中绘制一个困难的函数并通过二等分找到根

Question

eq <- function(x){ sum(exp(-Ln.M_Return[,1]/x)) / 168 - 1 } 
# Ln.M_Return[,1] is a vector of stock return, and its length is 168.

My first goal is to draw the plot for this function. However, there is always an error here. When I use "curve" function,

curve(eq(x))

it will appear an error message:

Error in curve(eq(x)) : 'expr'
In addition: Warning message:
In -Ln.M_Return[, 1]/x :
  longer object length is not a multiple of shorter object length

I have tried some other methods, like plot() and xyplot(), but nothing changed.

My main purpose is actually using bisection method to find the root of the function, which is the formula of Aumann-Serrano riskiness index. Thus, I choose to draw the plot first and approximate the location of the root. Then I will use some codes of bisection to find the root. Thanks for reading my question patiently!

Answer 1

Without having your Ln.M_Return , I created some random data. You should Vectorize your equation in the curve function like this:

Ln.M_Return <- runif(168,0,10)
Ln.M_Return <- as.data.frame(Ln.M_Return)

eq <- function(x){ sum(exp(-Ln.M_Return[,1]/x)) / 168 - 1 } 

g <- Vectorize(eq)

curve(g, from=1, to=100, , xlab="xvalue", ylab="yvalue", 
      col="blue", lwd=2)

Output: