Compute the inverse of function

Question

I want to compute and plot the inverse of given function f. I have the following code in R : ``` ############## Parameters ###### r1 <- 0.0125 r2 <- 0.0305 S1 <- 0.0400 S2 <- 0.0900 s1 <- sqrt(S1) s2 <- sqrt(S2) rho <- -0.45 ############## function f ########### f <- function(u) ((((u-r2)/(r1-r2))^2)*((s1)^2)) + (((1-((u-r2)/(r1-r2)))^2)*((s2)^2)) + (2*(((u-r2)/(r1-r2)))*(1-((u-r2)/(r1-r2)))*rho*s1*s2) f(r1) #### the values is equal to 0.04 ``` The problem is how to invert this function, I try to use the function uniroot but I was not able to solve my problem. this was the code that I used : ``` ######## First I define the function inverse : inverse = function (f, lower = 0, upper = 0.035) { function (y) uniroot((function (x) f(x) - y), lower = lower, upper = upper)[1]} ######## second I define the inverse of f using inverse f_inverse = inverse(function (u) f(u), 0.0, 0.035) ``` Thanks for your help,

Accepted Answer

So you are trying to invert a function that is not bijective. Look at `curve(f, 0, 0.035)` and `abline(h=0.04, col="red")` to see that. If you give `uniroot` proper bounds, the following will work:

```
f_inverse <- function(y, lower=0.0, upper=0.02) 
   uniroot((function (x) f(x) - y), lower = lower, upper = upper)$root

f_inverse(0.04)
## [1] 0.01250961

f_inverse(0.04, 0.02, 0.04) # to get the other root...
## [1] 0.02561455
```

EDIT: Careful, the 0.02 was really just a guess. To find the actual value, use `optimize(f, lower=0, upper=1)`. Then you can also plot the "inverse" function:

```
optim <- optimize(f, lower=0, upper=1)
seq1 <- seq(f(0), optim$objective, length=100)
inv1 <- sapply(seq1, f_inverse, lower=0, upper=optim$minimum)
seq2 <- seq(optim$objective, f(0.04), length=100)
inv2 <- sapply(seq2, f_inverse, lower=optim$minimum, upper=1)
plot(c(seq1, seq2), c(inv1, inv2), type="l")
```

On the other hand, this doesn't seem to have any advantages over `curve(f, 0, .04)`, which is much easier.

Compute the inverse of function

Solution

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