How do I solve a function for a given x?

Question

I have been browsing through quite some help pages, but I did not find the solution for my - probably - simple problem. I defined a function

funB <- function(x) (0.8042851 + 
    ((3.9417843-0.8042851)/(1+((x/0.4039609)^(-3.285016))))) 

and would like to solve it for a given x (say, x = 0.2). How do I do that? I have looked at uniroot() and polyroot() , but they did not seem to fit my function.

Answer 1

If you wanted to find the value of x such that funB(x) was equal to 0.2, you would do something like this:

funB <- function(x) (0.8042851 + 
    ((3.9417843-0.8042851)/(1+((x/0.4039609)^(-3.285016))))) 
target <- 0.2
uniroot(function(x) funB(x)-target, interval=c(-5,10))

but there's a problem. It's up to you to pick an interval value that brackets a root (ie funB(x)<0.2 for the lower value and >0.2 for the upper value, or vice versa. funB is NaN for x<0, 0.8042851 for x==0, and increasing for x>0 (try curve(funB, from=-5, to=100, n=1001) for example). So the solution you want (if I've guessed right about the meaning of your question) doesn't seem to exist.

note : in general a negative value raised to a negative power is NaN in R (even in cases where the answer "should" be defined, eg (-8)^(1/3) is the cube root of -8, which is -2...). If you're sure you know what you're doing you could replace (x/a)^b with sign(x)*(abs(x)/a)^b) ... (if you make this change, the function appears well-behaved for x>-0.4 and funB(x)-0.2 does have a root between -0.3 and -0.2... but I have no idea if this makes sense for your application or not )

Answer 2

Just to be sure that there is a root where you are expecting it, plot the graph of funB .

curve(funB)

Define an auxiliary function, f , taking an extra argument and solve this new function for a = <target_value> .

f <- function(x, a) funB(x) - a

uniroot(f, interval = c(0, 1e3), a = 2)

#$root
#[1] 0.3485097
#
#$f.root
#[1] -0.0001305644
#
#$iter
#[1] 12
#
#$init.it
#[1] NA
#
#$estim.prec
#[1] 6.103516e-05

Answer 3

Well, I guess I must like doing things the hard way. I just rearranged your function to find its inverse:

funC <- function(y) (((3.137499)/(y - 0.8042851) - 1)^(-1/3.285016)) * 0.4039609

So if I want to know when funB(x) == 3.7 I can do:

funC(3.7)
#> [1] 0.860193

and sure enough

funB(0.860193)
#> [1] 3.7

or indeed

funB(funC(1))
#> [1] 1

And as others have pointed out, x doesn't have a real value at funB(x) == 0.2 as you can see in this plot:

curve(funC, 0, 4)

Now, if you really want to know the complex root where funB(x) == 0.2 then you can modify funC like so:

funC <- function(y) (((3.137499)/(as.complex(y) - 0.8042851) - 1)^(-1/3.285016)) * 0.4039609

So now:

funC(0.2)
#> [1] 0.1336917+0.1894797i

And therefore the answer to your question is 0.1336917 +/- 0.1894797i

funB(complex(real = 0.133691691, imaginary = 0.1894797))
[1] 0.1999996+0i

Close enough.

Answer 4

funB <- function(x) (0.8042851 + ((3.9417843-0.8042851)/(1+((x/0.4039609)^(-3.285016)))))

# call the function with desired input
funB(0.2)

...and the output:

> funB(0.2)
[1] 1.087758
>