
Inverse function
I have been working on a project where I have a function F(x) which is closely related to a function G(x). The function G(x) has been well studied already and tools exist for using it. To make the equations I derive for F(x) useful I wanted to convert them into G(x) so that people can evaluate things for F(x) using the tools that already exist for G(x)
So far I derived their relationship
$\displaystyle r\cdot F(x)+ c= G(x)$
So for example if I had this equation to find an important number B
$\displaystyle B= F(5)F(y)$
I could turn it into
$\displaystyle B= \frac{G(5)c}{r}\frac{G(y)c}{r}$
$\displaystyle B= \frac{G(5)G(y)}{r}$
And use the tools that exist for G(x) to find B for any y
But I am stuck when it comes to their inverse functions. As an example of this I have derived the equation
$\displaystyle L=F^{1}(y0.5)$
I am confused about how I can use the relationship $\displaystyle r\cdot F(x)+ c= G(x)$ to get L in terms of the inverse function of $\displaystyle G$. I am not even sure if this is possible to do

Re: Inverse function
Hey Shakarri.
One method I think you should investigate is the rule of the derivative of an inverse function (in terms of its noninverse) and the chain rule of differentiation.
If you combine those together you should get a Taylor series for your function and if its in a simple enough form (or recognizable form), then you can substitute the appropriate elementary functions in.

Re: Inverse function
Hi !
if I well understand, you previously compute the values of c and r so that (G(x)c)/r be a good approximation for F(x).
Then, you want to compute x so that F(x)=y , given y. (i.e. the inverse function of F)
G(x)=c+r*F(x) = c+r*y
Hence : x = value of the inverse function of G, for the argument X=(c+r*y)
That is : x = F^(1){y} = G^(1){X} where X=c+r*y

Re: Inverse function
Ah thank you, I was wondering about that solution but I was really unsure about inverse functions. Anyway now you have derived it and it works with all the cases which I can think of with trivial solutions so it must be correct :)