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I messed up the equation last time I asked this - Can someone please solve this function for X?

$Y = \displaystyle 0.5\:a\:\text{erf}\left(\frac{x-b}{c\sqrt{2}}+.5\right)+d$

When I solve for Y with x=1 and

a =       1.412 b =       1.259 c =       1.003 d =      0.3016  

I get 0.5460

I want to plug 0.5460 into a formula and get 1 back.

I am aware that there exists a function called "inverse error function" (erfinv) where erfinv(erf(x)) = x but I still can't figure this one out.

  • 0
    possible duplic$a$te of [How can $I$ solve this e$q$uation (contains error function) ?](http://math.stac$k$exchange.com/questions/38524/how-can-i-solve-this-equation-contains-error-function)2011-05-12

2 Answers 2

1

As in the previous question, you just "unpack" it. $Y-d=0.5a\text{ erf} \left(\frac{x-b}{c\sqrt{2}}+.5\right)$
$2\frac{Y-d}{a}=\text{erf}\left(\frac{x-b}{c\sqrt{2}}+.5\right)$
$\text{erf}^{-1}\left(2\frac{Y-d}{a}\right)-.5=\frac{x-b}{c\sqrt{2}}$
$c\sqrt{2}\left[\text{erf}^{-1}\left(2\frac{Y-d}{a}\right)-.5\right]+b=x$

3

I do not know how familiar you are with basic algebra, so I will do it step by step. Please feel free to hurry the process!

$1$. Subtract $d$ from both sides. You should get

$Y-d=(0.5a)\text{erf}\left(\frac{x-b}{c\sqrt{2}}+ 0.5\right)$

$2$. Divide both sides by $0.5a$. You should get

$\frac{Y-d}{0.5a}=\text{erf}\left(\frac{x-b}{c\sqrt{2}}+ 0.5\right)$

Now for brevity let

$w=\frac{Y-d}{0.5a}$

So we have

$w=\text{erf}\left(\frac{x-b}{c\sqrt{2}}+ 0.5\right)$

$3$. Now apply $\text{erfinv}$ to both sides. This is where we use the fact that $\text{erfinv}(\text{erf}(u))=u$. We get

$\text{erfinv}(w)= \left(\frac{x-b}{c\sqrt{2}}+ 0.5\right)$

$4$. Subtract $0.5$ from both sides. We get

$\text{erfinv}(w)-0.5= \left(\frac{x-b}{c\sqrt{2}}\right)$

$5$. Multiply both sides by $c\sqrt{2}$. We get

$c\sqrt{2}(\text{erfinv}(w)-0.5)= {x-b}$

$6$. Finally, add $b$ to both sides, and because I like $x$ on the left of the $=$ sign, interchange the two sides. We get

$x= c\sqrt{2}(\text{erfinv}(w)-0.5) +b$