I have a problem, while solving some equations in Gams because there is another expression of so called error function then in my problem..
Error function which I am working with is: $$erf(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}dt$$
And in Gams we have similar function: $$errorf(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-\frac{1}{2}t^{2}}dt$$
If I want to express one function by the other one I have an expression:
$$errorf(x)=\frac{1+erf(\frac{x}{\sqrt{2 }})}{2}$$
$$erf(\frac{x}{\sqrt{2}})=2\cdot errorf(x)-1$$ Well I need to express $$erf(x)=..??$$
Can anyone give me some advices please how to solve this problem?
I tried the aproximation of function $$erf(x)=tanh(\sqrt\pi \cdot log(2)\cdot x))$$
which is great aproximation but I would like to solve this in another way.
The other problem comes in because I need to use this error function in transcendental equation $$x\cdot e^{x^{2}}\;\cdot erf(x )=konst$$ and in equation: $$T_{l}=T_{0}+({T_{m}-T_{0}})\frac{erf\left [ \frac{y}{2(\alpha _{l}t)^{\frac{1}{2}}} \right ]}{erf(x )}$$
$\endgroup$1 Answer
$\begingroup$What Gams is calling errorf is just the cumulative distribution function of the standard normal distribution, often denoted $\Phi(x)$. As you indicated, one can write
$$\text{errorf(x)} = \Phi(x) = \frac{1+\text{erf}(\frac{x}{\sqrt{2}})}{2}$$
You can solve for $\text{erf}(x)$ as a function of $\Phi(x)$ from this expression by defining $y=\frac{x}{\sqrt{2}}$. Then
$$\text{erf}(y) = 2\Phi(y\sqrt{2})-1 = 2\cdot \text{errorf}(y\sqrt{2})-1$$
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