I want to create a curve with the above properties, and which is odd in that it is continuous, and produces a negative version of itself for negative $x$. A picture may help give the general idea (but please note it's not accurate):
This is created by shifting the normal (probability) curve to centre on $x=1$, then subtracting another normal curve shifted to centre on $x=-1$, using this basic formula for a normal curve:
$$\frac{e^\frac{-x^2}{2}}{\sqrt (2\pi)}$$
I then stretched the curve vertically by $\frac{\sqrt (2\pi)}{1-\frac{1}{e^2}}$ to give the curve above.
Unfortunately, it's not quite what I am looking for.
- I want maxima and minima at $x=1$ and $x=-1$
- I want the value of the function at those points to be $1$ and $-1$
- I want the decay after that point to follow (at least very roughly) $\frac{1}{x}$ - i.e., far less rapid
Any suggestions?
The simplest function available that satisfies the given properties is a rational function: $$\frac{2x}{1+x^2}$$ That it is odd (and so has $f(0)=0$) comes from the $x$ in the numerator. That it decays at the rate of $\frac1x$ comes from comparing the degrees of the polynomials above and below (1 vs. 2). That it has a maximum of $f(1)=1$ can be verified by differentiating the function.