I'm looking for a concave down increasing-function, see the image in the right lower corner.
Basically I need a function f(x) which will rise slower as x is increasing. The x will be in range of [0.10 .. 10], so f(2x) < 2*f(x) is true. Also if
I would also like to have some constants which can change the way/speed the function is concaving.
Any suggestion on a formula?
5 Answers
$\begingroup$If you are restricted to a positive $x$, you could use
$$ f(x) = a \sqrt{x}, \quad a > 0 $$
which satisfies
$$ f(2 \cdot x) = a \sqrt{2 \cdot x} = \sqrt{2} \cdot a \sqrt{x} = \sqrt{2} \cdot f(x) < 2 \cdot f(x) $$
$\endgroup$ $\begingroup$\begin{equation} f(x) = -\frac{a}{x^b} \end{equation} where $a$ is a positive number and $b>0$.
$\endgroup$ $\begingroup$I think natural logarithm is a good function for your purpose. It's increasing and have low-rate slope, lower than your diagram. Let $f(x)=a+b\ln x$, then $f(2x)<2f(x)$ concludes that $\displaystyle\ln\frac{2}{x}<\frac{a}{b}$. In other hand for $0.1<x<10$ gives us $$-1.6<\ln\frac{2}{x}<3$$ so we must set $\displaystyle3<\frac{a}{b}$ and we have $0<3b<a$.
For example set $a=4$ and $b=1$ which satisfy our condition, so $f(x)=4+\ln x$ has the condition $f(2x)<2f(x)$ with below diagram in $[0.1,10]$:
Below graph shows $f(2x)<2f(x)$ in $[0.1,10]$:
$\log(x)$ comes to mind, but we also need to consider $f(2x)\lt2f(x)$. If we set $f(x)=\log(20x)$, then, since $x\gt0.1$, $$ \begin{align} f(2x) &=\log(40x)\\ &=\log(2)+\log(20x)\\ &=\log(2)+f(x)\\ &\lt f(x)+f(x)\\ &=2f(x) \end{align} $$
$\endgroup$ $\begingroup$Constant $a$ in
$$ 1- e^{-x/a}$$
changes climb rate as desired. ( Twice the function value more than function with double argument).
