Looking for a specific differentiable function

Question

I'm from computer science and not a mathematician, so i hope you can help me by finding a function with the following properties:

In principle it is about the following function: \begin{equation} f(x)= \begin{cases} x^2,\; if \; x < 0 \\ x, \;if \; x > 0 \end{cases} \end{equation}

so something like this: 1 My problem is that i need a continuous and differentiable function, which doesn't have to be perfect, but should approximate this function. The obligatory properties are: \begin{equation} f(0) = 0 \end{equation} \begin{equation} \min f(x) = 0 \end{equation} \begin{equation} if \; x < 0 \rightarrow f(x) \approx x^2 \end{equation} \begin{equation} if \; x > 0 \rightarrow f(x) \approx x \end{equation} and function must be continuous and differentiable everywhere.

It would be best, of course, if there were any possibility to approximate such a function, regardless of which function I want to have on the left and which I want to have on the right side, of the y-axis, as long as it meets all the requirements described above. So, for example also two quadratic function but with different exponents on both sides. But i don't think that such an general solution is possible ...

Answer 1

Let $s_h(x): \mathbb{R} \to (0;1)$ be the sigmoid function with parameter $h$: $$ s_h(x) = \frac{1}{1 + e^{-hx+20}} $$ Note that $\lim_{x \to - \infty} s_h(x) = 0$ and $\lim_{x \to + \infty} s_h(x) = 1$ and that $f(x)$ can be extended to be continuous for $x=0$ by choosing $f(0)=0$.

A possible function $g_h$ can be: $$ g_h(x) = (1 - s_h(x)) x^2 + s_h(x) x $$

Larger values of the parameter $h$ will lead to a narrower sigmoid function, meaning a narrower transition between $x^2$ and $x$.

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Edit: I changed the formula for $s_h$ (and added a missing minus sign in there). This way, the derivative $g'(0)$ is close to zero.

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Edit2: I got a better one:

Let $r_h(x): \mathbb{R} \to (0;1)$ be the sigmoid function with parameter $h$: $$ r_h(x) = 1 + \frac{1}{1 + e^{hx}} $$ Note that $\lim_{x \to - \infty} r_h(x) = 2$ and $\lim_{x \to + \infty} r_h(x) = 1$.

A possible function $y_h$ can be: $$ y_h(x) = |x|^{r_h(x)} $$ that has a minimum for $x=0$, $y_h(0)=0$.

Answer 2

Both sides must have derivative $0$ at the origin to guarantee differentiability at the origin. I think that means there's no way to avoid a corner there if $f(x) \approx x$ on the right.

You could try $$ f(x)= \begin{cases} |x|^\alpha& x < 0 \\ x^\beta & x > 0 \end{cases} $$ for any $\alpha$, $\beta > 1$ .

Here is a picture with $\alpha = 2.5$, $\beta = 1.2$. The second graph shows a magnification near the origin. The more you enlarge it there the more you see that there is no corner.