Family of curves that transitions smoth from linear to logarithmic

Question

I am looking for a family of curves that smoothly transitions from

$f(x)=x$

to

$f(x)=log(x+1)$

Answer 1

In general, you can combine any two functions $f_0(x)$ and $f_1(x)$ using a third (interpolating) function $w(y)$: $$f(x, y) = \biggl(1 - w(y)\biggr) f_0(x) + w(y) f_1(x) \tag{1a}\label{1a}$$ such that $$\begin{aligned} f(x, 0) &= f_0(x) \\ f(x, 1) &= f_1(x) \\ \end{aligned}$$ and $f(x, y)$ for $0 \lt y \lt 1$ are "combinations" of the two functions. The interpolating function $w(y)$ defines how the two functions get mixed.

Any function that fulfills $w(0) = 0$, $w(1) = 1$, $0 \lt w(y) \lt 1$ for $0 \lt y \lt 1$ will work, but some common (symmetric) ones are $$\begin{aligned} w_1(y) &= y \\ w_3(y) &= 3 y^2 - 2 y^3 \\ w_5(y) &= 10 y^3 - 15 y^4 + 6 y^5 \\ w_\lambda(y) &= \begin{cases} 0, & y \le 0 \\ 2^{\lambda - 1} y^\lambda, & 0 \lt y \le \frac{1}{2} \\ 1 - 2^{\lambda - 1} (1 - y)^\lambda, & \frac{1}{2} \lt y \lt 1 \\ 1, & y \ge 1 \\ \end{cases} \end{aligned} \tag{1c}\label{1c}$$ If you want replacing $y$ with $1-y$ to behave as if one swapped $f_0$ and $f_1$, then you do want a symmetric function, one that for $0 \le y \le 1$ fulfills $w(y) + w(1-y) = 1$.

Note that outside $0 \le y \le 1$, we actually do extrapolation, which may or may not make any sense at all. Again, it depends on the behaviour of the intepolating/extrapolating function $w(y)$.

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For example, using linear interpolation, $w(y) = y$, and $f_0(x) = x$, $f_1(x) = \log(x + 1)$, we get $$f(x, y) = (1 - y) x + y \log(x + 1)$$ and $f(x, 0) = x$, $f(x, 1) = \log(x + 1)$, and something in between for $0 \lt y \lt 1$ using $f(x, y)$.

Answer 2

Let's say we have function $g_1(x)$ that applies for all $x \le x_1$, and another function $g_2(x)$ that applies for all $x \ge x_2$, with $x_1 \le x_2$, and we want to combine these into a single function $f(x)$.

As described by Jean Marie in a comment to the question, we could have $g_1(x) = x$, $g_2(x) = \log (x + 1)$, with $x_1 = x_2 = 0$.

(Whenever there exists an $x$ for which $g_1(x) = g_2(x)$, that can be used as the shared boundary.)

The important detail here is smoothness. When $g_1(x) = g_2(x)$, we have $\mathbb{C}^0$ continuity at $x$. If their derivatives also match, we have $\mathbb{C}^1$ continuity at $x$. Again, as Jean Marie commented, the above two functions derivatives ($d g_1 / d x = 1$, $d g_2 / d x = 1/(x+1)$) match at $x = 0$ but the second derivatives ($d^2 g_1 / d x^2 = 0$, $d^2 g_2 / d x^2 = -1 / (x+1)^2$) do not, so we have $\mathbb{C}^1$ continuity here.

The combined function is usually defined as a piecewise function: $$f(x) = \begin{cases} x, & x \lt 0 \\ \log(x + 1), & x \ge 0 \\ \end{cases}$$

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When $x_1 \lt x_2$, we can "fill in" the gap using a third function, $g_3(x)$, such that it has desired smoothness at both $x_1$ and $x_2$. Very commonly, these functions are simple polynomials, because a polynomial of degree $N$ has $N+1$ constants that can be set by fitting.

Basically, we define $g_3$ as an $N$'th degree polynomial, $$g_3(x) = \sum_{i=0}^{N} C_i x^i$$ noting that its first derivative is then $$\frac{d g_3(x)}{d x} = \sum_{i=1}^{N} i C_i x^{i-1}$$ and second derivative $$\frac{d^2 g_3(x)}{d x^2} = \sum_{i=2}^{N} i (i-1) C_i x^{i-2}$$ and so on. We choose $N$ so that we have the exact number ($N+1$) of unknowns $C_i$, matching the number of smoothness requirements at $x_1$ and $x_2$. This forms a system of $N+1$ equations, which can be solved (at least numerically) using linear algebra (or any computer algebra system, like Maxima/wxMaxima, Sagemath, Matlab, Octave, Mathematica, Maple, etc.).

For example, let's say we have $g_1(x) = \frac{1}{2} x$ for $x \le 1$, and $g_2(x) = x$, for $x \ge 2$, and we want to join them with $\mathbb{C}^1$ continuity. This means $$\left\lbrace \begin{aligned} g_3(1) &= g_1(1) = \frac{1}{2} \\ g_3(2) &= g_2(2) = 2 \\ \left.\frac{d g_3(x)}{d x} \right\rvert_{x=1} &= \left.\frac{d g_1(x)}{d x} \right\rvert_{x=1} = \frac{1}{2} \\ \left.\frac{d g_3(x)}{d x} \right\rvert_{x=2} &= \left.\frac{d g_2(x)}{d x} \right\rvert_{x=2} = 1 \\ \end{aligned} \right.$$ and we have four unknowns, so $g_3(x)$ will be of third degree (a cubic): $$g_3(x) = C_0 + C_1 x + C_2 x^2 + C_3 x^3, \quad \frac{d g_3(x)}{d x} = C_1 + 2 C_2 x + 3 C_3 x^2$$ Fitting in the values we have a system of four equations and four unknowns, $$\left\lbrace \begin{aligned} g_3(1) &= C_0 + C_1 + C_2 + C_3 = \frac{1}{2} \\ g_3(2) &= C_0 + 2 C_1 + 4 C_2 + 8 C_3 = 2 \\ \left.\frac{d g_3(x)}{d x}\right\rvert_{x=1} &= C_1 + 2 C_2 + 3 C_3 = \frac{1}{2} \\ \left.\frac{d g_3(x)}{d x}\right\rvert_{x=2} &= C_1 + 4 C_2 + 12 C_3 = 1 \\ \end{aligned} \right.$$ which has a single solution, $C_0 = 4$, $C_1 = -9$, $C_2 = 7$, $C_3 = -3/2$, and therefore $$g_3(x) = -\frac{3}{2} x^3 + 7 x^2 - 9 x + 4$$ and the combined function is $$f(x) = \begin{cases} g_1(x) = \frac{1}{2} x, & x \le 1 \\ g_3(x) = -\frac{3}{2} x^3 + 7 x^2 - 9 x + 4, & 1 \lt x \lt 2 \\ g_2(x) = x, & x \ge 2 \\ \end{cases}$$

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Sometimes we really want $x_1 \gt x_2$, and have to "compromise" somehow for $x_2 \le x \le x_1$ instead.

This can be done using a "blending" or interpolation function $w(x)$, such that $w(x) = 0$ for $x \lt x_2$, $0 \le w(x) \le 1$ for $x_2 \le x \le x_1$, and $w(x) = 1$ for $x \ge x_1$. Normally, we want symmetry, so that $w(x) + w(1 - x) = 1$.

If we use an intermediate variable $y$ such that it is zero at $x = x_2$ and one at $x = x_1$, i.e. $$y = \frac{x - x_2}{x_1 - x_2} \quad \iff \quad 1 - y = \frac{x_1 - x}{x_1 - x_2} \quad \iff \quad x = x_2 + (x_1 - x_2) y$$ then there are a couple of commonly used intepolation polynomials that can be useful: $$\begin{aligned} w_1(y) &= y \\ w_3(y) &= y^2 (3 - 2 y) \\ w_5(y) &= y^3 (10 - 15 y + 6 y^2 ) \\ \end{aligned}$$ noting that $w_3(y)$ is pretty close to $\frac{1 - \cos \pi y}{2}$. Within this interval, $f(y) = (1 - w(y)) g_1(x) + w(y) g_2(x)$.

There are also interpolating functions like $$w_\lambda(y) = \begin{cases} 2^{\lambda - 1} y^\lambda, & y \le \frac{1}{2} \\ 1 - 2^{\lambda - 1} (1 - y)^\lambda, & y \gt \frac{1}{2} \\ \end{cases}$$ where $\lambda \gt 0$ determines the curvature, for when one needs a "tunable" interpolating function.

The combined function is then $$f(x) = \begin{cases} g_1(x), & x \le x_2 \\ g_1(x) w\left(\frac{x_1 - x}{x_1 - x_2}\right) + g_2(x) w\left(\frac{x - x_2}{x_1 - x_2}\right), & x_2 \lt x \lt x_1 \\ g_2(x), & x \ge x_1 \\ \end{cases}$$ assuming a symmetric interpolating function $w$ (i.e. $w(x) + w(1-x) = 1$ for $0 \le x \le 1$); and $$f(x) = \begin{cases} g_1(x), & x \le x_2 \\ g_1(x) \left(1 - w\left(\frac{x - x_2}{x_1 - x_2}\right)\right) + g_2(x) w\left(\frac{x - x_2}{x_1 - x_2}\right), & x_2 \lt x \lt x_1 \\ g_2(x), & x \ge x_1 \\ \end{cases}$$ for a general interpolating function $w$ for which $0 \le w(x) \le 1$ for $0 \le x \le 1$.

Obviously, the degree of smoothness, and even the exact behaviour of $f(x)$ in the interpolated interval, depends completely on the interpolating function $w$. This means that in practice, the combined function (and its derivatives, up to the degree of smoothness/continuity desired) is plotted for interpolation function candidates.

For "tunable" interpolating functions, the "tuning" variables (like $\lambda$ above) can be fitted to some desired condition; the more conditions, the more complex interpolating function is needed.