Strategies to build functions with given expansions at two different points

Question

Consider two non-constant real polynomials $f(x)$ and $g(x)$:
$$f=f_0 + f_1 (x-x_0) +...+f_N(x-x_0)^N $$ $$g=g_0 + g_1(x-x_1) +...+g_M(x-x_1)^M $$ where $f_0...f_N,g_0...g_M,x,x_0,x_1 \in \mathbb{R}$ and $M,N\in \mathbb{N}^+$.

I would like to know if there is any function $H(f,g)$ such that:

• the Taylor expansion of $h(x)=H(f(x),g(x))$ up to order $N$ around $x_0$ coincides with $f$.

• the Taylor expansion of $h(x)$ up to order $M$ around $x_1$ coincides with $g$.

I was thinking about using a sort of harmonic mean of $f$ and $g$ or transition functions built out of $f$ and $g$, but I am stuck. Moreover, I am not even sure that such a function exists if we restrict $H$ to be a rational function of two variables. For sure different strategies are possible, if you have any idea please tell me. Thank you in advance for any help/hint/idea!

Note: I am asking the function $h(x)$ to stem from $H(f,g)$, not $H(f,g,x)$. Therefore, a "trivial" construction like $\tilde{h}(x)= \phi_0(x)f(x)+\phi_1(x)g(x)$ for some appropriate $\phi_0(x)$ and $\phi_1(x)$ is not admissible. In other words, since it is required that $h(x)=H(f(x),g(x))$, we also have that $$ h'(x) = f'(x) \partial_f H + g'(x) \partial_g H \, . $$ The above property is not respected by the "trivial" $\tilde{h}(x)$, as $\tilde{h}'(x)$ is not linear in the derivatives $f'$ and $g'$.

Answer 1

Here is a partial solution when $|f(x_0)-g(x_0)| > \epsilon$, $|f(x_1)-g(x_1)| > \epsilon$, $|f(x_0)-f(x_1)| > \epsilon$, and $|g(x_0)-g(x_1)| > \epsilon$ all hold. Or to put it differently, pick

$\epsilon = \tfrac{1}{2} \min(|f(x_0)-g(x_0)|, |f(x_1)-g(x_1)|, |f(x_0)-f(x_1)|, |g(x_0)-g(x_1)|)$

and the assumption is that this number isn't 0. Under the stated assumption you can pick

$H(f,g) = \Theta(\epsilon - |f-f(x_0)|) f + \Theta(\epsilon - |g-g(x_1)|)g$

where $\Theta$ is the Heaviside step function or smooth approximation that is flat enough around $0$ to preserve the Taylor series to a given finite order.

Here is a counterexample showing that you need to assume something more: When $x_0 = x_1$, and $f_i \neq g_i$ for some $i \leq \min(M,N)$, the Taylor series of $h(x)$ at $x_0=x_1$ can obviously agree with at most one of the series for $f(x)$ and $g(x)$.

Answer 2

I assume there are some restrictions on where the agreement needs to arise since $h$ can't agree with $f$ and $g$ everywhere. So I propose a similar question: how can we construct a function $h$ such that

• on $x<a$, it agrees with $f$
• on $x>b$, it agrees with $g$
• $h$ is smooth (provided $f$ and $g$ are)

This idea came up in a video by EpsilonDelta on YouTube a while back, which I later implemented into a Desmos demo out of curiosity. Naturally, then, it should work just fine in the polynomial case, but can also work for smooth functions in general.

The process of interpolation works as follows:

• We define a sufficiently smooth "base interpolating" function of sorts. The video works with $$ \psi(x) = e^{-1/x} \cdot \mathbf{1}_{[0,1]}(x) $$ The video notes that we require $\psi$ monotone increasing and smooth, if we want the interpolation to be smooth. We also force $\psi(0) = 0$. These conditions allow us to get the useful properties for the next function, $\phi$.

• We then modify $\psi$ into a new function $\phi$, as below. The goal is to construct a function that is "like" $\psi$ (and in particular, inherits its starting value, monotonicity, and smoothness), but also has $\phi(1)=1$. Hence, $\phi$ will function as a sort of sigmoid-like "weight" between $f$ and $g$, telling how much of a value is contributed based on its distance from the boundaries of the domain (and based on the definition of the smooth "base" function). $$ \phi(x) = \frac{\psi(x)}{\psi(x)+\psi(1-x)} $$ This has some useful properties:

  • $\phi$ is smooth
  • $\phi(0) = 0$
  • $\phi(1) = 1$
  • $\phi$ is monotone increasing
  • $\phi$ is symmetric about $(\frac 1 2, \frac 1 2)$
  • If $\psi^{(n)}(0) = 0$ for $k = 1,2,\cdots,n$ the same is true of $\phi$

• To handle interpolation on $[a,b]$, we modify $\phi$ with the rule $$ \phi_{a,b}(x) = \phi \left( \frac{x-a}{b-a} \right) $$

• The interpolating function $h$ is then given by $$ h(x) = \left( 1 - \phi_{a,b}(x) \right) f(x) + \phi_{a,b}(x) g(x) $$

For instance, interpolating $\sin(5x)-x$ on $(-\infty,-1)$ with $\cos(7x)+2$ on $(3,\infty)$, the interpolation graphically looks as so: