Lagrange polynomial can be used to obtain an identity: $$(k+t)^n = \sum_{i=0}^n (k+d_i)^n \prod_{\substack{j=0\\ j\not=i}}^n \frac{t-d_j}{d_i-d_j},$$ which holds for any integer $n>0$, any real numbers $k,t$, and any distinct real numbers $d_0,\dots,d_n$. The key idea is that the difference between the left-hand and right-hand sides represents a polynomial of degree $\leq n$ in $t$, while it has $n+1$ zeros $d_0,\dots,d_n$. The fundamental theorem of algebra then implies that this polynomial is zero.
The same identity can restated in terms of function $f(t)=t$ as $$(\star)\qquad f(k+t)^n = \sum_{i=0}^n f(k+d_i)^n \prod_{\substack{j=0\\ j\not=i}}^n \frac{f(t-d_j)}{f(d_i-d_j)}.$$
Recently I learned that the identity $(\star)$ holds for (non-polynomial) function $f(t)=U_t=U_t(P,Q)$ being Lucas sequence (equivalently, $U_t=\frac{a^t-b^t}{a-b}$, where $a,b$ are the zeros of the characteristic polynomial, which enables real values for $t$). That is, $$U_{k+t}^n=\sum_{i=0}^{n} U_{k+d_i}^n\prod_{\substack{j=0\\ j\not=i}}^{n}\frac{U_{t-d_j}}{U_{d_i-d_j}},$$ where again $n>0$ is integer, $k,t$ are any real numbers, and $d_0,\dots,d_n$ are any distinct real numbers.
What would be the analog of the fundamental theorem of algebra here?
As pointed out by Fedor Petrov, the difference here represents a polynomial of degree $\leq n$ in $(a/b)^t$, and the same arguments hold.
What would be other interesting examples of $f(t)$, for which the identity $(\star)$ holds?
