# Two-point Taylor expansion with one assymptotic point?

According to this paper, a two-point Taylor expansion can be definied like this:

$$\text{Let }f\left(z\right)\text{ be an analytic function and }z_1 \text{and }z_2\in \mathbb{C}, z_1\neq z_2\text{.}\\ \text{The two-point Taylor expansion is defined as:}\\ P_n\left(z_1, z_2;z\right)=\sum_{k=0}^n{\left[a_k\left(z_1,z_2\right)\left(z-z_1\right)+a_k\left(z_2,z_1\right)\left(z-z_2\right)\right]\left(z-z_1\right)^k\left(z-z_2\right)^k}\\ \text{with coefficients}\\ a_0\left(z_1,z_2\right)=\frac{f\left(z_2\right)}{z_2-z_1}\\ \forall_{n>0}{: a_n\left(z_1,z_2\right)=\sum_{k=0}^n{\frac{\left(n+k-1\right)!}{k!\left(n-k\right)!}\frac{(-1)^{n+1}n f^\left(n-k\right)\left(z_2\right)+(-1)^{k}k f^\left(n-k\right)\left(z_1\right)}{n!\left(z_1-z_2\right)^{n+k+1}}}}$$ This is actually pretty nice. However, seeing that there are things like $z_1-z_2$, I'm not sure how you'd do an assymptotic expansion, where one of the two points is actually at infinity.

Is this possible? If so, how would you do it?

the formula above it is only correct for some function as Sin(x) but not for Sin(x+1) XE Sin[x] about x=-2 and x=2 gives $$\frac{1}{2} z \sin (2)+(z-2)^2 z (z+2)^2 \left(-\frac{\sin (2)}{256}-\frac{1}{128} 3 \cos (2)\right)+(z-2) z (z+2) \left(\frac{\cos (2)}{8}-\frac{\sin (2)}{16}\right)$$ that is correct but Sin[x+1] gives $$z \left(\frac{\sin (1)}{4}+\frac{\sin (3)}{4}\right)+(z-2)^2 z (z+2)^2 \left(\frac{3 (2 \sin (1)+2 \sin (3))}{1024}-\frac{\sin (1)}{128}-\frac{\sin (3)}{128}+\frac{1}{256} (-2 \cos (1)-\cos (3))+\frac{1}{256} (-\cos (1)-2 \cos (3))\right)+(z-2) z (z+2) \left(\frac{1}{32} (-\sin (1)-\sin (3))+\frac{\cos (1)}{16}+\frac{\cos (3)}{16}\right)+\frac{\sin (3)}{2}-\frac{\sin (1)}{2}-2 \left(\frac{3 (2 \sin (1)+2 \sin (3))}{2048}-\frac{\sin (1)}{128}+\frac{1}{256} (-2 \cos (1)-\cos (3))\right)+2 \left(\frac{3 (2 \sin (1)+2 \sin (3))}{2048}-\frac{\sin (3)}{128}+\frac{1}{256} (-\cos (1)-2 \cos (3))\right)-2 \left(\frac{1}{64} (-\sin (1)-\sin (3))+\frac{\cos (1)}{16}\right)+2 \left(\frac{1}{64} (-\sin (1)-\sin (3))+\frac{\cos (3)}{16}\right)$$ is incorrect and the correct series expansion is $$-\frac{1}{720} (z-2)^3 (z+2)^3 \sin (1)+\frac{1}{64} (z-2)^2 (z+2)^2 \sin (1)-\frac{1}{512} (z-2)^2 z (z+2)^2 \sin (1)-\frac{1}{64} (z-2)^2 (z+2)^2 \sin (3)-\frac{1}{512} (z-2)^2 z (z+2)^2 \sin (3)+\frac{1}{4} z \sin (1)+\frac{1}{4} z \sin (3)+\frac{1}{128} (z-2)^2 (z+2)^2 \cos (1)-\frac{3}{256} (z-2)^2 z (z+2)^2 \cos (1)-\frac{1}{128} (z-2)^2 (z+2)^2 \cos (3)-\frac{3}{256} (z-2)^2 z (z+2)^2 \cos (3)-\frac{1}{8} (z-2) (z+2) \cos (1)+\frac{1}{8} (z-2) (z+2) \cos (3)+(z-2) z (z+2) \left(-\frac{\sin (1)}{32}-\frac{\sin (3)}{32}+\frac{\cos (1)}{16}+\frac{\cos (3)}{16}\right)+\frac{\sin (3)}{2}-\frac{\sin (1)}{2}$$ other Example for the function $$e^{-x}$$ The formula of the paper using x=1 and x=-1 gives $$\left(\frac{1}{16} \left(\frac{2}{e}+e\right)+\frac{1}{16} \left(\frac{1}{e}+2 e\right)+\frac{3}{32} \left(\frac{2}{e}-2 e\right)-\frac{e}{16}+\frac{1}{16 e}\right) (z-1)^2 z (z+1)^2+\left(\frac{1}{4} \left(e-\frac{1}{e}\right)-\frac{e}{4}-\frac{1}{4 e}\right) (z-1) z (z+1)+\left(\frac{1}{2 e}-\frac{e}{2}\right) z+\frac{1}{8} \left(\frac{1}{e}-e\right)+\frac{1}{16} \left(\frac{2}{e}+e\right)+\frac{1}{8} \left(e-\frac{1}{e}\right)+\frac{1}{16} \left(-\frac{1}{e}-2 e\right)+\frac{5}{16 e}+\frac{13 e}{16}$$ which is wrong and cause I Know simple using plot and see the result but using the following $$\frac{1}{720} (z-1)^3 (z+1)^3+\frac{(z-1)^2 (z+1)^2}{8 e}+\frac{7 (z-1)^2 z (z+1)^2}{16 e}-\frac{1}{16} e (z-1)^2 z (z+1)^2+\frac{1}{4} e (z-1) (z+1)-\frac{(z-1) z (z+1)}{2 e}-\frac{(z-1) (z+1)}{4 e}+\frac{z}{2 e}-\frac{e z}{2}+\frac{1}{2 e}+\frac{e}{2}$$ plot it and see the result enter image description here