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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?

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One thing I have done is use two sets of expansions: one at, say, zero and one at infinity. So I would derive separate expansions and get two sets of coefficients; then save an extra parameter for matching the solutions to make a global approximation.

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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

f is the function ,St is taylor series , S112 as above and the red is the E^-x using the formula of the paper

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  • $\begingroup$ Why is their formula wrong? Why is yours right? $\endgroup$ – Antonio Vargas Apr 6 '18 at 10:50

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