The second order Taylor-expansion with Lagrange-Remainder of $e^u$ around $u=0$ for some $s_1$ such that $0<s_1<u$ is:
$$e^u=1+u+\frac{u^2}{2!}+\frac{e^{s_1}u^3}{3!}$$
If we now say: $u=x+1$ for some $s_2$ such that $0<s_2<x+1$ we get:
$$e^{x+1}=1+(x+1)+\frac{(x+1)^2}{2!}+\frac{e^{s_2}(x+1)^3}{3!}$$
If I understand it correctly the first series is centrered around $0$ while the second series is centered around $-1$.
Question: Are $s_1$ and $s_2$ equal? Why (not)? If not, can I express $s_2$ in terms of $s_1$?
Since $u=x+1$ and $u>0$ we get $e^{s_1}u^3=e^{s_2}(x+1)^3$, thus $e^{s_1}=e^{s_2}$ and therefore $s_1=s_2$