<s>[INST] Find Taylor's expansion around $ x=2$ for the function $f(x)=x^5+x^4+2x+1 $ 
Find Taylor's expansion around $ x=2$ for the function $f(x)=x^5+x^4+2x+1 $

My try:
$$f(2)=53$$
$$f^{(1)}(x)=5x^4+4x^3+2 \Rightarrow f^{(1)}(2)=114$$
$$f^{(2)}(x)=20x^3+12x^2 \Rightarrow f^{(2)}(2)=208$$
$$f^{(3)}(x)=60x^2+24x \Rightarrow f^{(3)}(2)=288$$
$$f^{(4)}(x)=120x+24 \Rightarrow f^{(4)}(2)=264$$
$$f^{(5)}(x)=120 \Rightarrow f^{(5)}(2)=120$$
$$f^{(n)}(x)=0, n\ge6 \Rightarrow f^{(n)}(2)=0$$That is why I have: $$f(x)=53+114(x-2)+\frac{208}{2!}(x-2)^2+\frac{288}{3!}(x-2)^3+\frac{264}{4!}(x-2)^4+\frac{120}{5!}(x-2)^5$$ I checked this answer in Mathematica and I get that this is true. However I also checked diagrams for $f(x)=x^5+x^4+2x+1 $ and $f(x)=53+114(x-2)+\frac{208}{2!}(x-2)^2+\frac{288}{3!}(x-2)^3+\frac{264}{4!}(x-2)^4+\frac{120}{5!}(x-2)^5$ and I get that there are the same when $x\ge2$.Why this diagrams are different? I think that these diagrams should be the same in all domain.
 [/INST] You have polynomials on both sides of $$x^5+x^4+2x+1=53+114(x-2)+\frac{208}{2!}(x-2)^2+\frac{288}{3!}(x-2)^3+\frac{264}{4!}(x-2)^4+\frac{120}{5!}(x-2)^5$$
Thus the two sides match on the entire range of $(-\infty, \infty).$ 
You do not have to worry about the interval of convergence of either side.  
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