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I learned that if you have a function, let's call it $f(x)$, and you want its Taylor expansion about a point "$a$" you can get the Maclaurin expansion of the function $f(x-a)$ and this will be equal with the Taylor expansion of $f(x)$ near $a$. I can't understand this and I didn't find any proof. How would you prove this?

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  • $\begingroup$ Welcome to the site ! - Just make $x=y+a$ and work around $y=0$. $\endgroup$ Commented Apr 2, 2020 at 12:35
  • $\begingroup$ I don't understand. By doing that the only thing that I get is the Maclaurin expansion of f(a). I want to show that the Maclaurin expansion of f(x-a) is equal to the Taylor expansion of f(x) when you have the center in a. $\endgroup$
    – Vladimir
    Commented Apr 2, 2020 at 13:31

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You have learned that the MacLaurin polynomial $$M^{(r)}_g(y):=\sum_{k=0}^r{g^{(k)}(0)\over k!}y^k$$ is the single polynomial $y\mapsto p(y)$ of degree $\leq r$ whose derivatives of order $\leq r$ at $y=0$ are all equal to the corresponding derivatives of $g$ at $0$. It is then an obvious conjecture that $$T^{(r)}_{f,a}(x)=\sum_{k=0}^r{f^{(k)}(a)\over k!}(x-a)^k$$ is the single polynomial $x\mapsto p(x)$ of degree $\leq r$ whose derivatives of order $\leq r$ at $x=a$ are all equal to the corresponding derivatives of $f$ at $a$.

Proof. It is easy to check that $${d^k\over dx^k}T^{(r)}_{f,a}(x)\biggr|_{x=a}=f^{(k)}(a)\qquad(0\leq k\leq r)\ .$$ If we had two different polynomials $p$ and $q$ producing the derivative values $f^{(k)}(a)$ $\>(0\leq k\leq r)$ then the new polynomials $$p_1(y):=p(a+y),\qquad q_1(y):=q(a+y)$$ would contradict the unicity statement of the MacLaurin polynomial.

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For notation Tf means taylor expansion of f and Mf Maclaurin expansion.

Basically Maclaurin series is evaluation of the Taylor series at x=0 and \begin{equation} \left.Mf(x)\right|_{x=0}=\sum_{n=0}^{\infty} c_{n}(x)^{n} \text { for } c_{n}=\frac{f(0)}{n !} \end{equation} \begin{equation} \left.Tf(x)\right|_{x=a}=\sum_{n=0}^{\infty} d_{n}(x-a)^{n} \text { for } d_{n}=\frac{f(a)}{n !} \end{equation} \begin{equation} \text { for }\left.Tf(x)\right|_{x=0}=f(0) \end{equation} Means when you evaluate Taylor expansion of f(x) at x=0 you will get the Maclaurin expansion.

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  • $\begingroup$ That's not what I meant. The think that I don't understand is why Mf(x-a)= Tf(x)|x=a $\endgroup$
    – Vladimir
    Commented Apr 2, 2020 at 13:07
  • $\begingroup$ I can't see any reason for this equation to hold you can try it with x^3 for simplicity it won't work. $\endgroup$
    – asd.123
    Commented Apr 2, 2020 at 13:42

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