# How do you prove the usage of Maclaurin expansion in order to get the Taylor expansion about a point.

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?

• Welcome to the site ! - Just make $x=y+a$ and work around $y=0$. Commented Apr 2, 2020 at 12:35
• 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. Commented Apr 2, 2020 at 13:31

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