Unfortunately I'm not talking about Taylor's theorem, but Taylor's polynomial. The problem stated is : Let $\displaystyle f$ be a function that can be derived at least $\displaystyle n$ times in $\displaystyle a$. Define with precision what is the Taylor's polynomial of order $\displaystyle n$ of $\displaystyle f$ in $\displaystyle a$. Prove that it exists and that it is unique.

My attempt : I define it : The Taylor's polynomial of order $\displaystyle n$ of $\displaystyle f$ in $\displaystyle a$ is the only polynomial $\displaystyle p(x)$ of degree lesser or equal to $\displaystyle n$ that satisfies $\displaystyle p^k(a)=f^k(a)$, $\displaystyle \forall 0\leq k \leq n$, with $\displaystyle k \in \mathbb{N}$ union $\displaystyle \text { { }0 \text {}}$. Now I have to prove that it exists and that it's unique. How can I prove that?

Looking to my notes, the proof is not done, but it says to write the polynomial this way : Let the polynomial be $\displaystyle p_{n,a}(x)=a_0+a_1(x-a)+a_2(x-a)^2+...+a_n(x-a)^n$, with $\displaystyle a_k=\frac{f^{k}(a)}{k!}$ and then after a few algebra with the derivatives of the polynomial, it says $\displaystyle p^k_{n,a}(a)=f^k(a)\Leftrightarrow k!a_k=f^k(a) \Leftrightarrow a_k=\frac{f^k(a)}{k!}$. The funny part is that now it says, $\displaystyle \Leftarrow$) shows the existence while $\displaystyle \Rightarrow$) shows the uniqueness.

I'm very confused. If I understood well, I must show thatifthe polynomial can be written such as above,then$\displaystyle a_k=\frac{f^{k}(a)}{k!}$ and thatif$\displaystyle a_k=\frac{f^{k}(a)}{k!}$,thenthe polynomial can be written as above. But I'm not sure, so I'm asking you. And in case of an affirmation, would you help me? I thought about induction but realized it was not worth it.