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Math Help - Taylor's polynomial, proof for existence and uniqueness

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    MHF Contributor arbolis's Avatar
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    Taylor's polynomial, proof for existence and uniqueness

    Unfortunately I'm not talking about Taylor's theorem, but Taylor's polynomial. The problem stated is : Let f be a function that can be derived at least n times in a. Define with precision what is the Taylor's polynomial of order n of f in a. Prove that it exists and that it is unique.
    My attempt : I define it : The Taylor's polynomial of order n of f in a is the only polynomial p(x) of degree lesser or equal to n that satisfies p^k(a)=f^k(a), \forall 0\leq k \leq n, with k \in \mathbb{N} union \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 p_{n,a}(x)=a_0+a_1(x-a)+a_2(x-a)^2+...+a_n(x-a)^n, with a_k=\frac{f^{k}(a)}{k!} and then after a few algebra with the derivatives of the polynomial, it says 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, \Leftarrow) shows the existence while \Rightarrow) shows the uniqueness.
    I'm very confused. If I understood well, I must show that if the polynomial can be written such as above, then a_k=\frac{f^{k}(a)}{k!} and that if a_k=\frac{f^{k}(a)}{k!}, then the 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.
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    Quote Originally Posted by arbolis View Post
    Unfortunately I'm not talking about Taylor's theorem, but Taylor's polynomial. The problem stated is : Let f be a function that can be derived at least n times in a. Define with precision what is the Taylor's polynomial of order n of f in a. Prove that it exists and that it is unique.
    If f is differenciable n times on an interval containing a then we define \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k to be its n-th Taylor polynomial at a. It of course exists since f^{(0)}(a),f^{(1)}(a),...,f^{(n)}(a) all exist by hypothesis. It is unique because we given a formula how to compute this polynomial.
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    MHF Contributor arbolis's Avatar
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    Thanks ThePerfectHacker, you gave me a very fast and nice way to solve the problem.
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