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Let $V$ denote the set of all Taylor polynomials of degree $\leq n$ for a fixed natural number $n$ (including the zero polynomial), regraded as real-valued functions of a real variable. Then is $V$ a (real) linear space? If so, how? If not, which axiom(s) fail(s) to hold?

My work:

If $f$ and $g$ are two functions having derivatives of order upto $n$ at a given point $x=x_0$, then the Taylor polynomials of $P_f$ and $P_g$ of $f$ and $g$, respectively, are given by $$ P_f(x) = \sum_{k=0}^n \frac{(x-x_0)^k}{k!} f^{(k)}(x_0)$$ and $$ P_g(x) = \sum_{k=0}^n \frac{(x-x_0)^k}{k!} g^{(k)}(x_0). $$

Then the sume $P_f + P_g$ is the Taylor polynomial $P_{f+g}$ of the sum $f+g$ and, for any real number $a$, the polynomial $aP_f$ is the Taylor polynomial of the function $af$.

So the set of all Taylor polynomials at a fixed point $x_0$ of real-valued functions of a real variable having derivatives of order upto $n$ at $x_0$ IS a real vector space.

What is the situation like in the general case?

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  • $\begingroup$ What general case? Taylor polynomials around different points? $\endgroup$ – Git Gud Jul 26 '14 at 10:36
  • $\begingroup$ Git Gud, exactly. $\endgroup$ – Saaqib Mahmood Jul 26 '14 at 10:41
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    $\begingroup$ Taylor polynomials are in particular regular polynomials and also every regular polynomial coincides with its taylor polynomial. So if you know that the set of regular polynomials forms a $\mathbb R$-vector space, everything follows smoothly. $\endgroup$ – Git Gud Jul 26 '14 at 10:45
  • $\begingroup$ @SaaqibMahmuud You mean an affine space? $\endgroup$ – Paracosmiste Jul 26 '14 at 10:46
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Every polynomial is a Taylor polynomial (of itself, for example), so the space in question is just the space of all polynomials of degree $\leq n$. Of course it is a linear space.

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