# Is Taylor polynomial of a function around a point $a$ the limit of an orthogonal projection of that function?

I've learned in my course of linear algebra that we can find the best approximation of a continuous real function $f$ defined in $[a-r,a+r]$ ($a,r$ real numbers with $r>0$), by a polynomial $p_r$ of degree $n$ by taking the orthogonal projection of that function over the subspace of polynomials of degree $n$. But there is a polynomial $T$ of degree $n$ that gives us a better approximation of $f$ for points that are close enough to $a$, namely its taylor polynomial. In other words, if we restrict $f$ to some small (by "small" I mean "small enough so that this statement becames true") neighborhood of $a$ of radius $\epsilon$ its Taylor polynomial $T$ restricted to that neighborhood is a better approximation than the restriction of $p_r$ to that same neighborhood.
If now we compare $p_{\epsilon}$ instead of $p_r$, that is the orthogonal projection of $f$ restricted to the neighborhood of $a$ of radius $\epsilon$ over the subspace of polynomials of degree $n$ in that neighborhood, $p_{\epsilon}$ should be a better "global approximation" for that restriction of $f$ than the restriction of $T$. But again we can choose a smaller neighborhood of $a$ so that the resctriction of $T$ beats the restriction of $p_{\epsilon}$.

So, does it make sense to think of the Taylor polynomial of degree $n$ of $f$ around the point $a$ as the limit of $p_{\epsilon}$ as $\epsilon \to 0$?

• There is a way to think of it in terms of direct limits (which is a more advanced, algebraic object). – Michael Burr Jan 4 '17 at 13:31

## 1 Answer

I'm being intentionally vague here to give some ideas without the lengthy background of abstract algebra.

Let $O_a$ be the set of analytic functions defined at $a$ where two functions are identified if they agree in a neighborhood of $a$. This is sometimes called the "germ" of functions at $a$.

Inside of $O_a$ let $m_a$ be the set of functions that vanish at $a$. In this case, $m_a$ is generated by $(x-a)$.

In this case (using abstract algebra), we can consider the ring $O_a/m_a$. The image of a function $f$ in this ring is the value of this function at $a$.

Consider the image of the function $f$ in $O_a/(m_a^2)$. This is, essentially, the first order Taylor expansion of $f$.

We can take more powers of zeros at $a$, i.e., $m_a^k$ and look at the image (think of this as your projection), this corresponds to the Taylor expansion of degree $k$. By taking what is called the inverse limit of these values, we can recover the Taylor expansion of $f$.

• Thanks for the explanation! I definitely need to read it again more carefully and look for some background material. – la flaca Jan 4 '17 at 13:45