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I am using Wikipedia to brush up on these key concepts. I will write what I believe I read and maybe an expert can put me back on track. First , they are all functions.

I get the idea that a Taylor polynomial is the same as a Taylor series other than maybe the polynomial is finite?

So that maybe a quadradic approximation is only the first two terms of a Taylor series?

And finally all Taylor series are analytic but going the other way I am not sure, since all analytic functions are locally given by a convergent power series.

Would the function having the real numbers as domain and the square as the output be considered "continuously differentiable" and it's Taylor series would have all it's terms 0 accept for the first two?

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    $\begingroup$ I'm not clear which function you're referring to in your final paragraph. Are you referring to the function $x \mapsto x^2$? If so, the function is continuously differentiable, analytic, and is literally its own Taylor series. $\endgroup$ – Theo Bendit Sep 16 '18 at 20:21
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For a given function $f(x)$ and a given point $x=a$, the Taylor polynomial of degree n is the polynomial $$ P(x) = f(a) + f'(a) (x-a) + f''(a)(x-a)^2 /{2!}+...+ f^{(n)} (a) (x-a)^n /{n!}$$ This polynomial has its value and up to $n^{th}$ derivatives match with the function $f(x).$

The Taylor series is an infinite series which may or may not have infinite non-zero terms and its terms are similar to terms of the Taylor polynomial.

The Taylor series may or may not converge to the function $f(x).$

A function is analytic at a given point if and only if it is equal to its Taylor series on a neighborhood of the given point.

A quadratic approximation is the first three terms of the Taylor polynomial of the function at the given point.

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