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I am struggling understand the linear approximation and Taylor.series. Could you give me a hint what are the derivatives of these functions?

$$a_2(x_1-x_0)^2 + a_3(x_1-x_0)^3?$$ If it’s stated that $x_1=x_0$.

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    $\begingroup$ what have you tried that you couldn't find the derivatives? $\endgroup$ – pointguard0 Jul 5 at 8:35
  • $\begingroup$ I mean, I can find a he derivatives, but I can’t get the whole intuition behind the Taylor Series concept. $\endgroup$ – Maria Lavrovskaya Jul 5 at 8:51
  • $\begingroup$ Derivatives with respect to which variable? Anyway, for intuition about Taylor series, you should watch 3Blue1Brown's video on that topic: youtube.com/watch?v=3d6DsjIBzJ4 $\endgroup$ – Hans Lundmark Jul 5 at 9:34
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Peano's theorem - Let $f : (a,\,b) \to \mathbb{R}$ and $x_0 \in (a,\,b)$. If $f$ is $n$ times derivable at $x_0$, the Taylor polynomial of order $n$ and centered at $x_0$, $$T_n(x) := \sum_{k = 0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$$ is the only polynomial of degree $\le n$ such that $$f(x) = T_n(x) + o\left((x-x_0)^n\right) \quad \text{for} \; x \to x_0$$ and also such that $$f^{(k)}(x_0) = T_n^{(k)}(x_0) \quad \text{for} \; k = 0,\,1,\,\dots,\,n.$$


Application example - Let $f : (-3,\,3) \to \mathbb{R}$ of law $f(x) := e^x$ and $x_0 = 0$, then it follows that:

$\hspace{3cm}$enter image description here

Now the "meaning" of "Taylor polynomials" should be obvious.

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