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The taylor series expansion about point $a$ for function $f(x)$ is given by $$f(x) \approx f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 +...$$

As more and more terms are added to the taylor polynomial, the interval of approximation over $f(x)$ increases as well.

For a taylor series expansion about point $x_0 + \epsilon$, shouldn't the expansion be $$f(x) \approx f(x_0 +\epsilon) + f'(x_0+\epsilon)(x-(x_0+\epsilon)) + ... $$

Why do they write it as $$f(x_0 + \epsilon) = f(x_0) + f'(x_0)\epsilon + \frac{1}{2}f''(x_0)\epsilon^2$$ This is taken from wolfram alpha

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  • $\begingroup$ They probably meant " The Taylor Series expanded about $x_0$ to evaluate small variations from $x_0$, $x=x_0+\epsilon$ ". $\endgroup$
    – Zalnd
    Nov 18 '20 at 2:45
  • $\begingroup$ so the taylor expansion is about point $x_0$ and we use the function approximation at point $x_0$ to estimate the value at $x_0 + \epsilon$ ? $\endgroup$
    – calveeen
    Nov 18 '20 at 2:46
  • $\begingroup$ Exactly. That way you obtain a function for $\epsilon$: since $x_0$ is a constant, the only variable in the expression of $f(x_0+\epsilon)$ is $\epsilon$ itself. $\endgroup$
    – Zalnd
    Nov 18 '20 at 2:52
  • $\begingroup$ so this function of $\epsilon$ allows me to obtain the value of $f(x_0 + \epsilon)$ for any small values of $\epsilon$ ? $\endgroup$
    – calveeen
    Nov 18 '20 at 3:03
  • $\begingroup$ Yes. In this context, another common notation is to write $\epsilon$ as $\Delta x$: $f(x_0+\Delta x) \equiv f(\Delta x) = f(x_0) + f'(x_0) \Delta x$ $\endgroup$
    – Zalnd
    Nov 18 '20 at 3:09
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It's kind of a weird mixed series obtained from the Taylor series. Start with the Taylor series about $x=x_0$, namely

$$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2}f''(x_0)(x-x_0)^2+ \dots$$

Now substitute $x-x_0=\varepsilon$ on the RHS, but leave the evaluations of $f$ and its derivatives on the RHS at $x_0$. Then on the LHS, write $f(x)=f(x_0+\varepsilon)$. This gives

$$f(x_0+\varepsilon)=f(x_0)+f'(x_0)\varepsilon+\frac{1}{2}f''(x_0)\varepsilon^2+ \dots$$

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