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The best linear approximation of a real function infinitely times differentiable is given by:

$$ f(x) = f(x_0) + f^{(1)}(x_0)(x-x_0) + \frac{f^{(2)}(\xi_x)}{2}(x-x_0)^2 = \\ f(x_0) + f^{(1)}(x_0)(x-x_0) + r_1(x) $$

What happens if I make $x_0 = g(x)$? you can assume is infinitely times differentiable. Is there a way to write such expansion? Is it gonna look like

$$ f(x) = f(g(x)) + f^{(1)}(g(x))(x-g(x)) + r_1(g(x)) $$

Or would it look like something $$ f(x) = f(g(x)) + f^{(1)}(g(x))g^{(1)}(x)(x-g(x)) + \hat{r}_1(g(x)) $$

I believe is the second one, but I'm not entirely sure. Is there a theorem I'm not aware of that maybe I should read through?

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    When you expand $f(x)$ it is assumed that $x_0$ is a fixed point. You cannot later replace $x_0$ by $g(x)$. You are free to consider $x_0$ as a free parameter, but not as a function of $x$. What you can do is to Taylor expand $f(g(x))$ about a point $x_0$ for which it becomes $$f(g(x)) = f(g(x_0)) + f'(g(x_0))g'(x_0)(x-x_0) + \ldots$$ – 2017-01-20
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    How about If I do something like I get $x$ and then I fix it so $g(x)$ would be fixed as well, couldn't I write the displacement $f(x) - f(g(x))$ as a taylor expansion where $g(x)$ is the expansion center? – 2017-01-20
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    You cannot fix $x$ and at the same time have it be a variable in the series (if $x$ is fixed then the resulting series would not be a function of $x$). You can expand the function $h(x) = f(x) - f(g(x))$ as a Taylor series, but this will have to be about a point $x_0$ not depending on $x$. Look at the definition of a [Taylor series](https://en.wikipedia.org/wiki/Taylor_series#Definition). – 2017-01-20

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