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I am going through the lecture notes of David Gross on dynamical systems.

In Section 1.1.3 on page 10, the first equation in the section is given below.

$$f(x_0 + t) = f(x_0) + f'(x_0) t + \mathcal{O(t^2)}$$

Could anyone help me to understand the intuition behind this particular equation? I assume $f'$ is the first derivative. Also, why the last term is $t^2$ and not some other function of $t$?

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You can also write this as $$ f(x_0+t)-f(x_0)=t\int_0^1f'(x_0+st)\,ds=f'(x_0)t+t^2\int_0^1(1-s)f''(x_0+st)\,ds $$ which is the Taylor expansion with integral remainder term.

As long as $f$ is twice continuously differentiable, the last integral is a continuous function in $x_0$ and $t$, so that indeed the term is $O(t^2)$.

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we have, $$ f'(x_0) = \frac{f(x_0 + t)-f(x_0)}{t} + \frac{O(t^2)}{t} $$

from definition of derivative and big O notation. The result follow.
Note : we can also replace $O(t^2)$ by $o(t)$

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  • $\begingroup$ thanks for the reply. I am familiar with the big O notation. My question is why it is $t^2$? $\endgroup$ Sep 3 '20 at 6:20

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