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Given we know the value of all order derivatives at a point $x_0$ for a given f(x).

As per my knowledge all the geometric properties like slope, curvature, convexity are functions of solely the derivatives of any order, like in vector calculus, for Serret-Frenet equations, all the parameters can be determined once we know the values of some of the higher derivatives at point and we can construct the function from $x_0$ up to $\infty$ using initial given values at $x_0$ alone however we know this is true only for analytic functions and a limited domain.

So what is the hidden intrinsic property of a function other than derivatives that prevents it from being analytic ?

Example: Take y=ln(1+x), i know the value of function and all derivatives at x=0, both I and nature will use these values to construct the function for x>0, yet, for my taylor series, it will diverge for x>1, still one curious fact is that as i keep on adding terms, function diverges to +$\infty$ for even terms and -$\infty$ for odd terms so it still keeps oscillating about original value and each successive term seems to correct the offshoot introduced by previous term.

essentially what prevents its convergence ?

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    $\begingroup$ It is the sum of its Taylor series – which is always the case on $\mathbf C$, but not necessarily on $\mathbf R$. $\endgroup$
    – Bernard
    Apr 19, 2019 at 20:12
  • $\begingroup$ @bernard i am unable to understand, kindly elaborate $\endgroup$
    – Kutsit
    Apr 19, 2019 at 20:22
  • $\begingroup$ I mean it is not necessary true that an indefinitely differentiable function $f$ is such that $\;f(x)=\sum_{i=0}^\infty \frac {f^{(i)}(x_0)}{i!}(x-x_0)^i$. Functions which satisfy this equality are called analytic. $\endgroup$
    – Bernard
    Apr 19, 2019 at 20:32

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