# What intrinsic property determines whether a function is analytic

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 ?

• It is the sum of its Taylor series – which is always the case on $\mathbf C$, but not necessarily on $\mathbf R$. Apr 19, 2019 at 20:12
• @bernard i am unable to understand, kindly elaborate Apr 19, 2019 at 20:22
• 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. Apr 19, 2019 at 20:32