A function is analytic only if the Taylor series associated to it about ‘x

_{0}’ is converging to the function in any neighborhood for every ‘x

_{0}’ to be in its domain.

**Let a function ‘f’ be real analytic on an open set say ‘D’ that lies on the real line. Then for any ‘x**_{0}’ in ‘D’, we can write as,

$f(x)=\sum_{n=0}^{\infty }a_{n}(x-x_{0})^{n} = a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^{2}+...$

Here, the coefficients a_{0}, a_{1},….., belong to the set of real numbers. Also this series will converge to some f(x) for ‘x’ lying in the neighborhood of ‘x_{0}’.

An analytic function is also a differentiable function which is infinitely differentiable alternatively, in such manner that its domain at any particular point ‘x

_{0}’ will converge to some f(x) at certain ‘x’ which lies in the neighborhood of ‘x

_{0}’ uniformly.

$f(x)$=$\sum_{n=0}^{\infty }$$\frac{f^{(n)}x_{0}}{n!}$$(x-x_{0})^{n}$Generally, the set of all real valued such functions are denoted by C

^{W}(D), on a given set ‘D’. We can define a function ‘f’ on a subset of the real line to be a real analytic one at some point ‘x’ only if there exist some neighborhood ‘D’ of ‘x’ in which ‘f’ is real and analytic.

Also, if ‘f’ is a differentiable equation which is infinitely differentiable on an open subset ‘D’ contained in ‘R’, then the conditions stated below are all equivalent.

**1)**. Function ‘f’ is analytically real.

**2)**. There will exist an extension of ‘f’ which is complex to an open set ‘G’ contained in ‘C’ which will contain ‘D’.

**3)**. Let us have a compact set ‘K’ contained in ‘D’. Then there exists a constant ‘C’ in a manner that for each ‘x’ belonging to K and for every ‘k’ (non negative integer), the given bound holds true.

$|\frac{d^{k}f(x) }{dx^{k}}|$ $\leq C^{k+1}k!$The FBI transform can be used to characterize the real analyticity of function ‘f’ for a given point say ‘x’.

**While complex analytic ones being completely equivalent to holomorphic functions can be characterized much more easily and simply.**