Solution to which problem in mathematics is only given/solved by Tylor Series which is otherwise either impossible or extremely hard to solve?

What is the real advantage of Tylor's series ?

closed as too broad by Dietrich Burde, Hans Lundmark, Lord Shark the Unknown, Dave, RRL Dec 7 at 0:44

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

This one is probably more on the the Physics side, but Perturbation Theory is definitely one of those problems. The idea is you know how to solve a system subject to an interaction $V_0$, and now want to know the solution if you change to $A$. To solve it, you expand

$$ A = A_0 + \epsilon + \epsilon A_1 + \frac{1}{2}\epsilon^2A_2 + \cdots $$

and iteratively find the solutions at each order. Applications of this include scattering theory, Feynman Diagrams, ...

The Taylor Series is an expansion of the power series into an infinite sum of terms. For example: $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$

The Taylor Series is used to get an approximate value of a function. The basic formula is f(x): $f(x) = f(a) +\frac{f'(a)}{1}(x-a) + \frac{f''(a)}{2!}(x-a)^{2} + \frac{f'''(a)}{3!}(x-a)^{3} + ... $

This formula is useful for

  1. Approximating definite integrals of functions that have no definite integrals.
  2. Understanding the growth of functions
  3. Solving differential equations

This website explains it in detail why the Taylor series works and how you get it.

Not the answer you're looking for? Browse other questions tagged or ask your own question.