Question

Why do we use x=0 to find f(0) in maclaurin series? why another numbers are not used?
and why f(0) is needed?

Answer 1

Answer to Question 50683, Math, Differential Calculus | Equations

Why do we use $x=0$ to find $f(0)$ in Maclaurin series? Why another numbers are not used? And why $f(0)$ is needed?

The answer to this question comes from the Taylor’s theorem. If the function is infinitely differentiated in the neighborhood of the point $x=x_{0}$, then it can be represented as

$f(x)=f(x_{0})+f^{\prime}(x_{0})\frac{(x-x_{0})}{1!}+f^{\prime\prime}(x_{0})\frac{(x-x_{0})^{2}}{2!}+..+f^{(k)}(x_{0})\frac{(x-x_{0})^{k}}{k!}+..$

This theorem is a generalization of Lagrange’s theorem

$f(x)-f(x_{0})=f^{\prime}(\xi)(x-x_{0})$

Thus, Maclaurin series ($x_{0}=0$) is only a particular case of Taylor series. Saying rigorously, if we use $x=x_{0}$, we must call it Taylor series (not Maclaurin series).

Actually, we can use any point $x=x_{0}$ to write an expansion, but only when this function at the point satisfies the Taylor’s theorem.