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Let \( f \) be a real function \(n \) times differentiable such as:
$$f(1)=1, \,\,\,f'(1)=\frac{1}{2}, \,\,\,f''(1)=\frac{1}{2}, \,\,\,f'''(1)=\frac{3}{4}, \,\,\,\cdots, f^{(n)}(1)=\frac{n!}{2^n}$$
a. Express the Taylor Series around \(x_0=1 \).

b. For what values of \( x \) does the previous series converge?

c. Find the value of \(f(0) \).

\(\displaystyle{\rm{a}.}\) The function \(\displaystyle{f}\) is n times differentiable and for each \(\displaystyle{n\in\mathbb{N}}\) holds :

\(\displaystyle{0<f^{(n)}(1)\leq 1}\) . Therefore, there is \(\displaystyle{\delta>0}\) such that :

\(\displaystyle{f(x)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(1)\,\left(x-1\right)^{n}}{n!}=\sum_{n=0}^{\infty}\left(\dfrac{x-1}{2}\right)^{n}\,\forall\,x\in\left(1-\delta,1+\delta\right)}\)

which is the Taylor series of \(\displaystyle{f}\) around \(\displaystyle{x_0=1}\) .

\(\displaystyle{\rm{b}.}\) The previous series is a geometric one and converges if, and only if,

\(\displaystyle{\left|\dfrac{x-1}{2}\right|<1\iff -1<\dfrac{x-1}{2}<1\iff -1<x<3}\) .

So, \(\displaystyle{f(x)=\sum_{n=0}^{\infty}\left(\dfrac{x-1}{2}\right)^{n}\,,x\in\left(-1,3\right)}\) .

\(\displaystyle{c.\,\,\,f(0)=\sum_{n=0}^{\infty}\left(\dfrac{-1}{2}\right)^{n}=\dfrac{1}{1-(-1/2)}=\dfrac{2}{3}}\)