Assume that the Taylor expansion $f(x)=\sum_{k=0}^n a_k x^k$$f(x)=\sum_{k=0}^\infty a_k x^k$ is convergent for some $|x|>1$. Then f$f$ can be extended in a natural way into the complex domain by writing $f(z)=\sum_{k=0}^n a_k z^k$$f(z)=\sum_{k=0}^\infty a_k z^k$ with $z$ complex and $|z|≤1$. So we may look at $f$ on the unit circle $|z|=1$. Consider $f$ as a function of the polar angle $\phi$ there, i.e., look at the function $F(\phi):=f(e^{i\phi})$. This function $F$ is $2\pi$-periodic, and its Fourier expansion is nothing else but $F(\phi)=\sum_{k=0}^n a_k e^{ik\phi}$$F(\phi)=\sum_{k=0}^\infty a_k e^{ik\phi}$ where the $a_k$ are the Taylor coefficients of the "real" function $x\mapsto f(x)$ we started with.
Assume that the Taylor expansion $f(x)=\sum_{k=0}^n a_k x^k$ is convergent for some $|x|>1$. Then f can be extended in a natural way into the complex domain by writing $f(z)=\sum_{k=0}^n a_k z^k$ with $z$ complex and $|z|≤1$. So we may look at $f$ on the unit circle $|z|=1$. Consider $f$ as a function of the polar angle $\phi$ there, i.e., look at the function $F(\phi):=f(e^{i\phi})$. This function $F$ is $2\pi$-periodic, and its Fourier expansion is nothing else but $F(\phi)=\sum_{k=0}^n a_k e^{ik\phi}$ where the $a_k$ are the Taylor coefficients of the "real" function $x\mapsto f(x)$ we started with.
Assume that the Taylor expansion $f(x)=\sum_{k=0}^\infty a_k x^k$ is convergent for some $|x|>1$. Then $f$ can be extended in a natural way into the complex domain by writing $f(z)=\sum_{k=0}^\infty a_k z^k$ with $z$ complex and $|z|≤1$. So we may look at $f$ on the unit circle $|z|=1$. Consider $f$ as a function of the polar angle $\phi$ there, i.e., look at the function $F(\phi):=f(e^{i\phi})$. This function $F$ is $2\pi$-periodic, and its Fourier expansion is nothing else but $F(\phi)=\sum_{k=0}^\infty a_k e^{ik\phi}$ where the $a_k$ are the Taylor coefficients of the "real" function $x\mapsto f(x)$ we started with.
Assume that the Taylor expansion $f(x)=\sum_{k=0}^n a_k x^k$ is convergent for some $|x|>1$. Then f can be extended in a natural way into the complex domain by writing $f(z)=\sum_{k=0}^n a_k z^k$ with $z$ complex and $|z|≤1$. So we may look at $f$ on the unit circle $|z|=1$. Consider $f$ as a function of the polar angle $\phi$ there, i.e., look at the function $F(\phi):=f(e^{i\phi})$. This function $F$ is $2\pi$-periodic, and its Fourier expansion is nothing else but $F(\phi)=\sum_{k=0}^n a_k e^{ik\phi}$ where the $a_k$ are the Taylor coefficients of the "real" function $x\mapsto f(x)$ we started with.