Any identity (involving addition, multiplication, substitution) between real/complex power series is an identity in the ring of formal power series.

I need help to prove the following principle-

Any identity between real or complex power series, involving addition, multiplication (possibly infinite sums and products), and substitution, is an identity in the ring of formal power series.

Suppose $$f(x)=\sum_{n \geq 0} a_nx^n$$ and $$g(x)=\sum_{n \geq 0} b_nx^n$$ be two power series in $$\mathbb{R}$$.

Also let the identity $$f(x)=g(x)$$ holds in $$\mathbb{R}$$.

Then we have to show that same identity $$f(X)=g(X)$$ i.e., $$\sum_{n \geq 0} a_nX^n=\sum_{n \geq 0} b_nX^n$$ holds in the ring of formal power series $$\mathbb{R}[[X]]$$.

Let $$h(x)=f(x)-g(x)=\sum_{n \geq 0} a_nx^n-\sum_{n \geq 0} b_nx^n=0$$.

We know that Taylor series of an analytic function is unique.

Does this conclude the proof?

Help me to prove the principle with any further requirements.

Hint: $$a_n = \dfrac{f^{(n)}(0)}{n!}$$ holds in the ring of analytic functions on $$\mathbb R$$ and in the ring of formal power series $$\mathbb{R}[[X]]$$.
• @@lhf, How does it work? I think it should be $c_n=a_n-b_n=\frac{h^{(n)}(0)}{n!}=0$. But since $h(x)=0$, we must have $c_n=0$ implying $a_n-b_n=0 \Rightarrow a_n=b_n$. Is not it? – M. A. SARKAR Jan 1 at 14:50
• ok now $a_n=b_n$ implies $\sum a_nX^n=\sum b_nX^n$. Is it? – M. A. SARKAR Jan 1 at 15:03