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 $\mathbb{R}[[X]]$.

  • $\begingroup$ @@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? $\endgroup$ – M. A. SARKAR Jan 1 at 14:50
  • $\begingroup$ @M.A.SARKAR, exactly $\endgroup$ – lhf Jan 1 at 15:01
  • $\begingroup$ ok now $a_n=b_n$ implies $ \sum a_nX^n=\sum b_nX^n$. Is it? $\endgroup$ – M. A. SARKAR Jan 1 at 15:03

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