-2
$\begingroup$

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.

Answer:

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @@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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.