# $\sum_{n=1}^{\infty}a_nx^n=\sum_{n=1}^{\infty}b_nx^n,\space x\in\mathbb{R}$

Suppose we have $$\sum_{n=1}^{\infty}a_nx^n=\sum_{n=1}^{\infty}b_nx^n,\space x\in\mathbb{R}$$ Can we say that $a_n=b_n$ for all $n\in \mathbb{N}$?

It is true in case of polynomial. I don't know that holds true for power series too. Note that we do not assume that $x\in$ Radius of convergence of those power series.

Secondly, if we assume that $x\in$ Radius of convergence of those power series, is the statement holds for that??

• What does the equality mean if the series don't converge? – Robert Israel Jun 24 '17 at 3:12
• If the equality is just for one single $x$, then you can look at $x=1$, $a_i=0$ for all $i\neq 1$, $a_1=1$, $b_i=0$ for all $i\neq2$, and $b_2=1$. Even for finitely many $x$ one can construct similar examples. The interesting thing is when it holds for infinitely many $x$ and those $x$ accumulate somewhere. – OR. Jun 24 '17 at 3:14
• @RobertIsrael thanks – MAN-MADE Jun 24 '17 at 3:34

If both series have nonzero radius of convergence $r_a$, $r_b$, then since the sums are analytic functions in the disk $|z| < \min(r_a, r_b)$, if they agree for all real $x$ in that disk (or even for some sequence of $x$ with a limit point in that disk) they must be equal as analytic functions, and then all $a_n = b_n$.

If both have radius of convergence $0$, they don't converge for any $x \ne 0$. Then the statement $\sum_n a_n x^n = \sum_n b_n x^n$ for $x \ne 0$ is true in the sense "undefined = undefined", and this certainly does not imply $a_n = b_n$.

hint

If $f (x)=g (x)$ for $x\in (-R,R)$ then

$f'(x)=g'(x)$ .

with $x=0$, we get $a_1=b_1$. now think $f''$.

If both series converges, and converges to the same function then yes, it follows that $a_n=b_n$ for all $n \in \mathbb{N}$.

Indeed, if $s(x):=\displaystyle \sum_{n=0}^{\infty} a_nx^n$ and $k \in \mathbb{Z}^+$ then by hypotesis $$s^{(k)} = \sum_{n=k}^{\infty} n(n-1)...(n-k+1)x^{n-k}=\sum_{n=k}^{\infty} a_n\frac{n!}{(n-k)!}x^{n-k}.$$ Putting $x=0$ in the last equality yields $$a_k=\frac{s^{(k)}(0)}{k!}.$$

Note: This holds if $x$ is in the interval of convergence, id est, if $x \in [-R,R]$, were $R$ is the radius of convergence.