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Is it true that if $x\in\mathbb{R}$ for two converging sums $$\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty b_n x^n,$$ then $$\forall {n≥0},a_n=b_n?$$ What if $x$ is not real but complex? Or any ring?

Gary
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Damian Albrecht
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    Certainly not true for one $x,$ but if there are infinitely many $x$ in a bounded region for which the sums are equal, this would be true. – Thomas Andrews Oct 09 '23 at 01:54
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    If the set of points $x \in \mathbb{C}$ for which equality holds has an accumulation point, then the coefficients are equal. – Kakashi Oct 09 '23 at 02:04
  • You can also do something like this by writing $$\sum_{n=0}^\infty (a_n - b_n) x^n = 0$$ Now RHS is zero and in LHS $x \ne 0, , \text{so} , (a_k - b_k)=0 , \forall , 0 \leq k \leq n$ – Lucky Chouhan Oct 09 '23 at 02:58
  • @LuckyChouhan Are you claiming this if it is true for one $x \ne 0$? That's clearly not the case. – Robert Israel Oct 09 '23 at 03:04
  • @RobertIsrael Actually I was saying "It is true for all x, I took the case of $x=0$ as a trivial. That's why I types $x \ne 0$ – Lucky Chouhan Oct 09 '23 at 03:06
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    @LuckyChouhan: the argument you wrote in your comment needs a proof. Essentially the problem can be reduced to following: if $\sum_{n\geq 0} c_nx^n=0$ for all real $x\in(-r, r) $ for some $r>0$ then we have $c_n=0$ for all non-negative integers $n$. Your comment performs this reduction step. But after this you need to prove the reduced version. The accepted answer here also has the same mistake. – Paramanand Singh Oct 09 '23 at 04:57
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    See this related thread https://math.stackexchange.com/q/3433492/72031 – Paramanand Singh Oct 09 '23 at 05:07
  • @ParamanandSingh thank you for correcting me and for that thread :) – Lucky Chouhan Oct 09 '23 at 07:44

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Suppose the sums $\sum_{n=0}^{\infty} a_n x^n$ and $\sum_{n=0}^{\infty} b_n x^n$ exit in $\mathbb R$. Then their difference, $\sum_{n=0}^{\infty} a_n x^n-\sum_{n=0}^{\infty} a_n x^n$ makes sense in $\mathbb R$. But the sums are equal, so \begin{align} & \sum_{n=0}^{\infty} a_n x^n-\sum_{n=0}^{\infty} b_n x^n=0 \\& \Rightarrow \sum_{n=0}^{\infty}(a_n-b_n)x^n=0 \cdot x^n, \end{align} which is an identity. Therefore, comparing both sides, we have $a_n-b_n=0$ implying $a_n=b_n$ for all $n$.

Edit: In general, just read the property "if $\sum_{n \geq 0} a_n (x-c)^n$ and $\sum_{n \geq 0} b_n (x-c)^n$ are two power with radius of convergence $r_1, r_2>0$ respectively, then the two power series are equal for all $x$ such that $|x-c|<r=\min{r_1, r_2}$ if and only if $a_n=b_n$ for all $n$. Now take $x \in \mathbb{R}$, the whole set and put $c=0$. You will get the desired result.

Other method: since the two series are equal in $\mathbb R$, which is a open set, they are infinitely differentiable. Taking the first differentiation of $\sum_{n \geq 0} a_n x^n=\sum_{n \geq 0} b_n x^n$, we have $a_0+\sum_{n \geq 1}n a_n x^{n-1}=b_0+\sum_{n \geq 1} nb_n x^{n-1}$, putting $x=0$, you get $a_0=b_0$. And continue the process to get $a_n=b_n$.

MAS
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  • How do you prove that there is no sequence $c_n=a_n-b_n$ such that $\sum_{n=0}^\infty c_n x^n = 0$? – Damian Albrecht Oct 09 '23 at 03:06
  • @DamianAlbrecht, Can you please clarify the question ? The 2nd equation says it is an identity, that is true for all $x \in \mathbb R$ and so you can compare both sides. – MAS Oct 09 '23 at 03:09
  • I'm asking how you deduced that there is no sequence $c_n$ such that $\sum_{n=0}^\infty c_n x^n = 0$ – Damian Albrecht Oct 09 '23 at 03:21
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    This is not a proof at all. Perhaps you have not really understood the question. The idea of comparing both sides (actually comparing coefficients of $x^n$ on both sides) is based on the result asked in question. – Paramanand Singh Oct 09 '23 at 04:52
  • A proper proof requires a bit of analysis. The key is that the values of $x$ for which the identity holds must have a limit point so that coefficients on both sides can be equated. – Paramanand Singh Oct 09 '23 at 04:54
  • @DamianAlbrecht, I assumed that the sun exit for all real numbers. – MAS Oct 09 '23 at 07:34
  • @ParamanandSingh, you really didn't understand the answer. I assumed the sum exit for all real numbers. So limit exits – MAS Oct 09 '23 at 07:35
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    Your updated answer is fine. This level of detail is necessary. I will reverse my down vote to an up vote. – Paramanand Singh Oct 09 '23 at 09:41
  • @DamianAlbrecht, the answer is same in $\mathbb C$ or any topologically complete ring. – MAS Oct 09 '23 at 11:22