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I have an equation of the form:

$$∑_{n=1}^{∞} a_{n}(x)=0 \tag{*}$$

where $a_{n}(x)$ are real non constant functions in $x$ for all $n≥1$. Assume that $a_{n}(x)≠0$ for all $n≥2$. Assume also that $a_{1}(x)=0$ for some $x=a$.

Then equation $(*)$ becomes $$∑_{n=2}^{∞} a_{n}(x)=-a_{1}(x)$$ My question is: Can we deduce that the only solution of $(*)$ is $x=a$. Any conterexample is very welcome.

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    What if $a_n$ are constant functions such that the series is zero? Then, every x would be a solution – 2017-01-29
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    @Blaza: This is not the case. I will change the question. – 2017-01-29
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    How about $a_1(x)=-\sin(x)$ and $$a_n(x)=\frac{(-1)^n}{(2n-3)!}x^{2n-3},\quad n\ge 2.$$ Then on the LHS we have Taylor series for $\sin(x)$ and equation (*) also holds for all $x \in \mathbb R$. – 2017-01-29
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    @ZoranLoncarevic the only problem I see with that is that the condition $a_n(x)\neq0$ isn't satisfied at zero. Maybe this can be circumvented, e.g. by using the domain $(0,\infty $ if that's allowed. – 2017-01-29
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    @Blaza We could also define functions $a_i(x)$ as above for $x \neq 0$, and set $a_n(0)$ to arbitrary non-zero values so that $$\sum_{n=0}^\infty a_n(0)=0.$$ – 2017-01-29
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    @ZoranLoncarevic: So a unicity of the solution $x$ is needed before solving that equation. – 2017-01-29
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    @E.J "*So a unicity of the solution is needed before solving that equation*" I don't know what do you mean by that. – 2017-01-29
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    @ZoranLoncarevic: For $x=a$ is the only solution, we must proves that the equation has only one solution. – 2017-01-29
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    @ZoranLoncarevic I agree completely. We could even set $a_n$ to be piecewise constant functions to get every x as a solution. We are getting a lot of counterexamples. I guess there should be some quite stronger conditions to get uniqueness of x=a. I'm not sure even if continuity of $a_n$ is enough. – 2017-01-29
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    @E.J Well, "to prove that equation has only one solution we must prove that the equation has only one solution" is certainly a tautology. – 2017-01-29
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    @ZoranLoncarevic: I mean that we must proves that the equation has one root and then we find it. – 2017-01-29
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    @E.J can you tell us more about what you have tried and where the confusion arises? Zoran Loncarevic gave a counterexample with $a_1=-\sin x$. Have you checked that that really is a counterexample? Then we can't prove that there is only one solution, because that is not true. – 2017-01-29

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