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I have an equation of the form: $$f(x)=\sum_{n=2}^{\infty}a_{n}(x)=x^2+ax-1$$ where $x$ is a real number. that is, $$x^2+ax-1-f(x)=0$$ Solutions are: $$x=\frac{1}{2}\sqrt{a^2+4f(x)+4}-\frac{a}{2}$$ and $$x=-\frac{a}{2}-\frac{1}{2}\sqrt{a^2+4f(x)+4}$$ provided that: $$a^2+4f(x)+4>0$$ I have two versions: Replacing $f(x)$ by $x^2+ax-1$ to get $$a^2+4f(x)+4=a^2+4(x^2+ax-1)+4= (a+2x)^2>0$$ or replacing $f(x)$ by $\sum_{n=2}^{\infty}a_{n}(x)$ to get $a^2+4f(x)+4=a^2+4+4\sum_{n=2}^{\infty}a_{n}(x)$ which is hard to decide if it is positive or not.

My question is: Is the first approach correct?

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    Grammar note: $f$ is a function, $f(x)$ is a number. Everywhere you've written $f$ alone, you mean $f(x)$. Or maybe you mean to introduce a new variable $y$ satisfying $y = f(x)$ and use $y$ everywhere. – 2017-01-06
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    @Hurkyl: Yes you are right. the equation is edited – 2017-01-06
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    You are running in circles. What exactly is the input to the problem, what the output? What is known about or demanded of the functions $a_n(x)$? – 2017-01-06
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    @LutzL: It is the term of the series defining $f$. – 2017-01-06
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    So this means that the $a_n$ as well as $a$ are the input? And you are looking for the solutions $x$ of the equation? Then your transformation leads to a fixed point formula where you now have to investigate if it is contractive. You could also use other methods as the bisection,secant or regula falsi methods to find roots in a more systematic way. – 2017-01-06
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    @LutzL: So, my first approach correct. – 2017-01-06
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    Yes, if something like convergence can be proven. Else this transformation is nice but useless. $-$ The condition $a^2+4f(x)+4\ge0$ only tells you that you need not look for solution where it is not satisfied. As nothing more concrete is provided about $f$, this is the most information you get out of that. – 2017-01-06
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    @LutzL: Thank you very much. – 2017-01-06

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