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I'm using this method:

First, write the polynomial in this form: $$a_nx^n+a_{n-1}x^{n-1}+......a_2x^2+a_1x=c$$ Let the LHS of this expression be the function $f(x)$. I'm gonna write the Taylor series of $f^{-1}(x)$ around x=0 and then put $x=c$ in it to get $f^{-1}(c)$ which will be the value of $x$.

Since, $f^{-1}(0)=0$ here, so we've got the first term of our Taylor series as $0$.

Now, the only thing that remains is calculating the derivatives of $f^{-1}(x)$ at $x=0$.

I'm using the fact that $$\frac{d(f^{-1}(x))}{dx}=\frac{1}{f'(f^{-1}(x))}$$

By differentiating this equation, we can get the second derivative of f−1(x) as: $$\frac{d^2(f^{-1}(x))}{dx^2}=-\frac{1}{(f'(f^{-1}(x)))^2}\cdot f''(f^{-1}(x))\cdot (f^{-1})'(x)$$

Similarly, we can get the other derivatives by further differentiation of this equation. Then we can evaluate all the derivatives at x=0 to get the Taylor series of $f^{-1}(x)$ and evaluate it at $x=c$ to get the value of $x$.

I don't know the formula of $f^{-1}(x)$ but I know the value of $f^{-1}(x)$ at $x=0$. After doing all the formulas, what I have to do in the end in evaluating that expression at $x=0$ and I've the value of $f^{-1}(x)$ at $x=0$. For example, $$f'^{-1}(x)=\frac{d(f^{-1}(x))}{dx}=\frac{1}{f'(f(^{-1}(x))}$$ $$=\frac{1}{n*a_{n}(f(^{-1}(x)))^{n-1}+(n-1)a_{n-1}(f(^{-1}(x)))^{n-2}+.............2a_2f^{-1}(x)+a_1}$$ which gives $$\frac{1}{a_1}$$ at $x=0$ since $$f^{-1}(0)=0$$ Similarly, $$\frac{d^2(f^{-1}(x))}{dx^2}=-\frac{1}{(f'(f^{-1}(x)))^2}\cdot f''(f^{-1}(x))\cdot (f^{-1})'(x)$$

We already have the value of the first derivative at $x=0$ so we can substitute that here, so $$\frac{d^{2}}{dx^{2}}(f^{-1}(x))=-\frac{1}{a_1^{2}}\cdot 2a_2\cdot \frac{1}{a_1}$$ $$=-\frac{2a_2}{a_1^{3}}$$ I think this process can be continued to get more derivatives by the product rule.

1.Is this method correct?

2.Can something be done to make it better and remove the limitations (if there are any)?

UPDATE: If I'm not wrong, then I think this method only works if none of the coefficients of the polynomial are zero. Is there some way to remove that limitation?

UPDATE: Oh, I just figured out that we can obtain the Taylor series of the inverse of a polynomial around any point $x=a$ by this method. I also just found out about the Lagrange inversion theorem. It was also about getting Taylor series of inverse functions. I didn't understand much but it was the same series as mine except the coefficients. Are the coefficients also the same? Have I been doing the same thing?

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    As regards the typesetting, you don't need to do anything special to make $'$ appear as a superscript. Just type f' in your mathematics and it will be displayed just fine: $f'$. – 2017-02-21
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    How can you not write prime as a superscript? It's just f'(x) and looks like $f'(x)$. You could alternatively use the notation $f^{(n)}(x)$ for the nth derivative – 2017-02-21
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    http://mathworld.wolfram.com/SeriesReversion.html – 2017-02-21
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    I now understand the OP's problem: you should use brackets, $(f^{-1})'$, if you want to typeset the derivative of the inverse of a function. $\LaTeX$ and MathJax will barf or give bad results for input along the lines of $f^{-1}'$. (And quite right too: $f^{{-1}{'}}$ and $\mbox{$f^{-1}$}'$ are both incomprehensible and grotesque.) – 2017-02-21
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    I am confused as to how you got $f^{'-1}(x)$ at the end of your formula for the second derivative ... google Lagrangian Inversion & look at the wikipedia page. https://en.wikipedia.org/wiki/Lagrange_inversion_theorem – 2017-02-21
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    @DonaldSplutterwit Sorry, I don't know advanced mathematics. But it's just the chain rule. I'm denoting the derivative of the inverse of $f(x)$ by $f'^{-1}(x)$ – 2017-02-21
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    $f'^{-1}$ can only mean the function that is inverse to the derivative of $f$ so your revised notation is worse than before. Moving on to the mathematics in your question, I suggest you try your method out on simple examples, e.g, $f(x) = 1 + x$. – 2017-02-21
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    @RobArthan Okay. I'll edit it. – 2017-02-21
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    @RobArthan Does my method seem correct? – 2017-02-21
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    @Dove Yeah right, the chain rule ... I get $(f^{-1}(x))^{'}$ which you calculated to be $1/f^{'}(f^{-1}(x))$ ? – 2017-02-21
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    @DonaldSplutterwit I had used a different notation to avoid too many braces. I had written in the top of the post then about the notation. Now, I've edited the post with proper notation. – 2017-02-21
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    @Dove: as I said, I suggest you try your method out on simple examples to see if it helps you to find roots: I don't see how it can. – 2017-02-21
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    *Style comment:* Don't use an asterisk $*$ for multiplication. Use either nothing or a raised dot $\cdot$ which is produced by \cdot . There's even $\times$ which is produced by \times . – 2017-02-22
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    \begin{eqnarray*} \frac{d^2 f^{-1}(x)}{dx^2}=\frac{-f^{''}(f^{-1}(x))}{(f^{'}(f^{-1}(x)))^3} \end{eqnarray*} – 2017-02-22
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    @DonaldSplutterwit Oh, Thanks. I always make elementary calculation mistakes. Thanks again. – 2017-02-22
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    @Dove look at the Lagrange inversion formula ... there are some absolutely delightful combinatorics hidden in these formulea ... your method does work, try it for a quadratic & you will discover the Catalan numbers ... good luck exploring & discovering. – 2017-02-22
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    @RobArthan Obviously, it does work for $f(x)=x+1$. By my method, the first derivative of its inverse at $x=0$ will come out to be 1. There are no further derivatives there. So, its Taylor series at $x=0$ will be (x-0)*1, which when evaluated at x=-1, gives -1 which is the zero of the polynomial. – 2017-02-22
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    @Dove: so keep going: how does your method work for quadratic $f$ or cubic $f$ or .... – 2017-02-22
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    @DonaldSplutterwit If I'm not wrong, I think this method only works if none of the coefficients of polynomial are zero. Is there some way to remove that limitation? And, I looked up Lagrange inversion formula. It said that it gives the taylor expansion for the inverse function. It is possible to get one series for polynomials also by that method? – 2017-02-22

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\begin{eqnarray*} f(x)=x+a_1x^2+a_2x^3+a_3x^4\cdots \end{eqnarray*} The inverse function needs to satisfy $f(f^{[-1]}(x))=x$ so \begin{eqnarray*} f^{[-1]}(x)=x-a_1x^2+(2a_1^2-a_2)x^3+(-5a_1^3+5a_1a_2-a_3)x^4 \cdots \end{eqnarray*}

Now set $a_2=0$ and $a_3=0$ etc... \begin{eqnarray*} f^{[-1]}(x)=x-a_1x^2+2a_1^2x^3-5a_1^3x^4 +14x^5\cdots \end{eqnarray*} This sum is also \begin{eqnarray*} f^{[-1]}(x)=\frac{1-\sqrt{1-4ax}}{2a} \end{eqnarray*} This is overkill to derive the formula for the solution of a quadratic but if you now allow $a_2$ to be non zero you will have a series solution for a cubic equation & so on.

Your method will work ... it is just Lagrangian Inversion in disguise !

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    Are the coefficients obtained by my expansion also the same as the ones in Lagrangian expansion? I didn't understand much about Lagrangian inversion theorem. Am I doing the same thing? – 2017-02-22