Quartic Formulation of tan(x) = x

Question

I want to convert the above equation into quartic form. The only thing I can think of is to use the Taylor series expansion of tan(x). Is there any other way, anyone can suggest to convert the above equation into quartic form.

Answer 1

If you want the roots of the equation $\tan(x)=x$, as said in comments, there is an inifinite number of solutions closer and closer to $(2n+1)\frac \pi 2$.

Consider instead that you look for the zero's of $$f(x)=\sin(x)-x \cos(x)$$ and develop as series around $\color{blue}{x_{(n)}=(2n+1)\frac \pi 2}$. $$f(x)=1+x_{(n)} (x-x_{(n)})+\frac{1}{2} (x-x_{(n)})^2-\frac{1}{6} x_{(n)} (x-x_{(n)})^3-\frac{1}{8} (x-x_{(n)})^4+O\left((x-x_{(n)})^{5}\right)$$

Using it for $n=4$ and solving the quartic would give as an approximate solution $14.0661939128333$ while the exact solution, obtained using Newton method, is $14.0661939128315$.

Going one step further, you could even use series reversion to get $$x_n=x_{(n)}-\frac{1}{x_{(n)}}-\frac{2}{3 x_{(n)}^3}-\frac{19}{24 x_{(n)}^5}-\frac{5}{8 x_{(n)}^7}+O\left(\frac{1}{x_{(n)}^9}\right)$$ Applied to the case where $n=4$, this would give $x_4=14.06619405$