<s>[INST] Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$. 
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.

Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no roots hence proving that above polynomial has $4$ real roots.
But using online calculators I get zero real roots. Where am I wrong?
 [/INST] We can compute the number of real roots using Sturm's Theorem.
$$
\begin{array}{rll}
\text{Sturm Chain}&+\infty&-\infty\\\hline
x^6+6x^5+30x^4+120x^3+360x^2+720x+720&+\infty&+\infty\\
6x^5+30x^4+120x^3+360x^2+720x+720&+\infty&-\infty\\
-5x^4-40x^3-180x^2-480x-600&-\infty&-\infty\\
-48x^3-432x^2-1728x-2880&-\infty&+\infty\\
45x^2+360x+900&+\infty&+\infty\\
384x+1920&+\infty&-\infty\\
-225&-225&-225
\end{array}
$$
There are $3$ changes of sign at $+\infty$ and $3$ changes of sign at $-\infty$. Thus, there are no real roots.
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