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?

Solution 1:

Let $E_n(x):=\sum_{k=0}^n\,\frac{x^k}{k!}$ for $n=0,1,2,\ldots$. We shall prove that $E_n(x)$ has no real roots if $n$ is even, and $E_n(x)$ has exactly one real root, which is simple, if $n$ is odd.

Suppose that $n$ is even. Clearly, $E_n(x)$ has no roots in $\mathbb{R}_{\geq 0}$. By Taylor's Theorem, we have $\exp(x)=E_n(x)+R_n(x)$, where the remainder term is given by $$R_n(x)=\int_0^x\,\frac{\exp^{(n+1)}(t)}{n!}\,(x-t)^n\,\text{d}t=\int_0^x\,\frac{\exp(t)}{n!}\,(x-t)^n\,\text{d}t\,.$$ If $x<0$, then $$R_n(x)=-\int_0^{|x|}\,\frac{\exp(-t)}{n!}\,|x+t|^n\,\text{d}t<0\,.$$ That is, $$E_n(x)=\exp(x)-R_n(x)>\exp(x)>0$$ for all $x<0$. That is, $E_n(x)$ has no negative roots either; i.e., $E_n(x)$ has no real roots.

If $n$ is odd, then $E'_n(x)=E_{n-1}(x)$ has no real roots. Thus, $E_n(x)$ can have at most one real root, due to Rolle's Theorem. Clearly, $E_n(x)$ has a real root, being a polynomial in $\mathbb{R}[x]$ of an odd degree. Consequently, $E_n(x)$ has exactly one real root, which is simple.

Solution 2:

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.

Solution 3:

let $y = 1+x/1!+x^2/2!+x^3/3! + \cdots + x^6/6! .$ it is clear that $y \ge 1$ for all $x \ge 0.$ we will show that $y(a) > 0$ for $a < 0$ and that will prove that $y$ is never zero.

pick an $a < 0.$ we have $$y' = y - x^6/6!, \space y(0) = 1.\tag 1$$

rearranging $(1)$ and multiplying by $e^{-x}$ gives $$ (ye^{-x})' = -x^6e^{-x}/6!.$$ integrating the last equation from $a$ to $0$ we get $$1-y(a)e^{-a}=-\int_a^0 x^6e^{-x}/6!\, dx\to y(a)e^{-a} = 1+\int_a^0 x^6e^{-x}/6!\, dx > 0$$

therefore $y(a) > 0$ and that concluded the claim that $y > 0$ for all $x.$