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?
 A: 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.
A: $$
\begin{align}
\sum_{i=1}^6 \dfrac {x^i} {i!} &=\dfrac 1 {720} \cdot (x^6+6x^5+30x^4+120x^3+360x^2+720x+720= \\
&=\dfrac 1 {720} \cdot \{x^4(x+3)^2+20x^2(x+3)^2+x^4+180x^2+720x+720\}
\end{align}
$$
It can be easily proved that $x^4+180x^2+720x+720 > 0$ by using the derivative. Therefore, there are no real roots.
A: 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 $x.$
A: More generally,
a Google search
for "partial sums of exponential series"
 turned up this:
https://www.math.washington.edu/~morrow/336_09/papers/Ian.pdf
This paper shows that,
in particular,
if
$s_n(z)
=\sum\limits_{k=0}^n \frac{z^k}{k!}
$,
then,
if
$p_n(z)
=s_n(nz)
$,
the zeroes of
$p_n(z)$
fall asymptotically
near the curve
$$\Gamma
=\{z: |ze^{1-z}| = 1,
|z| \le 1\}.
$$
This paper also has
this surprising characterization
of the exponential function:
Theorem 3.7. Suppose 
$f(z) 
=\sum\limits_{k=0}^{\infty} a_k z^k$
is an entire function. 
The following
two statements are equivalent:
(i) There is a positive number 
$c$ such that for each $n$,
 the function
$\sum\limits_{k=0}^{n} a_k z^k$
has no zeroes with norm less than $cn$.
(ii) The function $f$
 can be represented as 
$ae^{bz}
$.
A: First, rescale by $720$ to get integer coefficients: $$x^6+6x^5+30x^4+120x^3+360x^2+720x+720$$
Now repeated completion of binomial powers:
$$\begin{align}
&\phantom{{}={}}(x+1)^6+15x^4+100x^3+345x^2+714x+719\\
&=(x+1)^6+15(x+5/3)^4+95x^2+\frac{3926}{9}x+\frac{16288}{27}\\
\end{align}$$
You could complete the square again on this last quadratic and you will be left with a positive constant, or you can just compute its discriminant to see that the quadratic itself has no roots (and has a positive quadratic term, so is therefore positive). So $$(x+1)^6+(x+5/3)^4+q(x)$$ is positive.
A: Here I am giving a proof without using exponential series and Taylor's series. Let $E(x)= 1+ \frac{x}{1!} +\ldots+\frac{x^n}{n!}$. Then $E(x) = 0$ has no repeated real root. If not, let $a \in \Bbb{R}$ be a repeated real root for $E(x) =0$. Then $E(a) =0$ and $E'(a) = 0$. These two together imply that $a = 0$ but $0$ is not a root of $E(x)$.
Case 1 : When $n$ is odd.
If $n$ is odd then $E(x)= 0$ has at least one real root and such a root must be negative ($E(x)= 0$ does not have any positive real root as the change of sign of the coffecients is $0$). So there exist $b \in \Bbb{R}$ such that $b < 0$ and $E(b) = 0$.  Since $E(x)$ has no repeated real root by Rolle's Theorem between two distinct roots of $E(x) = 0$, there is a root of $E'(x) = 0$. Now I prove that $E'(x)$ does not have any real root. Note that $E'(x)$ is similar to $E(x)$ but with $n$ even.
Case 2 : $n$ is even, $n = 2m$
Then $E(x) = 1+ \frac{x}{1!}+\ldots+\frac{x^{2m}}{(2m)!}$. Then $E'(x) = 1 + \frac{x}{1!}+\ldots+\frac{x^{2m-1}}{(2m-1)!}$. Then there exist $b \in \Bbb{R}$ such that $b < 0$ and $E'(b) = 0$. Then $E''(b) = -\frac{b^{2m-1}}{ (2m-1)!}> 0$ and therefore at $x= b$,  $E(x)$ has a minimum value. But $E(b) = \frac{b^{2m}}{(2m)!}> 0$. Now $E(x)$ is a polynomial function which admits only minimum value (when $n$ is even) and the minimum value is always positive. So $E(x)$ is always positive (when $n$ is even). As a result, $E(x)$ has no real root if $n$ is even and exactly one real root if $n$ is odd.
A: 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.  
A: I think the notion that the fourth derivative having no real roots proves that the polynomial itself has four real roots is your problem.  Can you explain your reasoning a bit more?  I mean to say, $x^6+x^4+1$ clearly has no real roots (it is everywhere positive), but its fourth derivative $360x^2+24$ has no real roots either (it is likewise everywhere positive).
The polynomial in your problem does indeed have no real roots.
A: Previously, I posted an answer using Sturm Chains. That answer did not give any idea whether this was true for all even-ordered truncations of the Taylor Series for $e^x$.  Here is an answer that works for all even-ordered truncations of the Taylor Series for $e^x$.

Polynomial Product Solution
Define
$$
e_n(x)=\sum_{k=0}^n\frac{x^k}{k!}\tag1
$$
Then
$$
\begin{align}
e_n(x)\,e_n(-x)
&=\sum_{k=0}^n\frac{x^k}{k!}\sum_{j=0}^n(-1)^j\frac{x^j}{j!}\tag2\\
&=\sum_{k=0}^{2n}\sum_{j=\max(0,k-n)}^{\min(k,n)}(-1)^j\binom{k}{j}\frac{x^k}{k!}\tag3\\
&=\sum_{k=0}^{2n}\sum_{j=k-n}^n(-1)^j\binom{k}{j}\frac{x^k}{k!}\tag4\\
&=\sum_{k=0}^{2n}\sum_{j=0}^k\left[(-1)^n\binom{-1}{n-j}-(-1)^{k-n-1}\binom{-1}{k-n-1-j}\right]\binom{k}{j}\frac{x^k}{k!}\tag5\\
&=\sum_{k=0}^{2n}\left[(-1)^n\binom{k-1}{n}+(-1)^{k-n}\binom{k-1}{k-n-1}\right]\frac{x^k}{k!}\tag6\\
&=1+(-1)^n\sum_{k=1}^{2n}\left(1+(-1)^k\right)\binom{k-1}{n}\frac{x^k}{k!}\tag7\\[3pt]
&=1+(-1)^n\sum_{k=1}^n2\binom{2k-1}{n}\frac{x^{2k}}{(2k)!}\tag8
\end{align}
$$
Explanation:
$(2)$: apply $(1)$
$(3)$: swap order of summation, substitute $k\mapsto k-j$,
$\phantom{\text{(3):}}$ swap order of summation again
$(4)$: when $k\lt j\le n$, $\binom{k}{j}=0$, so set the upper limit to $n$
$\phantom{\text{(4):}}$ when $k-n\le j\lt 0$, $\binom{k}{j}=0$, so set the lower limit to $k-n$
$(5)$: $(-1)^n\binom{-1}{n-j}-(-1)^{k-n-1}\binom{-1}{k-n-1-j}=(-1)^j[k-n\le j\le n]$
$\phantom{\text{(5):}}$ and $\binom{k}{j}$ set the limits; we only need to sum $0\le j\le k$
$(6)$: Vandermonde's Identity
$(7)$: the $k=0$ term is $1$ and for $k\ge1$, $\binom{k-1}{k-n-1}=\binom{k-1}{n}$
$(8)$: terms for odd $k$ are $0$; substitute $k\mapsto2k$
For even $n$, $e_n(x)\,e_n(-x)\ge1$. Since $e_n(x)\ge1$ for $x\ge0$, $(8)$ says that
$$
\bbox[5px,border:2px solid #C0A000]{e_n(x)\gt0\quad\text{for even $n$ and all $x\in\mathbb{R}$}}\tag9
$$


Here is a solution that extends the answer by abel.

Differential Equation Solution
Define $e_n(x)$ as in $(1)$. Then
$$
e_n(t)-e_n(t)'=\frac{t^n}{n!}\tag{10}
$$
Solving via integrating factor:
$$
\begin{align}
e^x\int_x^\infty\frac{t^n}{n!}e^{-t}\,\mathrm{d}t
&=e^x\int_x^\infty\left(e_n(t)-e_n(t)'\right)e^{-t}\,\mathrm{d}t\tag{11}\\
&=e^x\int_x^\infty\left(-e_n(t)e^{-t}\right)'\,\mathrm{d}t\tag{12}\\[3pt]
&=e^x\left[-e_n(t)e^{-t}\right]_x^\infty\tag{13}\\[9pt]
&=e_n(x)\tag{14}
\end{align}
$$
For even $n$, it is easy to see that the left hand side of $(11)$ is positive for all $x$, so $(14)$ is also positive. That is,
$$
\bbox[5px,border:2px solid #C0A000]{e_n(x)\gt0\quad\text{for even $n$ and all $x\in\mathbb{R}$}}\tag{15}
$$
A: Let $f(x)= 1+x+\frac{x^2}{2!}+\ldots +\frac{x^6}{6!}$
Differentiating $f(x)$:
$$f'(x)= 1+x+\frac{x^2}{2!}+\ldots+\frac{x^5}{5!}$$
Consider $a$ as one real root for $f'(x)=0$, i.e. $f'(a)=0$. Obviously $a\ne 0$.
Now, $f(a)= 0+ \frac{a^6}{6!}>0$. This means that all the potential local minima of $f(x)$ are already positive. Hence, there are no real roots for $f(x)=0$.
