# 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?

• You have proven that the polynomial has at most four real roots. – Batominovski Aug 14 '15 at 19:25
• For a different approach: The form of the polynomial suggest that studying a function like $P_n(x) = e^{-x}[1 + x + \frac{x^2}{2!} + \ldots + \frac{x^n}{n!}]$ might simplify the analysis. In fact $\frac{dP_{n}(x)}{dx} = -\frac{x^{n}e^{-x}}{n!}$ so if $n$ is even the derivative is always negative. Finally since $\lim_{x\to\infty} P_n(x) = 0$ it follows that the polynomial cannot have any real roots. – Winther Aug 14 '15 at 19:37
• @EmmadKareem That would only mean there are no positive real roots, which is quite obvious. – Macavity Aug 16 '15 at 3:19
• @SujithZis derivative being zero is necessary for multiple roots, not sufficient. E.g. Think of $x^2+1$, the derivative has a zero, but the polynomial has no real roots. – Macavity Aug 16 '15 at 3:27
• @EmmadKareem You could also check for negative roots, but after replacing $x \mapsto -x$. However in this case it is inconclusive at it allows upto $6$ negative roots. – Macavity Aug 16 '15 at 4:18

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.

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.

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 $x.$

• (+1) This is a good answer, and it forced me to post a couple of other approaches. – robjohn Sep 15 '20 at 0:00
• I have appended a generalization of your answer to my second answer. I hope that is okay. – robjohn Sep 15 '20 at 15:39

\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.

• Note that $x^4+180x^2+720x+720=x^4+180(x+2)^2$. Hence, there is no need for derivatives. – Batominovski Aug 14 '15 at 21:15

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 $ae^{bz}$.

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.

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.

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.

• you better delete the comment. The answer is right, and I have made a small edit. The forum requires politeness in spite of downvotes – vidyarthi Sep 13 '20 at 17:30

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, $$\bbox[5px,border:2px solid #C0A000]{e_n(x)\gt0\quad\text{for even n and all x\in\mathbb{R}}}\tag{15}$$