Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.

Question

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?

Answer 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.

Answer 2

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.

Answer 3

$$ \begin{align} \sum_{i=1}^6 \dfrac {x^i} {i!} &=\dfrac 1 {720} \cdot (x^6+6x^5+30x^4+120x^3+360x^2+720x+720= \\ &=\dfrac 1 {720} \cdot \{x^4(x+3)^2+20x^2(x+3)^2+x^4+180x^2+720x+720\} \end{align} $$

It can be easily proved that $x^4+180x^2+720x+720 > 0$ by using the derivative. Therefore, there are no real roots.

Answer 4

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 $x.$

Answer 5

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} $.

Answer 6

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.

Answer 7

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.

Answer 8

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 $e^x$.

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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 $$ \bbox[5px,border:2px solid #C0A000]{e_n(x)\gt0\quad\text{for even $n$ and all $x\in\mathbb{R}$}}\tag9 $$

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Here is a solution that extends the answer by abel.

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

Answer 9

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 $f(x)=0$.