11
$\begingroup$

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?

$\endgroup$
10
  • 2
    $\begingroup$ You have proven that the polynomial has at most four real roots. $\endgroup$ – Batominovski Aug 14 '15 at 19:25
  • 5
    $\begingroup$ 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. $\endgroup$ – Winther Aug 14 '15 at 19:37
  • 1
    $\begingroup$ @EmmadKareem That would only mean there are no positive real roots, which is quite obvious. $\endgroup$ – Macavity Aug 16 '15 at 3:19
  • 1
    $\begingroup$ @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. $\endgroup$ – Macavity Aug 16 '15 at 3:27
  • 1
    $\begingroup$ @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. $\endgroup$ – Macavity Aug 16 '15 at 4:18
15
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $x.$

$\endgroup$
2
  • $\begingroup$ (+1) This is a good answer, and it forced me to post a couple of other approaches. $\endgroup$ – robjohn Sep 15 '20 at 0:00
  • $\begingroup$ I have appended a generalization of your answer to my second answer. I hope that is okay. $\endgroup$ – robjohn Sep 15 '20 at 15:39
4
$\begingroup$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ Note that $x^4+180x^2+720x+720=x^4+180(x+2)^2$. Hence, there is no need for derivatives. $\endgroup$ – Batominovski Aug 14 '15 at 21:15
2
$\begingroup$

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} $.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ you better delete the comment. The answer is right, and I have made a small edit. The forum requires politeness in spite of downvotes $\endgroup$ – vidyarthi Sep 13 '20 at 17:30
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.