# Real roots of the equation $1+\sum_{r=1}^{7}\frac{x^{r}}{r} = 0$

The number of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7} = 0$

$\bf{My\; Try}::$ Let $\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}$

Now $\displaystyle f^{'}(x) = 1+x+x^2+x^3+x^4+x^5+x^6$

and $\displaystyle f^{''}(x) = 1+2x+3x^2+4x^3+5x^4+6x^5$

$\displaystyle f^{'''}(x) = 0+2+6x+12x^2+20x^3+30x^4 = 2\left(1+3x+6x^2+10x^3+15x^4\right)$

Now I did not understand how can i solve it

Help Required

Thanks

Note that $$f'(x)=\sum_{k=0}^6 x^k=\frac{x^7-1}{x-1}.$$ Now, it is not too hard to see that $f'(x)>0$ for all $x\in\mathbb{R}$ (consider regions $x<-1$, $-1<x<0$, $0<x<1$ and $x>1$). So, there is a real root because the degree of the polynomial is odd, but there is only one because the function is monotonically increasing.

• This is not a valid answer if the only tag is "algebra-precalculus" ;-) But since the OP started differentiating, maybe we should just edit the tags... – Julien Dec 24 '13 at 3:45
• That's a good point. I hadn't looked at the tags. – Jonathan Dec 24 '13 at 3:53

Observe that

$(x - 1)f'(x) = (x - 1)(\sum_0^6 x^i) = x^7 - 1, \tag{1}$

and that the polynomial $x^7 - 1$ has exactly one real zero, $x = 1$. Thus the zeroes of $f'(x)$ must be the remaining zeroes of $x^7 - 1$, which are the six complex $7$-th roots of unity $e^{2 \pi i / 7}$ for $1 \le i \le 6$. This shows that $f'(x)$ has no real zeroes; and since $f'(1) = 7 > 0$, $f'(x) > 0$ for all $x \in \Bbb R$. Thus

$f(x) = \sum_0^7 (x^r /.r) \tag{2}$

is monotonically increasing everywhere. Being of odd degree, $f(x)$ has at least one real zero; being monotonically increasing, it can have no more.

Hope this helps. Good Yule to One and All,

and as always,

Fiat Lux!!!

Law of signs is covered in pre-calc/algebra. Using the standard law of signs, one has to show that there is exactly one real root between -1 and -2. Now I am rusty on how this is done (I am 60+ and I learnt it 40 years ago!) but goes something like this.

\begin{align} f(x) &= x^7/7 + x^6/x + x^5/5 + x^4/4 + x^3/3 + x^2/2 + x^1/1 + 1 && \hbox{No sign changes}\\ f(-x) &= -x^7/7 + x^6/x - x^5/5 + x^4/4 - x^3/3 + x^2/2 - x^1/1 + 1 && \hbox{7 sign changes}\\ \end{align} So we know that all the seven roots (if they exist) is to the left of $0$. Similarly $f(x-1)$ has no sign changes but $f(x-2)$ has 7 sign changes. So all the roots have to be between $-2$ and $-1$.

Most of the calculations are done using synthetic division and all this is covered in the precalc/algebra course. Hope someone can expand on this.

I will also dust off my precalc books and check if no one answers this