# no. of real roots of the equation $1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+…+\frac{x^7}{7} = 0$

The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$

$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}$

So $$\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$$ Then Differentiate

both side w . r to $x\;,$ We get $$\displaystyle f'(x)=1+x+x^2+x^3+..........+x^6$$

Now for max. and Minimum Put $$f'(x) = 0\Rightarrow 1+x+x^2+x^3+x^4+x^5+x^6 = 0$$

We can write $f'(x)$ as $$\displaystyle \left(x^3+\frac{x^2}{2}\right)^2+\frac{3}{4}x^4+x^3+x^2+x+1$$

So $$\displaystyle f'(x) = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[3x^4+4x^3+4x^2+4x+4\right]$$

So $$\displaystyle f'(x) = = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[\left(\sqrt{2}x^2+\sqrt{2}x\right)^2+(x^2)^2+2(x+1)^2+2\right]>0\;\forall x\in \mathbb{R}$$

So $f'(x) = 0$ does not have any real roots. So Using $\bf{LMVT}$ $f(x) = 0$ has at most one root.

In fact $f(x) = 0$ has exactly one root bcz $f(x)$ is of odd degree polynomial and it

will Cross $\bf{X-}$ axis at least one time.

My question is can we solve it any other way, i. e without using Derivative test.

Help me , Thanks

• Are you sure that no real roots for this polynomial? – Mhenni Benghorbal Feb 1 '15 at 13:09
• Could you make use of Descartes Rule of Signs? – Mufasa Feb 1 '15 at 13:11
• Can you please tell me what is LMVT – Snehil Sinha Feb 1 '15 at 13:11
• It should have a real root! – Mhenni Benghorbal Feb 1 '15 at 13:12
• It should have exactly one real root. – user 170039 Feb 1 '15 at 13:13

$f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$ So, as you note,

$f'(x)=1+x+x^2+x^3+..........+x^6$. Now consider $x$ in three ranges

1. For $x \ge 0$ then clearly $f'(x) >0$
2. For $-1 \le x < 0$, $f'(x)=(1+x) + (x^2+x^3) + (x^4+x^5) + x^6$. Each bracketed term is non-negative, and $x^6$ is positive, so $f'(x) >0$
3. For $x < -1$, $f'(x)=1 +(x + x^2) +(x^3 + x^4) +( x^5 + x^6)$. Again, each bracketed term is positive so $f'(x) >0$

From that conclude that $f(x)$ is a monotonic increasing function and can therefore have at most 1 zero.

Finally, as $x \to -\infty$, $f(x) \to -\infty$ and as $x \to +\infty$, $f(x) \to +\infty$ so that by continuity $f(x)$ must have at least one zero, and from the previous it therefore has exactly one zero.

The simplest way of seeing that it can have only one real root is IMO to look at the derivative. To that end we use the formula for a geometric sum $$f'(x)=1+x+x^2+\cdots+x^6=\frac{x^7-1}{x-1}$$ showing that $f'(x)>0$ whenever $x\neq1$, because the numerator and denominator both change signs only at $x=1$. Of course, $f'(1)=7>0$. So the function is everywhere increasing, and cannot have more than one zero. As you observed, being an odd degree polynomial it clearly has at least one zero.

Note that $(1+x+x^2+x^3+x^4+x^5+x^6)$ has $z, z^2,z^3,z^4,z^5,z^6$ as roots, thus f(x) should have atmost one real root (MVT)

Let

$$f(x)=1+\frac x1+\cdots+\frac{x^7}{7}$$ Assume that there are two different roots $a_1$ and $a_2, (a_1<a_2)$ of the equation $f(x)=0$ so by the mean value theorem we get for $a_3\in(a_1,a_2)$

$$0=f(a_1)-f(a_2)=(a_1-a_2)f'(a_3)\implies f'(a_3)=1+a_3+a_3^2+\cdots+a_3^6=0$$ and using the geometric sum

$$1+x+\cdots+x^6=\frac{1-x^7}{1-x},\quad x\ne1$$ we see that $a_3$ doesn't exist so it exists at most one root of $f(x)=0$ and since the polynomial is odd then it exists at least one root. Hence the unicity of the root.

An odd polynomial has at least one real root is due to the fact that complex roots come in conjugate that is if $\alpha$ is a complex root of a polynomial then $\overline{\alpha}$ is a root too.

• This is known to the OP, has a much simpler reason, and does not suffice to answer the question. – Did Feb 1 '15 at 15:51
• @Did: After I gave the answer it became known for the OP!! – Mhenni Benghorbal Feb 1 '15 at 16:42
• @Did: Try not to depreciate people's work? – Mhenni Benghorbal Feb 1 '15 at 16:42
• No, OP knew about the fact before any of your posts on this question. Note that the question contains the following text: "In fact $f(x)=0$ has exactly one root bcz f(x) is of odd degree polynomial..." This was posted at 13:03, and since there is no history of a revision, the latest it could have been edited was 13:08. Your first comment was at 13:09. – epimorphic Feb 1 '15 at 16:52
• "This answer has been upvoted for three times!" For once, I agree with the exclamation mark, the fact that this answer was upvoted three times being truly amazing. – Did Feb 1 '15 at 18:34