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

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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)

Answer 4

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.

Answer 5

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.