Find the number of real roots of the polynomial: $f(x)=x^{11}-14x+19$?

Question

Q) Find the number of real roots of the polynomial: $f(x)=x^{11}-14x+19$

Ans) First of all let me show my approach. I was trying to find the number of real roots of this polynomial by applying the Descarte's Rule Of Signs, which states that:

The maximum number of positive real roots of a polynomial equation $f(x)=0$ is the number of changes of sign in $f(x)$ and the maximum number of negative real roots of a polynomial equation $f(x)=0$ is the number of changes of sign in $f(-x)$.

Here $f(x)=x^{11}-14x+19$ and $f(-x)=-x^{11}+14x+19$.

Clearly, the no. of changes in sign in $f(x)$ is $2$ and the number of changes in sign in $f(-x)$ is $1$. Therefore we get the total no. of real roots to be $3$. But there is only 1 real root. I can't understand where is my mistake.

On a side note, why does Descarte's Rule Of Signs not work for the polynomial equation $f(x)=x^{2}-x+1$?

Answer 1

Descartes rule of signs immediately gives that $f(x)$ has a maximum of two positive roots (or none), and exactly one negative root.

As we do not have a fix on the number of positive roots, we need to investigate $f(x)$ a bit more. In this case it is easy to apply AM-GM and then Bernoulli's inequality (if you don't want to calculate the exponentiation), like so: $x^{11}+19 = $ $$x^{11}+10\cdot \frac{19}{10} \geqslant 11\left(\frac{19}{10} \right)^{\frac{10}{11}}x = 11\cdot\frac1{(1-\frac{9}{19})^{10/11}} > 11\cdot \frac1{1-\frac{90}{209}}=\frac{2299}{119} x > 14x,$$ hence there cannot be a positive root.

Answer 2

Recall Descartes' Rule of Signs states that the number of positive real roots of a polynomial equation is at most equal to the number of changes of signs of the coefficients when written in descending powers of $x$, or it is less than it by an even number. In this case, $f(x)$ $=$ $x^{11}-4x+19$, as you say, has 2 changes of sign, so it has at most 2 positive real roots, or less than that by an even number. It actually has $0$ positive real roots, and only has the one real root at $x \approx -1.394$

Answer 3

For this type of problem, I would forget Descarte's Rule of Signs, and simply explore the first and second derivatives of the function.

The way to determine how many real roots that the polynomial has is by determining how many times the continuous function crosses the $~x$-axis.

You can examine the first derivative to determine when the derivative is positive, and when it is negative. You can also examine the 2nd derivative to get a reasonable idea about how the first derivative behaves.

You can also spot check the value of the original function at two or three positive values for $~x,~~x = 0,~$ and two or three negative values of $~x.$

This, coupled with your examination of the behavior of the first derivative should allow you to draw valid conclusions about the behavior of the function itself.

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Find the number of real roots of the polynomial: $f(x)=x^{11}-14x+19$

$~f'(x) = 11x^{10} - 14.$
$~f''(x) = 110x^9.~$

This means that the second derivative is decreasing for $~x < 0,~$ and increasing for $~x > 0.~$

Further, you have that $~f'(0) = -14.~$
Therefore, you can immediately conclude that the first derivative is strictly negative, for all $~x < 0.~$

Therefore, you can immediately conclude that the function is strictly decreasing on the interval $~(- \infty, 0].$

Further, you have that $~f(0) = 19.~$
Therefore, you can immediately conclude that the function $~f(x)~$ will cross the $~x$-axis exactly once in the interval $~(-\infty,0].$

Therefore, the problem now reduces to examining the behavior of $~f(x)~$ and $~f'(x),~$ for $~x > 0.~$

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From examination of the second derivative, you know that $~f'(x)~$ is strictly increasing on $~(0,+\infty).$

You also know that $~f'(0) = -14,~$ and $~\displaystyle f'(x_0) = 0,~$
where $~x_0 = \displaystyle ~\sqrt[10]{\frac{14}{11}} \implies 1 < x_0 < 1.5.~$

This implies that $~f(x)~$ is strictly decreasing on the interval $~(0,x_0)~$ and strictly increasing on the interval $~(x_0,+\infty).~$

You also have that $~f(0) = 19, ~f(1) = 6, ~$ and $~f(2) ~$ is much greater than $~0.~$

Therefore, you know that as $~x~$ ranges from $~0~$ through $~2,~$ that $~f(x)~$ has the shape of a valley. So, the sole remaining question is: does the bottom of this valley [i.e. $~f(x_0)~$] occur below the $~x$-axis, on the $~x$-axis, or above the $~x$-axis?

Then:

• $~x^{10} = \dfrac{14}{11},~$ and $~1 < x < 1.5.$
  Therefore, $1 < ~x^{11} < 3.$

• Similarly $14 < 14x < 21.$

At this point, it seems that $~f(x_0) > 0~$ but the situation is a little too close to call, without a calculator. The calculator confirms that $~f(x_0) > 0.~$

Therefore, $~f(x)~$ never crosses the $~x$-axis in the interval $~(0,+\infty).$

Therefore, $~f(x)~$ has exactly one real root, and that root is negative.

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$\underline{\text{Addendum}}$

I do advocate using intermediate results, such as Descarte's Rule of Signs, to ease the work. That is, I do not advocate re-inventing the wheel.

However, for this particular problem, there were two compelling points:

• The problem asked for the exact number of real roots.

• The nature of the $~f(x)~$ polynomial immediately indicated that exploration of $~f'(x)~$ and therefore $~f''(x),~$ would be elementary.
  
  That is, $~f'(x)~$ is a $~10$-th degree polynomial of form $~Ax^{10} = B.~$ When I realized that, I imagined that this was intentional on the part of the problem composer, and that the problem solver was supposed to take advantage of this simplicity.

Answer 4

Consider the following equation $$x\bigg( 1 + \frac{x}{r} \bigg)^r = x\cdot e_r(x)= z$$ assuming $r \in Q$ and $z \in C$. The solution $x$ is the so-called Lambert-Tsallis function, $W_r(z)$, proposed by R. V. Ramos (please see some answers in my profile). Also, $[e_r(x)]^\alpha = e_{r\alpha}(x\alpha)$. As can be seen, $W_r(z)$ is a multivalued function and it has a lot of interesting properties.

Now, one can write your problem like

$$-14x\bigg(1 -\frac{x^{10}}{14}\bigg) = -19$$ $$x \cdot e_1 \bigg(-\frac{x^{10}}{14}\bigg) = \frac{19}{14}$$ lets raise to $\color{red}{r=10}$ $$x^re_r \bigg(-\frac{r}{14}x^{10}\bigg) = \bigg(\frac{19}{14}\bigg)^r$$ which produces $$\frac{W_r\bigg[ -\frac{r}{14}\bigg(\frac{19}{14}\bigg)^r\bigg]}{-\frac{r}{14}} = x^r$$

raising again to ^$\frac{1}{r}$ we have

$$ \color{red} {x = \frac{19/14}{1 + \frac{W_r(z)}{r} } }$$

with $z= -\frac{r}{14}\bigg(\frac{19}{14}\bigg)^r$

Properties:

1. $W_r(z)$ when $r \in Z_+$ has $r+1$ values. In your case $r=10$ so $x$ will have 11 solutions (we already know)
2. The real roots of $W_r(z)$ when $r$ is even will assume 3 branches (The Lambert function has 2 real branches and it looks like the picture below.

Results $W_{10}(z) = \{ 1.5265 + 2.4833i, 1.5265 - 2.4833i, -2.2131 + 7.3439i, -2.2131 - 7.3439i, -7.7444 + 9.9328i, -7.7444 - 9.9328i, -13.6206 + 9.2325i, -13.6206 - 9.2325i, -19.7383 + 0.0000i, -18.0793 + 5.5122i, -18.0793 - 5.5122i\} $ which produces $x = \{ 1.1252 - 0.2424i, 1.1252 + 0.2424i, 0.9224 - 0.8699i, 0.9224 + 0.8699i 0.2951 - 1.2993i, 0.2951 + 1.2993i, -0.4996 - 1.2740i, -0.4996 + 1.2740i, -1.3936 + 0.0000i, -1.1462 - 0.7820i, -1.1462 + 0.7820i\} $

In your problem, you can see that $z= -\frac{10}{14}\bigg(\frac{19}{14}\bigg)^{10} < z_b=-(\frac{r}{r+1})^{r+1} = -\frac{10}{11}^{11}$ so there is only one real root to $W_{10}(z)$ and, as a consequence, to $x$.