Riemann surface
matlab

1.3 Investigate the properties of the polynomial equation
$$f(x)=x^7+5 x^6+x^4-x^3+x^2-2=0,$$
by proceeding as follows.
(a) By writing the fifth-degree polynomial appearing in the expression for $f^{\prime}(x)$ in the form $7 x^5+30 x^4+a(x-b)^2+c$, show that there is in fact only one positive root of $f(x)=0$.
(b) By evaluating $f(1), f(0)$ and $f(-1)$, and by inspecting the form of $f(x)$ for negative values of $x$, determine what you can about the positions of the real roots of $f(x)=0$.

(a) We start by finding the derivative of $f(x)$ and note that, because $f$ contains no linear term, $f^{\prime}$ can be written as the product of $x$ and a fifth-degree polynomial:
\begin{aligned} f(x) & =x^7+5 x^6+x^4-x^3+x^2-2=0, \ f^{\prime}(x) & =x\left(7 x^5+30 x^4+4 x^2-3 x+2\right) \ & =x\left[7 x^5+30 x^4+4\left(x-\frac{3}{8}\right)^2-4\left(\frac{3}{8}\right)^2+2\right] \ & =x\left[7 x^5+30 x^4+4\left(x-\frac{3}{8}\right)^2+\frac{23}{16}\right] . \end{aligned}
Since, for positive $x$, every term in this last expression is necessarily positive, it follows that $f^{\prime}(x)$ can have no zeros in the range $0<x<\infty$. Consequently, $f(x)$ can have no turning points in that range and $f(x)=0$ can have at most one root in the same range. However, $f(+\infty)=+\infty$ and $f(0)=-2<0$ and so $f(x)=0$ has at least one root in $0<x<\infty$. Consequently it has exactly one root in the range.
(b) $f(1)=5, f(0)=-2$ and $f(-1)=5$, and so there is at least one root in each of the ranges $0<x<1$ and $-1<x<0$.

There is no simple systematic way to examine the form of a general polynomial function for the purpose of determining where its zeros lie, but it is sometimes

helpful to group terms in the polynomial and determine how the sign of each group depends upon the range in which $x$ lies. Here grouping successive pairs of terms yields some information as follows:
$x^7+5 x^6$ is positive for $x>-5$,
$x^4-x^3$ is positive for $x>1$ and $x<0$, $x^2-2$ is positive for $x>\sqrt{2}$ and $x<-\sqrt{2}$. Thus, all three terms are positive in the range(s) common to these, namely $-51$. It follows that $f(x)$ is positive definite in these ranges and there can be no roots of $f(x)=0$ within them. However, since $f(x)$ is negative for large negative $x$, there must be at least one root $\alpha$ with $\alpha<-5$.

1.5 Construct the quadratic equations that have the following pairs of roots:
(a) $-6,-3 ;$; (b) 0,4 ; (c) 2,$2 ;$; (d) $3+2 i, 3-2 i$, where $i^2=-1$.

Starting in each case from the ‘product of factors’ form of the quadratic equation, $\left(x-\alpha_1\right)\left(x-\alpha_2\right)=0$, we obtain:
(a) $(x+6)(x+3)=x^2+9 x+18=0$;
(b) $(x-0)(x-4)=x^2-4 x=0$;
(c) $(x-2)(x-2)=x^2-4 x+4=0$;
(d) $(x-3-2 i)(x-3+2 i)=x^2+x(-3-2 i-3+2 i)$
$$\begin{gathered} +\left(9-6 i+6 i-4 i^2\right) \ =x^2-6 x+13=0 \end{gathered}$$

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