some typos

quarto/precalc/polynomial_roots.qmd

@@ -523,7 +523,7 @@ A polynomial with real coefficients may or may not have real roots. The followin
 The study of polynomial roots is an old one. In $1637$ Descartes published a *simple* method to determine an upper bound on the number of *positive* real roots of a polynomial.
 
 
-> [Descartes' rule of signs](http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs): if  $p=a_n x^n + a_{n-1}x^{n-1} + \cdots a_1x + a_0$ then the number of  positive real roots is  either equal to the number of  sign differences between consecutive nonzero coefficients, or is  less than it by an even number. Repeated roots are counted separately.
+> [Descartes' rule of signs](http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs): if  $p=a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ then the number of  positive real roots is  either equal to the number of  sign differences between consecutive nonzero coefficients, or is  less than it by an even number. Repeated roots are counted separately.
 
 
 
@@ -553,7 +553,7 @@ N.(solve(j ~ 0, x))
 ### Cauchy's bound on the magnitude of the real roots.
 
 
-Descartes' rule gives a bound on how many real roots there may be. Cauchy provided a bound on how large they can be. Assume our polynomial is monic (if not, divide by $a_n$ to make it so, as this won't effect the roots). Then any real root is no larger in absolute value than $|a_0| + |a_1| + |a_2| + \cdots + |a_n|$, (this is expressed in different ways.)
+Descartes' rule gives a bound on how many real roots there may be. Cauchy provided a bound on how large they can be. Assume our polynomial is monic (if not, divide by $a_n$ to make it so, as this won't effect the roots). Then any real root is no larger in absolute value than $|a_0| + |a_1| + |a_2| + \cdots + |a_{n-1}| + 1$, (this is expressed in different ways.)
 
 
 To see precisely [why](https://captainblack.wordpress.com/2009/03/08/cauchys-upper-bound-for-the-roots-of-a-polynomial/) this bound works, suppose $x$ is a root with $|x| > 1$ and let $h$ be the bound. Then since $x$ is a root, we can solve $a_0 + a_1x + \cdots + 1 \cdot x^n = 0$ for $x^n$ as: