typos

quarto/precalc/polynomial_roots.qmd

@@ -366,7 +366,7 @@ solve(x^5 - x + 1)
 Related to `solve` is the specialized `roots` function for identifying roots, Unlike solve, it will identify multiplicities.
 
 
-For a polynomial with only one indeterminate the usage is straight foward:
+For a polynomial with only one indeterminate the usage is straight forward:
 
 
 ```{julia}
@@ -533,7 +533,7 @@ An immediate consequence, is that a polynomial whose coefficients are all non-ne
 Applying this to the polynomial $x^5 -x + 1$ we get That the coefficients have signs: `+ 0 0 0 - +` which collapses to the sign pattern `+`, `-`, `+`. This pattern  has two changes of sign. The number of *positive* real roots is either $2$ or $0$. In fact there are $0$ for this case.
 
 
-What about negative roots? Cleary, any negative root of $p$ is a positive root of $q(x) = p(-x)$, as the graph of $q$ is just that of $p$ flipped through the $y$ axis. But the coefficients of $q$ are the same as $p$, except for the odd-indexed coefficients ($a_1, a_3, \dots$) have a changed sign.  Continuing with our example, for $q(x) = -x^5 + x + 1$ we get the new sign pattern `-`, `+`, `+` which yields one sign change. That is, there *must* be a negative real root, and indeed there is, $x \approx -1.1673$.
+What about negative roots? Clearly, any negative root of $p$ is a positive root of $q(x) = p(-x)$, as the graph of $q$ is just that of $p$ flipped through the $y$ axis. But the coefficients of $q$ are the same as $p$, except for the odd-indexed coefficients ($a_1, a_3, \dots$) have a changed sign.  Continuing with our example, for $q(x) = -x^5 + x + 1$ we get the new sign pattern `-`, `+`, `+` which yields one sign change. That is, there *must* be a negative real root, and indeed there is, $x \approx -1.1673$.
 
 
 With this knowledge, we could have known that in an earlier example the graph of `p = x^7 - 16129x^2 + 254x - 1` – which indicated two positive real roots – was misleading, as there must be $1$ or $3$ by a count of the sign changes.
@@ -1001,7 +1001,7 @@ Now consider the case $p_0 > 0$. There are two possibilities either `pos(p)` is
 So there is parity between `var(p)` and `pos(p)`: if $p$ is monic and $p_0 < 0$ then both `var(p)` and `pos(p)` are both odd; and if $p_0 > 0$ both `var(p)` and `pos(p)` are both even.
 
 
-Descartes' rule of signs will be established if it can be shown that `var(p)` is at least as big as `pos(p)`. Supppose $r$ is a positive real root of $p$ with $p = (x-r)q$. We show that `var(p) > var(q)` which can be repeatedly applied to show that if $p=(x-r_1)\cdot(x-r_2)\cdot \cdots \cdot (x-r_l) q$, where the $r_i$s are the postive real roots, then `var(p) >= l + var(q) >= l = pos(p)`.
+Descartes' rule of signs will be established if it can be shown that `var(p)` is at least as big as `pos(p)`. Supppose $r$ is a positive real root of $p$ with $p = (x-r)q$. We show that `var(p) > var(q)` which can be repeatedly applied to show that if $p=(x-r_1)\cdot(x-r_2)\cdot \cdots \cdot (x-r_l) q$, where the $r_i$s are the positive real roots, then `var(p) >= l + var(q) >= l = pos(p)`.
 
 
 As $p = (x-c)q$ we must have the leading term is $p_nx^n = x \cdot q_{n-1} x^{n-1}$ so $q_{n-1}$ will also be `+` under our monic assumption. Looking at a possible pattern for the signs of $q$, we might see the following unfinished synthetic division table for a specific $q$:
@@ -1014,7 +1014,7 @@ As $p = (x-c)q$ we must have the leading term is $p_nx^n = x \cdot q_{n-1} x^{n-
   + - - - + - + + 0
 ```
 
-But actually, we can fill in more, as the second row is formed by multiplying a postive $c$:
+But actually, we can fill in more, as the second row is formed by multiplying a positive $c$:
 
 
 ```{verbatim}