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Update rational_functions.qmd

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some typos.
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Fang Liu 2023-04-17 15:14:24 +08:00
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quarto/precalc/rational_functions.qmd
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@ -239,7 +239,7 @@ bottom = x-1
quo, rem = divrem(top, bottom)
```
The graph of has nothing in common with the graph of the quotient for small $x$
The graph has nothing in common with the graph of the quotient for small $x$
```{julia}
@ -265,7 +265,7 @@ $$
\frac{(x-2)^3\cdot(x-4)\cdot(x-3)}{(x-5)^4 \cdot (x-6)^2}.
$$
By looking at the powers we can see that the leading term of the numerator will the $x^5$ and the leading term of the denominator $x^6$. The ratio is $1/x^1$. As such, we expect the $y$-axis as a horizontal asymptote:
By looking at the powers we can see that the leading term of the numerator will the $x^5$ and the leading term of the denominator $x^6$. The ratio is $1/x^1$. As such, we expect the $x$-axis as a horizontal asymptote:
#### Partial fractions
@ -378,7 +378,7 @@ plot(𝒇, -5, 5, ylims=(-20, 20))
This isn't ideal, as the large values are still computed, just the viewing window is clipped. This leaves the vertical asymptotes still effecting the graph.
There is another way, we could ask `Julia` to not plot $y$ values that get too large. This is not a big request. If instead of the value of `f(x)` - when it is large - -we use `NaN` instead, then the connect-the-dots algorithm will skip those values.
There is another way, we could ask `Julia` to not plot $y$ values that get too large. This is not a big request. If instead of the value of `f(x)` - when it is large - we use `NaN` instead, then the connect-the-dots algorithm will skip those values.
This was discussed in an earlier section where the `rangeclamp` function was introduced to replace large values of `f(x)` (in absolute values) with `NaN`.
@ -621,7 +621,7 @@ a, b = 4, 6
pq = 𝐩 // one(𝐩)
x = variable(pq)
d = Polynomials.degree(𝐩)
numerator(lowest_terms( (x + 1)^2 * pq((a*x + b)/(x + 1))))
numerator(lowest_terms( (x + 1)^d * pq((a*x + b)/(x + 1))))
```
---
@ -630,7 +630,7 @@ numerator(lowest_terms( (x + 1)^2 * pq((a*x + b)/(x + 1))))
Now, why is this of any interest?
Mobius transforms are used to map regions into other regions. In this special case, the transform $\phi(x) = (ax + b)/(x + 1)$ takes the interval $[0,\infty]$ and sends it to $[a,b]$ ($0$ goes to $(a\cdot 0 + b)/(0+1) = b$, whereas $\infty$ goes to $ax/x \rightarrow a$). Using this, if $p(u) = 0$, with $q(x) = (x-1)^d p(\phi(x))$, then setting $u = \phi(x)$ we have $q(x) = (\phi^{-1}(u)+1)^d p(\phi(\phi^{-1}(u))) = (\phi^{-1}(u)+1)^d \cdot p(u) = (\phi^{-1}(u)+1)^d \cdot 0 = 0$. That is, a zero of $p$ in $[a,b]$ will appear as a zero of $q$ in $[0,\infty)$ at $\phi^{-1}(u)$.
Mobius transforms are used to map regions into other regions. In this special case, the transform $\phi(x) = (ax + b)/(x + 1)$ takes the interval $[0,\infty]$ and sends it to $[a,b]$ ($0$ goes to $(a\cdot 0 + b)/(0+1) = b$, whereas $\infty$ goes to $ax/x \rightarrow a$). Using this, if $p(u) = 0$, with $q(x) = (x+1)^d p(\phi(x))$, then setting $u = \phi(x)$ we have $q(x) = (\phi^{-1}(u)+1)^d p(\phi(\phi^{-1}(u))) = (\phi^{-1}(u)+1)^d \cdot p(u) = (\phi^{-1}(u)+1)^d \cdot 0 = 0$. That is, a zero of $p$ in $[a,b]$ will appear as a zero of $q$ in $[0,\infty)$ at $\phi^{-1}(u)$.
The Descartes rule of signs applied to $q$ then will give a bound on the number of possible roots of $p$ in the interval $[a,b]$. In the example we did, the Mobius transform for $a=4, b=6$ is $15 - x - 11x^2 - 3x^3$ with $1$ sign change, so there must be exactly $1$ real root of $p=(x-1)(x-3)(x-5)$ in the interval $[4,6]$, as we can observe from the factored form of $p$.
@ -640,7 +640,8 @@ Similarly, we can see there are $2$ or $0$ roots for $p$ in the interval $[2,6]$
```{julia}
mobius_transformation(𝐩, 2,6)
p = fromroots([1,3,5]) # (x-1)⋅(x-3)⋅(x-5) = -15 + 23*x - 9*x^2 + x^3
mobius_transformation(p, 2,6)
```
This observation, along with a detailed analysis provided by [Kobel, Rouillier, and Sagraloff](https://dl.acm.org/doi/10.1145/2930889.2930937) provides a means to find intervals that enclose the real roots of a polynomial.
@ -662,7 +663,6 @@ Applying these steps to $p$ with an initial interval, say $[0,9]$, we would have
```{julia}
#| hold: true
p = fromroots([1,3,5]) # (x-1)⋅(x-3)⋅(x-5) = -15 + 23*x - 9*x^2 + x^3
mobius_transformation(p, 0, 9) # 3
mobius_transformation(p, 0, 9//2) # 2
mobius_transformation(p, 9//2, 9) # 1 (and done)