|
|
Post by mondo on Sept 24, 2022 19:39:20 GMT
Hello,
at the bottom of page 123 author encourages us to decompose $M(z)$ into the set of equations shown the beginning of the next page. My question is how to "decompose" $M(z)$ to get it? It seems to not feasible to me.
Also, on the same page author also says "Note that if $(ad - bc) = 0$ then $M(z)$ is an uninteresting constant mapping sending every point $z$ to the same image point $\frac{a}{c}$". How is it possible? If $(ad - bc) = 0$ then the transformation marked by (iii) is just $z -> 0$. This $\frac{a}{c}$ is part of (iv) but this is a distinct transformation. So why was it related?
|
|
|
|
Post by Admin on Sept 24, 2022 21:46:33 GMT
Mondo Hello, at the bottom of page 123 author encourages us to decompose $M(z)$ into the set of equations shown the beginning of the next page. My question is how to "decompose" $M(z)$ to get it? It seems to not feasible to me. This is feasible. I don't think you understand how to combine transformations. Try finding out and think again. Vasco |
|
|
|
Post by mondo on Sept 25, 2022 15:58:14 GMT
Mondo Hello, at the bottom of page 123 author encourages us to decompose $M(z)$ into the set of equations shown the beginning of the next page. My question is how to "decompose" $M(z)$ to get it? It seems to not feasible to me. This is feasible. If we assume I have a freedom to zero some of the constants $a,b,c,d$ then yes. Otherwise I don't see how we can get say $z -> z + \frac{d}{c}$ from $M(z)$. Where are $a,b$ constants? I don't think you understand how to combine transformations. Try finding out and think again. If I assume a composition of (iv) and (iii) then $z-> \frac{ad-bc}{c^2}z + \frac{a}{c}$ and then if $(ad-bc) = 0$ this is indeed a mapping $z -> \frac{a}{c}$ as author wrote. However we can as well do composition of (iii) and (i) to get $z -> \frac{d}{c}$. But this one is not mentioned in the book - is that overlooked or can't be done for whatever reason? |
|
|
|
Post by Admin on Sept 25, 2022 16:33:38 GMT
If we assume I have a freedom to zero some of the constants $a,b,c,d$ then yes. Otherwise I don't see how we can get say $z -> z + \frac{d}{c}$ from $M(z)$. Where are $a,b$ constants? We don't. It's the other way round: we get $M(z)$ from $z -> z + \frac{d}{c}$. I don't think you understand how to combine transformations. Try finding out and think again. If I assume a composition of (iv) and (iii) then $z-> \frac{ad-bc}{c^2}z + \frac{a}{c}$ and then if $(ad-bc) = 0$ this is indeed a mapping $z -> \frac{a}{c}$ as author wrote. However we can as well do composition of (iii) and (i) to get $z -> \frac{d}{c}$. But this one is not mentioned in the book - is that overlooked or can't be done for whatever reason? We do these four transformations in order.Vasco |
|
|
|
Post by mondo on Sept 26, 2022 7:50:48 GMT
If we assume I have a freedom to zero some of the constants $a,b,c,d$ then yes. Otherwise I don't see how we can get say $z -> z + \frac{d}{c}$ from $M(z)$. Where are $a,b$ constants? We don't. It's the other way round: we get $M(z)$ from $z -> z + \frac{d}{c}$. If I assume a composition of (iv) and (iii) then $z-> \frac{ad-bc}{c^2}z + \frac{a}{c}$ and then if $(ad-bc) = 0$ this is indeed a mapping $z -> \frac{a}{c}$ as author wrote. However we can as well do composition of (iii) and (i) to get $z -> \frac{d}{c}$. But this one is not mentioned in the book - is that overlooked or can't be done for whatever reason? Vasco "As a first step towards making sense of (1), let us decompose M(z) into the following sequence of transformations" - so it sounds to me like I should take $M(z)$ and produce/decompose these four transformations from it. We do these four transformations in order. Ok, its not said explicitly but I think it can be guessed. |
|
|
|
Post by Admin on Sept 26, 2022 8:46:44 GMT
Mondo
The way I would tackle it is to apply the first then the second and so on and you should then get $M(z)$
1. $z\mapsto z+\frac{d}{c}$
2. Now substitute this expression for $z$ into (ii) and so on
You should get $M(z)$ after substitution (iv).
Then you should see the answer to your other question about what happens when $(ad-bc)=0$
Vasco
|
|
|
|
Post by mondo on Sept 26, 2022 16:51:53 GMT
Yes now it makes perfect sense - I got $M(z)$ from these four transformation by $f_4(f_3(f_2(f_1)))$. Thank you Vasco.
|
|
