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This is in continuation of my earlier post here.

On pg.#7-8, question #7 is about an alternate model of growth (i.e., proportional growth, herewith referred as P) and its comparison with the uniform growth model (herewith referred as U) used to form one rectangle from another by shrinking or expanding $x,y$ coordinates. The different parts of the question are attempted below, but am clueless about the (c), (d) parts:

(a) Asks to compare P with U in question #5(a) -(d). (There is an erratum there, refers to question #7 rather than question #5). The U model is covered in a previous post here. Below will attempt using the P model:

a) Describe all the points in the lh-plane that represent the rectangles proportionally grown from the rectangle $(3,2)$.
-> $(3*i, 2*i), i\in \mathbb{R}, i\ne 0$
b) Given the rectangle $(3,2)$ and the rectangle $(5,3)$, can you find a rectangle that can be proportionally grown from each of them? Explain.
-> Need find if possible for real $s,t$ s.t. $(i) 3s = 5t\implies \frac{s}{t} = \frac53; 2s = 3t\implies \frac{s}{t} = \frac32$. No, it is not possible.
c) If one rectangle is proportionally grown from another, can they be similar? Describe all such pairs of rectangles.
-> Need show that some real $s$ and given point $(a,b), \frac{a}{b} = \frac{as}{bs}$. It can be seen that it is an axiom for any value of $s$.
d) Is it possible to grow (or shrink) uniformly a square from any rectangle? Explain.
-> Need show that for a point $(a,b)$ with $a\ne b$, there exists a real $s$ s.t. $as= bs$. No, it is always false.

(b) What is the relation between proportional growth and aspect ratio?
No change in aspect ratio.

(c) Consider two rectangles (two points in the lh-plane, for example, A & B).
What if you are allowed to use two growth processes—one proportional and one uniform—in sequence? Can you always grow one rectangle from another when you are allowed to use at most two steps?

(d) Under which conditions on the dimensions of two rectangles, you can grow one rectangle from another by using both the uniform and proportional growth? What can be said about the number of times you must switch from one type of growth to another one? Does the order in which you apply them matter?

jiten
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1 Answers1

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For part $(a)$, you might like to impose the condition that $i > 0$.

Guide for part $(c)$,

A $P_p$ operation maps a point from $(x,y)$ to $(px,py)$ where $p>0$.

A $U_u$ operations maps a point from $(x,y)$ to $(x+u, y+u)$.

$U_u(P_p(x,y))=U_u(px,py)=(px+u, py+u)$

Given positive numbers, $x,y,r$ and $s$, can we solve for $px+u=r$, $py+u=s$.

$$\begin{bmatrix} x & 1 \\ y & 1\end{bmatrix}\begin{bmatrix} p \\ u\end{bmatrix} = \begin{bmatrix} r \\ s\end{bmatrix} $$

Suppose $x \ne y$, then we have $$\begin{bmatrix} p \\ u\end{bmatrix} = \frac{1}{x-y}\begin{bmatrix} 1 & -1 \\ -y & x\end{bmatrix}\begin{bmatrix} r \\ s\end{bmatrix} $$

Try to pick positive $x,y,r,s$ such that $p$ is negative and hence it can't be done in the order of $U$ and then $P$.

For the other order, note that we have $$P_q(U_v(x,y))=P_q(x+v,y+v)=(qx+vq, qy+vq)=U_{qv}(P_q(x,y))$$

Siong Thye Goh
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  • Sorry for delay, had been to other topics. For part (a), modified O.P. to have non-zero real $i$. Hope am correct! In part (c), have a pre-defined sequence of $2$ steps - first P, then U. Constraint needed is to have dimensions $(x,y,r,s)$ with positive values s.t. $p\lt 0$ to make reverse impossible. (i) Request set of such values of dimensions. (ii) Second part of (c) is also not clear, please help. (iii) (d) asks first reverse of your stated constraint for (c). (iv) (d)-second part asks for # times to switch models, which to me is having no logical basis. (v) (d)-third part is vague to me. – jiten May 24 '18 at 07:40
  • Have modified your answer's last part (I interpreted it as for part (d)) by elaborating the condition stated by you : For the other order, note that we have $$P_q(U_v(x,y))=P_q(x+v,y+v)=(qx+vq, qy+vq)=U_{qv}(P_q(x,y))$$ i.e., $$qx+vq = r, qy+vq=s\implies x= y+\frac{(r-s)}{q}$$ – jiten May 24 '18 at 07:49
  • it is not clear to me whether we need $i \ne 0$ or $i > 0$. – Siong Thye Goh May 24 '18 at 17:08
  • Thanks, but if you think chat is needed to solve the rest of the questions/sub-questions; then would create one. – jiten May 24 '18 at 19:26
  • Please visit the chat room : https://chat.stackexchange.com/rooms/77956/shape-of-algebra-1 – jiten May 24 '18 at 19:36
  • I am feeling hopeless with the problems in sec 1.1 itself of the book. Please help. Chat may help explore me a lot, even if no answer is clearly visible. Your matrix approach may lead to some solution, may be my linear algebra (i.e., the matrix part) is weak, so am unable to understand its further applicability. I hope that the problem will add a lot to the understanding of solution approach using Linear Algebra. At least give me some more advice on where to study such problems. – jiten May 24 '18 at 20:15
  • In chat, get curve connecting peaks (max. A for given P) given by: $A= \frac{P^2}{16}=x^2$, by finding derivative of $A=(\frac{P}2-x)x\implies dA/dx = \frac{P}2 - 2x\implies x=y=\frac{P}4$. It is eqn. of curve joining peaks (for various $P$) of curves & those curves open down, say for $P=4, A= -x^2 +2x$, but $dA/dx$ curve opens upwards. Also, the book's mapping between point and $AP$ model is on page 28. of chapter 1 (freely accessible at : https://www.worldscientific.com/doi/suppl/10.1142/7810/suppl_file/7810_chap01.pdf). Request vetting of the above. – jiten May 26 '18 at 09:11
  • $A= \frac{P^2}{16}$ is indeed the solution that you are looking for. – Siong Thye Goh May 27 '18 at 03:10
  • In generating functions, on pg. 12 (as in image at: https://i.stack.imgur.com/NSvbp.png) of notes at : https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/MIT6_042JF10_chap12.pdf, could not understand the significance of why only the first fraction with numerator as $x$ is referred to for taking $F'(x)$, while the second fraction with numerator $x^2$ is not. Posted it as question (image: https://i.stack.imgur.com/EiGcl.png) , but deleted due to negative votes. – jiten May 31 '18 at 16:52
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    your question looks fine besides it lacks working. just include the working so that others don't have to repeat the work and it should be fine. (not saying that you are wrong). in fact intuitively your solution makes more sense. i have yet to read the original document. – Siong Thye Goh May 31 '18 at 17:02
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    It's just bad use of notation, they use $F$ to refer to two things. The $F$ from the part that you are confused with refers to another function $G(x)=\frac{1}{(1-x)^4}$. Learn to cope with negative vote, one negative vote is just a warning to improve your post. I get negative votes regularly too. Editing in the future is better than deleting, now we are hiding a post that could have helped others. – Siong Thye Goh May 31 '18 at 17:26