matrix transformation to generate a sequence.

[lljj]

PROBLEM
A graphic designer, Ben, wants to create an animation in which a sequence of squares is created by composition of successive enlargements and translations. The first four frames of the animation are shown in greater detail in the drawing in the attachment.

The width of each successive square is one half of the adjacent larger square. Let the sequence be $$U_o,U_1,U_2,…$$ , and the first square $U_0$ has a width of 4 cm.

(a) Find an expression for the width of $U_n$ in centimeters.

Ben decides to generate the squares using the transformation $$\begin{pmatrix} x_n \\ y_n \end{pmatrix}=A_n \begin{pmatrix} x_o \\ y_o \end{pmatrix} + b_n$$, where $A_n$ is a 2x2 matrix that represents an enlargement, $b_n$ is a 2x1 column vector that represents a translation, $(x_0,y_0)$ is a point in $U_0$ and $(x_n,y_n)$ is its image in $U_n$.

(b) (i) Write down $A_1$.

(ii) Write down $A_n$ in terms of n.

(c) (i) By considering the case where $(x_0,y_0)$ is (0,0), state the coordinates $(x_1,y_1)$, of its image in $U_1$.

(ii) Hence, find $b_1$.

(ii) Show that $$b_n = \begin{pmatrix} 8(1-2^{-n}) \\ 8(1-2^{-n}) \end{pmatrix}$$
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ANSWER

I have problem solving question (c) (iii). I tried to use induction to prove the identity, but could not do it. Please help. Thanks.

My answers for the other questions:

(a) This is a GP with first term 4 and common ratio 0.5, so $4(0.5)^{n}$

(b) (i) $$\begin{pmatrix} 0.5 & 0 \\ 0 & 0.5 \end{pmatrix}$$
(ii) $$\begin{pmatrix} 0.5^{n} & 0 \\ 0 & 0.5^{n} \end{pmatrix}$$
(c) (i) (4,4)

(ii) $$\begin{pmatrix} 4 \\ 4 \end{pmatrix}$$

 

[MaxWong]

Note that $$\begin{pmatrix}x_n \\y_n\end{pmatrix} = A_1 \begin{pmatrix}x_{n-1} \\y_{n-1}\end{pmatrix}+b_1$$Replace all n by n-1 and then substituting into the right hand side gives $$\begin{pmatrix}x_n \\y_n\end{pmatrix} = A_2 \begin{pmatrix}x_{n-2} \\y_{n-2}\end{pmatrix}+(A_1 + I)b_1$$where $I$ is the identity matrix. Repeat that again until we get $$\begin{pmatrix}x_n \\y_n\end{pmatrix} = A_n \begin{pmatrix}x_{0} \\y_{0}\end{pmatrix}+(A_{n-1}+A_{n-2}+\cdots+A_2+A_1+I)b_1$$Then $b_n = (A_{n-1}+A_{n-2}+\cdots+A_2+A_1+I)b_1$.

Now, note that $$(A_{n-1}+A_{n-2}+\cdots+A_2+A_1+I)(I-A_1)=I-A_n$$(To see why, recall the analogous formula in algebra: ($1 + x + x^2 + \cdots + x^{n-1})(1 - x) = 1-x^n$, where $x$ is a real number and $n$ is a positive integer.)

Then, $$A_{n-1}+A_{n-2}+\cdots+A_1+I=(I-A_n)(I-A_1)^{-1} = \begin{pmatrix}1 - 2^{-n}&0\\0&1-2^{-n}\end{pmatrix}\begin{pmatrix}0.5&0\\0&0.5\end{pmatrix}^{-1} = \begin{pmatrix}1 - 2^{-n}&0\\0&1-2^{-n}\end{pmatrix}\begin{pmatrix}2&0\\0&2\end{pmatrix} =\begin{pmatrix}2(1-2^{-n})&0\\0&2(1-2^{-n})\end{pmatrix}$$
Using this, we have $$b_n = \begin{pmatrix}2(1-2^{-n})&0\\0&2(1-2^{-n})\end{pmatrix}\binom{4}4 = \binom{8(1-2^{-n})}{8(1-2^{-n})}$$