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Proof, EXT2 EQ-Bank 34

Consider a sequence of rectangles with side lengths \(a_{ n }\) and \(b_{ n }\).

The first rectangle has  \(a_1=2\)  and  \(b_1=1\).

For integers  \(n \geq 1,\ \ a_{n+1}=\dfrac{a_n+b_n}{2}\)  and  \(b_{n+1}=\dfrac{2}{a_{n+1}}.\)

  1. Show that every rectangle in the sequence has an area of 2 square units.   (1 mark)

  2. Use the relationship between the arithmetic mean and the geometric mean to prove that  \(a_n \geq \sqrt{2}\)  for any integer  \(n \geq 1\).   (2 marks)

  3. Use mathematical induction to prove that  \(a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2})\)  for any integer  \(n \geq 1\).    (4 marks)

  4. Use the squeeze theorem to show that the rectangles approach a square as \(n\) approaches infinity.   (2 marks)

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a.    \(\text{Since}\ \ a_1 b_1=2\ \ \text{and}\ \ a_{n+1} b_{n+1}=a_{n+1} \times \dfrac{2}{a_{n+1}}=2\)

\(\Rightarrow\ \text{Each rectangle has area 2.}\)
 

b.    \(a_1=2\ \ \Rightarrow\ \ a_1 \geq \sqrt{2}\ \ \text{(true for 1st rectangle)}\)

\(\text{AM/GM inequality:}\ \ \dfrac{a_n+b_n}{2} \geq \sqrt{a_nb_n} \)

\(a_nb_n=2\ \ \text{(from part (a))}\)

\(\dfrac{a_n+b_n}{2} \geq \sqrt{2}\ …\ (1)\)

\(\text{Since}\ \ a_{n+1}=\dfrac{a_n+b_n}{2}:\)

\(\ a_{n+1} \geq \sqrt{2}\ \ \ \text{(using (1) above)}\)

\(\therefore a_n \geq \sqrt{2}\)
  

c.    \(\text{Prove}\ \ a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}),\ \ \text{for}\ \ n \geq 1\)

\(\text{If}\ \ n=1:\)

\(\ a_1-\sqrt{2}=2-\sqrt{2} \leq \dfrac{1}{2^{0}}(2-\sqrt{2})\).

\(\Rightarrow\ \text{True for}\ \ n=1.\)
 

\(\text{Assume true for}\ \ n=k:\)

\(a_k-\sqrt{2} \leq \dfrac{1}{2^{k-1}}(2-\sqrt{2})\ …\ (1) \)

\(\text{Prove true for}\ \ n=k+1:\)

\(\text{i.e.}\ \ a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})\)

\(\text{By definition,} \ \ a_{k+1}=\dfrac{1}{2}\left(a_k+b_k\right)\)

\(a_{k+1}-\sqrt{2}\) \(=\dfrac{1}{2}\left(a_k-\sqrt{2}\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) \)  
  \(\leq \dfrac{1}{2}\left(\dfrac{1}{2^{k-1}}(2-\sqrt{2})\right)+\dfrac{1}{2}\left(b_k-\sqrt{2}\right)\ \ \text{(see (1) above)}\ \)  

 
\(\text{Since}\ \ a_k \geq \sqrt{2}\ \ \text{and}\ \ a_kb_k=2:\)

\(\ b_k \leq \sqrt{2}\ \ \text{and}\ \ b_k-\sqrt{2} \leq 0\)

\(a_{k+1}-\sqrt{2} \leq \dfrac{1}{2^k}(2-\sqrt{2})+\dfrac{1}{2}\left(b_k-\sqrt{2}\right) \leq \dfrac{1}{2^k}(2-\sqrt{2})\)

\(\Rightarrow\ \text{True for}\ \ n=k+1.\)

\(\therefore\ \text{Since true for}\ \ n=1,\ \text{by PMI, true for integers}\ \ n \geq 1.\)
  

d.    \(\text {Combining parts (b) and (c):}\)

\(0 \leq a_n-\sqrt{2} \leq \dfrac{1}{2^{n-1}}(2-\sqrt{2}).\)

\(\text{As}\ \ n \rightarrow \infty, \ \dfrac{1}{2^{n-1}} \rightarrow 0\)

\(\Rightarrow a_n-\sqrt{2} \rightarrow 0\ \ \text{(by squeeze theorem)}\)

\(\text{Since rectangles have an area = 2:}\)

\(\text{As}\ \ n \rightarrow \infty,\ b_n \rightarrow \sqrt{2}\)

\(\text{i.e. rectangles approach a square.}\)