Recursive Induction (1 Viewer)

Intilegience · Feb 2, 2024

A sequence is defined by an+1 = 1/2 (an + 2/an) where a1=1 and n is a positive integer

Use induction to show that (an - √2)/an + √2) = ((1 - √2)/(1+√2))^2^(n-1)

liamkk112 · Feb 2, 2024

im just writing the sequence in latex: $a_{n+1} = \frac{1}{2}(a_n + \frac{2}{a_n})$

we want to prove the statement that $\frac{a_n -\sqrt{2}}{a_n +\sqrt{2}} = (\frac{1-\sqrt{2}}{1+\sqrt{2}})^{2^{n-1}}$ for all positive integers n

base case: n = 1
LHS = $\frac{a_n - \sqrt{2}}{a_n + \sqrt{2}} = \frac{1-\sqrt{2}}{1+\sqrt{2}}$ = RHS, so the statement is true for n = 1

assumption: assume that the statement is true for n = k ,where k is a positive integer:
$\frac{a_k -\sqrt{2}}{a_k +\sqrt{2}} = (\frac{1-\sqrt{2}}{1+\sqrt{2}})^{2^{k-1}}$
prove that the statement is true for n = k + 1:
$\frac{a_{k+1} -\sqrt{2}}{a_{k+1} +\sqrt{2}} = (\frac{1-\sqrt{2}}{1+\sqrt{2}})^{2^{k}}$
LHS = $\frac{a_{k+1} -\sqrt{2}}{a_{k+1} +\sqrt{2}}$
= $\frac{ \frac{1}{2}(a_k + \frac{2}{a_k}) -\sqrt{2}}{ \frac{1}{2}(a_k + \frac{2}{a_k}) +\sqrt{2}}$ by the recursive formula
= $\frac{a_k^2 - 2\sqrt{2}a_k + 2}{a_k^2 + 2\sqrt{2}a_k + 2}$ (multiplied numerator and denominator by 2a_k)
= $\frac{(a_k -\sqrt{2})^2}{(a_k + \sqrt{2})^2}$
= $(\frac{a_k -\sqrt{2}}{a_k + \sqrt{2}})^2$ by index laws
= $((\frac{1-\sqrt{2}}{1+\sqrt{2}})^{2^{k-1}})^2$ by the assumption
= $(\frac{1-\sqrt{2}}{1+\sqrt{2}})^{2^{k-1}\times 2}$
= $(\frac{1-\sqrt{2}}{1+\sqrt{2}})^{2^{k}}$
= RHS as required

hence by mathematical induction the statement is true for all positive integers n

Recursive Induction (1 Viewer)

Intilegience

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liamkk112

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