Find-3-2-3-2-3-2-

Question Number 183342 by Shrinava last updated on 25/Dec/22

Find : 3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{\dots}}} = ?

Answered by Red1ight last updated on 25/Dec/22

3 - \frac{2}{x} = x

x^{2} - 3 x + 2 = 0

x = \frac{3 \pm \sqrt{9 - 8}}{2}

x = 2, x = 1

Commented by mr W last updated on 25/Dec/22

3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{\dots}}} = 1

3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{\dots}}} = 2

3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{\dots}}} = 3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{\dots}}}

\Rightarrow 1 = 2 ?

Answered by Frix last updated on 25/Dec/22

a_{1} \in R ∖ {0}

a_{n + 1} = 3 - \frac{2}{a_{n}}

a_{\infty} = {\begin{cases} 1; a_{1} = 1 \\ 2; a_{1} \neq 1 \end{cases}

Because we cannot forward calculate

x = 3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{\dots}}} = a_{\infty} we cannot decide

whether x = 1 or x = 2

Commented by Shrinava last updated on 25/Dec/22

Answer : 2

Commented by Frix last updated on 25/Dec/22

Then try to prove x \neq 1

Answered by mahdipoor last updated on 25/Dec/22

A_{1} = 3 - \frac{2}{i_{1}} = \frac{3 i_{1} - 2}{i_{1}} \Rightarrow i_{1} = 3 - \frac{2}{i_{2}} \Rightarrow i_{1} i_{2} = 3 i_{2} - 2

A_{2} = \frac{3 i_{1} i_{2} - 2 i_{2}}{i_{1} i_{2}} = \frac{7 i_{2} - 6}{3 i_{2} - 2} \Rightarrow i_{2} = 3 - \frac{2}{i_{3}} \Rightarrow

A_{3} = \frac{7 i_{2} i_{3} - 6 i_{3}}{3 i_{2} i_{3} - i_{3}} = \frac{15 i_{3} - 14}{7 i_{3} - 6}

\Rightarrow A_{n} = \frac{{a i}_{n} - b}{{c i}_{n} - d} \Rightarrow A_{n + 1} = \frac{(3 a - b) i_{n + 1} - 2 a}{{a i}_{n + 1} - b}

a_{1} = k, b_{1} = k - 1 \Rightarrow a_{2} = 2 k + 1, b_{2} = 2 k

a_{3} = 4 k + 3, b_{3} = 4 k + 2 \Rightarrow a_{4} = 8 k + 7, b_{4} = 8 k + 6

\Rightarrow\Rightarrow

a_{n} = 2^{n - 1} k + (2^{n - 1} - 1) b_{n} = a_{n} - 1

\Rightarrow\Rightarrow k = 3

A_{n} = \frac{(2^{n + 1} - 1) i - (2^{n + 1} - 2)}{(2^{n} - 1) i - (2^{n} - 2)}

l i m A_{n} n \to \infty = \frac{(2 - \frac{1}{2^{n}}) i - (2 - \frac{2}{2^{n}})}{(1 - \frac{1}{2^{n}}) i - (1 - \frac{2}{2^{n}})} =

\frac{2 i - 2}{i - 1} = 2 = 3 - \frac{2}{3 - \frac{2}{3 - \frac{2}{\dots}}} (i \neq \infty)

Commented by Frix last updated on 25/Dec/22

A_{n} = \frac{(2^{n + 1} - 1) i - (2^{n + 1} - 2)}{(2^{n} - 1) i - (2^{n} - 2)} =

= 2 + \frac{i - 2}{(2^{n + 1} - 1) i - 2 (2^{n} - 1)}

\Rightarrow

i = 1 \Rightarrow A_{n} = 1 \forall n \in N

i = 2 \Rightarrow A_{n} = 2 \forall n \in N

i \neq 1 \land i \neq 2 \Rightarrow \lim_{n \to \infty} A_{n} = 2

As you can see I^{'} m still not convinced .

Commented by Frix last updated on 25/Dec/22

I {don}^{'} t think numbers of the shape

x = m + \frac{n}{m + \frac{n}{m + \frac{n}{\dots}}} with m, n \in Z

can generally be given a value . We can

always solve for x \in C :

x = m + \frac{n}{x} \Rightarrow x = \frac{m \pm \sqrt{m^{2} + 4 n}}{2}

But how to prove the value of this example ?

x = 3 - \frac{3}{3 - \frac{3}{3 - \frac{3}{\dots}}} \overset{? ? ?}{=} \frac{3}{2} \pm \frac{\sqrt{3}}{2} i

You tell me adding, substracting, dividing

by real numbers gives complex numbers ?

Commented by mahdipoor last updated on 26/Dec/22

n o, i i s n o t \sqrt{- 1}, i i s r e a l n o m b e r a n d

n o t i m p o r t a n d h o w m a n y o f i,

w e k n o w n \to \infty, l i m A_{n} = 2 (\frac{i - 1}{i - 1}) = c

i f i \neq 1 \Rightarrow c = 2

i f i = 1 f o r n = \infty \Rightarrow h o p \Rightarrow \frac{2}{1} = 2 = c