Question-4232

Question Number 4232 by Yozzii last updated on 03/Jan/16

Commented by Yozzii last updated on 04/Jan/16

I^{'} m e x p e r i e n c i n g d i f f i c u l t y

t o s o l v e t h e 2 n d p a r t w i t h o u t u s e o f a t h e o r e m

I p o s t e d s o m e t i m e a g o, a n d

g e n e r a t i n g f u n c t i o n s .

Commented by prakash jain last updated on 04/Jan/16

a_{n} = \frac{1 + a_{n - 1}^{2}}{a_{n - 2}}, a_{n} = 3 a_{n - 1} - a_{n - 2}, a_{0} = 1, a_{1} = 1

a_{2} = \frac{1 + 1^{2}}{1} = 2, 3 a_{1} - a_{0} = 3 \times 1 - 1 = 2

a s s u m e a_{n} = 3 a_{n - 1} - a_{n - 2} i s t r u e f o r s o m e n .

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

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

t h e r e s u l t i s v a l i d f o r a l l n b y i n d u c t i o n .

g i v e n \frac{1 + a_{n - 1}^{2}}{a_{n - 2}} = a_{n} \Rightarrow 1 - a_{n} a_{n - 2} = - a_{n - 1}^{2}

Answered by prakash jain last updated on 04/Jan/16

a_{n} = 3 a_{n - 1} - a_{n - 2}

a_{n} = x^{n}

x^{n} = 3 x^{n - 1} - x^{n - 2}

x^{2} - 3 x + 1 = 0

x = \frac{3 \pm \sqrt{5}}{2}

a_{n} = k_{1} x_{1}^{n} + k_{2} x_{2}^{n}

Commented by Yozzii last updated on 07/Jan/16

C a n y o u a c q u i r e a_{n} i n t h e f o r m

i n d i c a t e d i n t h e q u e s i o n ?

Commented by Rasheed Soomro last updated on 15/Jan/16

W h y a_{n} = x^{n} ?

I f a_{n} = x^{n} \Rightarrow a_{1} = x

w h e r e a s a_{1} = 1 i s g i v e n .

Commented by prakash jain last updated on 16/Jan/16

To solve a linear difference equation . You

try to find a suitable solution to eliminate

n .

Commented by Rasheed Soomro last updated on 17/Jan/16

T h α n X! I need to study about difference equation .