Define-the-sequence-a-n-by-the-recurrence-equation-a-n-1-pa-n-qa-n-1-n-1-where-p-q-C-0-and-a-0-a-1-C-Find-a-n-in-terms-of-n-

Question Number 3565 by Yozzii last updated on 15/Dec/15

D e f i n e t h e s e q u e n c e {a_{n}} b y t h e

r e c u r r e n c e e q u a t i o n

a_{n + 1} = {p a}_{n} + {q a}_{n - 1} (n ⩾ 1)

w h e r e p, q \in C - {0} a n d

a_{0} = α, a_{1} = β

α, β \in C .

F i n d a_{n} i n t e r m s o f n .

Answered by prakash jain last updated on 16/Dec/15

a_{0} = α

a_{1} = β

a_{n} = {p a}_{n - 1} + {q a}_{n - 2}

x^{2} - p x - q = 0 \Rightarrow x = \frac{p \pm \sqrt{p^{2} + 4 q}}{2}

a s s u m e p^{2} + 4 q > 0

a_{n} = k_{1} {(\frac{p + \sqrt{p^{2} + 4 q}}{2})}^{n} + k_{2} {(\frac{p - \sqrt{p^{2} + 4 q}}{2})}^{n}

k_{1} + k_{2} = α \Rightarrow k_{2} = α - k_{1}

k_{1} (\frac{p + \sqrt{p^{2} + 4 q}}{2}) + k_{2} (\frac{p - \sqrt{p^{2} + 4 q}}{2}) = β

k_{1} (\frac{p + \sqrt{p^{2} + 4 q}}{2}) - k_{1} (\frac{p - \sqrt{p^{2} + 4 q}}{2}) + α (\frac{p - \sqrt{p^{2} + 4 q}}{2}) = β

k_{1} \sqrt{p^{2} + 4 q} = β - α (\frac{p - \sqrt{p^{2} + 4 q}}{2})

k_{1} = \frac{β}{\sqrt{p^{2} + 4 q}} - \frac{α p}{2 \sqrt{p^{2} + 4 a}} + \frac{α}{2}

k_{2} = \frac{α}{2} - \frac{β}{\sqrt{p^{2} + 4 q}} + \frac{α p}{2 \sqrt{p^{2} + 4 a}}

a_{n} = k_{1} {(\frac{p + \sqrt{p^{2} + 4 q}}{2})}^{n} + k_{2} {(\frac{p - \sqrt{p^{2} + 4 q}}{2})}^{n}

Commented by Yozzii last updated on 15/Dec/15

A g e n e r a t i n g f u n c t i o n a p p r o a c h c a n

b e u s e d ?

Commented by prakash jain last updated on 16/Dec/15

Thanks for the hint . I {didn}^{'} t think about it

yet .

Commented by Rasheed Soomro last updated on 17/Dec/15

W h e r e h a s x^{2} - p x - q = 0 c o m e ?

Commented by Rasheed Soomro last updated on 17/Dec/15

I s t b i s^{'} g e n e r a t i n g f u n c t i o n a p p r o a c h^{'} ?

I {d i d n}^{'} t u n d e r s t a n d .

Commented by Yozzii last updated on 17/Dec/15

N o . T h i s s o l u t i o n i s b a s e d o n a

t h e o r e m o n d i f f e r e n c e e q u a t i o n s

o f t h e f o r m

{t a}_{n + 1} + {r a}_{n} + {c a}_{n - 1} = 0 .

Commented by Yozzii last updated on 17/Dec/15

L e t t, r, c b e c o n s t a n t s, t \neq 0, i n t h e

r e c u r r e n c e r e l a t i o n

{t a}_{n + 1} + {r a}_{n} + {c a}_{n - 1} = 0 (n ⩾ 1)

d e f i n i n g t h e s e q u e n c e {a_{n}} w i t h

a_{0} = f a n d a_{1} = h .

C o n s i d e r t h e a u x i l i a r y e q u a t i o n

t λ^{2} + r λ + c = 0

w h o s e r o o t s a r e α a n d β .

T h e g e n e r a l t e r m a_{n} o f t h e s e q u e n c e

i s t h e n g i v e n b y

a_{n} = A α^{n} + B β^{n} (n ⩾ 0)

i f α \neq β .

I f α = β \neq 0, a_{n} = (A n + B) α^{n} (n ⩾ 0) .

T h e v a l u e s o f A a n d B a r e u n i q u e l y

d e t e r m i n e d b y t h e v a l u e s o f a_{1} a n d a_{0} .

Commented by Yozzii last updated on 17/Dec/15

{C a n}^{'} t p^{2} + 4 q < 0 ?

Commented by prakash jain last updated on 17/Dec/15

p^{2} + 4 q can be less than 0 r e a l o r c o m p l e x .

I just solved for one case > 0 .

As you mentioned if p^{2} + 4 q = 0, both roots

are equal say r

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

f o r u n e q u a l r o o t s p r e v i o u s r e s u l t i s c o r r e c t .