is-there-any-generale-form-for-this-sequense-u-n-1-au-n-b-cu-n-d-u-m-k-I-need-u-n-in-terms-of-n-i-have-try-to-derrive-it-for-a-long-time-but-i-cant-

Question Number 207477 by AliJumaa last updated on 16/May/24

i s t h e r e a n y g e n e r a l e f o r m f o r t h i s s e q u e n s e

{\begin{cases} u_{n + 1} = \frac{{a u}_{n} + b}{{c u}_{n} + d} \\ u_{m} = k \end{cases}

I n e e d u_{n} i n t e r m s o f n i h a v e t r y t o d e r r i v e i t f o r a l o n g t i m e b u t i c a n t

Commented by Frix last updated on 17/May/24

u_{n + 1} = \frac{a}{c} - \frac{\frac{a d - b c}{c^{2}}}{u_{n} + \frac{d}{c}}

A = \frac{a}{c} \land B = \frac{a d - b c}{c^{2}} \land C = \frac{d}{c}

u_{n + 1} = A - \frac{B}{u_{n} + C}

Still this looks rather nasty \dots

Commented by Tinku Tara last updated on 17/May/24

a s s u m i n g a, c \neq 0

b / a = j

d / c = k

f (n + 1) = \frac{f (n) + j}{f (n) + k}

Wolfram alpha gives a solution

Commented by Tinku Tara last updated on 17/May/24

Commented by Tinku Tara last updated on 17/May/24

https://m.wolframalpha.com/input?i=f%28n%2B1%29%3D%28%28f%28n%29%2Bj%29%2F%28f%28n%29%2Bk%29%29&lang=en

Commented by AliJumaa last updated on 17/May/24

A l l t h a n k s t o y o u

Commented by Tinku Tara last updated on 17/May/24

Maybe someone can find steps

to get the solution given by wolfram

alpha

Commented by AliJumaa last updated on 17/May/24

i w a n t t o t h a n k y o u f o r t h e l i n k

i^{'} v e t r i e d t h i s b e f o r e a n d i t w a s s o c o m p l e c a t e d

Commented by mr W last updated on 20/May/24

i h a v e s o l v e d t h i s i n m y o w n w a y,

s e e Q 207615

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