simplify-S-n-x-1-x-2-1-x-4-1-x-2-n-2-find-lim-n-S-n-x-if-x-lt-1-

Question Number 33719 by prof Abdo imad last updated on 22/Apr/18

s i m p l i f y S_{n} (x) = (1 + x^{2}) (1 + x^{4}) \dots . (1 + x^{2^{n}})

2) f i n d {l i m}_{n \to + \infty} S_{n} (x) i f ∣ x ∣< 1 .

Commented by prof Abdo imad last updated on 23/Apr/18

w e h a v e (1 + x^{2}) (1 + x^{4}) = 1 + x^{4} + x^{2} + x^{6}

= 1 + x^{2} + x^{4} + x^{6} = \frac{1 - x^{8}}{1 - x^{2}} i f x^{2} \neq 1

l e t s u p p o s e t h a t S_{n} (x) = \frac{1 - x^{2^{n + 1}}}{1 - x^{2}}

S_{n + 1} (x) = (1 + x^{2}) (1 + x^{4}) \dots . . (1 + x^{2^{n + 1}})

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

= \frac{1 - {(x^{2^{n + 1}})}^{2}}{1 - x^{2}} = \frac{1 - x^{2^{n + 2}}}{1 - x^{2}} t h e r e s u l t i s t r u e a t t e r m

n + 1 a n d w e m u s t s t u d y t b e c a s e x = \bar{+} 1

2) w e h a v e p r o v e d t h a t

S_{n} (x) = \frac{1 - x^{2^{n + 1}}}{1 - x^{2}} s o f o r ∣ x ∣< 1 {l i m}_{n \to + \infty} S_{n} (x) = \frac{1}{1 - x^{2}}

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18

(1 + x^{2}) (1 + x^{4}) = 1 + x^{2} + x^{4} + x^{6}

(1 + x^{2}) (1 + x^{4}) (1 + x^{8}) = (1 + x^{2} + x^{4} + x^{6}) (1 + x^{8})

= 1 + x^{2} + x^{4} + x^{6} + x^{8} + x^{10} + x^{12} + x^{14}

S_{n} (x) = (1 - x^{2 n}) / (1 - x^{2})

s o t h e v a l u e o f S_{n} (x) i s 1 / (1 - x^{2}) w h e n l i m i t n

t e n d s t o i n f i n i t y

\underset{x \to 0}{i}

Commented by prof Abdo imad last updated on 22/Apr/18

i t s n o t c o r r e c t b e c a u s e \frac{1 - x^{2 n}}{1 - x^{2}}

= \frac{1 - {(x^{2})}^{n}}{1 - x^{2}} = \frac{(1 - x^{2}) (1 + x^{2} + x^{4} + \dots + x^{2 n - 2})}{1 - x^{2}}

= 1 + x^{2} + x^{4} + \dots + x^{2 n - 2} \neq S_{n}