If-x-lt-1-then-x-1-x-2-1-x-4-1-x-8-1-x-16-1-is-equal-to-

Question Number 24286 by Joel577 last updated on 15/Nov/17

If ∣ x ∣ < 1 then

(x + 1) (x^{2} + 1) (x^{4} + 1) (x^{8} + 1) (x^{16} + 1) \dots . .

is equal to

Answered by mrW1 last updated on 15/Nov/17

l e t P_{n} = (x + 1) (x^{2} + 1) (x^{4} + 1) (x^{8} + 1) (x^{16} + 1) \dots . . (x^{2^{n}} + 1)

(x - 1) P_{n} = (x - 1) (x + 1) (x^{2} + 1) (x^{4} + 1) (x^{8} + 1) (x^{16} + 1) \dots . . (x^{2^{n}} + 1)

(x - 1) P_{n} = (x^{2} - 1) (x^{2} + 1) (x^{4} + 1) (x^{8} + 1) (x^{16} + 1) \dots . . (x^{2^{n}} + 1)

(x - 1) P_{n} = (x^{4} - 1) (x^{4} + 1) (x^{8} + 1) (x^{16} + 1) \dots . . (x^{2^{n}} + 1)

\dots \dots

(x - 1) P_{n} = (x^{2^{n}} - 1) (x^{2^{n}} + 1)

(x - 1) P_{n} = x^{2^{n + 1}} - 1

\Rightarrow P_{n} = \frac{1 - x^{2^{n + 1}}}{1 - x}

\lim_{n \to \infty} P_{n} = \frac{1}{1 - x}

\Rightarrow (x + 1) (x^{2} + 1) (x^{4} + 1) (x^{8} + 1) (x^{16} + 1) \dots . . = \frac{1}{1 - x}

Commented by math solver last updated on 16/Nov/17

sir, can you explain your last step

when n tending to infinity ?

Commented by mrW1 last updated on 16/Nov/17

w e h a v e P_{n} = \frac{1 - x^{2^{n + 1}}}{1 - x}

w h e n n \to \infty,

2^{n + 1} \to \infty

s i n c e ∣ x ∣< 1

x^{2^{n + 1}} \to 0

P_{n} \to \frac{1 - 0}{1 - x} = \frac{1}{1 - x}

Commented by math solver last updated on 16/Nov/17

okk, i {didn}^{'} t read ∣ x ∣ < 1 .

Commented by Joel577 last updated on 16/Nov/17

t h a n k y o u v e r y m u c h

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