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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)..... is equal to

$$\mathrm{If}\:\mid{x}\mid\:<\:\mathrm{1}\:\mathrm{then} \\ $$ $$\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{8}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{16}} \:+\:\mathrm{1}\right)..... \\ $$ $$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Answered by mrW1 last updated on 15/Nov/17

let P_n =(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)(x^(16) + 1).....(x^2^n +1) (x−1)P_n =(x−1)(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)(x^(16) + 1).....(x^2^n +1) (x−1)P_n =(x^2 −1)(x^2 + 1)(x^4 + 1)(x^8 + 1)(x^(16) + 1).....(x^2^n +1) (x−1)P_n =(x^4 −1)(x^4 + 1)(x^8 + 1)(x^(16) + 1).....(x^2^n +1) ...... (x−1)P_n =(x^2^n − 1)(x^2^n +1) (x−1)P_n =x^2^(n+1) −1 ⇒P_n =((1−x^2^(n+1) )/(1−x)) lim_(n→∞) P_n =(1/(1−x)) ⇒(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)(x^(16) + 1).....=(1/(1−x))

$${let}\:{P}_{{n}} =\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{8}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{16}} \:+\:\mathrm{1}\right).....\left({x}^{\mathrm{2}^{{n}} } +\mathrm{1}\right) \\ $$ $$\left({x}−\mathrm{1}\right){P}_{{n}} =\left({x}−\mathrm{1}\right)\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{8}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{16}} \:+\:\mathrm{1}\right).....\left({x}^{\mathrm{2}^{{n}} } +\mathrm{1}\right) \\ $$ $$\left({x}−\mathrm{1}\right){P}_{{n}} =\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{8}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{16}} \:+\:\mathrm{1}\right).....\left({x}^{\mathrm{2}^{{n}} } +\mathrm{1}\right) \\ $$ $$\left({x}−\mathrm{1}\right){P}_{{n}} =\left({x}^{\mathrm{4}} −\mathrm{1}\right)\left({x}^{\mathrm{4}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{8}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{16}} \:+\:\mathrm{1}\right).....\left({x}^{\mathrm{2}^{{n}} } +\mathrm{1}\right) \\ $$ $$...... \\ $$ $$\left({x}−\mathrm{1}\right){P}_{{n}} =\left({x}^{\mathrm{2}^{{n}} } −\:\mathrm{1}\right)\left({x}^{\mathrm{2}^{{n}} } +\mathrm{1}\right) \\ $$ $$\left({x}−\mathrm{1}\right){P}_{{n}} ={x}^{\mathrm{2}^{{n}+\mathrm{1}} } −\mathrm{1} \\ $$ $$\Rightarrow{P}_{{n}} =\frac{\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{x}} \\ $$ $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{P}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$ $$\Rightarrow\left({x}\:+\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{4}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{8}} \:+\:\mathrm{1}\right)\left({x}^{\mathrm{16}} \:+\:\mathrm{1}\right).....=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$

Commented bymath solver last updated on 16/Nov/17

sir, can you explain your last step when n tending to infinity?

$$\mathrm{sir},\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{your}\:\mathrm{last}\:\mathrm{step} \\ $$ $$\mathrm{when}\:\mathrm{n}\:\mathrm{tending}\:\mathrm{to}\:\mathrm{infinity}? \\ $$

Commented bymrW1 last updated on 16/Nov/17

we have P_n =((1−x^2^(n+1) )/(1−x)) when n→∞, 2^(n+1) →∞ since ∣x∣<1 x^2^(n+1) →0 P_n →((1−0)/(1−x))=(1/(1−x))

$${we}\:{have}\:{P}_{{n}} =\frac{\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } }{\mathrm{1}−{x}} \\ $$ $${when}\:{n}\rightarrow\infty, \\ $$ $$\mathrm{2}^{{n}+\mathrm{1}} \rightarrow\infty \\ $$ $${since}\:\mid{x}\mid<\mathrm{1} \\ $$ $${x}^{\mathrm{2}^{{n}+\mathrm{1}} } \rightarrow\mathrm{0} \\ $$ $${P}_{{n}} \rightarrow\frac{\mathrm{1}−\mathrm{0}}{\mathrm{1}−{x}}=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$

Commented bymath solver last updated on 16/Nov/17

okk, i didn′t read ∣x∣ < 1.

$$\mathrm{okk},\:\mathrm{i}\:\mathrm{didn}'\mathrm{t}\:\:\mathrm{read}\:\mid\mathrm{x}\mid\:<\:\mathrm{1}. \\ $$

Commented byJoel577 last updated on 16/Nov/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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