An algebra problem on sequences

Algebra Level 3

Let the sequence { a i } i = 0 ∞ \left \{ a_i \right \}_{i=0}^\infty be defined by a 0 = 1 2 a_0 = \frac 12 and a n = 1 + ( a n − 1 − 1 ) 2 a_n = 1 + (a_{n-1} - 1)^2 .

Find ∏ i = 0 ∞ a i \displaystyle \prod_{i=0}^\infty a_i .


The answer is 0.6667.

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Vijay Simha by Vijay Simha , 47, USA

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1 solution

Brian Moehring
Mar 1, 2017

Define b n = a n − 1 b_n = a_n - 1 for each n n . Then by induction, we may see b n = ( b n − 1 ) 2 = ⋯ = ( b 0 ) 2 n = ( − 1 2 ) 2 n b_n = \left(b_{n-1}\right)^2 = \cdots = (b_0)^{2^n} = \left(-\frac{1}{2}\right)^{2^n} and then by another induction, ∏ i = 0 n a i = ∏ i = 0 n ( 1 + ( − 1 2 ) 2 i ) = ∑ k = 0 2 n + 1 − 1 ( − 1 2 ) k = 1 − ( − 1 2 ) 2 n + 1 1 − ( − 1 2 ) . \prod_{i=0}^n a_i = \prod_{i=0}^n \left(1 + \left(-\frac{1}{2}\right)^{2^i}\right) = \sum_{k=0}^{2^{n+1}-1} \left(-\frac{1}{2}\right)^k = \frac{1 - \left(-\frac{1}{2}\right)^{2^{n+1}}}{1 - \left(-\frac{1}{2}\right)}.

Letting n → ∞ n\rightarrow\infty gives ∏ i = 0 ∞ a i = lim ⁡ n → ∞ 1 − ( − 1 2 ) 2 n + 1 1 − ( − 1 2 ) = 1 3 / 2 = 2 3 ≈ 0.6667 \prod_{i=0}^\infty a_i = \lim_{n\rightarrow\infty} \frac{1 - \left(-\frac{1}{2}\right)^{2^{n+1}}}{1 - \left(-\frac{1}{2}\right)} = \frac{1}{3/2} = \frac{2}{3} \approx \boxed{0.6667}

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Can u explain the second line ?

Kushal Bose - 4 years, 3 months ago

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As clarification, are you talking about this? ∏ i = 0 n ( 1 + ( − 1 2 ) 2 i ) = ∑ k = 0 2 n + 1 − 1 ( − 1 2 ) k \prod_{i=0}^n \left(1 + \left(-\frac{1}{2}\right)^{2^i}\right) = \sum_{k=0}^{2^{n+1}-1} \left(-\frac{1}{2}\right)^k

Brian Moehring - 4 years, 3 months ago

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How didi you get this.Can u show some steps

Kushal Bose - 4 years, 3 months ago

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@Kushal Bose – As I mentioned, it's a fairly direct, though not immediate, induction on n n .

As a base case, you may either set n = − 1 n=-1 , in which both sides equal 1 1 , or you may, if you don't like empty products, set n = 0 n=0 and note both side equal 1 2 \frac{1}{2} .

Then as an inductive step suppose n > − 1 n>-1 [or n > 0 n>0 ] and note that, using the inductive hypothesis, ∏ i = 0 n ( 1 + ( − 1 2 ) 2 i ) = ∏ i = 0 n − 1 ( 1 + ( − 1 2 ) 2 i ) ⋅ ( 1 + ( − 1 2 ) 2 n ) = ∑ k = 0 2 n − 1 ( − 1 2 ) k ⋅ ( 1 + ( − 1 2 ) 2 n ) = ∑ k = 0 2 n − 1 ( − 1 2 ) k + ∑ k = 0 2 n − 1 ( − 1 2 ) k + 2 n = ∑ k = 0 2 n − 1 ( − 1 2 ) k + ∑ k = 2 n 2 n + 1 − 1 ( − 1 2 ) k = ∑ k = 0 2 n + 1 − 1 ( − 1 2 ) k . \begin{aligned}\prod_{i=0}^n \left(1 + \left(-\frac{1}{2}\right)^{2^i}\right) &= \prod_{i=0}^{n-1} \left(1 + \left(-\frac{1}{2}\right)^{2^i}\right) \cdot \left(1 + \left(-\frac{1}{2}\right)^{2^n}\right) \\ &= \sum_{k=0}^{2^{n}-1} \left(-\frac{1}{2}\right)^k \cdot \left(1 + \left(-\frac{1}{2}\right)^{2^n}\right) \\ &= \sum_{k=0}^{2^{n}-1} \left(-\frac{1}{2}\right)^k + \sum_{k=0}^{2^{n}-1} \left(-\frac{1}{2}\right)^{k+2^n} \\ &= \sum_{k=0}^{2^{n}-1} \left(-\frac{1}{2}\right)^k + \sum_{k=2^n}^{2^{n+1}-1} \left(-\frac{1}{2}\right)^k \\ &= \sum_{k=0}^{2^{n+1}-1} \left(-\frac{1}{2}\right)^k.\end{aligned}

Brian Moehring - 4 years, 3 months ago

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@Brian Moehring – Ok Now its fine Thanks

Kushal Bose - 4 years, 3 months ago

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