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Math Help - double sequence

  1. #1
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    double sequence

    Let  d be a real number. For each integer  m \geq 0 , define a sequence  \{a_{m}(j) \} ,  j = 0,1,2, \ldots by the condition  a_{m}(0) = d/2^{m} , and  a_{m}(j+1) = (a_{m}(j))^{2}+2a_{m}(j), \ j \geq 0 . Find  \lim_{n \to \infty} a_{n}(n) .

    Basically I found that the sequence  a_{n}(n) monotone increasing. But the offsets are becoming smaller and smaller. It looks like a taylor series.

    Any ideas about how to get the limit?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by particlejohn View Post
    Let  d be a real number. For each integer  m \geq 0 , define a sequence  \{a_{m}(j) \} ,  j = 0,1,2, \ldots by the condition  a_{m}(0) = d/2^{m} , and  a_{m}(j+1) = (a_{m}(j))^{2}+2a_{m}(j), \ j \geq 0 . Find  \lim_{n \to \infty} a_{n}(n) .

    Basically I found that the sequence  a_{n}(n) monotone increasing. But the offsets are becoming smaller and smaller. It looks like a taylor series.

    Any ideas about how to get the limit?
    While playing with the first terms of the sequence I noticed that a_n(n)=\left( a_n(0)+1\right)^{2^n}-1 is true for n\leq2. Using induction it can be shown that it is in fact true for any non-negative integer n.

    Induction hypothesis \mathcal{P}_n\,: for all non-negative integer p, \,a_p(n)=\left(a_p(0)+1\right)^{2^n}-1

    Let's show \mathcal{P}_0 : let p be a non-negative integer. One has a_p(0)=a_p(0)+1-1=(a_p(0)+1)^{2^0}-1 so \mathcal{P}_0 is true.

    Let's assume that there exists a non-negative integer n such that \mathcal{P}_n is true. Let p be a non-negative integer.

    \begin{aligned}<br />
a_p(n+1)&=a_p(n)^2+2a_p(n)\\<br />
&=\left[a_p(n)+1\right]^2-1\\<br />
&=\left[\left(a_p(0)+1\right)^{2^n}-1+1\right]^2-1 \text{ since } a_p(n)=\left(a_p(0)+1\right)^{2^n}-1\\<br />
&=\left[a_p(0)+1 \right]^{2^{n+1}}-1.\end{aligned}

    This shows \mathcal{P}_{n+1} and we have, for any non-negative integers n and p, \,a_p(n)=\left(a_p(0)+1\right)^{2^n}-1. In particular, for n=p : a_n(n)=\left( a_n(0)+1\right)^{2^n}-1=\left( \frac{d}{2^n}+1\right)^{2^n}-1. Now, what is \lim_{n\to\infty}\left( 1+\frac{d}{2^n}\right)^{2^n}-1 ?
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  3. #3
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     e^{d}-1 .
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by particlejohn View Post
     e^{d}-1 .
    It seems good to me, I got the same result.
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  5. #5
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    basically \lim_{n\to\infty}\left( 1+\frac{d}{2^n}\right)^{2^n}-1 behaves as  \lim_{n \to \infty} \left(1+\frac{d}{n} \right)^{n} -1 right? Hence we get  e^{d}-1 .
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by particlejohn View Post
    basically \lim_{n\to\infty}\left( 1+\frac{d}{2^n}\right)^{2^n}-1 behaves as  \lim_{n \to \infty} \left(1+\frac{d}{n} \right)^{n} -1 right?
    Yes.
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  7. #7
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    Quote Originally Posted by particlejohn View Post
    basically \lim_{n\to\infty}\left( 1+\frac{d}{2^n}\right)^{2^n}-1 behaves as  \lim_{n \to \infty} \left(1+\frac{d}{n} \right)^{n} -1 right? Hence we get  e^{d}-1 .
    Yes. This is a subsequence of  \lim_{n \to \infty} \left(1+\frac{d}{n} \right)^{n} -1 . Since this sequence converges to e^d - 1. Thus, any subsequence converges to e^d - 1.
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