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Thread: double sequence

  1. #1
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    double sequence

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

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

    Basically I found that the sequence $\displaystyle 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 $\displaystyle a_n(n)=\left( a_n(0)+1\right)^{2^n}-1$ is true for $\displaystyle n\leq2$. Using induction it can be shown that it is in fact true for any non-negative integer $\displaystyle n$.

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

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

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

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

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