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Which is the general formula of this sequence?

$$ x_0 = -1$$

$$ x_{n+1} = ((-1)^n*X_n)/2^n$$

What baffles me more is the sign which is like this: $--++--++--++--++\cdots$

I've been wondering how the sign could change like that, any ideas?

Gtoyos
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  • The sign of the next number depends on the parity of $n$ and the sign of the current number, therefore this pattern. – Peter Aug 28 '17 at 22:37
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    $ x_{n+2} = (-1)^{n+1} \cdot \frac{x_{n+1}}{2^{n+1}} = (-1)^{2n+1}\cdot \frac{x_n}{2^{2n+1}}=-\frac{x_n}{2^{2n+1}},$, so the sign changes every two terms. – dxiv Aug 28 '17 at 22:47
  • Yes. But i don't know which is the pattern. How the succession does --++--++--++.... – Gtoyos Aug 28 '17 at 22:47
  • @Gtoyos Do you mean $\displaystyle (-1)^{\lfloor n/2 \rfloor+1},$? – dxiv Aug 28 '17 at 22:49
  • I edited it. sorry – Gtoyos Aug 28 '17 at 22:50
  • But the point is how the sign could be --++--++--++--. – Gtoyos Aug 28 '17 at 22:51
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    how the sign could be The sign is what it is based on the recurrence relation. Sorry, don't know what you are asking, or what kind of answer you expect. – dxiv Aug 28 '17 at 22:52
  • I want to know how the general formula must be in order to the sign to be --++--++--++. For example the general formula X_n = (-1)^n the sign is +-+-+-+-+. In this case, what is the general formula for a succession which it's sign is --++--++--++ – Gtoyos Aug 28 '17 at 22:56
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    @Gtoyos As I wrote in a previous comment, $\displaystyle (-1)^{\lfloor n/2 \rfloor + 1}$ will generate the sequence $-1,-1,,$ $+1,+1,,$ $-1,-1,,$ $+1,+1,,$ $\cdots$ – dxiv Aug 28 '17 at 23:00
  • Your question is not clear. Are you asking for a proof that the signs follow the pattern they appear to? If so, then I suggest proving it by induction. Do it by hand for the first four terms and then work by induction using the remainder on division by $4$. – lulu Aug 28 '17 at 23:08
  • I am looking for a general formula. @dxiv but, if you try with x =1 the result will be (-1)^1.5. That is an imaginary number! I am only working with natural numbers. Sorry for not stating it beforehand. – Gtoyos Aug 28 '17 at 23:12
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    @Gtoyos $\lfloor x \rfloor$ is the greatest integer function, and is always an integer. In particular $\lfloor 1/2 \rfloor = 0,$. – dxiv Aug 28 '17 at 23:14
  • But (-1)^1.5 is not -i? I'm working with "high school math" – Gtoyos Aug 28 '17 at 23:17
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    @Gtoyos There is *no* "1.5". For $n=1$ you get $(-1)^{\lfloor n/2 \rfloor +1}=(-1)^{0 +1}=-1,$. – dxiv Aug 28 '17 at 23:19
  • I don't understand what ⌊n/2⌋ means (greatest integer function). Is the integrer part? Like ⌊3/2⌋ = 1? – Gtoyos Aug 28 '17 at 23:23
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    @Gtoyos Did you click the link I posted two comments ago? Yes, it is the integer part. – dxiv Aug 28 '17 at 23:24
  • Just read it. Thank you so much, you really have patience. Is there any way i can show you my gratitude (upvote, reputation or something?) I'm new in this site – Gtoyos Aug 28 '17 at 23:26
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    @Gtoyos Glad to hear you sorted it out. I posted a writeup of my comments as an answer, it is entirely up to you to accept and/or upvote it as you see fit. – dxiv Aug 28 '17 at 23:40

3 Answers3

2

Let $\,a_n=(-1)^{\lfloor n/2 \rfloor+1}\,$, then by direct computation $a_0=a_1=-1\,$, $a_2=a_3=1\,$, and:

$$ a_{n+4}=(-1)^{\lfloor (n+4)/2 \rfloor+1} = (-1)^{\lfloor n/2 + 2 \rfloor+1}=(-1)^2 \cdot (-1)^{\lfloor n/2 \rfloor+1}=(-1)^{\lfloor n/2 \rfloor+1} = a_n $$

Therefore the sequence is periodic with period $\,4\,$, and so the first four values $\,-1,-1,1,1\,$ keep repeating indefinitely.

dxiv
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1

You can easly prove by induction that $\forall n\in\mathbb{N},x_n\neq0$.

Now let $n\ge 1$. We have $\forall k\in\{0,..,n-1\},x_{k+1}=\left(\dfrac{-1}{2}\right)^kx_k$. Thus \begin{aligned} \prod_{k=0}^{n-1}x_{k+1}&=\prod_{k=0}^{n-1}\left(\dfrac{-1}{2}\right)^k x_k\\ \prod_{k=1}^nx_k&=\left(\dfrac{-1}{2}\right)^{\left(\sum_{k=0}^{n-1}k\right)}\prod_{k=0}^{n-1}x_k. \end{aligned}

You can eliminate $\prod_{k=1}^{n-1}x_k$ and we have$\sum_{k=0}^{n-1}k=\dfrac{(n-1)n}{2}$. Therefore \begin{aligned} x_{n+1}&=\left(\dfrac{-1}{2}\right)^{\frac{(n-1)n}{2}}x_0\\ &=-\left(\dfrac{-1}{2}\right)^{\frac{(n-1)n}{2}}. \end{aligned}

Hence a general formula that gives your described sign is $(-1)^{\frac{(n-1)n}{2}}$. Note that even if this goes beyond your math level, you can prove the formula we got for $x_n$ by induction.

Scientifica
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Let $n=2k $ be even. Then $(-1)^n=1$ and $x_{n+1}$ will have the same sign as $x_n$.

Let $n=2k+1$ be odd. Then $(-1)^n=-1$ and $x_{n+1} $ will be the opposite sign as $x_n $.

Therefore $x_{n+2}$ will always be the opposite sign as $x_{n} $. Because if $n $ is even then $x_{n+1}$ will have the same sign, and $n+1$ is odd so $x_{n+1}$ has the opposite sign. If $n$ is odd then $x_{n+1}$ has the opposite sign, and $n+1$ is even so $x_{n+1} $ also has the opposite sign.

And $x_{n+4} $ will have the same sign as $x_{n+4} $ will have the opposite of $x_{n+2} $ which is the opposite of $x_n$.

So $x_0 = -$ by definition.

$x_1=-$ because 0 is even so sign stays the same.

$x_2=+$ because $1$ is odd so sign changes.

$x_3=+$ because $2$ is even so sign stays the same

$x_4=-$ and pattern repeats forever as pattern repeats four every fourth term.

fleablood
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