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How to put mathematically sequence that changes sign like:

$n = 0\quad f = 1$

$n = 1 \quad f = -1$

$n = 2 \quad f = -1$

$n = 3 \quad f = 1$

$n = 4 \quad f = -1$

$n = 5 \quad f = -1$

$n = 6 \quad f = 1$ .....

In the form of (-1)^(something) or similar analytical expression.

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  • $\begingroup$ Is there closed form analytical expression ? $\endgroup$
    – Anonymous
    Jun 7, 2016 at 18:39
  • $\begingroup$ @Anonymous That is the closed form expression for the sequence $\{ 1, -1, 1, -1, \dots \}$, which is not your sequence. $\endgroup$
    – Axoren
    Jun 7, 2016 at 21:06
  • $\begingroup$ This is easy if the use of modulus is permitted. $\endgroup$
    – PM 2Ring
    Jun 8, 2016 at 7:10

11 Answers 11

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This formula works: $$f(n)=-\frac13+\frac43\cos\left(\frac{2\pi}3n\right)$$ Here's a graph of what that looks like.

Since the sum of any three consecutive values is $-1$, it satisfies this recurrence relation: $$f(x+2)=-f(x+1)-f(x)-1$$ It also satisfies the recurrence relation $f(x+3)=f(x)$, though that's less interesting.

(Geometrically, think of a circle of radius $4/3$ centered on the point $(-1/3,0)$. This circle passes through the three equidistant points $(1,0)$, $(-1,2/\sqrt3)$, and $(-1,-2/\sqrt3)$ (they form the corners of an equilateral triangle inscribed in the circle). The function $f$ alternately takes the $x$-coordinate of each of these points.)

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\begin{align} a_1 &= 1 \\ a_2 &= -1 \\ a_n &= a_{n-1}a_{n-2} \end{align}

Using a product recurrence relation, we have

$$ a_n = a_2^{F_{n-1}}a_1^{F_{n-2}} $$

Where $F_k$ is the $k$'th Fibonacci number. Given a closed form for $F_k$, we can simplify $a_n$.

$$ F_k = \frac{(1 + \sqrt{5})^k - (1 - \sqrt{5})^k}{2^k\sqrt{5}} $$

$$ a_n = (-1)^{F_{n-1}} = (-1)^{\frac{(1 + \sqrt{5})^{n-1} - (1 - \sqrt{5})^{n-1}}{2^{n-1}\sqrt{5}}} $$

This was solved so that the starting index of the sequence was $1$. If you want to start at $0$, we simply shift the index by $+1$.

$$ a_n = (-1)^{F_{n}} = (-1)^{\frac{(1 + \sqrt{5})^{n} - (1 - \sqrt{5})^{n}}{2^{n}\sqrt{5}}} $$


Using Euler's Formula, we can get a far more (or less) elegant solution:

$$ a_n = (-1)^{F_n} = \cos(\pi F_n) + i\sin(\pi F_n) = e^{i\pi F_n} $$

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    $\begingroup$ LOL. (+1) ${}{}$ $\endgroup$
    – Micah
    Jun 8, 2016 at 0:41
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    $\begingroup$ Upon closer inspection, $$ a_n = (-1)^{F_n} = e^{i\pi F_n} = \cos(\pi F_n) + i\sin(\pi F_n) $$ $\endgroup$
    – Axoren
    Jun 8, 2016 at 4:29
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$$ {1 \over 3}\left[4\cos\left({2n\pi \over 3}\right) - 1\right]\,,\qquad n = 0,1,2,\ldots $$

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Another possibility (closer to the $(-1)^k$ that you claim to have played with) is: $$ (-1)^{1+\left\lfloor\frac{2(k-1)}{3}\right\rfloor} $$

(and I won't be surprised if I managed to get one of the shifts wrong).

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  • $\begingroup$ This seems like imaginary. $\endgroup$
    – Anonymous
    Jun 7, 2016 at 23:35
  • $\begingroup$ @Anonymous - that $\lfloor \cdot \rfloor$ is the floor function - the greatest integer $\le$ to what is inside. So the exponent is always an integer, and the value is always $1$ or $-1$. However, it could be simplified to $(-1)^{\left\lfloor \frac{2k+1}3\right\rfloor}$ $\endgroup$ Jun 8, 2016 at 3:02
  • $\begingroup$ I chose not to simplify the expression as I believed having both shifts made it a little clearer how I got from $(-1)^{\lfloor2k/3\rfloor}$ which I believe is a quite natural thing to try. I should probably have written something about that. $\endgroup$ Jun 8, 2016 at 6:50
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Here's a negative result: $f(n)$ cannot be written in the form $$ f(n) = (-1)^{g(n)} $$ where $g$ is a polynomial function.

The relevant condition on $g$ is that its values form the period 3 sequence

$$ g(n) \equiv 0, 1, 1, 0, 1, 1, \ldots \pmod{2} $$

The difference sequence of a sequence is defined by $\Delta h(n) = h(n+1) - h(n)$. We can compute the differences of $g$:

$$ \Delta g(n) \equiv 1, 0, 1, 1, 0, 1, \ldots \pmod{2} $$ $$ \Delta^2 g(n) \equiv 1, 1, 0, 1, 1, 0, \ldots \pmod{2} $$ $$ \Delta^3 g(n) \equiv 0, 1, 1, 0, 1, 1, \ldots \pmod{2} $$

and so forth. If $g$ were a polynomial function, then by taking enough differences we would arrive at the zero sequence; but that clearly cannot happen.

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  • $\begingroup$ This proves is that $g(n)$ cannot be a polynomial. An analytic function is not limited to this, it can be any function described by a convergent power series. Polynomials are such functions, but not all such functions are polynomials. $\endgroup$
    – Axoren
    Jun 8, 2016 at 18:36
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with $r = -\frac12 + \frac{\sqrt{3}}{2} i$

$$a_k = \frac{2(r^k + \overline{r^k} ) -1}{3} $$

should work.

I'm sorry for posting a wrong answer before.

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  • $\begingroup$ That doesn't work, the desired sequence is $1,-1,-1,1,-1,-1,1,-1,-1, \cdots $ and your sequence is $1,-1,-1, 1, 1,-1,-1, 1,1,-1,-1, 1,\dots $ $\endgroup$
    – zhw.
    Jun 7, 2016 at 19:40
  • $\begingroup$ r=(-1+1i*sqrt(2))/2; k=[0 1 2 3 4];(2*(r.^k+(conj(r.^k)))-1)/3 ans = 1.0000 -1.0000 -0.6667 0.5000 -0.9167 Matlab $\endgroup$
    – Anonymous
    Jun 7, 2016 at 20:35
  • $\begingroup$ @Anonymous r = (-1 + sqrt(3)*i)/2, sorry $\endgroup$ Jun 7, 2016 at 21:04
  • $\begingroup$ awesome thanks! $\endgroup$
    – Anonymous
    Jun 7, 2016 at 23:34
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The sign changes every 3 steps. So basically you've got a + if $n = 0$ and then $n = 3$ and $n = 6$ etc. Therefore it's an alternating sequence given as such: $a_n = 1$ if $n$ mod $3 = 0 $ and $-1$ otherwise.

Edit: The answer using the floor function was wrong as pointed by Erick Wong. Here is another way to write down the general term of the sequence : $$a_n = (-1)^{\mathbb{1}_{n \ mod \ 3 \ \neq \ 0}}.$$

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  • $\begingroup$ Is there closed form analytical expression ? $\endgroup$
    – Anonymous
    Jun 7, 2016 at 18:42
  • $\begingroup$ I've edited my answer. $\endgroup$
    – user242756
    Jun 7, 2016 at 18:46
  • $\begingroup$ @bgsk Your exponent certainly looks like it has period $4$, not $3$. $\endgroup$
    – Erick Wong
    Jun 7, 2016 at 21:47
  • $\begingroup$ I see. Thank you for the correction. $\endgroup$
    – user242756
    Jun 7, 2016 at 22:03
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The function $\lceil x \rceil - \lfloor x \rfloor$ is $1$ except for integers, where it vanishes. If we scale it to miss only $3$-integers and put as exponent onto $-1$: $$ (-1)^{\lceil n/3 \rceil - \lfloor n/3 \rfloor} $$ or similarly, but with the floor function only: $$ (-1)^{\lfloor -n/3 \rfloor + \lfloor n/3 \rfloor} $$ we get your desired sequence for $n=0$, $1$, $2\ldots$

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$$\ (-1)^{n^2\,\rm mod\,3}\ $$

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If you allow yourself other roots of unity, you can use the discrete fourier transform.

I presume you want the sequence to have period $3$. So, we use the cube root of unity $\zeta = \frac{-1 + i \sqrt{3}}{2}$. The DFT $\hat{f}$ is given by

$$ \hat{f}(n) = \sum_{k=0}^2 f(k) \zeta^{kn} $$

which we can tabulate as

  • $\hat{f}(0) = -1$
  • $\hat{f}(1) = 2$
  • $\hat{f}(2) = 2$

Then, we can recover $f$ by

$$ f(n) = \frac{1}{3} \sum_{k=0}^2 \hat{f}(k) \zeta^{-kn} $$

which works out to

$$ f(n) = \frac{1}{3} \left(-1 + 2 \zeta^n + 2 \zeta^{-n} \right) $$

which can be expressed in a variety of forms; e.g. since $\zeta = \exp(2 \pi i / 3)$, we have

$$ \cos\left( \frac{2 \pi n}{3} \right) = \frac{\zeta^n + \zeta^{-n}}{2} $$

and thus

$$ f(n) = \frac{1}{3} \left( -1 + 4 \cos\left( \frac{2 \pi n}{3} \right) \right) $$

More generally, you could solve this by picking any three linearly independent functions with period $3$, then solving a system of equations to recover what coefficients you need on them to construct $f$; the DFT is just a direct way to obtain such a solution when the three functions are $1$, $\zeta$, and $\zeta^2$.

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A meta-answer: if you already have the sequence $0,1,1,0,1,1\ldots$, then you may take the power of $-1,$ but you actually don't need exponentiation at all (similarly for other periodical sequences): \begin{align} a_n &=1 - 2(n^2 \bmod 3)\\ &=1 - 2n^2 + 6\lfloor n^2/3\rfloor\\ &=1 - 2\cdot \lvert (n-1) {\bmod} 3 -1\rvert\\ &= 1 - 2(\lceil n/3 \rceil - \lfloor n/3 \rfloor) \\ &= 1- 2\cdot \mathbf{1}_{\{n: \, 3\not\mid n\}}\\ &=\ldots \end{align} which I think is neater. Also: $$ a_n= (-1)^{2k+1 \bmod 3} $$

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