# Write the general term of the periodic sequence $1$, $-1$, $-1$, $1$, $-1$, $-1$, $1$, ..., as $(-1)^{g(n)}$ or other closed form

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.

• Is there closed form analytical expression ? Jun 7 '16 at 18:39
• @Anonymous That is the closed form expression for the sequence $\{ 1, -1, 1, -1, \dots \}$, which is not your sequence. Jun 7 '16 at 21:06
• This is easy if the use of modulus is permitted. Jun 8 '16 at 7:10

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.)

\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:

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

• LOL. (+1) ${}{}$ Jun 8 '16 at 0:41
• Upon closer inspection, $$a_n = (-1)^{F_n} = e^{i\pi F_n} = \cos(\pi F_n) + i\sin(\pi F_n)$$ Jun 8 '16 at 4:29

$${1 \over 3}\left[4\cos\left({2n\pi \over 3}\right) - 1\right]\,,\qquad n = 0,1,2,\ldots$$

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).

• This seems like imaginary. Jun 7 '16 at 23:35
• @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 $(-1)^{\left\lfloor \frac{2k+1}3\right\rfloor}$ Jun 8 '16 at 3:02
• 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. Jun 8 '16 at 6:50

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.

• 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. Jun 8 '16 at 18:36

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.

• That doesn't work, the desired sequence is $1,-1,-1,1,-1,-1,1,-1,-1, \cdots$ and your sequence is $1,-1,-1, 1, 1,-1,-1, 1,1,-1,-1, 1,\dots$
– zhw.
Jun 7 '16 at 19:40
• 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 Jun 7 '16 at 20:35
• @Anonymous r = (-1 + sqrt(3)*i)/2, sorry Jun 7 '16 at 21:04
• awesome thanks! Jun 7 '16 at 23:34

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 : $$a_n = (-1)^{\mathbb{1}_{n \ mod \ 3 \ \neq \ 0}}.$$

• Is there closed form analytical expression ? Jun 7 '16 at 18:42
– user242756
Jun 7 '16 at 18:46
• @bgsk Your exponent certainly looks like it has period $4$, not $3$. Jun 7 '16 at 21:47
• I see. Thank you for the correction.
– user242756
Jun 7 '16 at 22:03

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 $n=0$, $1$, $2\ldots$

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

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 $\zeta^2$.

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: $$a_n= (-1)^{2k+1 \bmod 3}$$