# Help with Sequences Problem: Expressing in Terms of Variable Integer k

• Mentallic

#### Mentallic

Homework Helper
I want to express

$$...-2\pi+\theta,\theta, 2\pi-\theta, 2\pi+\theta, 4\pi-\theta...$$

in terms of a variable integer k.

e.g. ...-x,0,x, 2x, 3x... = kx, k E Z

So I was thinking expressing it as so: $$2k\pi \pm \theta$$
but I believe it can be expressed in another way to avoid the $\pm$, using (-1)k somehow.

How can this be done?

Mentallic said:
I want to express

$$...-2\pi+\theta,\theta, 2\pi-\theta, 2\pi+\theta, 4\pi-\theta...$$

in terms of a variable integer k.

e.g. ...-x,0,x, 2x, 3x... = kx, k E Z

So I was thinking expressing it as so: $$2k\pi \pm \theta$$
but I believe it can be expressed in another way to avoid the $\pm$, using (-1)k somehow.

How can this be done?

The example sequence you gave seems to indicate you want "two plusses", "a minus", "a plus", and "a minus". Is this a mistake? If you want the sequence to alternate between plus and minus for theta and still use every integer k for $2\pi$ you may be able to use something like
$\pi(k+1)k + (-1)^k \theta[/tex] Yeah I'm not sure about your example, but I think this will do what you want. $$\lfloor k/2 \rfloor 2\pi + (-1)^k \theta$$ My example was just trying to illustrate what I meant, I didn't mean for it to be similar to my problem I think you might have a typo: $$\pi(k+1)k$$ ? Anyway, I solved it. $$...-\theta,\theta,2\pi-\theta,2\pi+\theta,4\pi-\theta...=\frac{\pi}{2}\left(2k+1\right)+(-1)^k\left(\theta-\frac{\pi}{2}\right), k \in Z$$ edit: I was wrong Last edited: uart, what does that symbol you used mean? uart said: $$\lfloor k/2 \rfloor$$ Mentallic said: I think you might have a typo: $$\pi(k+1)k$$ ? Try insert integers in [itex](k+1)k$ and see what you get. It corresponds to the integer truncation uart used in his post.

Mentallic said:
Anyway, I solved it.

$$...-\theta,\theta,2\pi-\theta,2\pi+\theta,4\pi-\theta...=(-1)^k(\pi-\theta)+2k\pi, k \in Z$$
For $k=0$ you then get $\pi - \theta$. Is that included in your sequence?

Mentallic said:
uart, what does that symbol you used mean?

It's called "floor(x)" and it means the "largest integer less than or equal to x"

filiplarsen said:
Try insert integers in $(k+1)k$ and see what you get. It corresponds to the integer truncation uart used in his post.
But k(k+1) grows at a quadratic rate, not linear.
For k=0,1,2... k(k+1)=0,2,6,12,20,30...
which grows too rapidly.

filiplarsen said:
For $k=0$ you then get $\pi - \theta$. Is that included in your sequence?
Yes sorry, I fixed up the error.

uart said:
It's called "floor(x)" and it means the "largest integer less than or equal to x"
Aha, like how computers truncate decimal expansions rather than rounding off by default Mentallic said:
But k(k+1) grows at a quadratic rate, not linear.
For k=0,1,2... k(k+1)=0,2,6,12,20,30...
which grows too rapidly.

Oh my, you are quite right. What I really was trying to do was to give the same sequence as uart also gave, but some part of my brain must have been a bit too creative with trying to get rid of the integer truncation while another part must have been sleeping instead of catching the nonsense. I do apologise.