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

[Mentallic]

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?

 

[Filip Larsen]

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]

 

[uart]

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

 

[Mentallic]

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

 

[Mentallic]

uart, what does that symbol you used mean?

\lfloor k/2 \rfloor

 

[Filip Larsen]

I think you might have a typo: \pi(k+1)k ?

Try insert integers in (k+1)k and see what you get. It corresponds to the integer truncation uart used in his post.

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?

 

[uart]

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"

 

[Mentallic]

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.

For k=0 you then get \pi - \theta. Is that included in your sequence?

Yes sorry, I fixed up the error.

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

 

[Filip Larsen]

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.