Sequences problem help

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  • #1
Mentallic
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Main Question or Discussion Point

I want to express

[tex]...-2\pi+\theta,\theta, 2\pi-\theta, 2\pi+\theta, 4\pi-\theta...[/tex]

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: [tex]2k\pi \pm \theta[/tex]
but I believe it can be expressed in another way to avoid the [itex]\pm[/itex], using (-1)k somehow.

How can this be done?
 

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  • #2
Filip Larsen
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I want to express

[tex]...-2\pi+\theta,\theta, 2\pi-\theta, 2\pi+\theta, 4\pi-\theta...[/tex]

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: [tex]2k\pi \pm \theta[/tex]
but I believe it can be expressed in another way to avoid the [itex]\pm[/itex], 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 [itex]2\pi[/itex] you may be able to use something like
[itex]\pi(k+1)k + (-1)^k \theta[/tex]
 
  • #3
uart
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Yeah I'm not sure about your example, but I think this will do what you want.

[tex]\lfloor k/2 \rfloor 2\pi + (-1)^k \theta [/tex]
 
  • #4
Mentallic
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My example was just trying to illustrate what I meant, I didn't mean for it to be similar to my problem :smile:
I think you might have a typo: [tex]\pi(k+1)k[/tex] ?

Anyway, I solved it.

[tex]...-\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[/tex]

edit: I was wrong :biggrin:
 
Last edited:
  • #5
Mentallic
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uart, what does that symbol you used mean?
[tex]\lfloor k/2 \rfloor[/tex]
 
  • #6
Filip Larsen
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I think you might have a typo: [tex]\pi(k+1)k[/tex] ?
Try insert integers in [itex](k+1)k[/itex] and see what you get. It corresponds to the integer truncation uart used in his post.

Anyway, I solved it.

[tex]...-\theta,\theta,2\pi-\theta,2\pi+\theta,4\pi-\theta...=(-1)^k(\pi-\theta)+2k\pi, k \in Z[/tex]
For [itex]k=0[/itex] you then get [itex]\pi - \theta[/itex]. Is that included in your sequence?
 
  • #7
uart
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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"
 
  • #8
Mentallic
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Try insert integers in [itex](k+1)k[/itex] 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 [itex]k=0[/itex] you then get [itex]\pi - \theta[/itex]. 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 :smile:
 
  • #9
Filip Larsen
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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.
 

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