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Infinite Series help

by beth192
Tags: infinite, series
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beth192
#1
Aug2-08, 11:31 PM
P: 4
My Calculus teacher posed this question to a recent class, and asked us (previous students) if we could figure it out(just for fun). I am stumped. The question is to find the general formula that represents the infinite series (1, -1, -1, 1,-1, -1, 1....) I am assuming it uses trig graphs, maybe something like sinx + cos2x... but even that doesn't really work. any ideas would be appreciated. thanks
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Defennder
#2
Aug2-08, 11:56 PM
HW Helper
P: 2,616
You could draw it out on a graph for values of x=1, x=2, x=3 and so on. Then perhaps find the Fourier series for the square wave.
futurebird
#3
Aug3-08, 12:31 AM
P: 276
It's not fancy but this would work:

For [tex]n \in \mathbb{N}^{+}[/tex],

[tex]S_{n} = [/tex]
1 if n = 1
1 if [tex]3|(n-1)[/tex]
-1 otherwise.

Were there any other constraints on how you could set it up?

beth192
#4
Aug3-08, 12:32 AM
P: 4
Infinite Series help

Not really, but we never got to far into series in calc two, so i am guessing it had to be pretty simple, just no recursive series or anything like that. The teacher keeps saying its not very complicated... thanks for the help y'all
futurebird
#5
Aug3-08, 12:38 AM
P: 276
I was having some issues with LaTeX. What's covered by "or anything like that" ?
spideyunlimit
#6
Aug3-08, 03:20 AM
P: 61
signum (sec(x))
spideyunlimit
#7
Aug3-08, 03:22 AM
P: 61
quadrant wise.....
in first quadrant cos is positive so sec is positive too and signum will give 1.
in 2nd and 3rd quadrant cos is negative so sec is also negative and signum will give -1 in both second and third quadrants.
then in 4th quadrant, positive so +1.
futurebird
#8
Aug3-08, 03:28 AM
P: 276
Quote Quote by spideyunlimit View Post
quadrant wise.....
in first quadrant cos is positive so sec is positive too and signum will give 1.
in 2nd and 3rd quadrant cos is negative so sec is also negative and signum will give -1 in both second and third quadrants.
then in 4th quadrant, positive so +1.
Wait, but the pattern is (1, -1, -1, 1,-1, -1, 1....)

not

1, -1, -1, 1, 1, -1, -1, ...
Defennder
#9
Aug3-08, 03:43 AM
HW Helper
P: 2,616
That doesn't work because it isn't tailored to discrete values of x=1,2,3,4. And if you're using sec x you might as well use cos x instead. The series must go 1,-1,-1,1,-1,-1,1..., not alternate between -1 and 1 or we could use (-1)^(2x-1) instead.
spideyunlimit
#10
Aug3-08, 04:22 AM
P: 61
i said sec x instead of cosx because else you get 0 too, but anyways, right! You won't get the afore said series :|
Hmmm, Let me think.
beth192
#11
Aug3-08, 10:20 AM
P: 4
Hey guys, the best i've come up with so far employees the use of sinx + cos2x... this gives you 1,1,-2,1,1,-2 at 4pie/6 intervals... then uses this (-1) ^ of sinx +cos2x .... i just have to shift the graph. My teacher said he did it two other ways though... any other ideas would be appreciated ( futurebird, i'm not really sure... those were his words : ( ) thanks everyone


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