Stompy

I need to find a formula for the general term of the sequence using:

{5, 1, 5, 1, 5, 1 ... }

I've thought about this problem in the back of my head as I'm working here... and I can't figure this one out.

The only thing I can think of is

If (-1)^n = -1 then 5
if (-1)^n = 1 then 1

...but none of our answers have ever been conditional like that and I don't think that's what he's lookin for.

These problems suck. There's absolutely no method or formula you can use to find out the function, yet they'll give you 5 of these on a 15 minute test an expect you to just mysteriously come up with some wacky equation to fit it.

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phukraut

Well, you haven't mentioned what rules we are allowed to define the function, but here's one: if n>=0, then

a_n = 5 Mod(n+1,2) + Mod(n,2),

where, Mod(x,y) is the remainder of x/y.

phukraut

Here's another (n>=0):

a_n = 1 + 2 (1 - (-1)^{n+1}).

Stompy

It doesn't say, but I think he said they all start at 1.

That answer seems a bit.. like something we haven't done yet. Being a programmer, I know mod, but we've never once had a problem in math that's needed a mod (nor is it discussed in that chapter anywhere).

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Pip

Or a_n=3-2*cos(n*pi) if the first term is a_1

Stompy

Crap, someone just emailed me the answer: 5^([-1^n]*.5 + .5)

There is NO way in hell I would've been able to figure that out... I really hope we don't have these on the test. The time I've spent actually w/ pen and paper trying to figure this out has been well over 20 minutes, and spending 20 minutes on ONE problem is crazy.

[edit]
No, that won't work. I'd have to be {1, 5, 1, 5 ...}

...it would if it was n+1 though!

Damn this is frustrating.

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Stompy

Is there a way that you can use to quickly come up with the formula? Or... how'd you know that so quickly?

Yeah, it's obvious now that I look at that, but sitting here staring at this problem... drew blanks.

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phukraut

so change it to n+1? I can't see why that should be a hindrance. Or just factor out (-1) from above and you can have it to the power of n.

Stompy

I did after I noticed it was shifted to the left (or right) by 1

The face was more or less a "oh god, dumb mistake" type face.

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quicksteal

Series problems in calc. 2 always seemed like toy problems to me...by that I mean that there's usually some mathematical pattern that you're supposed to realize, like 5^0 = 1. So basically, if you're lucky enough to find the pattern just by staring at it, then you can figure out the series. Usually, it's related to either a power series, or a sin or cos series. That's not very much help, for me it just clicked one day (luckily the day before my test).

Stompy

Thanks for the input and help, everyone!

Yeah, these seem like.. not really math problems. More like riddles or brain teasers.

At least with other problems like hydrostatic pressure or finding the area between two polar equations has a definite method to em. These seem more or less hit and miss... either you're lucky to find the answer, or you're not. IMO, not something you should be graded on.

Everything else calc related is very very easy to me, but this... almost all the sequence patterns we had to figure out I just looked at and went "...". It'd be different if it was something simple like n+3/(2^n+1), but we've had some equations that were just... I don't know how he honestly expects us to figure half that stuff out!

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Pip

Cos(n*pi) is my best friend for repetetive series. There are a few types of patterns and corresponding formulas that you just have to learn to recognize. Actually I sucked at series when I took that course, but in later courses I had to figure them out. Anyway, good luck with the exam.

filtherton

It's amazing what one can do when one has too.

Mattbastard

Actually I think it's 5^abs(cos((n*pi)/2)) abs=absolute value

If the series starts at n=0, the first is 5.

But if n=2, the answer is 5^(-1), or 1/5, hence the absolute value for the exponent.

I just ran it up to 10 and that seems to be the series. BTW I just did series' last week.

stingc

Um, ok. Math is not supposed to be about memorizing methods. That's what you have engineering courses are for. Given your complaint about spending 20 minutes on a problem, I guess you've only been given mechanical problems.

Anyways, I'll try to give you some idea of how to come up with these things on your own. If you have a sequence with an alternating pattern, (-1)^n is your friend (this is the same as cos(n pi)). Notice that (1+(-1)^n)/2={0,1,0,1,...}. This is very useful if you have one sequence nested in another.

You could therefore write your series as 1*(1+(-1)^n)/2+5*(1+(-1)^(n+1))/2. This simplifies to 3-2(-1)^n. You could also have come up with this directly by noting that 5 and 1 are equivalently 3+/-2.

For a more complicated example, take {1,4,3,16,5,36,...}. The odd terms are given by n, and the even terms by n^2. One way to write the general term is n*(1+(-1)^(n+1))/2 + n^2*(1+(-1)^n)/2. Of course this can be simplified and written in different ways, but the point is that you can separate out the even and odd terms very easily.

Also, if I were your prof, I wouldn't mark you off for using conditionals. It's easier that way. I'm not guaranteeing that he'll be the same though!

Stompy

We learned a few tricks here and there (like those you mentioned), but luckily he's not giving us any of these problems on any of his tests/finals.

He more or less acknowledged them as "brain teasers" and said it'd be a waste of time to test us on them. Instead he's gonna focus on the other aspects of series, like proving whether it converges/diverges along with the sum, etc.

Other than that, the whole section is pretty much a mirror of limits from Calc 1, so it's an easy A from this point on since this is our last chapter!

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