Question

Find an explicit formula and a recurrence formula for the following sequence. {1, -1/2, 3, -1/4, 5...}. I've been at it for awhile and just can't seem to come up with the correct formulas.

Answer 1

Yeah this is tricky looking. What sorta patterns have you found and tried out? I can help show you how to adjust them to work I think that'd be easier for me to explain and better for you too

Answer 2

so far I've only found complicated ones that haven't worked, and I've erased them (which I realize now was not a good idea) but the last one I have is \[\frac{n}{-n} * \frac{1}{n} * (-(n)^{2})\]

Answer 3

Are there any strategies to coming up with the formulas for these?

Answer 4

Yeah it's sorta hard to figure out a strategy for guessing these, but I like to kind of lay out what I know as best as I can. It might help to list stuff.

So when I see

1, -1/2, 3, -1/4, 5...

I see, ok it's kinda like

1, 2, 3, 4, 5, ...

And that much I know I can write as n. So maybe I can try to alter this to get it to work.

It looks like there are two other things going on here, we're getting this thing to flip flop between n and 1/n AND it's also flip flopping between + and -.

So I'd try to focus on this part next. I think + vs - flip-flopping is easier personally so let's try to figure out how to make this sequence:

1, -1, 1, -1, 1, ...

Then when we find it, we can figure out how to attach it the sequence 1,2,3,4,5,...

Answer 5

Yeah, I was actually just thinking about that as well. So a -(1/n) would almost work, except it would give me the inverse of each index. I feel like that is a good start though?

Answer 6

Yeah definitely!

Answer 7

starting the index from \(n=1\),

\(a_{2n+1} = 2n+1\)
\(a_{2n} = -\dfrac{1}{2n} \)

combine them and get
\(a_n = (1+(-1)^n)(\dfrac{-1}{n}) + (1-(-1)^n)\dfrac{2n+1}{2} \)

Answer 8

Also, are you familiar with writing fractions as exponents like this?

\[\frac{1}{n} = n^{-1}\]

I think this will make the alternating part easier for you to figure out since you probably know that a number itself is raised to the power 1:

\[n=n^1\]

So I'll just leave these here to give you some ideas to think about, try to figure it out

Answer 9

Yes, I'm familiar with that way of writing fractions, it definitely makes things like integrating a lot easier. Thanks, I'll try it out

Answer 10

i think the key thing here in writing an explicit formula for interleaved sequences is in noting below little trick:

\( 1+(-1)^n\) is \(0\) when \(n\) is odd and it is \(2\) when \(n\) is even

Answer 11

I'm trying to see how that works in the previous equation you posted above. Wouldn't that make it equal to zero any time that exponent is odd?

Answer 12

First of all, do you see why breaking the given sequence into two subsequences is easy here ?

Answer 13

Consider below two subsequences :

1) sequence formed by taking only odd indexed terms
2) sequence formed by taking only even indexed terms

Answer 14

Yes, that part made sense to me

Answer 15

Is it easy to write explicit formulas for each of those sequences indpendently ?

Answer 16

what are the explicit formulas ?

Answer 17

I think \[a _{2n+1} = n\]
and
\[a _{2n} = \frac{-1}{n}\]

Answer 18

looks we have a small problem with indexing
other than that you look good!

Answer 19

let me rephrase like this :

when \(n=2k-1\) is odd, the term is same as the index itself : \(n\)

when \(n=2k\) is even, the term is \(-\dfrac{1}{n}\)

Answer 20

see if that looks okay

Answer 21

Oh, ok. Yes I think I see how for the odd mind was 2k+1 when it should be 2k-1 right?

Answer 22

Yes, next see how you can make use of the \(\dfrac{1+(-1)^n}{2}\) thingy

Answer 23

Notice again, above expression is \(0\) when \(n\) is odd, and it is \(1\) when \(n\) is even

Answer 24

So it would be 1/2 when n is odd and 1 when n is even, right?

Answer 25

Nope, whats the value of (-1)^n when n is odd ?

Answer 26

whats the value of 1 + (-1)^n when n is odd ?

Answer 27

Oh, my bad, I see it's zero when n is odd