Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
Pages: 1
#1 2008-10-31 00:35:00
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
Shift of Geometric Series
Just curious, can anyone come up with the answer to the series sum:
(0<q<1)?
Last edited by George,Y (2008-10-31 01:08:18)
X'(y-Xβ)=0
Offline
#2 2008-10-31 00:38:20
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
Re: Shift of Geometric Series
Oh I got it. Just deduct the first term and get substraction.
X'(y-Xβ)=0
Offline
#3 2008-10-31 01:04:28
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
Re: Shift of Geometric Series
Last edited by George,Y (2008-10-31 02:28:34)
X'(y-Xβ)=0
Offline
#4 2008-10-31 01:24:43
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
Re: Shift of Geometric Series
This is to compute the weighed average life of maturity by discount value for Annuity.
An Annuity gives you A amount of every year.
Discount "all" of them to now gets you A/(1+r)+A/(1+r)[sup]2[/sup]+...=A/r
And the weighed average life is (1A/(1+r)+2A/(1+r)[sup]2[/sup] +...)/(A/r) = (1+r)/r
Last edited by George,Y (2008-10-31 01:25:57)
X'(y-Xβ)=0
Offline
#5 2008-10-31 01:55:02
- TheDude
- Member
- Registered: 2007-10-23
- Posts: 361
Re: Shift of Geometric Series
Your answer works for n = 1, but it doesn't appear to be correct for larger values of n. For example, for n = 2 we should get S(2) = 2q^2 + q = q(2q + 1). Here is what I get by using your formula:
It's pretty plain to see at this point that (q^3 - 2q^2 + 1) / (q^2 - 2q + 1) does not equal 2q + 1. You're close, but I think you've made a mistake in your algebra somewhere.
Wrap it in bacon
Offline
#6 2008-10-31 02:25:15
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
Re: Shift of Geometric Series
How about now? I found the term q[sup]n+1[/sup] needs a coefficient n. And thank you for pointing that out.
X'(y-Xβ)=0
Offline
#7 2008-10-31 02:34:13
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
Re: Shift of Geometric Series
Now it seems correct.
X'(y-Xβ)=0
Offline
#8 2008-10-31 02:37:40
- TheDude
- Member
- Registered: 2007-10-23
- Posts: 361
Re: Shift of Geometric Series
Alright, looks good now George.
How did you figure out that it needs a factor of n though? I tried following your work and thought it needed a factor of n + 1, which is obviously incorrect.
Last edited by TheDude (2008-10-31 02:38:52)
Wrap it in bacon
Offline
#9 2008-10-31 23:31:22
- George,Y
- Member
- Registered: 2006-03-12
- Posts: 1,379
Re: Shift of Geometric Series
The term (j-1)q[sup]j[/sup] when j=n+1, or the term iq[sup]i[/sup] when i=n
You see, I did the index shifting j=i+1, and I did the shifting of summation interval correspondingly, which may cause some confusion.
X'(y-Xβ)=0
Offline
Pages: 1