Solving a recursive relation ....

SaTaN · 09-01-2003, 12:19 AM

I am having problems solving this recursive relation .

Can someone help me with this

a(n) = -2n.a(n-1) +3n(n-1).a(n-2)

a(0) =1, a(1) = 2

Now I need to express a(n) as a function of n and not in terms of
a(n-1) and a(n-2) ;;

Skyline · 09-01-2003, 12:24 AM

Do you mean simplifying it?

Whitehat · 09-01-2003, 12:24 AM

I don't talk to evil beings. Satan, go away.

Skyline · 09-01-2003, 12:37 AM

reply?

MasterC · 09-01-2003, 01:02 AM

Quote:

Originally posted by Whitehat
I don't talk to evil beings. Satan, go away.

Although this literally had me laughing out loud, and couldn't type for a minute, I think in the future you may wanna reconsider posting such EXTREMELY humerous things until the person has their issue solved, even if it means resisting the greatest of urges

Cool

SaTaN · 09-01-2003, 01:33 AM

Quote:

Originally posted by Skyline
Do you mean simplifying it?

Yeah a(n) should be expressed as a function of only n
and not as function of a(n-1),a(n-2)......

1.e , a(n) should be of format

x. n ^ y + z . w^n + ......

where , a^b stands for a raised to the power b

SaTaN · 09-01-2003, 01:40 AM

Quote:

Originally posted by Whitehat
I don't talk to evil beings. Satan, go away.

This is what I found somewhere else in this forum and maybe you haven't seen it

Quote:

When you respond (as the first responder) to a thread with 0 replies it loses more than half it's viewers. Several people browse this board on quite slow connections, so they tend to glimpse over the forum, find a post they can respond to with 0 replies (meaning they don't have to read through a swarm of other posts) and post. If you respond with things like "Me to" or simple phrases like such, then those members will likely go without ever seeing the post. It really is a dis-service to the originator.

If you want to help , no one will stop you from doing it . But ,
please don't stop others from helping me.

Whitehat · 09-01-2003, 09:56 AM

Gotcha

I will try to be more helpful next time instead of being Mr. Funnyman

Peace.

SaTaN · 09-01-2003, 10:34 AM

Sorry WhiteHat , if I were a bit rude ......I hope you don't mind

What's it guys ??? Is it that NoOne does MATHS over here .
coz not even one ray of light in the right direction ..

Whitehat · 09-01-2003, 11:00 AM

It's cool. You were fine

I think not a lot of people do math because they don't need it. Like me....I was never a Compsci major. I was a marketing/business major. Didn't need a lot of math...

schatoor · 09-01-2003, 11:33 AM

Hi there. I see you are from Hyderabad. Is that in Pakistan or India?
But never mind that. This recrucive formula seems kinda hard. Do you know what a Z transform is? It's the descrete counterpart of the fourier transform. Aplie a Z transform on both sides and see where you get. After simplyfing the resulting equation, aply an inverse Z transform to get a(n).
Tell me if you understand what i'm saying. It's quite possible you don't know that a Z transform is, or you call it differently.

SaTaN · 09-01-2003, 11:41 AM

I am frm India .

I dunno what a Z-transform is .....

I use "Discrete Mathematics and Applications -- Kenneth H. Rosen"

Atleast tell me which book and where am I supposed to look for this

Thanx

quietguy47 · 09-01-2003, 12:09 PM

Maybe this will help.
http://mathworld.wolfram.com/Z-Transform.html
I don't know math(very well) but I know google.com

schatoor · 09-01-2003, 04:38 PM

No, if you don't know what a Z transform is, you probebly don't need to solve it that way.
I find this problem quite hard to solve because it's not linear. I don't understand how to solve somthing like that. I looked in my books and on the net I tried solving it my self. I can't seem to crack it.
What I tried doing was guessing an form of an solution and filling in the parameters. For instance, I looked if there were solutions of the form a(n)=r^n, where r is an arbitrary parameter. But no matter who you choose r, the equations just doesn't solve, because filling things in you equation:

Code:

a(n) = -2n*a(n-1)+3n(n-1)*a(n-2) <=>
r^n = -2n*r^(n-1)+3n(n-1)*r^(n-2) <=>
r^n = r^n*(-2n*r^-1+3*n(n-1)*r^-2)

from this it is clear that 

-2n*r^-1+3*n(n-1)*r^-2 = 1 for every n={...,-2,-1,0,1,2,....}

But that's not possible

I also tried it with a(n)=n^k*r^n with k an r arbitrary constants, but no luck

I hope you can crack the case. Good luck. And do tell me how it's supposed to be solved.

endorphinjunkie · 09-01-2003, 10:21 PM

Try substitution,

say, b = n + 2 and solve for b. when you know b, you will know n.