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#1 2012-11-06 02:20:34

steewan
Guest

Recurrence Relations and Backtracking

My teacher gave me this problem and I can't figure it out.

Jn= J(n-1) + n, j=2

I've gotten:
j1 = 1
j2 = 2 + 2
j3 = 2+3+2
j4 = 2+4+3+2
J5 = 2+5+4+3+2

Did I do this right so far and is if so, where do I go from here?

#2 2012-11-06 02:44:55

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 14,809

Re: Recurrence Relations and Backtracking

What does j=2 mean?


“Here lies the reader who will never open this book. He is forever dead.

“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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#3 2012-11-06 02:56:04

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 81,392

Re: Recurrence Relations and Backtracking

Hi steewan;

Every recurrence must always have an initial condition. It appears that yours is J(0)=1. In which case J(2) is incorrect.

Also you are not computing that recurrence correctly.

where do I go from here?

Depends on what you are being asked to do. Were you asked to solve the recurrence? To plot it? To determine it's limit?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#4 2012-11-06 11:12:12

scientia
Member
Registered: 2009-11-13
Posts: 222

Re: Recurrence Relations and Backtracking

steewan wrote:

My teacher gave me this problem and I can't figure it out.

Jn= J(n-1) + n, j1=2

I've gotten:
j1 = 2
j2 = 2 + 2
j3 = 2+3+2
j4 = 2+4+3+2
J5 = 2+5+4+3+2

Did I do this right so far and is if so, where do I go from here?


It's quite obvious that
. Try verifying it by mathematical induction.

Last edited by scientia (2012-11-06 11:12:49)

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