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Recurrence relation with Square roots in each term..

Posted on 2004-03-22
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Last Modified: 2011-08-18
Hi experts,
    How to solve a recurrence relation that have Square roots in each term...

    I can solve the normal recurrence relation problem (that hav no square roots)

    Give me some idea...

    Q will be like this

   ROOT(Rn) = ROOT(Rn-1)+ 2.ROOT(Rn-2)  with initial condition R0=R1=1

Rosh :)
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Question by:Roshan Davis
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Expert Comment

by:ozo
ID: 10655843
Rn = Sn²
where
Sn = (Sn-1) + 2×(Sn-2)
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Author Comment

by:Roshan Davis
ID: 10655884
Thankx for ur comment :)

Calculated associated equation
square(X) = X + 2
and itz roots are (x-2)(x+1)

C1 = u(-1)+v(2)
C2 = u.square(-1)+ v. square(2)

And reached the result

Cn = (-1/3)(-1)^n + (1/3)(2)^n

this is the Recurrence relation for "Sn = (Sn-1) + 2×(Sn-2)"

Is this correct?

Rosh :)
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Expert Comment

by:ozo
ID: 10655919
or in closed form
Rn = (((-1)^n+2^(n+1))/3)^2
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LVL 84

Expert Comment

by:ozo
ID: 10655927
Sorry, didn't see you had solved it
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Author Comment

by:Roshan Davis
ID: 10655941
But answers were different...?
Can u show me some steps...
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Author Comment

by:Roshan Davis
ID: 10655984
Hi, you can see the question here http://www.coolgoose.com/sites/puttalu/
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Expert Comment

by:ozo
ID: 10656001
They were different because you used C1=C2=1, while I used R0=R1=1
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Author Comment

by:Roshan Davis
ID: 10656009
So, for this Q http://www.coolgoose.com/sites/puttalu/ which is correct ?
Rosh :)
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Accepted Solution

by:
ozo earned 125 total points
ID: 10656073
for r0=r1=1
Rn = (((-1)^n+2*2^n)/3)^2
is correct
((-1)^0 + 2*2^0)/3 = 1
((-1)^1 + 2*2^1)/3 = 1
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