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Simplifying Linear Second Order Recurrence sequences...

Posted on 2011-03-13
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I'm not sure how to write the correct subscript notation for this expression, but I've got a sequence

Un = (some expression)

and need to prove an identity, where the LHS has Un+1 Un-1 - (Un)^2

(where n, n+1 and n-1 are all subscripts of U)

By multiplying out the expression substituting in n, n-1 and n+1 I have been able to cancel out a number of terms, but have some left that I can't simplify.

In the expression below there are no subscripts, everything is superscript:

9^(n-1).(-2)^(n+1) + (-2)^(n-1).9^(n+1) - 2(9^n (-2)^n)

is it possible to simplify this any further?


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Question by:purplesoup
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by:phoffric
ID: 35124693
>> LHS has Un+1     Un-1 - (Un)^2
                           ^
                           |
             are you missing an operator here?


>> 9^(n-1).(-2)^(n+1) + (-2)^(n-1).9^(n+1) - 2(9^n (-2)^n)
>> is it possible to simplify this any further?
   You can do some simplifying

Rules:
a^(b+c) = a^b a^c
a^(b-c) = a^b a^(-c) =  a^b / a^c

For example, 9^(n+1) = 9^n * 9    and    (-2)^(n-1) = (-2)^n / (-2)^(1)

You should end up with some common factors.
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by:TommySzalapski
ID: 35125459
I work out the equation and end up with a 121 (which is 11^2) and a bunch of -18s. That doesn't look like it'll work out well. If you get the same thing and it isn't right, then maybe something is amiss earlier in the equation.
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Author Comment

by:purplesoup
ID: 35128709
Sorry I still can't finish it. Applying the above rules I get this.

a^(b+c) = a^b a^c
a^(b-c) = a^b a^(-c) =  a^b / a^c

I get the following:

9^(n-1).(-2)^(n+1) + (-2)^(n-1).9^(n+1) - 2(9^n (-2)^n)

= (9^n.(-2)^n.(-2))/9 + ((-2)^n.9^n.9)/(-2) - 2.9^n.(-2)^n

= 9^n.(-2)^n (-2/9 - 9/2 -2)

= 9^n.(-2)^n (59/18)

???

Where did I go wrong?
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Expert Comment

by:phoffric
ID: 35129008
= 9^n.(-2)^n (-2/9 - 9/2 -2)
= 9^n.(-2)^n (-121/18)


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Author Comment

by:purplesoup
ID: 35129102
doh - sorry.

Is 9^2 . (-2)^2 = (-18)^n ??

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by:phoffric
ID: 35129135
>> Is 9^2 . (-2)^2 = (-18)^n ??
LHS has no n
RHS has n
So I'm not sure what you mean unless you were trying to solve for n.

>> 9^2 . (-2)^2 = 9*9 * (-2)*(-2) = 81 * 4 = 324
but something tells me that this is not what you are looking for.
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by:purplesoup
ID: 35129155
Sorry I typed it wrong, I was trying to simplify, I meant

Is 9^n . (-2)^n = (-18)^n ??

so the final expression would be

= (-18)^n (-121/18)

= -121. (-18)^(n-1)
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phoffric earned 500 total points
ID: 35129576
(ab)^n = a^n  b*n

so,  (-18)^n  =  (-2 * 9)^n  = (-2)^n 9^n

===

9^n.(-2)^n (-121/18) = (-18)^n (-121/18) = -121* (-18)^(n-1)


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