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How to convert this recursive formula into implicit formula

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Hi u/Raxreedoroid,

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You can form a vector space with 2^n and f(n) as the span. Then the matrix ((2,0)(1,3)) represents the transformation from (2^n ,f(n)) to (2^n+1 ,f(n+1)) You can diagonalize and get that ((2,0),(1,3))^n (1,f(0)) = (2^n ,3^n -2^n +f(0)*3^n ) So your f(n) is equal to 3^n -2^n +f(0)3^n
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I recently learned about generating functions, so I’m going to use one here.

Let F(x) = sum from n=1 to infinity of f_n * x^n / n!

If you want these calculations, let me know, but it should be pretty easy to show that F’(x) = 3F(x) + e^(2x)

F’(x)*e^(-3x) - 3*F(x)*e^(-3x) = e^(-x)

d/dx (F(x) * e^(-3x)) = e^(-x)

F(x) * e^(-3x) = -e^(-x) + c

F(x) = -e^(2x) + ce^(3x)

It also shouldn’t be too hard to show that the nth derivative of F at x=0 is the same as f_n

d^n F / dx^n = -2^n * e^(-2x) + c3^n * e^(3x)

At x=0

f_n = -2^n + c*3^n

We can rewrite this in terms of f_0

f_0 = -1 + c

c = f_0 + 1

f_n = -2^n + (f_0 + 1)3^n
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I recently learned this, so if someone can correct me please do:

The associated homogeneous polynomial (is that how it's called in english?) is r - 3 = 0, so 3\^n is the solution to the homogenous.

Then you can use this trick for the particular:

* p = 2\^(n-1)k (k constant)
* so you replace everything
* 2\^(n-1)k  - 3\*2\^(n-2)k = 2\^(n-1)
* 2k - 3/2k = 1 --> k = 2
* p = 2\^(n-1)\*2 = 2\^n

So now you know that every solution is: fn = C3\^n + 2\^n (C constant).

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