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7nadin3 [17]
3 years ago
6

Determine the exact formula for the following discrete models:

Mathematics
1 answer:
marshall27 [118]3 years ago
5 0

I'm partial to solving with generating functions. Let

T(x)=\displaystyle\sum_{n\ge0}t_nx^n

Multiply both sides of the recurrence by x^{n+2} and sum over all n\ge0.

\displaystyle\sum_{n\ge0}2t_{n+2}x^{n+2}=\sum_{n\ge0}3t_{n+1}x^{n+2}+\sum_{n\ge0}2t_nx^{n+2}

Shift the indices and factor out powers of x as needed so that each series starts at the same index and power of x.

\displaystyle2\sum_{n\ge2}2t_nx^n=3x\sum_{n\ge1}t_nx^n+2x^2\sum_{n\ge0}t_nx^n

Now we can write each series in terms of the generating function T(x). Pull out the first few terms so that each series starts at the same index n=0.

2(T(x)-t_0-t_1x)=3x(T(x)-t_0)+2x^2T(x)

Solve for T(x):

T(x)=\dfrac{2-3x}{2-3x-2x^2}=\dfrac{2-3x}{(2+x)(1-2x)}

Splitting into partial fractions gives

T(x)=\dfrac85\dfrac1{2+x}+\dfrac15\dfrac1{1-2x}

which we can write as geometric series,

T(x)=\displaystyle\frac8{10}\sum_{n\ge0}\left(-\frac x2\right)^n+\frac15\sum_{n\ge0}(2x)^n

T(x)=\displaystyle\sum_{n\ge0}\left(\frac45\left(-\frac12\right)^n+\frac{2^n}5\right)x^n

which tells us

\boxed{t_n=\dfrac45\left(-\dfrac12\right)^n+\dfrac{2^n}5}

# # #

Just to illustrate another method you could consider, you can write the second recurrence in matrix form as

49y_{n+2}=-16y_n\implies y_{n+2}=-\dfrac{16}{49}y_n\implies\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}\begin{bmatrix}y_{n+1}\\y_n\end{bmatrix}

By substitution, you can show that

\begin{bmatrix}y_{n+2}\\y_{n+1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n+1}\begin{bmatrix}y_1\\y_0\end{bmatrix}

or

\begin{bmatrix}y_n\\y_{n-1}\end{bmatrix}=\begin{bmatrix}0&-\frac{16}{49}\\1&0\end{bmatrix}^{n-1}\begin{bmatrix}y_1\\y_0\end{bmatrix}

Then solving the recurrence is a matter of diagonalizing the coefficient matrix, raising to the power of n-1, then multiplying by the column vector containing the initial values. The solution itself would be the entry in the first row of the resulting matrix.

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