Using generation functions solve the following difference equation
Solution 1
Thanks for the hint was able to solve it. This was my solution.
$$ a_{n+1} - 3a_{n+2} + 2a_n = 7n ; n\geq0; a_0 = -1; a_1 = 3. $$
$$ (\frac{a{(z)} - a_{0} + za_1}{z^2}) - 3(\frac{a{(z)} - a_{0}}{z^2}) + 2a(z) = \frac{7z}{(1 - z)^2}; a_0 = -1; a_1 = 3. $$
Substitute values in:
$$ (\frac{a{(z)} + 1 - 3z}{z^2}) - 3(\frac{a{(z)} + 1}{z^2}) + 2a(z) = \frac{7z}{(1 - z)^2}; a_0 = -1; a_1 = 3. $$
Expand and multiply through by $z^2$
$$ a(z) + 1 -3z - 3a(z)z -3z + 2a(z)z^2 = \frac{7z^3}{(1 - z)^2} $$
$$ a(z)[1 -3z + 2z^2] = \frac{7z^3}{(1 - z)^2} + 6z - 1 $$
$$ a(z) = \frac{13z^3 - 13z^2 + 8z -1}{(z-1)^2(2z-1)}$$
By Partial Fraction decomposition:
$$ a(z) = \frac{-11}{(2z-1)} + \frac{12}{(z-1)} + \frac{7}{(z-1)^2} + \frac{7}{(z-1)^3}$$
By Extracting coefficients we get: $$ a(z) = \frac{-7[n(n+1)]}{2} + 11 * 2^n - 12$$
Solution 2
Define the generating function $A(z) = \sum_{n \ge 0} a_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$, Recognize, e.g.: \begin{align} \sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2} \\ \sum_{n \ge 0} z^n &= \frac{1}{1 - z} \\ \sum_{n \ge 0} n z^n &= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \end{align} Solve the result for $A(z)$, write as partial fractions (a computer algebra system helps), and read off the coefficients by using geometric series or: $$ (1 + u)^{-m} = \sum_{k \ge 0} \binom{-m}{k} u^k = \sum_{k \ge 0} \binom{k + m - 1}{m - 1} (-1)^k u^k $$
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Comments
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Daniel about 1 year
Using generation functions solve the following difference equation
$$ a_{n+1} - 3a_{n+2} + 2a_n = 7n ; n\geq0; a_0 = -1; a_1 = 3. $$
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5xum over 9 yearsWhat have you done so far and where are you stuck?
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ShreevatsaR over 9 yearsThe same question was posted an hour ago by the same user, who's since deleted it for unclear reasons.
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Claude Leibovici over 9 yearsWhy did you delete the previous post ?
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Daniel over 9 years@ClaudeLeibovici It was a mistake I deleted it and didn't know how to get it back. I'm new to this
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Mhenni Benghorbal over 9 years
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Did over 9 years"It was a mistake" Hmmm... and you also chose to disregard the advice given to you there that "You can post your initial steps, and also what final answer you got" (advice repeated here by another user, and similarly disregarded).
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Anant over 9 yearsTypo: $(1+u)^\color{red}{m}$
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vonbrand over 9 years@Anant, thank you. Fixed.
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Daniel over 9 yearsThanks for the hint was able to solve it. This was my solution. @vonbrand
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vonbrand over 9 yearsThat's how it is done. Did you check the solution?I believe you left out the term with the square.