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Stuck... Find solutions of recursive equations using generating functions. $$ x_{n+2} = 14x_{n+1} - 49x_n + n7^n, n\ge 0\\ x_0 = 1\\ x_1=14 $$

What I tried: $$ x_{n} = 14x_{n-1} - 49x_{n-2} + (n-2)7^{n-2}, n\ge 2\\ a_{n} = 14a_{n-1} - 49a_{n-2} + (n-2)7^{n-2}\\ F(x) = \sum_{n=0}a_nx^n = 1 + \sum_{n=1}a_nx^n\\ F(x) - 1 = \sum_{n=1}a_nx^n$$ \begin{align} F(x) &= 1 + 14x + \sum_{n=2}^\infty (14a_{n-1} - 49a_{n-2} + (n-2)7^{n-2})x^n\\ &= 1 + 14x + 14\sum_{n=2}^\infty a_{n-1}x^n -49\sum_{n=2}^\infty a_{n-2}x^n + \sum_{n=2}^\infty (n-2)7^{n-2}x^n \\ &= 1 + 14x + 14x\sum_{n=2}^\infty a_{n-1}x^{n-1} -49x^2\sum_{n=2}^\infty a_{n-2}x^{n-2} + x^2\sum_{n=2}^\infty (n-2)7^{n-2}x^{n-2} \\ &= 1 + 14x + 14x\sum_{n=1}^\infty a_{n}x^{n} -49x^2\sum_{n=0}^\infty a_{n}x^{n} + x^2\sum_{n=0}^\infty n7^{n}x^{n} \\ &= 1 + 14x + 14x(F(x) - 1) - 49x^2F(x) + x^2\sum_{n=0}^\infty n7^{n}x^{n} \end{align}

And there is a problem. Is this correct? If yes, how to transform $ x^2\sum_{n=0}^\infty n7^{n}x^{n} $ into something like $x^2F(x)$?

It's a method which our teacher showed us to get generating function, but he didn't show us how make it with inhomogeneous.

RobPratt
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KanekiSenpai
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1 Answers1

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This looks correct so far. To proceed, note that $$\sum_{n \ge 0} n z^n = \sum_{n \ge 1} n z^n = z \sum_{n \ge 1} n z^{n-1} = z \frac{d}{dz}\sum_{n \ge 1} z^n = z \frac{d}{dz}\frac{z}{1-z} = \frac{z}{(1-z)^2}$$

RobPratt
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  • Or, by the generalized binomial theorem, $\frac{1}{(1-z)^2} = (1-z)^{-2} = \sum_{n=0}^\infty \binom{-2}{n} (-z)^n = \sum_{n=0}^\infty (n+1) z^n$ from which you can manipulate into the desired sum - and the reverse will eventually be immensely useful in expressing the solution of $F(z)$ as a series again. In general, $(1-z)^{-a} = \sum_{n=0}^\infty \binom{n+a-1}{a-1} z^n$ for $a$ a nonnegative integer. – Daniel Schepler May 15 '20 at 21:36
  • Thanks! One question. $x^2\sum_{n=0}^\infty n7^{n}x^{n} = x^27x\sum_{n=0}^\infty n(7x)^{n-1} =x^27x(\sum_{n=0}^\infty (7x)^n)' = x^27x(\frac{7x}{1-7x})' = x^27x\frac{1}{(1-7x)^2} = \frac{7x^3}{(1-7x)^2} $ It's correct? – KanekiSenpai May 15 '20 at 22:57
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    Yes, that's correct, or you can just substitute $z=7x$ directly: $x^2\sum_{n\ge 0} n (7x)^n=x^2 (7x)/(1-7x)^2$. – RobPratt May 15 '20 at 23:17