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I'm trying to solve the following recurrence relation (Find closed formula) using generating functions:

$f(n)=10f(n-1)-25f(n-2)$, $f(0)=0$, $f(1)=1$

I'm having a small difficulty at the end and can use a nudge in the right direction.

My solution

Define $$g(x)=\sum_{n=0}^{\infty}f(n)x^n = \sum_{n=2}^{\infty}f(n)x^n+x=10\sum_{n=2}^{\infty}f(n-1)x^n-25\sum_{n=2}^{\infty}f(n-2)x^n+x =$$

$$=10x\sum_{n=2}^{\infty}f(n-1)x^{n-1}-25x^2\sum_{n=2}^{\infty}f(n-2)x^{n-2}+x =$$

$$=10x\sum_{n=1}^{\infty}f(n)x^n-25x^2\sum_{n=0}^{\infty}f(n)x^n+x$$

But since the first element of $f(n)$ is $0$, then $$10x\sum_{n=1}^{\infty}f(n)x^n=10x\sum_{n=1}^{\infty}f(n)x^n+0=10x\sum_{n=0}^{\infty}f(n)x^n$$

So overall:

$$g(x)=10x\sum_{n=0}^{\infty}f(n)x^n-25x^2\sum_{n=0}^{\infty}f(n)x^n+x=10xg(x)-25x^2g(x)+x$$

To sum up:

$$g(x)=10xg(x)-25x^2g(x)+x$$

Solve for $g(x)$ and get:

$$g(x)=\frac{x}{(x-\frac{1}{5})^2}$$

But what do I do now? How can I extract $f(n)$ from this information?

Oria Gruber
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    Use the general binomial theorem (http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem) to get a series for $(x-\frac15)^{-2})$. – Macavity Jan 14 '14 at 17:39

3 Answers3

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Hint: The binomial theorem says $$ \begin{align} (1-x)^{-2} &=\sum_{k=0}^\infty\binom{-2}{k}(-x)^k\\ &=\sum_{k=0}^\infty\binom{k+1}{k}x^k\\ &=\sum_{k=0}^\infty(k+1)x^k\\ \end{align} $$ and $$ \frac{x}{(x-\frac15)^2}=\frac{25x}{(1-5x)^2} $$

robjohn
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An orderly way of attacking such problems is as follows: Define $F(z) = \sum_{n \ge 0} f(n) z^n$, write the recurrence so there are no substractions in indices: $$ f(n + 2) = 10 f(n + 1) - 25 f(n) $$ Multiply by $z^n$, sum over $n \ge 0$, recognize: \begin{align} \sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\ \sum_{n \ge 0} f(n + 2) z^n &= \frac{F(z) - f(0) - f(1) z}{z^2} \\ \end{align} to get: $$ \frac{F(z) - z}{z^2} = 10 \frac{F(z)}{z} - 25 F(z) $$ Maxima spits out the partial fraction expansion: $$ F(z) = \frac{1}{5 (1 - 5 z)^2} - \frac{1}{5 (1 - 5 z)} $$ Using the generalized binomial theorem, as $(-1)^n \binom{-2}{n} = \binom{n + 2 - 1}{1} = n + 1$: $$ f(n) = \frac{(n + 1) 5^n}{5} - \frac{5^n}{5} = n 5^{n - 1} $$

vonbrand
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Hint: The Taylor series for $g(x)$ is $\sum_{n=0}^{\infty}{g^{(n)}(0)x^{n}/n!}$ so we get that $f(n) = g^{(n)}(0)/n!$

neelp
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