Given:
$$\sum_{n = 0}^{\infty} a_nx^n = f(x)$$
where:
$$a_{n+2} = a_{n+1} - \frac{1}{4}a_n$$
is the recurrence relationship for $a_2$ and above ($a_0$ and $a_1$ are also given).
Is there a nice closed form to this pretty recurrence relationship?
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1$$a_{n+2}x^{n+2} = x.a_{n+1}x^{n+1} - x^2\frac{1}{4}a_nx^n$$ – sciona Jan 22 '15 at 18:32
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@PedroTamaroff What is $\varphi$ and $\varphi^{-1}$? – bzm3r Jan 22 '15 at 18:37
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@PedroTamaroff: here the characteristic polynomial has a double root. – Jack D'Aurizio Jan 22 '15 at 18:37
2 Answers
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Hint:
Solving your recurrence relationship gives $$ a_n=c_12^{-n}+c_2n2^{-n}, $$ where $c_1$ and $c_2$ can be determined by $a_0$ and $a_1$.
You will end up calculating the sums $$ \sum_{n=0}^{+\infty} \Bigl(\frac{x}{2}\Bigr)^n $$ and $$ \sum_{n=0}^{+\infty} n\Bigl(\frac{x}{2}\Bigr)^n $$ Do you know how to do that? Ask to fill in details where necessary...
mickep
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Thanks for the hint based answer! I am very new to series, so I think I know how to do the first sum, but maybe not the second -- however, I will look up how to do so and come back to you :) – bzm3r Jan 22 '15 at 18:41
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As Jack writes, the characteristic equation is $x^2-x+1/4=0$ which has the double root $x=1/2$. Then (see here) one get that $a_n=c_1(1/2)^n+c_2n(1/2)^n=c_12^{-n}+c_2n2^{-n}$. – mickep Jan 22 '15 at 19:08
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Since the characteristic polynomial of the recurrence $$ a_{n+2}=a_{n+1}-\frac{1}{4}a_n \tag{1}$$ is $p(x) = x^2-x+\frac{1}{4} = \left(x-\frac{1}{2}\right)^2$, we have that $$\left(2-x\right)^2\, f(x) = ax+b\tag{2}$$ where $a,b$ depend on the initial conditions. By decomposing $\frac{ax+b}{(2-x)^2}$ into partial fractions, we get: $$ a_n = 2^{-(n+2)} (b+(2 a+b) n). \tag{3}$$
Jack D'Aurizio
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