Solving Recurrence Relation Homework

pupeye11
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Homework Statement



The sequence f_n is defined by f_0=1, f_1=2 and f_n=-2f_{n-1}+15f_{n-2} when n \geq 2. Let

<br /> F(x)= \sum_{n \geq 2}f_nx^n<br />

be the generating function for the sequence f_0,f_1,...,f_n,...

Find polynomials P(x) and Q(x) such that

<br /> F(x)=\frac{P(x)}{Q(x)}<br />

The Attempt at a Solution



<br /> f_n+2f_{n-1}-15f_{n-2}=0<br />

So since we know that F(x)=f_0+f_1x+f_2x^2+...+f_nx^n+...

<br /> F(x)=f_0+f_1x+f_2x^2+...+f_nx^n+...<br />

<br /> 2xF(x)=2f_0x+2f_1x^2+...+2f_{n-1}x^n+...<br />

<br /> -15x^2F(x)= -15f_0x^2-...-15f_{n-2}x^n-...<br />

Summing these I get

<br /> (1+2x-15x^2)F(x)=f_0+(f_1+2f_0)x+(f_2+2f_1-15f_0)x^2+...+(f_n+2f_{n-1}-15f_{n-2})x^n<br />

After some algebra and substituting f_0=1, f_1=2 I get

<br /> F(x)=\frac{1+4x}{1+2x-15x^2}<br />

So

<br /> P(x)=1+4x<br />

and

<br /> Q(x)=1+2x-15x^2<br />

Is this correct?
 
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looks reasonable to me, you can always check by diferntiating you function a few times
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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