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Recurrence Relation |
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| Jun15-10, 08:51 PM | #1 |
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Recurrence Relation
1. The problem statement, all variables and given/known data
The sequence [tex]f_n[/tex] is defined by [tex]f_0=1, f_1=2[/tex] and [tex]f_n=-2f_{n-1}+15f_{n-2}[/tex] when [tex]n \geq 2[/tex]. Let [tex] F(x)= \sum_{n \geq 2}f_nx^n [/tex] be the generating function for the sequence [tex]f_0,f_1,...,f_n,...[/tex] Find polynomials P(x) and Q(x) such that [tex] F(x)=\frac{P(x)}{Q(x)} [/tex] 3. The attempt at a solution [tex] f_n+2f_{n-1}-15f_{n-2}=0 [/tex] So since we know that [tex]F(x)=f_0+f_1x+f_2x^2+...+f_nx^n+...[/tex] [tex] F(x)=f_0+f_1x+f_2x^2+...+f_nx^n+... [/tex] [tex] 2xF(x)=2f_0x+2f_1x^2+...+2f_{n-1}x^n+... [/tex] [tex] -15x^2F(x)= -15f_0x^2-...-15f_{n-2}x^n-... [/tex] Summing these I get [tex] (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 [/tex] After some algebra and substituting [tex]f_0=1, f_1=2[/tex] I get [tex] F(x)=\frac{1+4x}{1+2x-15x^2} [/tex] So [tex] P(x)=1+4x [/tex] and [tex] Q(x)=1+2x-15x^2 [/tex] Is this correct? |
| Jun21-10, 06:59 AM | #2 |
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Recognitions:
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looks reasonable to me, you can always check by diferntiating you function a few times
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