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Need to solve:

$$2a_{n-2} = a_n + a_{n-1}$$

with: $a_0 = 0$ and $a_1=1$

I get:

$$f(x) = \frac{2x^3-x^2-x}{2x^2-x-1}$$ so I tried to scompose the denominator and I get:

$$f(x) = \frac{2x^3-x^2-x}{(x-1)(x+\frac{1}{2})}$$ now I think I have to use partial fraction but in this case I do not see any $x^3$ coming out so how should I procede?

$$\frac{A}{x-1}+\frac{B}{x+\frac{1}{2}}$$

rtybase
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Christian Giupponi
  • 687

2 Answers2

1

I think you committed a typo in calculations. From: $$a_n=2a_{n-2}-a_{n-1}$$ we should have: $$f(x)=\sum\limits_{n=0}\color{blue}{a_{n}}x^{n}= x+\sum\limits_{n=2}a_nx^n= x+\sum\limits_{n=2}\left(2a_{n-2}-a_{n-1}\right)x^n=\\ x+2x^2\left(\sum\limits_{n=2}a_{n-2}x^{n-2}\right)-x\left(\sum\limits_{n=2}a_{n-1}x^{n-1}\right)=\\ x+2x^2\left(\sum\limits_{n=0}a_{n}x^{n}\right)-x\left(\sum\limits_{n=1}a_{n}x^{n}\right)=x+2x^2f(x)-xf(x)$$ and $$\color{red}{f(x)=\frac{x}{1+x-2x^2}}=\frac{1}{3(1-x)}-\frac{1}{3(1+2x)}=\\ \sum\limits_{n=0}\frac{1}{3}x^n-\sum\limits_{n=0}\frac{1}{3}(-2)^nx^n= \sum\limits_{n=0}\color{blue}{\frac{1-(-2)^n}{3}}x^n$$ and (also agrees with wolfram) $$a_n=\frac{1-(-2)^n}{3}$$

rtybase
  • 16,907
0

There's an error in the factorisation of the denominator. You should get $$f(x) = \frac{2x^3-x^2-x}{(x-1)(2x+1)}.$$ Now this isn't a proper rational function (i.e. the degree of the numerator is not $<$ degree of the denominator). So (in theory) you have to perform the Euclidean division of the numerator by the denominator first.

However, observe that $2x^3-x^2-x=x(2x^2-x-1)$, so the ‘fraction’ simplifies to $\color{red}x$.

Bernard
  • 175,478