Hello, I have the following: $\displaystyle A(x)=(x+1) + \frac{1}{2x^{2}-3x+1}$ How do I find a sequence $\displaystyle a_{0},a_{1},a_{2},...$ such that $\displaystyle A(x)=\sum_{i=0}^{\infty} a_{i}x^{i}$ Thanks.
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Originally Posted by surjective Hello, I have the following: $\displaystyle A(x)=(x+1) + \frac{1}{2x^{2}-3x+1}$ How do I find a sequence $\displaystyle a_{0},a_{1},a_{2},...$ such that $\displaystyle A(x)=\sum_{i=0}^{\infty} a_{i}x^{i}$ Thanks. $\displaystyle \frac{1}{2x^{2}-3x+1}=\frac{1}{(1-x)(1-2x)}=(1-x)^{-1}(1-2x)^{-1}$ You could now expand using the binomial theorem.
Could you elaborate a bit. It's not quite clear. I mean, when expanded how do I include x+1 ?? Assistance would be appreciated!!!
Originally Posted by surjective Could you elaborate a bit. It's not quite clear. I mean, when expanded how do I include x+1 ?? Assistance would be appreciated!!! $\displaystyle (1-x)^{-1}(1-2x)^{-1}=(1+x+x^2+...)(1+2x+4x^2+...)$ After simplifying the above expression, simply add $\displaystyle x+1$ to obtain the required power series.
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