Find a power series of $f(x)=\frac{x}{(1-x)(1-x^2)}$ around the center $0$ and determine its interval of convergence.

Question

Can anyone give me a detailed step to get the power series representation for $f(x)=\frac {x}{(1-x)(1-x^2)}$?

Answer 1

We can use this:

$$\frac { 1 }{ 1-x } \sum _{ n=0 }^{ \infty }{ { a }_{ n }{ x }^{ n } } =\sum _{ n=0 }^{ \infty }{ \left( { a }_{ 0 }+{ a }_{ 1 }+\cdots +{ a }_{ n } \right) { x }^{ n } } $$

Expand $\frac {1}{1-x^2}$,

$$\frac { 1 }{ 1-{ x }^{ 2 } } =\sum _{ n=0 }^{ \infty }{ { x }^{ 2n } } =\sum _{ k=0 }^{ \infty }{ \frac { 1+{ \left( -1 \right) }^{ k } }{ 2 } { x }^{ k } } $$

Therefore, $$\frac { 1 }{ 1-x } \sum _{ k=0 }^{ \infty }{ \frac { 1+{ \left( -1 \right) }^{ k } }{ 2 } { x }^{ k } } =\sum _{ k=0 }^{ \infty }{ \left( k+1 \right) { \left( { x }^{ 2k }+{ x }^{ 2k+1 } \right) } } $$

Finally multiply a $x$, we get,

$$\frac { 1 }{ 1-x } \frac { x }{ 1-{ x }^{ 2 } } =\sum _{ k=0 }^{ \infty }{ \left( k+1 \right) { \left( { x }^{ 2k+1 }+{ x }^{ 2k+2 } \right) } }=\sum _{ n=1 }^{ \infty }{ n{ \left( { x }^{ 2n-1 }+{ x }^{ 2n } \right) } } $$

The even term and odd term of the expansion is kind of different. I don't know if I write like above is ok.

That's the most difficult part, how to express it. The expansion is in this pattern:

$$x+x^2+2x^3+2x^4+3x^5+3x^6+4x^7+4x^8+...$$

Answer 2

Hint:

Multiply the power series expansions of $\dfrac1{1-x}$ and $\dfrac1{1-x^2}$. You'll get \begin{alignat}{4} 1+x&+x^2+x^3&&+x^4+x^5&&+x^6+x^7&&+x^8+\dotsm \\ &+x^2+x^3&& +x^4+x^5&&+x^6 +x^7&&+x^8+\dotsm \\ &&&+x^4+x^5&&+x^6+x^7&&+x^8+\dotsm \\ &&&&&+x^6+x^7&&+x^8+\dotsm\\ &&&&&&&+x^8+\dotsm \end{alignat} Can you see a pattern and prove it?

Answer 3

$$\frac{x}{(1-x)(1-x^2)}=\left(\frac{2x}{1-x^2}+\frac{x}{2(1-x)^2}\right)$$ we have $${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\tag 1$$ let $x\rightarrow x^2$ and then multiply by $x$ $${\frac {x}{1-x^2}}=\sum _{n=0}^{\infty }x^{2n+1}$$

now differentiate the equ.1 and multiply by $x$ to get $${\displaystyle {\frac {x}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n}}=\sum _{n=0}^{\infty }(n+1)x^{n+1}$$ so

$$\left(\frac{2x}{1-x^2}+\frac{x}{2(1-x)^2}\right)=\sum _{n=0}^{\infty }2x^{2n+1}+\frac{1}{2}(n+1)x^{n+1}$$