Power series of $\frac{x^{3}-2}{x^{2}+1}$

Question

I have to find the power series of the function $$f(x)=\frac{x^3-2}{x^{2}+1}$$ centered at $a=1$. I tried to write $f$ as $$f(x)=(x^3-2)\cdot\frac{1}{x^{2}+1}$$ and then, find the power series of $\displaystyle\frac{1}{x^{2}+1}$ centered at $a=1$. I want to know if this is the best way to do it.

Answer 1

We are looking for a Taylor series expansion of $f(x)$ at $x=1$, i.e. we are looking to find $a_n$ in \begin{align*} f(x)=\frac{x^3-2}{x^2+1}=\sum_{n=0}^\infty a_n (x-1)^n\tag{1} \end{align*}

In order to do so we start with a partial fraction expansion of $\frac{1}{x^2+1}$ and continue with expanding the linear terms at $x=1$.

We obtain \begin{align*} \color{blue}{\frac{1}{x^2+1}}&=-\frac{i}{2}\,\frac{1}{x-i}+\frac{i}{2}\frac{1}{x+i}\\ &=-\frac{i}{2}\,\frac{1}{(x-1)+1-i}+\frac{i}{2}\,\frac{1}{(x-1)+1+i}\\ &=-\frac{i}{2(1-i)}\,\frac{1}{1+\frac{x-1}{1-i}}+\frac{i}{2(1+i)}\,\frac{1}{1+\frac{x-1}{1+i}}\tag{2}\\ &=-\frac{i}{2(1-i)}\sum_{n=0}^\infty(-1)^n\frac{1}{(1-i)^n}(x-1)^n\\ &\qquad\quad+\frac{i}{2(1+i)}\sum_{n=0}^\infty(-1)^n\frac{1}{(1+i)^n}(x-1)^n\tag{3}\\ &=-\frac{i}{2}\sum_{n=0}^\infty(-1)^n\left(\frac{1+i}{2}\right)^{n+1}(x-1)^n\\ &\qquad\quad+\frac{i}{2}\sum_{n=0}^\infty (-1)^n\left(\frac{1-i}{2}\right)^{n+1}(x-1)^n\tag{4}\\ &=-\frac{i}{2}\sum_{n=0}^\infty (-1)^n\left(\frac{1}{\sqrt{2}}\right)^{n+1}e^{\frac{i(n+1)\pi}{4}}(x-1)^n\\ &\qquad\quad+\frac{i}{2}\sum_{n=0}^\infty(-1)^n\left(\frac{1}{\sqrt{2}}\right)^{n+1}e^{\frac{i(n+1)\pi}{4}}(x-1)^n\tag{5}\\ &\,\,\color{blue}{=\sum_{n=0}^\infty(-1)^n\left(\frac{1}{\sqrt{2}}\right)^{n+1}\sin\left(\frac{(n+1)\pi}{4}\right)(x-1)^n}\tag{6}\\ \end{align*}

Comment:

• In (2) we are ready to do a geometric series expansion.

• In (3) we are essentially finished with the example since we have found a representation according to (1). It has to be multiplied with $x^3-2$ but this is minor stuff. Nevertheless we can considerably simplify the expression, or better transform the series to better see what's going on.

• In (4) we rationalize the denominator as in $\frac{1}{1-i}=\frac{1+i}{(1-i)(1+i)}=\frac{1+i}{2}$.

• In (5) we use the identity $\frac{1+ i}{\sqrt{2}}=e^{\frac{i\pi}{4}}$ and apply De Moivre's formula. We do the same with $\frac{1-i}{\sqrt{2}}$.

• In (6) we use Euler's formula $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$.

Next step is to expand $x^3-2$ at $x=1$. We obtain \begin{align*} \color{blue}{x^3-2}&=((x-1)+1)^3-2\\ &\,\,\color{blue}{=(x-1)^3+3(x-1)^2+3(x-1)-1} \tag{7} \end{align*}

With (6) and (7) we can write $f(x)$ as

\begin{align*} \color{blue}{f(x)}&\color{blue}{=\frac{x^3-2}{x^2+1}}\\ &=\left[(x-1)^3+3(x-1)^2+3(x-1)-1\right]\\ &\qquad\quad\cdot\sum_{n=0}^\infty\frac{(-1)^n}{2^{\frac{n+1}{2}}}\sin\left(\frac{(n+1)\pi}{4}\right)(x-1)^n\\ &=\sum_{n=3}^\infty\left(\frac{(-1)^{n-3}}{2^{\frac{n-2 }{2}}}\sin\left(\frac{(n-2)\pi}{4}\right)\right)(x-1)^n\\ &\qquad\quad+3\sum_{n=2}^\infty\left(\frac{(-1)^{n-2}}{2^{\frac{n-1 }{2}}}\sin\left(\frac{(n-1)\pi}{4}\right)\right)(x-1)^n\\ &\qquad\quad+3\sum_{n=1}^\infty\left(\frac{(-1)^{n-1}}{2^{\frac{n}{2}}}\sin\left(\frac{n\pi}{4}\right)\right)(x-1)^n\\ &\qquad\quad-\sum_{n=0}^\infty\left(\frac{(-1)^{n}}{2^{\frac{n+1 }{2}}}\sin\left(\frac{(n+1)\pi}{4}\right)\right)(x-1)^n\tag{8}\\ &\,\,\color{blue}{-\frac{1}{2}+2(x-1)-\frac{1}{4}(x-1)^2}\\ &\qquad\quad\,\,\color{blue}{+\sum_{n=3}^\infty\left(-\frac{1}{\sqrt{2}}\right)^{n+1}\left(2\sqrt{2}\sin\left(\frac{(n-2)\pi}{4}\right)\right.}\\ &\qquad\qquad\quad\color{blue}{\left.-6\sin\left(\frac{(n-1)\pi}{4}\right)+3\sqrt{2}\sin\left(\frac{n\pi}{4}\right)+\sin\left(\frac{(n+1)\pi}{4}\right)\right)}\tag{9}\\ &\cdots\\ &=-\frac{1}{2}+2(x-1)-\frac{1}{4}(x-1)^2-\frac{1}{4}(x-1)^3+\frac{3}{8}(x-1)^4\\ &\qquad\quad-\frac{1}{4}(x-1)^5+\frac{1}{16}(x-1)^6+\frac{1}{16}(x-1)^7-\frac{3}{32}(x-1)^8+\cdots \end{align*}

Comment:

• In (8) we multply out and shift the index $n$ to always have terms $(x-1)^n$.

• In (9) we explicitely calculate the terms for $n=0,1,2$ and collect the terms with $n\geq 3$ in the series. Here we stop the calculation by noting that the sine terms can be further simplified. This is indicated in the last line calculated with Wolfram Alpha which show nice regularities of the coefficients .

Answer 2

If you make $x=y+1$ $$\frac{x^3-2}{x^{2}+1}=\frac{y^3+3 y^2+3 y-1}{y^2+2 y+2}=1+y-\frac{y+3}{y^2+2 y+2}$$ If you want a truncated series, continue with the long division.

Looking at the coefficient of the first terms, I suppose that their definition is not the simplest.

Edit

Working $$\frac{y+3}{y^2+2 y+2}=\sum_{n=0}^\infty a_n\,y^n$$ the coefficients are given by $$a_n=2^{-(n+2)} \left((3-i) (-1-i)^n+(3+i) (-1+i)^n\right)$$ which are all real. Replacing $y$ by $(x-1)$, we then have $$\frac{x^3-2}{x^{2}+1}=1+(x-1)-\sum_{n=0}^\infty a_n\,(x-1)^n=-\frac 12+2(x-1)-\sum_{n=2}^\infty a_n\,(x-1)^n$$