Finding Taylor polynomial of the right degree and checking the remainder

Question

There are few things which I can't quite grasp while trying to find Taylor polynomial of composite functions and their products. This will be lenghty for such an easy topic so i apologize in advance. I will use this problem as an example:

Find Taylor polynomial centered at $0$ of the fourth degree of the function:$$f(x) = \frac{1+x+x^2}{1-x+x^2}$$

I work with remainder in Peano form, defined to be $\omega(x)= \frac{R_n}{(x-a)^n}$ and $R_n = f(x)-P_n(x)$, where $P_n$ is Taylor polynomial of n-th degree and $a$ is where the polynomial is centered at. By Taylor theorem we know for $P_n$ this holds: $\lim_{x\to a}\omega(x) = 0$ and any given polynomial with this property is Taylor polynomial.

So since they're basically two multiplied functions I tried to make use of the Taylor expansion for $(1+x)^\alpha$ like this:$$(1+x+x^2)(1+(-x+x^2))^{-1}=(1+x+x^2)\bigg(\sum_{k=0}^{n}{\alpha\choose{k}}(-x+x^2)^k+\omega(-x+x^2)(-x+x^2)^n\bigg)$$ Where in $\omega(-x+x^2)$ the argument of $\omega(x)$ tends to zero as x tends to zero, so by limit of composite function: $\lim_{x\to0}\omega(-x+x^2)=0$. The remainder is multiplyied by $(-x+x^2)^n$. (Sorry for that notation. They use it at our uni and I'm not sure whether it's standard or not)

What degree should I plug into the sum? The problem asks for polynomial of degree $4$ but it's multiplied by polynomial of second degree. My initial thought was that I can expand the sum to second degree and I will get polynomial of 4th degree. It obviously didn't work. Can someone explain why I can't do this? The only explanation I came up with was that the remainder won't satisfy this limit: $$lim_{x\to\infty}\frac{R_4}{x^4}$$ so by Taylor theorem I can't tell if it's Taylor polynomial or not.

How does this multiplication of remainder change it? Is it okay to multiply it like this? From the expansion above:$$\omega(-x+x^2)(-x+x^2)^n+\omega(-x+x^2)(-x^2+x^3)^n+\omega(-x+x^2)(-x^3+x^4)^n$$ Does this whole term need to tend to zero when divided by $x^4$ by Taylor theorem? That would explain why I need to plug at least $n=4$, for $n<4$ the $\omega(-x+x^2)(-x+x^2)^n$ would be $\frac{0}{0}$.

Answer 1

To get the Taylor expansion around $x=0$, you should expand the terms of the form $(-x+x^2)^k$ as a sum of powers of $x$, adding equal powers up to order $4$. There are easier ways to get the Taylor expansion.

1. Long division. You divide $1+x+x^2$ by $1-x+x^2$ as polynomials, but in increasing orders of powers. The first term in the quotient will be $1$. Then compute $$ 1+x+x^2-1\times(1-x+x^2)=2\,x. $$ Dividing $2\,x$ by $1-x+x^2$, we find that the next term in the expansion will be $2\,x$. Compute now $$ 2\,x-(2\,x)\times(1+x+x^2)=-2\,x^2-2\,x^3. $$ Keep on going until you get to $x^4$.
2. Undetermined coefficients. Let $$ \frac{1+x+x^2}{1-x+x^2}=a_0+a_1\,x+a_2\,x^2+a_3\,x^3+a_4\,x^4+\dots $$ Then $$ 1+x+x^2=(1-x+x^2)\times(a_0+a_1\,x+a_2\,x^2+a_3\,x^3+a_4\,x^4+\dots)\tag{*} $$ Multiply the right hand side keeping only powers $\le4$. I wild it up to $x^2$: $$ a_0+(a_1-a_0)x+(a_2-a_1+a_0)x^2+\dots $$ Identifying coefficients on both sides of (*) we get the equations \begin{align} 1&=a_0\\ 1&=a_1-a_0\\ 1&=a_2-a_1+a_0\\ 0&=\dots \end{align} This allows to computebthe values of the coefficients $a_k$ recursively.