How to use "results from partial fractions"?

Question

Let ${a_n}$ be a sequence whose corresponding power series $A(x)=\sum_{i\geq 0}a_ix^i$ satisfies

$$A(x)=\frac{6-x+5x^2}{1-3x^2-2x^3}$$

The denominator can be factored into $(1-2x)(1+x)^2$. Using results from partial fractions, it can be shown that there exists constants $C_1,C_2,C_3$ such that

$$A(x)=\frac{C_1}{1-2x}+\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$$

Determine these constants and find $a_5$ using this new expression.

What exactly are "results from partial fractions" and how should I use them?

Answer 1

Hints:

Solve the resulting polynomial equation

$$A(x)=\frac{6-x+5x^2}{(1-2x)(1+x)^2}=\frac{C_1}{1-2x}+\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}\implies$$

$$5x^2-x+6=C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)$$

For example, if you substitute $\,x=-1\,$ in both sides , you get

$$12=3C_3\implies C_3=4$$

Now $\,x=\frac12\,$ :

$$\frac{27}4=\frac94C_1\implies C_1=3\;,\ldots\text{etc.}$$

Answer 2

We describe what we do once we have the coefficients $C_1,C_2,C_3$. It all comes from the expansion $$\frac{1}{1-t}=1+t+t^2+t^3+\cdots.\tag{$1$}$$ Putting $t=2x$, we find that $\frac{C_1}{1-2x}$ has expansion $$C_1+2C_1 x+4C_1x^2+8C_1x^3+\cdots.$$ Putting $t=-x$, we can in a similar way get the power series expansion of $\frac{C_2}{1+x}$.

For $\frac{C_3}{(1+x)^2}$ we need an additional idea. Look at Equation $(1)$, and differentiate both sides. We get $$\frac{1}{(1-t)^2}=1+2t+3t^2+4t^3+\cdots.$$ Put $t=-x$, and multiply by $C_3$.

Now that we have the three power series expansion, we can read off the coefficient of $x^n$ in their sum.

Details: Almost all the work in finding the $C_i$ has been done by DonAntonio. Bringing $\frac{C_1}{1-2x}=\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$ to the common denominator $(1-2x)(1+x)^2$, we find that the numerator is $C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)$. Thus, identically, we must have $$C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)=6-x+5x^2.$$ There are various ways to find the constants. Put $x=-1$. We get $C_3(3)=12$, so $C_3=4$. Put $x=1/2$. We get $C_1(9/4)=27/4$, so $C_1=3$. Finally, the constant term on the left is $C_1+C_2+C_3$, while on the right it is $6$. Since $C_1+C_3=7$, we get $C_2=-1$. So our original function is equal to $$\frac{3}{1-2x}-\frac{1}{1+x}+\frac{4}{(1+x)^2}.\tag{$2$}$$

Finally, we compute the coefficient $a_5$ of $x^5$. By the discussion in the main answer, the coefficient of $x^5$ in the expansion of $\frac{1}{1-2x}$ is $2^5$.

The coefficient of $x^5$ in the expansion of $\frac{1}{1+x}$ is $(-1)^5$.

The expansion of $\frac{1}{(1-t)^2}$ was obtained by differentiating $1+t+t^2+\cdots$. so the $t^5$ term is $6t^5$. Putting $t=-x$ we get that the coefficient of $x$ is $6(-1)^5$. Putting things together, and remembering our $C_i$, we get $$a_5=(3)(2^5)+(-1)(-1)^5+(4)(6)(-1)^5.$$

Answer 3

"Results from partial fractions" presumably refers to the following theorem about partial fraction decompositions.

Theorem: Let $\frac{P(x)}{Q(x)}$ be a rational function such that $\deg P < \deg Q$ and such that $Q(x)$ factors as $\prod (1 - r_i x)^{m_i}$ where the $r_i$ are distinct. Then there exist unique constants $c_{i, j}$ such that

$$\frac{P(x)}{Q(x)} = \sum_{i, j : 1 \le j \le m_i} \frac{c_{i, j}}{(1 - r_i x)^j}.$$

The special case of this result when $Q$ has distinct roots is probably familiar but the case of repeated roots is a little more subtle.

In any case, you don't need to know this; the result has already been used for you.

Answer 4

The main conclusion that drives the method of partial fractions is that any rational function of the form $\rm p(x)/q(x)$ where $\rm q$ factors into irreducibles as $\rm q(x)=w_1(x)^{e_1}\cdots w_n(x)^{e_n}$ can be represented in either of the equivalent forms

$$\rm \frac{p(x)}{q(x)}=a_0(x)+\frac{a_1(x)}{w_1(x)^{e_1}}+\cdots+\frac{a_n(x)}{w_n(x)^{e_n}}$$

$$\rm =a_0(x)+\left[\frac{b_{11}}{w_1(x)}+\frac{b_{12}}{w_1(x)^2}+\cdots+\frac{b_{1e_1}}{w_1(x)^{e_1}}\right]+\cdots+\left[\frac{b_{n1}}{w_n(x)}+\frac{b_{n2}}{w_n(x)^2}+\cdots+\frac{b_{ne_n}}{w_n(x)^{e_n}}\right]$$

for some polynomials $\rm a_i(x)$ with $\rm \deg a_i(x)<e_i\deg w_i(x)$ for each $\rm i=1,\cdots,n$ and constants $\rm b_{ij}$.

To use this theorem in practice, you set the desired equality and then solve for the unknown coefficients using the appropriate techniques from linear algebra.

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Given some abstract algebra and field theory as prerequisites, there is a very nice algebraic way of phrasing this theorem. The connection is similar to how a low-tech version of Sun-Ze (better known as the Chinese Remainder Theorem) describes existence and uniqueness of solutions to systems of congruences and the algebraic version describes a ring decomposition.

Let $F$ be a field and $T$ an indeterminate. Let $\mathsf{Irr}$ be the set of monic irreducibles in $F[T]$. We denote by $F(T)_\pi$ and $F[T]_\pi$ the ring completions with respect to the $(\pi(T))$-adic topology. Then

$$\frac{F(T)}{F[T]}\cong\bigoplus_{\pi\in\mathsf{Irr}}\frac{F(T)_\pi}{F[T]_\pi}$$

as additive quotient groups. There is also a number field analogue (specializing to $\bf Q$):

$$\frac{\bf Q}{\bf Z}\cong\bigoplus_{p~\text{prime}}\frac{{\bf Q}_p}{{\bf Z}_p},$$

which amounts to the $p$-primary decomposition of it as an abelian group; the direct summands are called the Prufer $p$-groups ${\bf Z}_{p^\infty}$ and can be alternately characterized as ${\bf Z}[p^{-1}]/{\bf Z}=\varinjlim{\bf Z}/p^n{\bf Z}$.