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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?

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"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.

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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 $\,x=\frac12\,$ :

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

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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 $$a_5=(3)(2^5)+(-1)(-1)^5+(4)(6)(-1)^5.$$

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  • $\begingroup$ Could you finish out the problem? It seems dumb that I still can't get it. $\endgroup$ – ithisa May 23 '13 at 11:43
  • $\begingroup$ You mean the original numerical problem? Did you find %C_2$? It may be for several hours, will be busy. $\endgroup$ – André Nicolas May 23 '13 at 11:56
  • $\begingroup$ Yesssssssssssss. $\endgroup$ – ithisa May 23 '13 at 11:56
  • $\begingroup$ Sorry, pressed enter key inadvertently. Can you please amswer the revised question, about $C_2$? $\endgroup$ – André Nicolas May 23 '13 at 12:00
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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.


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 ${\bf Z}[p^{-1}]/{\bf Z}=\varinjlim{\bf Z}/p^n{\bf Z}$.

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