Very long Taylor expansion/partial fraction decomposition

[jamesb1]

Homework Statement

I want to express the following expression in its Taylor expansion about x = 0:

$$
F(x) = \frac{x^{15}}{(1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^5)}
$$

The Attempt at a Solution

First I tried to rewrite the function in partial fractions (its been quite a while since I've last covered these).

So far I have:
\begin{align*}
F(x) = &\frac{A}{1-x} &&+ \\
&\frac{Bx + C}{1-x^2} &&+ \\
&\frac{Dx^2 + Ex + F}{1-x^3} &&+ \\
&\frac{Gx^3 + Hx^2 + Ix + J}{1-x^4} &&+ \\
&\frac{Kx^4 + Lx^3 + Mx^2 + Nx + O}{1-x^5}
\end{align*}

Then:

\begin{align*}
x^{15} = &A(1-x^2)(1-x^3)(1-x^4)(1-x^5) &&+ \\
&(Bx + C)(1-x)(1-x^3)(1-x^4)(1-x^5) &&+ \\
&(Dx^2 + Ex + F)(1-x)(1-x^2)(1-x^4)(1-x^5) &&+ \\
&(Gx^3 + Hx^2 + Ix + J)(1-x)(1-x^2)(1-x^3)(1-x^5) &&+ \\
&(Kx^4 + Lx^3 + Mx^2 + Nx + O)(1-x)(1-x^2)(1-x^3)(1-x^4)
\end{align*}

As you can see, this does not seem to make sense. For example, in the equation above I'm getting $1=0$ when $x=1$.

What am I doing wrong?

 

[Ray Vickson]

Homework Statement

I want to express the following expression in its Taylor expansion about x = 0:

$$
F(x) = \frac{x^{15}}{(1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^5)}
$$

The Attempt at a Solution

First I tried to rewrite the function in partial fractions (its been quite a while since I've last covered these).

If I were doing it I would proceed very differently. If
D(x) = (1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^5)
we can write
\frac{1}{D(x)} = \sum_a x^a \, \sum_b x^{2b} \, \sum_c x^{3c} \, \sum_d x^{4d} \, \sum_e x^{5e},
where $a,b,c,d,e$ run independently over $0,1,2,3,\ldots$. We can re-write this as
\frac{1}{D(x)} = \sum_{n=0}^{\infty} c_n x^n,
and, of course,
\frac{x^{15}}{D(x)} = \sum_{n=0}^{\infty} c_n x^{15+n}
Here, $c_n$ is the cardinality of the set
\{ (a,b,c,d,e): a + 2b + 3 c + 4d + 5e = n \}
(in other words, the number of $(a,b,c,d,e)$ bundles satisfying the sum condition).

 

[SammyS]

Consider $\displaystyle \ \frac{x}{1-x}\cdot\frac{x^2}{1-x^2}\cdot\ \dots\ \cdot\frac{x^5}{1-x^5}\ .$

 

[jamesb1]

If I were doing it I would proceed very differently. If
D(x) = (1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^5)
we can write
\frac{1}{D(x)} = \sum_a x^a \, \sum_b x^{2b} \, \sum_c x^{3c} \, \sum_d x^{4d} \, \sum_e x^{5e},
where $a,b,c,d,e$ run independently over $0,1,2,3,\ldots$. We can re-write this as
\frac{1}{D(x)} = \sum_{n=0}^{\infty} c_n x^n,
and, of course,
\frac{x^{15}}{D(x)} = \sum_{n=0}^{\infty} c_n x^{15+n}
Here, $c_n$ is the cardinality of the set
\{ (a,b,c,d,e): a + 2b + 3 c + 4d + 5e = n \}
(in other words, the number of $(a,b,c,d,e)$ bundles satisfying the sum condition).

Thank you for your answer! I actually tried that approach before (of expanding using the Maclaurin series for $(1-x)^{-1}$. I then couldn't continue after that as I never knew about this notion of the $c_n$ coefficient. Could you elaborate on how $c_n$ happens to be the cardinality of that set please?

 

[SammyS]

Consider $\displaystyle \ \frac{x}{1-x}\cdot\frac{x^2}{1-x^2}\cdot\ \dots\ \cdot\frac{x^5}{1-x^5}\ .$

Along with this, use the Maclaurin series for $\displaystyle \ \frac{x}{1-x} \ .$

 

[Ray Vickson]

Thank you for your answer! I actually tried that approach before (of expanding using the Maclaurin series for $(1-x)^{-1}$. I then couldn't continue after that as I never knew about this notion of the $c_n$ coefficient. Could you elaborate on how $c_n$ happens to be the cardinality of that set please?

It is just elementary algebra, but to help you get started I will go through one case of a smaller example. Consider the smaller example
S = \frac{1}{(1-x)(1-x^2)} = \underbrace{(1 +x + x^2 + x^3 + \cdots)}_{=F_1}\; \underbrace{(1 + x^2 + x^4 + x^6 + \cdots)}_{=F_2}.
When we expand out the product we will obtain a series of powers of $x$. How can we find the term $c_5 x^5$, consisting of all terms in $x^5$? These are made up of $x^a$ from factor $F_1$ and $x^{2b}$ from factor $F_2$. We have:
\begin{array}{lcl}<br /> c_5 x^5 &amp; = x^5 \times x^0 = x^{5 + 0 \cdot 2} &amp; (a = 5, b = 0) \\<br /> &amp; + x^3 \times x^2 = x^{3 + 1 \cdot 2} &amp; (a=3, b=1) \\<br /> &amp; + x^1 \times x^4 = x^{1 + 2 \cdot 2} &amp; (a = 1, b = 2)\\<br /> &amp;= 3 x^5 &amp; \end{array}
There are exactly three combinations of $a, b \in \{0,1,2,\ldots \}$ that give $a + 2b = 5$.

 

[vela]

So far I have:
\begin{align*}
F(x) = &\frac{A}{1-x} &&+ \\
&\frac{Bx + C}{1-x^2} &&+ \\
&\frac{Dx^2 + Ex + F}{1-x^3} &&+ \\
&\frac{Gx^3 + Hx^2 + Ix + J}{1-x^4} &&+ \\
&\frac{Kx^4 + Lx^3 + Mx^2 + Nx + O}{1-x^5}
\end{align*}

Then:

\begin{align*}
x^{15} = &A(1-x^2)(1-x^3)(1-x^4)(1-x^5) &&+ \\
&(Bx + C)(1-x)(1-x^3)(1-x^4)(1-x^5) &&+ \\
&(Dx^2 + Ex + F)(1-x)(1-x^2)(1-x^4)(1-x^5) &&+ \\
&(Gx^3 + Hx^2 + Ix + J)(1-x)(1-x^2)(1-x^3)(1-x^5) &&+ \\
&(Kx^4 + Lx^3 + Mx^2 + Nx + O)(1-x)(1-x^2)(1-x^3)(1-x^4)
\end{align*}

As you can see, this does not seem to make sense. For example, in the equation above I'm getting $1=0$ when $x=1$.

What am I doing wrong?

As Ray and SammyS have suggested, partial fractions isn't the best way to go here. But I'll point out some problems with what you did in case you're still curious. First, the degree of the numerator is equal to the degree of the denominator. You want the degree of the numerator to be strictly less than the degree of the denominator. That's easy enough to deal with. Just pull the factor of $x^{15}$ off to the side for now.

The second problem is that you can't use any old factorization of the denominator. You have to factor the denominator into a product of irreducible polynomials:
$$(1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^5) = -(x-1)^5(x+1)^2(x^2+1)(x^2+x+1)(x^2+ b_+ x + 1)(x^2 + b_- x + 1)$$ where $b_\pm = \frac{1 \pm \sqrt 5}{2}$. As you can see, partial fractions looks to be tedious and messy.

 

[jamesb1]

It is just elementary algebra, but to help you get started I will go through one case of a smaller example. Consider the smaller example
S = \frac{1}{(1-x)(1-x^2)} = \underbrace{(1 +x + x^2 + x^3 + \cdots)}_{=F_1}\; \underbrace{(1 + x^2 + x^4 + x^6 + \cdots)}_{=F_2}.
When we expand out the product we will obtain a series of powers of $x$. How can we find the term $c_5 x^5$, consisting of all terms in $x^5$? These are made up of $x^a$ from factor $F_1$ and $x^{2b}$ from factor $F_2$. We have:
\begin{array}{lcl}<br /> c_5 x^5 &amp; = x^5 \times x^0 = x^{5 + 0 \cdot 2} &amp; (a = 5, b = 0) \\<br /> &amp; + x^3 \times x^2 = x^{3 + 1 \cdot 2} &amp; (a=3, b=1) \\<br /> &amp; + x^1 \times x^4 = x^{1 + 2 \cdot 2} &amp; (a = 1, b = 2)\\<br /> &amp;= 3 x^5 &amp; \end{array}
There are exactly three combinations of $a, b \in \{0,1,2,\ldots \}$ that give $a + 2b = 5$.

Wow, now that is quite interesting and it makes so much sense. Looks like I need to revise some combinatorics! Thank you again.