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Representing Functions by Series · Tutorial 783 of 998

Quotients by Normalizing the Denominator

Normalize the denominator first, then use the geometric series and careful multiplication to find a valid power-series representation of the quotient.

Advanced 9 min read

What You'll Learn

  • Factor a denominator into a nonzero constant times one minus a function.
  • Identify where the geometric expansion is valid by checking the condition that the inner function has magnitude below one.
  • Multiply the resulting reciprocal series by a polynomial numerator and compute its coefficients.
  • Verify a quotient expansion by multiplying the proposed series by the original denominator.
  • Distinguish the region guaranteed by the geometric expansion from the full domain of the rational function.

Why Normalize the Denominator?

A geometric series is a useful starting point for building power series for rational functions. The central identity is \(1/(1-u)=\sum_{n=0}^{\infty}u^n\), valid when \(|u|<1\). A denominator that does not already have the form \(1-u\) can often be rewritten to have it, at the cost of factoring out a constant.

After that normalization, the reciprocal has a geometric expansion. Multiplying by the numerator then gives a series for the quotient. The key tasks are to keep track of the factored constant, state the region where the expansion converges, and multiply the series correctly.

Definition: To normalize a denominator \(D(x)\), write it in the form \(D(x)=C(1-h(x))\), where \(C\) is a nonzero constant. Wherever \(|h(x)|<1\), its reciprocal has the geometric representation \(\frac{1}{D(x)}=\frac{1}{C}\sum_{n=0}^{\infty}h(x)^n\).

The Geometric Expansion After Normalization

The condition \(|h(x)|<1\) is the convergence condition for the geometric series. In particular, it is not enough to recognize a geometric-looking expression: the value of \(h(x)\) must satisfy this strict inequality at the point under consideration. Also, the factor \(1/C\) remains in the reciprocal. Omitting it changes every coefficient.

The geometric identity itself follows from the finite geometric sum. For \(u\ne 1\), the partial sum through the \(N\)th power satisfies the following identity:

$$ \sum_{n=0}^{N}u^n=\frac{1-u^{N+1}}{1-u}. $$

If \(|u|<1\), then \(u^{N+1}\) tends to zero as \(N\) tends to infinity. Taking the limit of both sides gives \(\sum_{n=0}^{\infty}u^n=1/(1-u)\). Substituting \(u=h(x)\) proves the normalized reciprocal formula wherever \(|h(x)|<1\).

Important condition: The geometric expansion gives a representation on the set where \(|h(x)|<1\). That set is guaranteed to be a region of convergence, but it need not be the entire domain of the rational function. A rational function can be defined at points where this particular geometric expansion does not converge.

Multiplying by a Numerator

Suppose the numerator is a polynomial \(P(x)\), and the denominator is normalized as \(C(1-h(x))\). In the region where \(|h(x)|<1\), multiply the reciprocal series by \(P(x)/C\). If \(P\) is a polynomial in \(x\) and \(h(x)\) is linear in \(x\), collecting like powers of \(x\) gives a power series. The coefficients can be found by adding the finite contributions that produce each power.

For example, if \(P(x)=p_0+p_1x\) and the reciprocal series has coefficients \(a_n\), then the product has constant coefficient \(p_0a_0\); its coefficient of \(x^k\), for \(k\geq 1\), is \(p_0a_k+p_1a_{k-1}\). This is ordinary multiplication of series: a term of degree \(k\) can come only from terms whose degrees add to \(k\).

More generally, if the numerator already has a power-series representation, multiply the two series where both converge. Inside their common interval of convergence, power series converge absolutely, so their Cauchy product is valid; the coefficient of a given power is the finite sum of products whose indices add to that power. The examples below use polynomial numerators so that the coefficient calculations remain explicit.

Worked Examples

Worked Example: Expanding a Linear Quotient

Represent \(\frac{2+x}{3-x}\) as a power series centered at \(x=0\), and state the region where the geometric expansion applies.

First factor \(3\) from the denominator:

$$ 3-x=3\left(1-\frac{x}{3}\right). $$

Thus \(C=3\) and \(h(x)=x/3\). The condition \(|h(x)|<1\) becomes \(|x|<3\). In that region,

$$ \frac{2+x}{3-x} = \frac{2+x}{3}\sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n. $$

Now collect coefficients. The constant term is \(2/3\). For \(k\geq 1\), the coefficient of \(x^k\) receives one contribution from \(2\sum (x/3)^n/3\) and one from \(x\sum (x/3)^n/3\):

$$ \frac{2}{3}\frac{1}{3^k}+\frac{1}{3}\frac{1}{3^{k-1}} = \frac{2}{3^{k+1}}+\frac{1}{3^k} = \frac{5}{3^{k+1}}. $$

Therefore the representation is

$$ \frac{2+x}{3-x} = \frac{2}{3}+\sum_{k=1}^{\infty}\frac{5}{3^{k+1}}x^k, \qquad |x|<3. $$

To check it, use the geometric identity to sum the series before collecting coefficients:

$$ \frac{1}{3}\left(2+x\right)\sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n = \frac{1}{3}(2+x)\frac{1}{1-x/3} = \frac{2+x}{3-x}. $$

This recovers the original quotient on the stated region.

Worked Example: A Denominator with a Plus Sign

Find a power series for \(\frac{1-x}{2+x}\) centered at zero, using normalization to apply the geometric series.

Factor \(2\) from the denominator and write the result as one minus a function:

$$ 2+x=2\left(1+\frac{x}{2}\right) =2\left(1-\left(-\frac{x}{2}\right)\right). $$

Here \(C=2\) and \(h(x)=-x/2\). The geometric condition is \(|x|<2\). Hence

$$ \frac{1-x}{2+x} = \frac{1-x}{2}\sum_{n=0}^{\infty}\left(-\frac{x}{2}\right)^n. $$

The first terms are obtained by multiplying the numerator through:

$$ \frac{1}{2}\left(1-\frac{x}{2}+\frac{x^2}{4}-\frac{x^3}{8}+\cdots\right) - \frac{x}{2}\left(1-\frac{x}{2}+\frac{x^2}{4}-\cdots\right) = \frac{1}{2}-\frac{3}{4}x+\frac{3}{8}x^2-\frac{3}{16}x^3+\cdots. $$

For \(k\geq1\), the coefficient of \(x^k\) is the contribution from the first series minus the contribution shifted by the factor \(x\):

$$ \frac{1}{2}\left(-\frac{1}{2}\right)^k - \frac{1}{2}\left(-\frac{1}{2}\right)^{k-1} = \frac{(-1)^k3}{2^{k+1}}. $$

Thus

$$ \frac{1-x}{2+x} = \frac{1}{2}+\sum_{k=1}^{\infty}\frac{(-1)^k3}{2^{k+1}}x^k, \qquad |x|<2. $$

For a check, sum the uncollected geometric expression:

$$ \frac{1-x}{2}\cdot\frac{1}{1-(-x/2)} = \frac{1-x}{2}\cdot\frac{1}{1+x/2} = \frac{1-x}{2+x}. $$

The sign in the inner function matters: the plus sign in the denominator produces alternating powers in the reciprocal series.

Worked Example: Multiplying by a Quadratic Numerator

Expand \(\frac{1+x+x^2}{5-2x}\) around zero and find a formula for its coefficients.

Normalize the denominator:

$$ 5-2x=5\left(1-\frac{2x}{5}\right). $$

Here \(C=5\), \(h(x)=2x/5\), and \(|h(x)|<1\) means \(|x|<5/2\). Therefore

$$ \frac{1+x+x^2}{5-2x} = \frac{1+x+x^2}{5}\sum_{n=0}^{\infty}\left(\frac{2x}{5}\right)^n. $$

The coefficient of \(x^0\) is \(1/5\). The coefficient of \(x^1\) is \(1/5+2/25=7/25\). The coefficient of \(x^2\) is \(1/5+2/25+4/125=39/125\). For \(k\geq2\), the three possible contributions come from the terms \(1\), \(x\), and \(x^2\) in the numerator:

$$ c_k = \frac{1}{5}\left[ \left(\frac{2}{5}\right)^k+ \left(\frac{2}{5}\right)^{k-1}+ \left(\frac{2}{5}\right)^{k-2} \right] = \frac{39}{125}\left(\frac{2}{5}\right)^{k-2}. $$

Consequently,

$$ \frac{1+x+x^2}{5-2x} = \frac{1}{5}+\frac{7}{25}x+ \sum_{k=2}^{\infty}\frac{39}{125}\left(\frac{2}{5}\right)^{k-2}x^k, \qquad |x|<\frac{5}{2}. $$

To verify the coefficients, multiply the proposed series by \(5-2x\). The constant coefficient becomes \(5(1/5)=1\). The coefficient of \(x\) becomes \(5(7/25)-2(1/5)=7/5-2/5=1\). The coefficient of \(x^2\) becomes \(5(39/125)-2(7/25)=39/25-14/25=1\). For each \(k\geq3\), the coefficient is \(5c_k-2c_{k-1}=0\), because \(c_k=(2/5)c_{k-1}\). The product is therefore exactly \(1+x+x^2\), as required.

Common Pitfall: Losing the Factored Constant

A frequent error is to expand the denominator after recognizing its variable part but forget the constant that was factored out. For instance, the incorrect calculation

$$ \frac{2+x}{3-x} \ne (2+x)\sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n $$

is missing the factor \(1/3\). The error is visible even at \(x=0\): the left side is \(2/3\), while the proposed right side is \(2\). The correct normalization is \(3-x=3(1-x/3)\), so the reciprocal is \(\frac{1}{3}\sum_{n=0}^{\infty}(x/3)^n\), and the quotient is \(\frac{2+x}{3}\sum_{n=0}^{\infty}(x/3)^n\).

A second check is to multiply the proposed series by the original denominator. If the product does not give the numerator, a factor or coefficient has been lost. It is also important not to claim that the geometric series converges wherever the rational function is defined: the expansion requires the stricter condition \(|h(x)|<1\).

Why This Method Matters

Normalizing a denominator turns many quotients into a familiar geometric expansion. It provides a direct way to obtain coefficients without first solving for them recursively, and the resulting series can be checked by multiplication. The method also makes the convergence restriction explicit: it comes from the normalized variable part of the denominator.

When applying the method, keep the steps distinct. First identify \(C\) and \(h(x)\); next solve \(|h(x)|<1\); then write the reciprocal with its factor \(1/C\); finally multiply by the numerator and collect coefficients. These checks help distinguish a correct series representation from a formally plausible but invalid one.

Check Your Understanding

Use denominator normalization in each question, and state the condition under which the geometric expansion applies.

  1. Write \(4-x\) as \(C(1-h(x))\). What inequality on \(x\) guarantees convergence of the reciprocal expansion?
  2. Find the first four terms of a power series for \(\frac{1}{2-x}\) centered at zero.
  3. In \(\frac{1+x}{3+ x}\), what are the constant \(C\) and function \(h(x)\) after normalization?
  4. Explain why a factor \(1/C\) must appear when expanding the reciprocal of \(C(1-h(x))\).
  5. For a proposed quotient series, what multiplication check can confirm that its coefficients are correct?