Partial Frac. Decomp. I tried both B and Bx+C. Which is correct?

Question

Disclaimer: I am not a student trying to get free internet homework help. I am an adult who is learning Calculus. I am deeply grateful to the members of this community for their time.

I am not sure why you need to use B vs. Bx+C. I tried both, and got 2 different answers. In the past, the only explanation I got to why you need to use Bx+C is that just using B would not give an answer. Well, this time both ways yielded an answer. Why is one incorrect??

Answer 1

A simple sanity check shows that your second solution is incorrect: $$\frac{1}{x^2+x+1} - \frac{1}{x-1} = \frac{x-1 - x^2 -x - 1}{(x-1)(x^2+x+1)} = \frac{-x^2-2}{x^3-1}\neq \frac{x^2-2x-2}{x^3-1}$$

It is easiest to explain why your second approach does not work using the language of linear algebra. Let $P_2(\Bbb R)$ be the vector space of all polynomials of degree $\leq 2$. In addition, let: $$f(x) = x^2+x+1$$ $$g(x) = x - 1$$ $$h(x) = x^2 - x$$ In your second approach, you are attempting to find a linear combination of $f(x)$ and $g(x)$ that yields $x^2-2x-2$. But this is impossible because $x^2-2x-2$ is not in the span of $f(x)$ and $g(x)$. Thus, there are no constants $A,B$ such that $Af + Bg = x^2 - 2x - 2$. Using $Bx+C$ instead of $B$ works because it will give you three polynomials that span all of $P_2(\Bbb R)$ (from your first approach, you can see that these polynomials are $f(x)$, $g(x)$, and $h(x)$). Hence, you are guaranteed to be able to find constants $A,B,C$ such that $Af + Bg + Ch = x^2-2x-2$.

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Note also that when you solved for $A,B$ in your second approach, you were essentially constructing a parabola that matches $x^2-2x-2$ on exactly two points. There are infinitely many such parabolas. A parabola is uniquely defined by three points, so if you want to make sure that whatever you are solving for ends up being equal to $x^2-2x-2$, you must confirm that there are three points that both of them pass through. Obviously, you cannot force that to happen using your second approach, since $A(x^2+x+1) + B(x-1)$ can be completely defined using only two points. So that approach will fail most of the time.

Answer 2

Why you need to use $Bx+C$ instead of just using $B$?

Since you know the trick of "let $x=1$" to find coefficient $A$ in $\dfrac{A}{x-1}$, I'll give you another answer (for fun) using the same substitution trick, but applied on each complex root $r_i$ of the denominator $Q(z)$.

It would be more instructive to work on a general setting, so that calculation mistakes won't hinder you from getting the whole idea.

To find $A_i$ in $\require{action}$

$$\bbox[5pt, border: 1pt solid red] {\toggle {\frac{P(z)}{Q(z)} = \sum_{j = 1}^{\deg Q} \frac{A_j}{z-r_j}, \\ \bbox[3pt]{\text{Click to multiply it by } z-r_i}} {\frac{P(z)}{\displaystyle \prod\limits_{\substack{j=1 \\ j\ne i}}^{\deg Q} (z-r_j)} = A_i + \sum_{\substack{j=1 \\ j\ne i}}^{\deg Q} \frac{A_j}{z-r_j} (z-r_i),} \endtoggle} $$

we substitute $r_i$ into the above clicked equality, so that

$$\bbox[4pt, border: 1pt solid blue]{ A_i = \frac{P(r_i)}{\displaystyle \prod\limits_{\substack{j=1 \\ j\ne i}}^{\deg Q} (r_i-r_j)}. } \tag{*} \label1$$

If $Q$ has real coefficients, the complex roots of $Q$ must occur in conjugate pair(s), so you may add up the corresponding terms to see why you need $Bx+C$ instead of $B$ in the denominator.

$$\frac{A_k}{z-r} + \frac{A_\ell}{z-\bar{r}} = \toggle {\frac{\mbox{degree one polynomial in } z}{\mbox{degree two polynomial in } z}} {\frac{(A_k+A_\ell)z - A_k \bar{r} - A_\ell r}{z^2 - 2 \mathrm{Re}(r)z + |r|^2}} \endtoggle \tag{click for details}$$

Well, this time both ways yielded an answer. Wh[ich] is one incorrect?

Looking at the degree of real factors of the denominator $Q(x)$, we know that the second approach is incorrect.

Exercise: Calculate $A$ in the question body using \eqref{1}.

Let $\omega$ be the 3^rd root of unity. $Q(x) = x^3-1 = (x-1)(x-\omega)(x-\omega^2)$. Note that $\omega^2+\omega+1=0$, $\omega^2 = \bar{\omega}$ and $\omega^3=1$. Let $\beta$ and $\gamma$ correspond to $\omega$ and $\omega^2$ respectively. \begin{align} A &= \frac{1-2-2}{1+1+1} = -1 \\ \beta &= \frac{\omega^2 - 2\omega - 2}{(\omega-1)(\omega-\omega^2)} \\ &= \frac{-\omega-1 - 2\omega - 2}{-\omega^3 + 2\omega^2 - \omega} \\ &= \frac{-3\omega-3}{-1 + 2(-\omega-1) - \omega} = 1 \\ \gamma &= \frac{\bar\omega^2 - 2\bar\omega - 2}{(\bar\omega-1)(\bar\omega-\omega)} \\ &= \frac{\overline{{\omega^2 - 2\omega - 2}}}{\overline{(\omega-1)(\omega-\omega^2)}} = \bar\beta = 1 \end{align} \begin{align} \frac{Bx+C}{x^2+x+1} &= \frac{\beta}{x-\omega} + \frac{\gamma}{x-\omega^2} = \frac{2x-\omega-\omega^2}{x^2+x+1} = \frac{2x+1}{x^2+x+1} \end{align}