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Concerning partial fraction decomposition I can proof the following theorem:

For two polynomials $f(x),g(x) \ne 0$ with coefficients in $\mathbb C$ one can find the following unique representation:

$$\frac{f(x)}{g(x)}=\sum_{i=1}^{m}\sum_{j=1}^{n_i}\frac{a_{ij}}{(x-\alpha_i)^{j}}+p(x)$$

where $\alpha_1,\dots,\alpha_m$ are zeros of $g(x)$, $n_1,\dots,n_m$ their multiplicities, $p(x)$ is another polynomial and $a_{ij} \in \mathbb C, \forall i,j$.

Now I read in lecture notes that if $f(x)$ and $g(x)$ have coefficients in $\mathbb R$, then one can represent $\frac{f(x)}{g(x)}$ as a linear combination (with real coefficients) of terms of the following types:

  1. $x^k, k \in \mathbb N_{\ge0}$;

  2. $\frac{1}{(x-\alpha)^k}, k \in \mathbb N_{\ge0}$ where $\alpha$ is a real zero of $g(x)$;

  3. $\frac{1}{\big((x-\mathcal Re(\alpha))^2+\mathcal Im(\alpha)\big)^k},k \in \mathbb N_{\ge0}$ where $\alpha$ is a complex zero of $g(x)$;

  4. $\frac{x-Re(\alpha)}{\big((x-\mathcal Re(\alpha))^2+\mathcal Im(\alpha)\big)^k},k \in \mathbb N_{\ge0}$ where $\alpha$ is a complex zero of $g(x)$;

I do not fully see why it should be the case, but some aspects are clear to me.

Specifically, elements of the type $x^k$ come from the polynomial $p(x)$, elements of the type $\frac{1}{(x-\alpha)^k}$ one can also clearly see in the representation given in the theorem above.

To get elements of the types 3 and 4 one apparently needs to add two elements which are also present in the representation given by the theorem:

$$\frac{a}{(x-\overline \alpha)^k}+\frac{b}{(x-\alpha)^k}=\frac{b(x-\overline \alpha)^k+a(x- \alpha)^k}{\big((x-\mathcal Re(\alpha))^2+\mathcal Im(\alpha)^2\big)^k}$$

So I am not sure how can I turn $b(x-\overline \alpha)^k+a(x- \alpha)^k$ into $1$ or $x-\mathcal Re(\alpha)$.

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  • $\begingroup$ I'm sure you meant $\left((x - \mathcal{Re}(\alpha))^2+(\mathcal{Im}(\alpha))^2\right)^k$ !!! $\endgroup$ – Gribouillis Aug 19 '17 at 20:03
  • $\begingroup$ Indeed, thanks. $\endgroup$ – Sergey Zykov Aug 20 '17 at 11:50
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In the double sum, if you take all the terms corresponding to $(x-\alpha)^ {-j}$ and $(x-\overline{\alpha})^{-j}$, then this partial sum can be written $$S = \frac{A(x)}{(x-\alpha)^n(x-\overline{\alpha})^n}$$ where $A$ is a polynomial of degree $<2n$. You can divide $A(x)$ by $(x-\alpha)(x-\overline{\alpha})$, $$A(x) = A_1(x)(x-\alpha)(x-\overline{\alpha}) + b_n x + c_n$$ so that $$S = \frac{A_1(x)}{(x-\alpha)^{n-1}(x-\overline{\alpha})^{n-1}} + \frac{b_n x + c_n}{(x-\alpha)^n(x-\overline{\alpha})^n}$$ and $A_1$ has a degree $< 2(n-1)$. By induction, you get the result.

To prove that $A(x)$ has real coefficients, note that when $f$ and $g$ have real coefficients, then for all $x$ such that $g(x)\not = 0$ $$\overline{\left(\frac{f(\overline{x})}{g(\overline{x})}\right)} = \frac{f(x)}{g(x)}$$ Applying this to the complex decomposition yields $$\frac{f(x)}{g(x)} = \sum\sum \frac{\overline{a_{i j}}}{x-\overline{\alpha_i}} + \overline{p(\overline{x})}$$ The uniqueness of the decomposition shows that $p$ has real coefficients and also that the quantity $S = S(x)$ above satisfies $\overline{S(\overline{x})} = S(x)$. It follows that $$A(x) = S(x)(x-\alpha)^n(x-\overline{\alpha})^n$$ satisfies $\overline{A(\overline{x})} = A(x)$. Hence it is a polynomial with real coefficients.

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  • $\begingroup$ How do you proof that $b_n=1$ and $c_n = -\mathcal Re(\alpha)$ or $b_n=0$? $\endgroup$ – Sergey Zykov Aug 19 '17 at 13:36
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    $\begingroup$ You don't prove it because it is false! Your result say that $\frac{f}{g}$ is a linear combination of such terms. It means that $b_n$ and $c_n$ can be any (real) numbers. Now, try to prove that $b_n, c_n$ are real numbers, because it's a very important part of the theorem. $\endgroup$ – Gribouillis Aug 19 '17 at 13:40
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    $\begingroup$ OK, I see, basically if $b_n=0$, then I have $c_n\frac{1}{(x-\alpha)^n(x-\overline{\alpha})^n}$ which is a term of type 3, and if $b_n \ne 0$, then $\frac{b_n x + c_n}{(x-\alpha)^n(x-\overline{\alpha})^n}=b_n(c_n/b_n+\mathcal Re (\alpha))\frac{ 1}{(x-\alpha)^n(x-\overline{\alpha})^n} + b_n\frac{ x - \mathcal Re (\alpha)}{(x-\alpha)^n(x-\overline{\alpha})^n}$, which are term of type 3 and 4. $\endgroup$ – Sergey Zykov Aug 19 '17 at 13:52
  • $\begingroup$ as for $b_n$ and $c_n$, they are real because $A(x)$ and $(x-\alpha)(x-\overline \alpha)$ arу real, but I do not see yet why $A(x)$ must be real. $\endgroup$ – Sergey Zykov Aug 19 '17 at 14:07
  • $\begingroup$ If you know how to prove that $b_n$ and $c_n$ are real, then you know why $A(x)$ is real, because $A(x) = \sum_{i\ge1}(b_i x + c_i)(x-\alpha)^{n-i}(x-\overline{\alpha})^{n-i}$ $\endgroup$ – Gribouillis Aug 19 '17 at 15:04
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Because your original polynomials are real, you should be able to show that $a$ and $b$ are complex conjugates of each other. This means your expression is real.

Then when you have the expression you found, you can expand the numerator in terms of powers of the quadratic in the denominator to split it into fractions of the kind you want.

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