Partial fractions in complex analysis: Difference between revisions

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{{Short description|Way of writing a meromorphic function}}
In [[complex analysis]], a '''partial fraction expansion''' is a way of writing a [[meromorphic function]] ''<math>f(z)''</math> as an infinite sum of [[rational functions]] and [[polynomials]]. When ''<math>f(z)''</math> is a rational function, this reduces to the usual [[partial fractions|method of partial fractions]].

==Motivation==

By using [[polynomial long division]] and the partial fraction technique from algebra, any rational function can be written as a sum of terms of the form ''1 /<math display=inline>\frac{1}{(az + b)^{^k}''} + ''p(z)''</math>, where ''<math>a''</math> and ''<math>b''</math> are complex, ''<math>k''</math> is an integer, and ''<math>p(z)''</math> is a polynomial . Just as [[polynomial factorization]] can be generalized to the [[Weierstrass factorization theorem]], there is an analogy to partial fraction expansions for certain meromorphic functions.

A proper rational function, i.e. (one for which the [[degree of a polynomial|degree]] of the denominator is greater than the degree of the numerator,) has a partial fraction expansion with no polynomial terms. Similarly, a meromorphic function ''<math>f(z)''</math> for which <math>|''f(z)''|</math> goes to 0 as ''<math>z''</math> goes to infinity at least as quickly as <math display=inline>|''\frac{1/}{z''}|,</math> has an expansion with no polynomial terms.

==Calculation==

Let ''<math>f(z)''</math> be a function meromorphic in the finite complex plane with [[pole (complex analysis)|poles]] at ''&lambda;<submath>1</sub>''\lambda_1, ''&lambda;₂''\lambda_2, ...,</math> and let (''&Gamma;<submath>1</sub>''(\Gamma_1, ''&Gamma;₂''\Gamma_2, ...)</math> be a sequence of simple closed curves such that:

* The origin lies inside each curve ''&Gamma;<submath>k\Gamma_k</submath>''

* No curve passes through a pole of ''<math>f''</math>

* ''&Gamma;<submath>k\Gamma_k</submath>'' lies inside ''&Gamma;<submath>\Gamma_{k+1}</submath>'' for all ''<math>k''</math>

* <math>\lim_{k\rightarrow \infty} d(\Gamma_k) = \infty</math>, where ''<math>d(&Gamma;<sub>k\Gamma_k)</submath>)'' gives the distance from the curve to the origin

* one more condition of compatibility with the poles <math>\lambda_k</math>, described at the end of this section

Suppose also that there exists an integer ''<math>p''</math> such that

:<math>\lim_{k\rightarrow \infty} \oint_{\Gamma_k} \left|\frac{f(z)}{z^{p+1}}\right| |dz| < \infty</math>

Suppose also that there exists an integer ''p'' such that

Writing <math>\operatorname{PP}(''f(z)''; ''z = &lambda;<sub>k\lambda_k)</submath>'') for the [[principal part]] of the [[Laurent expansion]] of ''<math>f''</math> about the point ''&lambda;<submath>k\lambda_k</submath>'', we have

:<math>\lim_{k\rightarrow \infty} \oint_{\Gamma_k} \left|\frac{f(z)}{z^{p+1}}\right| dz < \infty</math>

Writing PP(''f(z)''; ''z = &lambda;_k'') for the [[principal part]] of the [[Laurent expansion]] of ''f'' about the point ''&lambda;_k'', we have

:<math>f(z) = \sum_{k=0}^{\infty} \operatorname{PP}(f(z); z = \lambda_k),</math>

if ''<math>p = -1'',</math>. andIf if ''<math>p > -1''</math>, then

:<math>f(z) = \sum_{k=0}^{\infty} (\operatorname{PP}(f(z); z = \lambda_k) + c_{0,k} + c_{1,k}z + \cdots + c_{p,k}z^p),</math>

where the coefficients ''c<submath>c_{j,k}</submath>'' are given by

:<math>c_{j,k} = \operatorname{Res}_{z=\lambda_k} \frac{f(z)}{z^{j+1}}</math>

&lambda;<submath>0\lambda_0</submath> should be set to 0, because even if ''<math>f(z)''</math> itself does not have a pole at 0, the [[residue (complex analysis)|residue]]s of ''<math display=inline>\frac{f(z)/}{z<sup>^{j+1}}</supmath>'' at ''<math>z'' = 0</math> must still be included in the sum.

Note that in the case of &lambda;<sub>0</submath>\lambda_0 = 0</math>, we can use the Laurent expansion of ''<math>f(z)''</math> about the origin to get

:<math>f(z) = \frac{a_{-m}}{z^m} + \frac{a_{-m+1}}{z^{m-1}} + \cdots + a_0 + a_1 z + \cdots</math>

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:<math>\sum_{j=0}^p c_{j,k}z^j = a_0 + a_1 z + \cdots + a_p z^p</math>

so that the polynomial terms contributed are exactly the [[regular part]] of the Laurent series up to ''z<supmath>z^p</supmath>''.

For the other poles ''&lambda;<submath>k\lambda_k</submath>'' where ''<math>k'' &\ge; 1</math>, ''1/z<supmath display=inline>\frac{1}{z^{j+1}}</supmath>'' can be pulled out of the [[residue (complex_analysiscomplex analysis)|residue]] calculations:

:<math>c_{j,k} = \frac{1}{\lambda_k^{j+1}} \operatorname{Res}_{z=\lambda_k} f(z)</math>

:<math>\sum_{j=0}^p c_{j,k}z^j = [\operatorname{Res}_{z=\lambda_k} f(z)] \sum_{j=0}^p \frac{1}{\lambda_k^{j+1}} z^j</math>

* To avoid issues with convergence, the poles should be ordered so that if &lambda;<submath>k\lambda_k</submath> is inside &Gamma;<submath>n\Gamma_n</submath>, then &lambda;<submath>j\lambda_j</submath> is also inside &Gamma;<submath>n\Gamma_n</submath> for all ''<math>j'' < ''k''</math>.

To avoid issues with convergence, the poles should be ordered so that if &lambda;_k is inside &Gamma;_n, then &lambda;_j is also inside &Gamma;_n for all ''j'' < ''k''.

==Example==

The simplest examples of meromorphic functions with an infinite number of poles are the non-[[entire function|entire]] trigonometric functions,. soAs takean the functionexample, <math>\tan(''z''). tan(''z'')</math> is meromorphic with poles at ''<math display=inline>(n + \frac{1/}{2})&\pi;''</math>, ''<math>n'' = 0, &plusmn;\pm 1, &plusmn;\pm 2, ...</math> The contours ''&Gamma;<submath>k\Gamma_k</submath>'' will be squares with vertices at ''&plusmn;&<math>\pm \pi; k &plusmn;\pm &\pi;ki'' k i</math> traversed counterclockwise, ''<math>k'' > 1</math>, which are easily seen to satisfy the necessary conditions.

[[Image:ContourSquares.PNG|right]]

The simplest examples of meromorphic functions with an infinite number of poles are the non-entire trigonometric functions, so take the function tan(''z''). tan(''z'') is meromorphic with poles at ''(n + 1/2)&pi;'', ''n'' = 0, &plusmn;1, &plusmn;2, ... The contours ''&Gamma;_k'' will be squares with vertices at ''&plusmn;&pi;k &plusmn; &pi;ki'' traversed counterclockwise, ''k'' > 1, which are easily seen to satisfy the necessary conditions.

On the horizontal sides of ''&Gamma;<submath>k\Gamma_k</submath>'',

:<math>z = t \pm \pi k i,\ \ t \in [-\pi k, \pi k],</math>

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:<math>|\tan(z)|^2 = \frac{\sin^2(t)\cosh^2(\pi k) + \cos^2(t)\sinh^2(\pi k)}{\cos^2(t)\cosh^2(\pi k) + \sin^2(t)\sinh^2(\pi k)}</math>

<math>\sinh(''x'') < \cosh(''x'')</math> for all real ''<math>x''</math>, which yields

:<math>|\tan(z)|^2 < \frac{\cosh^2(\pi k)(\sin^2(t) + \cos^2(t))}{\sinh^2(\pi k)(\cos^2(t) + \sin^2(t))} = \coth^2(\pi k)</math>

For ''<math>x'' > 0</math>, <math>\coth(''x'')</math> is continuous, decreasing, and bounded below by 1, so it follows that on the horizontal sides of ''&Gamma;<submath>k\Gamma_k</submath>'', <math>|\tan(''z'')| < \coth(''&\pi;'')</math>. Similarly, it can be shown that <math>|\tan(''z'')| < 1</math> on the vertical sides of ''&Gamma;<submath>k\Gamma_k</submath>''.

With this bound on <math>|\tan(''z'')|</math> we can see that

:<math>\oint_{\Gamma_k} \left|\frac{\tan(z)}{z}\right| dz \le \operatorname{length}(\Gamma_k) \max_{z\in \Gamma_k} \left|\frac{\tan(z)}{z}\right| < 8k \pi \frac{\coth(\pi)}{k\pi} = 8\coth(\pi) < \infty.</math>

(TheThat is, the maximum of <math display=inline>|\frac{1/''}{z''}|</math> on ''&Gamma;<submath>k\Gamma_k</submath>'' occurs at the minimum of <math>|''z''|</math>, which is ''<math>k&\pi;'')</math>.

Therefore ''<math>p'' = 0</math>, and the partial fraction expansion of <math>\tan(''z'')</math> looks like

:<math>\tan(z) = \sum_{k=0}^{\infty} (\operatorname{PP}(\tan(z); z = \lambda_k) + \operatorname{Res}_{z=\lambda_k} \frac{\tan(z)}{z}).</math>

The principal parts and [[residue (complex analysis)|residue]]s are easy enough to calculate, as all the poles of <math>\tan(''z'')</math> are simple and have residue -1:

:<math>\operatorname{PP}(\tan(z); z = (n + \frac{1}{2})\pi) = \frac{-1}{z - (n + \frac{1}{2})\pi}</math>

:<math>\operatorname{Res}_{z=(n + \frac{1}{2})\pi} \frac{\tan(z)}{z} = \frac{-1}{(n + \frac{1}{2})\pi}</math>

We can ignore ''&lambda;<submath>0</sub>''\lambda_0 = 0</math>, since both <math>\tan(''z'')</math> and <math display=inline>\frac{\tan(''z'')/''}{z''}</math> are analytic at 0, so there is no contribution to the sum, and ordering the poles ''&lambda;<submath>k\lambda_k</submath>'' so that ''&lambda;<submath display=inline>1</sub>''\lambda_1 = ''&\frac{\pi;''/}{2}, ''&lambda;₂''\lambda_2 = \frac{-''&\pi;''/}{2}, ''&lambda;₃''\lambda_3 = \frac{3''&\pi;''}{2}</2math>, etc., gives

:<math>\tan(z) = \sum_{k=0}^{\infty} \left[\left(\frac{-1}{z - (k + \frac{1}{2})\pi} - \frac{1}{(k + \frac{1}{2})\pi}\right) + \left(\frac{-1}{z + (k + \frac{1}{2})\pi} + \frac{1}{(k + \frac{1}{2})\pi}\right)\right]</math>

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===Infinite products===

Because the partial fraction expansion often yields sums of ''<math display=inline>\frac{1/(}{a+bz)''}</math>, it can be useful in finding a way to write a function as an [[infinite product]]; integrating both sides gives a sum of logarithms, and exponentiating gives the desired product:

:<math>\tan(z) = -\sum_{k=0}^{\infty} \left(\frac{1}{z - (k + \frac{1}{2})\pi} + \frac{1}{z + (k + \frac{1}{2})\pi}\right)</math>

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:<math>\tan(z) = 2\sum_{k=0}^{\infty} \sum_{n=0}^{\infty} \frac{2^{2n+2}}{(2k + 1)^{2n+2}\pi^{2n+2}} z^{2n + 1},</math>

which shows that the coefficients ''a<submath>na_n</submath>'' in the Laurent (Taylor) series of ''<math>\tan(z)''</math> about ''<math>z'' = 0</math> are

:<math>a_{2n+1} = \frac{T_{2n+1}}{(2n+1)!} = \frac{2^{2n+3}}{\pi^{2n+2}} \sum_{k=0}^{\infty} \frac{1}{(2k + 1)^{2n+2}}</math>

:<math>a_{2n} = \frac{T_{2n}}{(2n)!} = 0,</math>

where ''T<submath>nT_n</submath>'' are the [[tangent numbers]].

Conversely, we can compare this formula to the Taylor expansion for <math>\tan(''z'')</math> about <math>z = 0</math> to calculate the infinite sums:

:<math>\tan(z) = z + \frac{1}{3}z^3 + \frac{2}{15}z^5 + \cdots</math>