User:Pelliott11/sandbox

Complex Roots
Shouldn't this article include something about how to deal with terms that have complex/imaginary roots? Damn complex roots screwed me over on the exam.... —Preceding unsigned comment added by 203.128.4.54 (talk • contribs)


 * There is such a section: it's the one on irreducible quadratic factors. Having non-real complex roots is the same as having irreducible quadratic factors. Michael Hardy (talk) 18:52, 6 November 2008 (UTC)

Z-domain
This page needs some talk of PFD of transfer functions in the z-domain&mdash;they're hard. jScott 00:01, 2005 Mar 18 (UTC)

I don't know about partial fraction decompositions of, specifically, transfer functions, but I agree that the article in its present form is missing a lot of the standard material. Michael Hardy 01:46, 18 Mar 2005 (UTC)


 * I know my edits today have left the article a bit oddly organized, but I'll be back with much more later. Michael Hardy 19:37, 14 Apr 2005 (UTC)

Ahmad Hamidi seorang yang pandai.
This doesn't sound like it belongs in this article. I'm not sure what it means, but if it does belong in the article please put it back in. I'm going to delete it until someone says otherwise.-The Lab Rat


 * You're right; the previous editor put in some fanciful stuff and then fixed all of it except that. Michael Hardy 23:49, 18 May 2005 (UTC)
 * That is Malay, which means 'Ahmad Hamidi is a smart person'. :)  -x42bn6 07:45, 1 September 2005 (UTC)

shouldn't be merged
The sentence
 * This article or section should be merged with partial fraction decomposition over the reals.

isn't this POV ? The fact that people mix up both subjects is not enough. I think it is a good idea to distinguish
 * the algebraic view of partial fractions (associated to a given denominator polynomial), and problems of decomposition of an element of R(X) into those (for different types of rings R)
 * the analysis point of view, over the reals (and, not to forget, over the complex numbers, useful also for the real case), with methods for this, including limits, derivatives, changes of variables,...

There is enough to be said about both subject to justify two separate pages, of course with tight links especially in the first sections of both pages. &mdash; MFH: Talk 18:28, 26 May 2005 (UTC)

There should be some instruction on how to handle partial fraction decomposition over Z.

Don't agree with the merging
The two topics are not the same thing, merging would be ideal but i've left an internal link to the decomposition topic instead since this seems quite neat. since it's been up for months and nobody seems to have agreed with the merging i'm removing the suggestion, feel free to revert (giving reasons) however

General procedures
This page should have less example, and more general explanation. Each type of Partial fraction expansion should have its own algorithem given, and an example later - i.e. we shouldn't use examples to explain the algorithem - examples go after explanation to support it - not to replace it. Fresheneesz 19:47, 21 April 2006 (UTC)


 * I am new to the discussion. There is a Vedic algorithm or procedure that can be done mentally! You set up the partial fractions with numerators, A, B, C, D, etc. When the denominator contains powers of a binomial you set up partial fractions of the descending powers in the denominator and the relation of the numerators helps to solve for the numerators.


 * I scanned the article. Only parts would be comprehensible to an algebra student. I will add a simple procedure with no large gaps and no long solution process. Also, should we not mention that this procedure is used in integral calculus to break the Gordian Knots of some fractional algebraic expressions into simple fractional expressions that are easy to integrate as the sums and differences of the natural log? Larry R. Holmgren 20:04, 4 March 2007 (UTC)

irreducible???
Why must the denominator be irreducible, it looks to me as if most of the examples use the fact that the denominator is reducible to go on with the partial fraction expansion. What gives? Fresheneesz 20:39, 21 April 2006 (UTC)
 * Oh - its saying that the result of the PFE leaves an irrucible denominator. However, I don't think this is true either.
 * Under the header An irreducible quadratic factor in the denominator, it shows such an example. Another example is the expansion of (5s^3 - 3 s^2 + 2s - 1) / (S^4 + s^2). Fresheneesz 20:47, 21 April 2006 (UTC)
 * Nevermind - I get it finally... I'll fix up the page to make it more clear what it means. Fresheneesz 20:51, 21 April 2006 (UTC)

must have irreducible polynom in denom???
Does a partial fraction expansion have to have a power of an irreducible polynomial in the denominator? If I had started this article, I would have said that "partial fraction expansion" is the act of changing a fraction A(x)/B(x) into the sum of multiple fractions. Is this general definition not correct? Fresheneesz 21:01, 21 April 2006 (UTC)

the new "simple example"
Don't you think that the newly-added "simple example" is kinda dumb? I mean, you can simply turn (x) to (x+a-a), and then (x+a)/(x+a) -a/(x+a) = 1-a/(x+1). There's no reason to mess around with partial fraction decomposition here....


 * I think the simple examples used here (and elsewhere) are valuable. They may be obvious to you and I, but for someone meeting partial fractions for the first time it will be a valuable first step before they meet the more complicated examples. DavidMcKenzie 10:46, 26 February 2007 (UTC)

Partial Fractions Algorithm
Separation of a fractional algebraic expression into partial fractions is the reverse of the process of combining fractions by converting each fraction to the lowest common denominator (LCM) and adding the numerators. This separation is accomplished by a mental, one-line Vedic formula called the Paravartya Sutra. Case one has fractional expressions where factors in the denominator are unique. Case two has fractional expressions where some factors may repeat as powers of a binomial.

In integral calculus we would want to write a fractional algebraic expression as the sum of its partial fractions in order to take the integral of each simple fraction separately. Once the original denominator, D0, has been factored we set up a fraction for each factor in the denominator. We may use a subscripted D to represent the denominator of the respective partial fractions which are the factors in D0. Letters A, B, C, D, E, and so on will represent the numerators of the respective partial fractions.

We calculate each respective numerator by (1) calculating the Paravartya value of the denominator (which is the value of the variable making that factor zero) and (2) then substituting this value into the original expression but ignoring that factor in the denominator. Each Paravartya value for the variable is the value which gives an undefined value to the expression since we do not divide by zero.

General formula:
 * $$\frac{lx^2 + mx + n}{(x-a)(x-b)(x-c)}$$ = $$\frac{A}{(x-a)}$$ + $$\frac{B}{(x-b)}$$ + $$\frac{C}{(x-c)}$$

Here, a, b, c, l, m, and n are given integer values.

Where x=a and A = $$\frac{la^2 + ma + n}{(a-b)(a-c)}$$;

and where x=b and B = $$\frac{lb^2 + mb + n}{(b-c)(b-a)}$$;

and where x=c and C = $$\frac{lc^2 + mc + n}{(c-a)(c-b)}$$.

CASE ONE
Factorize the expression in the denominator. Set up a partial fraction for each factor in the denominator. Apply the Paravartya Sutra to solve for the new numerator of each partial fraction.

Example

 * $$\frac{3x^2 + 12x + 11}{(x+1)(x+2)(x+3)}$$ = $$\frac{A}{x+1}$$+$$\frac{B}{x+2}$$+$$\frac{C}{x+3}$$

Set up a partial fraction for each factor in the denominator. With this framework we apply the sutra to solve for A, B, and C by mental math.

1. D1 is x+1, set it equal to zero. This gives the Paravartya value for A when x = -1.

2. Next, substitute this value of x into the fractional expression, but without D1.

3A. Put this value down as the value of A. Proceed similarly for B and C.

3B. For Paravartya B use x = -2.

3C. For Paravartya C use x = -3.

Thus, to solve for A: use x = -1.
 * $$\frac{3x^2 + 12x + 11}{(x+2)(x+3)}$$ = $$\frac{3 -12 +11}{(1)(2)}$$ = $$\frac{2}{2}$$ = $$1$$ = A

Thus, to solve for B: use x = -2.
 * $$\frac{3x^2 + 12x + 11}{(x+1)(x+3)}$$ = $$\frac{12 -24 +11}{(-1)(1)}$$ = $$\frac{-1}{(-1)}$$ = +1 = B

Thus, to solve for C: use x = -3.
 * $$\frac{3x^2 + 12x + 11}{(x+1)(x+2)}$$ = $$\frac{27 -36 +11}{(-2)(-1)}$$ = $$\frac{2}{(+2)}$$ = +1 = C

Thus,
 * $$\frac{3x^2 + 12x + 11}{(x+1)(x+2)(x+3)}$$ = $$\frac{1}{x+1}$$+$$\frac{1}{x+2}$$+$$\frac{1}{x+3}$$

CASE TWO
When factors of the denominator include powers of one expression we (1) Set up a partial fraction for each unique factor and each lower power of D; (2) We set up an equation showing the relation of the numerators if all were converted to the LCD. From the equation of numerators we solve for each numerator, A, B, C, D, and so on. This equation of the numerators is an absolute identity, true for all values of x. So, we may select any value of x and solve.

Example

 * $$\frac{3x + 5}{(1-2x)^2}$$ = $$\frac{A}{(1-2x)^2}$$ + $$\frac{B}{(1-2x)}$$

Here, we set up a partial fraction for each descending power of the denominator. Then we solve for the numerators, A and B. As the Paravartya value for A and B will be the same, x = ½, we need an additional relation in order to solve for both. To write the relation of numerators the second fraction needs another factor of (1-2x) to convert it to the LCD, giving us 3x + 5 = A + B(1-2x).

To solve for A: Set the denominator of the first fraction to zero, 1-2x = 0. This is the Paravartya value for A, x = ½. When we substitute this value, x = ½, into the relation of numerators we have 3(1/2) + 5 = A + B(0). Solving for A gives us A = 3/2 + 5 = 13/2. Numerator A equals six and one-half.

To solve for B: Since the equation of the numerators, 3x + 5 = A + B(1-2x), is true for all values of x, pick a value for x and use it to solve for B. We have solved for the value of A above, A = 13/2. We may use it to solve for B.

We may pick x = 0, use A = 13/2, and solve for B. 3x + 5 =   A  + B(1-2x) 0 + 5 = 13/2 + B(1+0) 10/2 = 13/2 + B   -3/2 = B

We may pick x = 1. Then solve for B. 3x + 5 =   A  + B(1-2x) 3 + 5 = 13/2 + B(1-2) 8  = 13/2 + B(-1) 16/2 = 13/2 - B      B = -3/2

We may pick x = -1. Solve for B. 3x + 5 =   A  + B(1-2x) -3 + 5 = 13/2 + B(1+2) 4/2 = 13/2 + 3B -9/2 =  3B -3/2 =  B

Hence,
 * $$\frac{3x + 5}{(1-2x)^2}$$ = $$\frac{13/2}{(1-2x)^2}$$ + $$\frac{-3/2}{(1-2x)}$$

Or
 * $$\frac{3x + 5}{(1-2x)^2}$$ = $$\frac{13}{2(1-2x)^2}$$ $$-$$ $$\frac{3}{2(1-2x)}$$

Technique Three
A third technique is an analysis of the expanded relation of the numerators. Just match up the x-terms of each degree. Then one can see or solve for the missing numerator.


 * $$\frac{3x + 5}{(1-2x)^2}$$ = $$\frac{A}{(1-2x)^2}$$ + $$\frac{B}{(1-2x)}$$

Converting each fraction to the LCD we have the relation in the numerators: 3x+5 = A + B(1-2x). As A and B are constants, we can expand and match up the constant terms and the x-terms. The values of A and B will then be apparent. 3x+5 = A + B(1-2x) 3x+5 = A+B -2Bx Hence, 5   = A+B 3x  = -2Bx Therefore, 3   = -2B -3/2 = B And, 5   = A -3/2 5+3/2 = A  13/2 = A Hence,
 * $$\frac{3x + 5}{(1-2x)^2}$$ = $$\frac{13/2}{(1-2x)^2}$$ + $$\frac{-3/2}{(1-2x)}$$

Merge with Partial fraction decomposition

 * See also the merge proposal in the next section

This merge was proposed before but opposed with the argument that it is a good idea to distinguish the algebraic view of partial fractions and the analysis point of view. While that may be true: --Lambiam Talk 07:40, 18 June 2007 (UTC)
 * 1) the articles do a good job of making that distinction;
 * 2) the distinction, to the extent it is needed, can well be made within a single article;
 * 3) there is no relationship between the distinction and the names of the articles – or, in fact, it is, rather, the reverse, as the article entitled "Partial fraction" talks a whole lot more about decomposition, even going into algorithms for computing the decomposition (most of which can be adapted to any Euclidean ring);
 * 4) other articles referring to the process of partial fraction decomposition understandably have links to Partial fraction decomposition, while links to Partial fraction would usually be more appropriate in the context (see, e.g., Rational function);
 * 5) what is said in the "P.f. decomposition" article should really also be said in the other article;
 * 6) the "P.f. decomposition" article is a stub; to flesh it out to a full-fledged article would result in a lot of duplication.

I've struck out the preceding comment, since the earlier merge proposal appear to have involved a different article, namely Partial fraction decomposition over the reals. I think the latter article largely overlaps with the present one and should also be merged, but that is another proposal. --Lambiam Talk 07:47, 18 June 2007 (UTC)

Support Get everything organized in one place, if a split is needed later (I doubt it will be) then so be it.--Cronholm144 08:48, 18 June 2007 (UTC)

Merge with Partial fraction decomposition over the reals

 * See also the merge proposal in the preceding section

This merge was proposed before but opposed with the argument that it is a good idea to distinguish the algebraic view of partial fractions and the analysis point of view. While that may be true: --Lambiam Talk 08:00, 18 June 2007 (UTC)
 * 1) the articles don't do a good job of making that distinction;
 * 2) the distinction, to the extent it is needed, can well be made within a single article;
 * 3) the article named Partial fraction decomposition over the reals would be an orphan if it was not for a "See also" link here;
 * 4) most of what is said in the other article is also said here; there is a lot of overlap and duplication.

Support the merger might take a little bit of effort, both articles need organization and I worry that merging the two will just confuse things more. So this merger needs to be done carefully. --Cronholm144 08:46, 18 June 2007 (UTC)

Comment: one reason why there may be some confusion now of "theory" and "practice" is the addition of a great deal of how-to and algorithmic material. While this may well be useful to some readers, it leads to a more important question of organisation that just the merge proposed. Logically, the existence of the decomposition sought is always prior to finding it by some undetermined-coefficient method. You can get round that either by saying (in an example) that verification is immediate by addition of fractions, or (more problematically) proving some algorithm correct. But in any case an encyclopedic article should not be dominated by tricks for hand calculation: examples, yes. The content of algebra books from the nineteenth century is not the most fruitful material to start off with. Charles Matthews (talk) 13:00, 1 January 2009 (UTC)

Comment If these are to be merged, the merged article should be titled partial fraction. Michael Hardy (talk) 18:23, 1 January 2009 (UTC)


 * In fact the merge has already happened. Charles Matthews (talk) 23:07, 1 January 2009 (UTC)

What does Indian culture have to do with the subject?
I don't see how the "Parāvartya Sūtra" section is anything but a special case of partial fraction decomposition, or how "calculating the Parāvartya value of the denominator (which is the value of the variable making that binomial factor equal to zero)" is anything other than finding a zero of the denominator.

Maybe I'm missing some historical background of its development, or haven't read the section well enough, but it looks to me like someone is trying to attach Indian terminology to standard algebra techniques, and that seems out of place on the English Wikipedia. If there is some historical Indian basis for partial fraction development, I appologize and suggest creating a History section instead.

--BlueGuy213 (talk) 02:19, 13 November 2008 (UTC)

Yes, a history section on the early results about partial fractions would be interesting; in particular about the Indian contributes to the problem (in the western mathematical education we learn so little about the impressive progresses made by Indian and Chinese mathematicians. Today English is the universally aknowledged language of Maths, but it is important to recall that it so only since the last 60 years; in long and important periods the common language of Math was Greek, then Arabic, then Latin, with probably no preferred language, or maybe German, from the French revolution to the end of World War II). But outside a historical section I agree that it is unopportune introducing an exotic term where the terminology is already fixed. --PMajer (talk) 14:03, 28 December 2008 (UTC)

Indian mathematics is very rich and has an immense history. That history should be explained by someone competent to do it real justice. Patriotic alpha myth stuff is already in plentiful supply in the Anglophone nations, and there is no need of importation of more of it from India, even as a corrective to the home-grown version. Changing familiar terms to unfamiliar ones is is simply irritating, in the absence of a solidly argued case for the superiority or priority of the new terms. As it happens, there are reputable Indian mathematicians who argue that the collection of ideas that has been purveyed under the name Vedic mathematics is, in fact, Neither Vedic Nor Mathematics (Alma Teao Wilson (talk) 22:21, 10 July 2010 (UTC)).

Okay, I've ripped out that stuff. The result is still clunky, however, and needs further editing. (Alma Teao Wilson (talk) 23:12, 10 July 2010 (UTC))


 * Actually, I was thinking that the techniques as presented--not just the names attached to them--were entirely redundant with the "normal" techniques I had just learned in Precalculus Algebra (here in the US). I wasn't looking to replace the Hindi names with "Heaviside cover-up method", I was looking to remove the "redundacy" (as I perceived it).  After following the link in Heaviside cover-up method, which I note is from a Christian college website (see their main page), it seems I was mistaken as to this.


 * But then, if the method is genuinely useful--as, I gather, a way to abbreviate the paperwork involved in partial fraction decomposition--then I see no reason why Oliver Heaviside should get more credit for it than Hindu mathematicians (especially if they both developed it independently). Your link to a website that refers to "Hindutva terrorism" here may not be an NPOV source, either.  (I don't know, I haven't looked into this deeply.)


 * I won't be making any changes to the article myself, since I certainly do not know much about the more advanced methods and applications of partial fractions decomposition, nor their history. But I thought this information might be useful to someone who does know, and I still believe (like PMajer apparently does) that a History section discussing all contributors' known work would be nice.  --BlueGuy213 (talk) 05:40, 5 August 2010 (UTC)

Where's Taylor?
A major issue was lacking in this very long article: the useful formula for the coefficients of a PFD via Taylor polynomial expansion. I have therefore added a small section on the subject with a sketch of the proof. It seems to me that this article is not bad locally, but still contains things repeated several times in different places; sooner or later it should be rectified--PMajer (talk) 20:30, 31 December 2008 (UTC)

External link error
Moved from Talk:Partial fraction/Comments, but it'll have a better chance of being seen here. Astronaut (talk) 08:14, 2 April 2009 (UTC)

the link at the bottom of the page ("Article on solving partial fractions") seems to be dead. the page it links to ("http://www.wikinotes.hosted.hostmax.co.uk/wiki/Partial_Fractions") gives this as it's html output:

[an error occurred while processing this directive]

— Preceding unsigned comment added by 98.97.132.70 (talk) 23:45, 1 April 2009

Connection with matrix canonical form. Rational canonical form
Connection with the rational canonical form of a matrix, and therefore with finitely generated modules over PIDs, is missing. Also missing in the rational canonical form page.  franklin  14:38, 16 December 2009 (UTC)

Symbolic Characterization Has Errors
The characterization at the beginning of the article:


 * In symbols, one can use partial fraction expansion to change a rational function in the form


 * $$ \frac{f(x)}{g(x)} $$


 * where ƒ and g are polynomials, into a function of the form


 * $$ \sum \frac{a_n}{h_n(x)} $$


 * where hn are polynomials that are factors of g(x), and are in general of lower degree.

is not correct. In addition, I don't know what "are in general of lower degree" means. (A good rule for clear writing: When making a comparison, make sure that what is being compared to what is clear.)

I think this is what was intended:


 * In symbols, one can use the partial fraction expansion technique to change a rational function in the form


 * $$ \frac{f(x)}{g(x)} $$


 * where ƒ and g are polynomials, into a function of the form


 * $$ \sum \frac{a_n(x)}{h_n(x)} $$


 * where an and hn are polynomials, the hns are factors of g, and each an is of lower degree than the corresponding hn.

SDCHS (talk) 00:19, 28 May 2010 (UTC)

Complete rewrite
I think this article should be completely rewritten. Its arrangement is just a mess&mdash;what article first gives you examples then explain what it is? And it seems to me the examples are too many.--Netheril96 (talk) 10:39, 25 October 2010 (UTC)


 * The rewrite should probably take place here rather than in user space. This way it is more inclusive and transparent.  Moreover, changes can more easily be discussed on the talk page.  Also, I see that major changes have taken place here after the rewrite started.  I am going to remove the template, since we should not have people simultaneously editing two versions of the article.   Sławomir Biały  (talk) 11:28, 5 November 2010 (UTC)


 * If you think pages in userspace are inappropriate, I can move it to somewhere like Partial fraction/draft. This article need not just revisions but complete rewrite instead. To do that, we need to start from almost scratch because this article contains too much redundant and unencylopediac information and before we finish the main article should remain intact.--Netheril96 (talk) 13:28, 5 November 2010 (UTC)


 * Major edits are already underway here, though. I think the article should be left where it is while this happens.  Sławomir Biały  (talk) 14:20, 5 November 2010 (UTC)


 * Fine. I'll wait until you guys finish the trunk of the new article.--Netheril96 (talk) 14:57, 5 November 2010 (UTC)


 * I think it's down to a basic trunk. I don't think the article requires a complete rewrite anymore, although there is certainly still plenty of room for improvement.   Sławomir Biały  (talk) 12:37, 7 November 2010 (UTC)

Moved from article
I've moved this from the article. It was out of place where it was. It could go into an "applications" section, but that would need to be written, probably after the article has been cleaned up. Sławomir Biały (talk) 12:32, 7 November 2010 (UTC)

=== Use in deriving the logistic general equation ===

In many beginning calculus courses, partial fractions are introduced as a way to derive the general equation for a logistic function.

Logistic functions model a population which grows until it reaches a limit. The rate of change for the function is proportional (constant k) to both the population reached (P) and the fraction of the total carrying capacity (M) remaining. Thus the differential equation is:


 * $$\frac{dP}{dt}=kP\left({M-P \over M}\right)$$

Separating variables:


 * $${dP \over P(M-P)} = {k \, dt\over M}\, $$

The partial fraction expansion is


 * $${1 \over P(M-P)} = {A \over P} + {B \over M-P}$$

implying


 * $$1=A(M-P) + BP \,$$

for all values of P except for P = 0 and for P = M. But when the equation is true for all other values, it is also true for these limiting values. P → 0 gives


 * $$1=AM \,$$

and P &rarr; M gives


 * $$1=BM\,$$

So


 * $$A=B={1\over M}\,$$

and the final partial fraction expansion is


 * $${1 \over P(M-P)} = {1 \over MP} + {1 \over M(M-P)}.$$

The differential equation is then


 * $${dP \over MP} + {dP \over M(M-P)} = {k \, dt \over M} $$


 * $${dP \over P} + {dP \over (M-P)} = k \, dt $$


 * $${dP \over P} - {d(M-P) \over (M-P)} = k \, dt $$


 * $$d(\ln P) - d(\ln (M-P)) = k \, dt $$


 * $$d\left(\ln{P \over M-P}\right) = k \, dt $$

Integrating:


 * $$\ln{P \over M-P} = kt + C $$


 * $${P \over M-P} = e^{kt + C} $$


 * $${M-P \over P} = e^{-kt - C} $$


 * $${M \over P} - 1 = e^{-kt - C} $$


 * $$P = {M \over 1+e^{-kt - C}} $$

Setting t to zero and infinity gives


 * $$P_0 = {M \over {1+e^{- C}}} $$


 * $$P_\infty = M $$

and finally


 * $$P= {1 \over {1\over P_{\infty}} -\left({1 \over P_{\infty}}-{1 \over P_0}\right)e^{-kt}}$$

Moved from article 2
I moved the following inscrutable and out-of-place section as well. It can be restored if there are strong feelings about it, but my sense is that the article is better off without it. Sławomir Biały (talk) 12:32, 7 November 2010 (UTC)

== The role of the Taylor polynomial ==

The partial fraction decomposition of a rational function can be related to Taylor's theorem as follows. Let


 * $$P(x), Q(x), A_1(x),\dots, A_r(x)$$

be real or complex polynomials; assume that


 * $$\textstyle Q=\prod_{j=1}^{r}(x-\lambda_j)^{\nu_j},$$

that


 * $$\textstyle\deg(P)<\deg(Q)=\sum_{j=1}^{r}\nu_j,$$

and that


 * $$\textstyle\deg A_j<\nu_j\text{ for }j=1,\dots,r.$$

Define also


 * $$\textstyle Q_i=\prod_{j\neq i}(x-\lambda_j)^{\nu_j}=\frac{Q}{(x-\lambda_i)^{\nu_i}} \text{ for }i=1,\dots,r.$$

Then we have


 * $$\frac{P}{Q}=\sum_{j=1}^{r}\frac{A_j}{(x-\lambda_j)^{\nu_j}}$$

if, and only if, for each $$\textstyle i$$ the polynomial $$\textstyle A_i(x)$$ is the Taylor polynomial of $$\textstyle\frac{P}{Q_i}$$ of order $$\textstyle\nu_i-1$$ at the point $$\textstyle\lambda_j$$:


 * $$A_i(x):=\sum_{k=0}^{\nu_i-1} \frac{1}{k!}\left(\frac{P}{Q_i}\right)^{(k)}(\lambda_i)\ (x-\lambda_i)^k.

$$

Taylor's theorem (in the real or complex case) then provides a proof of the existence and uniqueness of the partial fraction decomposition, and a characterization of the coefficients.

Sketch of the proof: The above partial fraction decomposition implies, for each 1 ≤ i ≤ r, a polynomial expansion


 * $$\frac{P}{Q_i}=A_i + O((x-\lambda_i)^{\nu_i})\qquad $$, as $$x\to\lambda_i;$$

so $$\textstyle A_i$$ is the Taylor polynomial of $$\textstyle\frac{P}{Q_i}$$, because of the unicity of the polynomial expansion of order $$\textstyle\nu_i-1$$, and by assumption $$\textstyle\deg A_i<\nu_i$$.

Conversely, if the $$\textstyle A_i$$ are the Taylor polynomials, the above expansions at each $$\textstyle\lambda_i$$ hold, therefore we also have


 * $$P-Q_i A_i = O((x-\lambda_i)^{\nu_i})\qquad $$, as $$x\to\lambda_i,$$

which implies that the polynomial $$\textstyle P-Q_iA_i$$ is divisible by $$\textstyle (x-\lambda_i)^{\nu_i}.$$

For $$\textstyle j\neq i$$ also $$\textstyle Q_jA_j$$ is divisible by $$\textstyle (x-\lambda_i)^{\nu_i}$$, so we have in turn that $$\textstyle P- \sum_{j=1}^{r}Q_jA_j$$ is divisible by $$\textstyle Q$$. Since $$\textstyle\deg\left( P- \sum_{j=1}^{r}Q_jA_j \right) < \deg(Q)$$ we then have $$\textstyle P- \sum_{j=1}^{r}Q_jA_j=0$$, and we find the partial fraction decomposition dividing by $$\textstyle Q$$

Discussion
Hi Sławomir. I noticed your post just now, when I have already restored the section... Can you explain better your objections? --pm a 23:08, 20 November 2010 (UTC)


 * This approach is basically already redundant with the much more succinct section on computing the partial fraction decomposition via residues. I think the only substantial difference is that that section uses the Laurent polynomial rather than the Taylor polynomial, but it's the same underlying idea.  Anyway, I have moved it toward the end of the article for now, but I believe it should be possible to condense the entire section into a remark at the end of the residues section, or if not, then the residues section can be tweaked to encompass this point of view.   Sławomir Biały  (talk) 15:34, 21 November 2010 (UTC)

Well, I'm sorry to disagree. It seems we shall have to discuss a little; I hope you too will take this as a good opportunity of improvement.

Actually, as you can check, the residue approach in the article is not really "much more succint"; even because it contains no proofs, just some sentences, and so is the linked article. In fact, referring to residues this way may even turn out to be somehow misleading. Usually one computes a residue of a rational function by means of a partial fraction decomposition (as it has been said few lines above in the introduction, in a sentence about motivations), not the opposite. So, a naive reader may be confused, and may even believe that the machinery of the residue calculus be needed in order to compute explicitly a PFD. The limit formula ($$a_{ij}=\lim$$ etc), as it is, is little more than a tautology (how should one compute the limit?). In contrast, I'd like you to observe that the Taylor formula approach consists in a neat statement, with a self-contained proof of the existence and unicity of the decomposition. Also, please notice how more neat is the characterization of the numerator $$A_j(x)$$ in terms of a Taylor polynomial, collecting all terms $$ a_{ij}$$ corresponding to the pole $$\lambda_j,$$ instead of a statement and a formula about every single one of them. As to the Laurent polynomials or series, I preferred not to refer to them there, for the sake of simplicity.

Incidentally, note that the Taylor section currently has the only proof in the whole article, (thus, we can't say it's redundant). My suggestion is to put the Taylor theorem approach immediately after the division algorithm (they are quite linked indeed). The connections with the calculus of residued comes more naturally as an application, and as such I would find its natural position at the end of the article. On a minor side, things that we may try to improve in the rest of the article, IMO, are: here and there the style is a bit too much hand-waving for an encyclopedia; and three examples seems a bit more than enough. --pm a 20:17, 21 November 2010 (UTC)


 * Some of the run-up to the statement of the decomposition theorem in the very first section might be considered a "proof". (It's true that not all details are examined, but it would be quite easy to turn this into a completely rigorous proof.)  I'll have more to say on this subject later, but I need to go right now.  Sławomir Biały  (talk) 21:46, 21 November 2010 (UTC)


 * Yes, I agree on this. In fact, it is also not bad as it is, i.e. an introductory informal proof may be nice. --pm a 12:37, 22 November 2010 (UTC)


 * The limit sign is not tautological. For example, $$(x-a)\frac{1}{x-a}$$ is not defined at point a, and neither is its derivative. The limit sign is to ensure that the example always equals 1, even at points it normally does not exist. This should be no problem in calculation, but must be specified for the sake of rigor. And I totally disagree with "Usually one computes a residue of a rational function by means of a partial fraction decomposition (as it has been said few lines above in the introduction, in a sentence about motivations), not the opposite.". I have never come across a book that performs partial fraction before computing residues and neither does Residue.--Netheril96 (talk) 02:13, 22 November 2010 (UTC)


 * Yes, I said "little more than tautological", meaning: that formula, as it is written, has little content as to how to compute the residue in practice, and has also little or no content as to the issue of the existence of the decomposition. It assumes the existence of a decomposition, then identifies the coefficients as residues or limits. This is OK to me, but that was already contained in the Taylor section, and even better, in that that section also gives the existence of the decomposition, and adds a useful characterization of the components. Moreover, referring to the residues may result a little misleading unless one explains why they are necessary, and how to use them to make the computation and/or to prove the existence of a decomposition (at the moment, a  naive reader may think that it is required doing a path integration).  Also, I wrote "usually" meaning that e.g. computing the residue of 1/x(x+1) at x=0 one first makes an algebraic preparation writing it in the form a/x + b/(x+1) .  You disagree, that's ok, but you did not conclude the reasons of your objections. Could you explain what should be the connection between residue calculus and PFD in your opinion/experience? I do not think we disagree too much. --pm a  12:21, 22 November 2010 (UTC)

This is getting offtopic. The current state of the exposition is unsatisfactory. The ideal scenario, as I see it, is for the Taylor section and residues section to get merged, possibly under a more general heading. The Taylor section should be reduced and summarized (more explanation, fewer symbols). Ideally, a formal proof should not even be required; reference to the unicity of Taylor expansion should be sufficient if it is presented correctly. Sławomir Biały (talk) 01:17, 23 November 2010 (UTC)

OK, let's try to be essential. IMO there is a reason to keep the Taylor approach separated from a residue calculus: the Taylor formula is material of elementary courses (say, even good high schools), while the residue calculus is not (at least, a bit less). But a merging is possible, if conveniently done. As to the proof, my opinion is that short, self-contained proofs are a most valuable content of the maths articles of wikipedia. It is also thank to these quick references, that wikipedia gathers highly qualified and experienced mathematicians among the users. Anyway, either we discuss in a constructive way, or if you prefer, you can do the job yourself (I am sure you will do it at your best) and postpone a discussion at the end. Of course, if you prefer not to discuss at all, we may ask for other people opinion, or just put on the table our respective curricula (stupid things like published articles in the top worldwide maths journals, academic and visiting positions in the top worldwide maths institutions etc). Best, --pm a 10:22, 23 November 2010 (UTC)


 * I don't think there is any need to compare credentials. I trust that you are more than qualified to edit an article on elementary calculus topics (and much more).  I hope that this trust is mutual.  My concern over the section is that it delves into too much detail for something that is already proven by other means that are more elementary (using pure algebra) or more elegant (using residues) methods elsewhere.  There is, at any rate, consensus that proofs generally don't belong in articles.  Personally, I prefer to talk about the proof and say roughly how it goes in plain words rather than to go into details.  It's not a big deal though.  I'll try to simplify the treatment.   Sławomir Biały  (talk) 19:18, 3 April 2011 (UTC)


 * Sławomir, yes, there isn't any need to speak of credentials --but read the whole sentence that I wrote. I just had the impression that you refused the discussion from the beginning (maybe you think discussing couldn't be productive and just turn into a waste of time?). Look as you started:
 * ''I moved the following inscrutable and out-of-place section as well. It can be restored if there are strong feelings about it, but my sense is that the article is better off without it.


 * Anyway I appreciated that you made an effort to improve the form. To be honest, it does not seem better than before, but maybe there is still room for improving ;-) --pm a 20:50, 16 April 2011 (UTC)

Issue with notation
The recent edit highlights an issue with the notation in this article. Specifically, the divisor polynomial is denoted by Q(x) as in rational function, whereas in polynomial long division it is denoted by D(x), with Q(x) denoting the quotient polynomial. Can the notation be improved so that we get closer to P(x) = D(x)Q(x) + R(x)? Isheden (talk) 10:37, 15 February 2012 (UTC)

Example 4 might be wrong
Hello everyone, when comparing Example 4 with the results given by Wolfram Alpha, i noticed that the result is quite different. Since i'm absolutely no expert on this subject, could someone review this and correct the example? 134.28.202.206 (talk) 15:07, 10 March 2012 (UTC)
 * Wolfram Alpha gives the following partial fraction expansion: $$\frac{i}{z-i} - \frac{i}{z+i} - \frac{3}{2(z+1)} + \frac{3}{2(z-1)}$$ Isheden (talk) 20:00, 10 March 2012 (UTC)
 * Starting with the expression $$\frac{z^2+5}{(z^2-1)(z^2+1)}$$, the partial fraction expansion is $$-\frac{2}{z^2+1}-\frac{3}{2(z+1)}+\frac{3}{2(z-1)}$$. Isheden (talk) 20:38, 10 March 2012 (UTC)
 * It turned out that the numerator should have been $$z^2-5$$ instead of $$z^2+5$$. I've fixed it and expanded the example to clarify what idea it is based on. Isheden (talk) 21:49, 10 March 2012 (UTC)

Example 5 has bad reasoning leading to correct result.
The first example of incorrect reasoning is the following:
 * Therefore:


 * $$\lim_{x \to 1} \frac{1}{(x - 1)(x^2 + x + 1)} = \lim_{x \to 1}\frac{A}{x - 1},$$


 * and thus:


 * $$A = \lim_{x \to 1}{\frac{1}{x^2 + x + 1}} = \frac{1}{3}$$

This simply does not follow. For example, $$\lim_{x \to 0}{\frac{1}{x}} = \lim_{x \to 0}{\frac{2}{x}}$$. But $$ 1 \neq 2$$. From the "equality" of two limits which are equal by virtue of both being equal to infinity, we can conclude absolutely nothing!

The second case of bad reasoning in Example 5 is the following:


 * Therefore:


 * $$ \lim_{x \to \infty} \frac{A + B}{x} = \lim_{x \to \infty} \frac{1}{(x - 1)(x^2 + x + 1)} =0,$$


 * which implies A + B = 0 and so

From $$ \lim_{x \to \infty} \frac{A + B}{x} =0$$, one cannot conclude that   A + B = 0.

For example: $$ \lim_{x \to \infty} \frac{2}{x} =0$$, But $$ 2 \neq 0$$.

I believe I have a line of reasoning that corrects both of these problems. However, there are two problems with it:
 * 1) It is more complex and difficult to follow.
 * 2) It is my original work, and therefore not eligible to be put in Wikipedia.

Basically, the idea is instead of considering and comparing two separate expressions which are assumed to be equal, one considers the ratios of those expressions which should be 1. Someone please help! Pelliott11 (talk) 00:51, 22 May 2017 (UTC)

Why not simply say that it is the sum of all the principal parts of the Laurent expansions around each pole?
This is also the most efficient way to compute this, the ugly formula involving derivatives mentioned in the section about the residue method is not the most practical way to compute series expansions. Also note that if the degree of the numerator is larger than that of the numerator then the point at infinity is a pole, the principal part of the Laurent expansion around infinity is then precisely the polynomial you would get if you perform a division to reduce the degree of the numerator to below that of the denominator. Count Iblis (talk) 23:48, 5 April 2013 (UTC)
 * This could be said somewhere in the article, but this article should be understandable to people that know only polynomials and rational fractions and ignore the theory of analytic functions (poles, Laurent expansion, ....). Nevertheless, this article is badly written, and several sections, in particular that on the residue method, need to be completely rewritten. About your assertion on the ugly formula, I suppose that it is the formula for aij. I agree with you, that the general formula is not useful. But in case of a simple root of the denominator, the formula $$a_{i1}=P(\alpha)/Q'(\alpha)$$ is the simplest way to compute the "logarithmic part". D.Lazard (talk) 09:22, 6 April 2013 (UTC)
 * Yes, but then because we're dealing with rational functions only, you can formulate and prove this using only elementary means (a sketch of the proof is that if you subtract from a rational function its complete singular behavior for each singularity, the result doesn't have any singular behavior left anymore and must therefore be a polynomial. Then if the original rational function tends to zero at infinity, that polynomial must be identical to zero). So, one has the same result that all you need to do is expand around each singularity and that's something every first year student ought to be capable of. Count Iblis (talk) 16:15, 6 April 2013 (UTC)

Assessment comment
Substituted at 02:27, 5 May 2016 (UTC)

Decomposition over the Complex Numbers
I think the article is lacking proper explanation and examples on how to perform partial fraction decomposition over the Complex Numbers. In particular, Example 2 is not at all clear on how the values of D and E were derived from the initial decomposition. — Preceding unsigned comment added by Frikdt (talk • contribs) 08:49, 23 December 2016 (UTC)
 * There is nothing special with complex numbers, except that every denominator of the decomposition is a linear polynomial or a power of a linear polynomial. Nevertheless, I have edited example 2, for clarification. Note also that if the denominator of the input fraction is a product of distinct linear polynomials, the residue method is almost alway the simplest method for computing the decomposition. D.Lazard (talk) 09:38, 23 December 2016 (UTC)