Talk:Partial fraction

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[edit] Complex Roots

Shouldn't this article include something about how to deal with terms that have complex/imaginary roots? Damn complex roots fucked me over on the exam....

[edit] Z-domain

This page needs some talk of PFD of transfer functions in the z-domain—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)

[edit] 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)

[edit] 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. — MFH: Talk 18:28, 26 May 2005 (UTC)


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

[edit] 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

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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)

[edit] 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[1]. 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)}.[2]

[edit] 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.

[edit] 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}[3]

[edit] 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.[4]

[edit] 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.[5]

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)}


[edit] 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)}

[edit] References

  1. ^ Vedic Mathematics: Sixteen Simple Mathematical Formulae from the Vedas, by Swami Sankaracarya (1884-1960), Motilal Banarsidass Indological Publishers and Booksellers, Varnasi, India, 1965; reprinted in Delhi, India, 1975, 1978. 367 pages.
  2. ^ Page 188, Vedic Mathematics
  3. ^ Page 186, Vedic Mathematics
  4. ^ Pages 188-189, Vedic Mathematics
  5. ^ Page 189, Vedic Mathematics
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