आंशिक भिन्न

बीजगणित में, आंशिक भिन्न प्रसार (partial fraction expansion) एक विधि है जो किसी परिमेय भिन्न के अंश या हर के डेग्री (degree) को कम करने के काम आती है।

सांकेतिक रूप में, निम्नलिखित परिमेय भिन्न को आंशिक भिन्नों में तोड़ा जा सकता है-

 

जहाँ ƒ और g बहुपद (polynomials) है। इसके आंशिक भिन्न निम्नवत होंगे-

 

जहाँ gj (x) बहुपद हैं और ये g(x) के गुणखण्ड हैं। 

उदाहरण -

को आंशिक भिन्नों में बदलकर निम्नलिखित प्रकार से भी लिखा जा सकता है-

 विधि 
माना दिया हुआ भिन्न  है तो:

विधि 1

जब दिये हुए भिन्न के हर को  जैसे रैखिक गुणनखण्ड हो सकें ; जहाँ n >=1

विधि 2

जब दिये हुए भिन्न के हर का रैखिक गुणनखण्ड न हो बल्कि  जैसे द्विघात गुणखण्ड हो (जहाँ n >= 1) :

उदाहरण
 उदाहरण १ 

 

Here, the denominator splits into two distinct linear factors:

 

so we have the partial fraction decomposition

 

Multiplying through by x2 + 2x − 3, we have the polynomial identity

 

Substituting x = −3 into this equation gives A = −1/4, and substituting x = 1 gives B = 1/4, so that

 

 उदाहरण २ 

 

After long-division, we have

 

Since (−4)2 − 4×8 = −16 < 0, the factor x2 − 4x + 8 is irreducible, and the partial fraction decomposition over the reals has the shape

 

Multiplying through by x3 − 4x2 + 8x, we have the polynomial identity

 

Taking x = 0, we see that 16 = 8A, so A = 2. Comparing the x2 coefficients, we see that 4 = A + B = 2 + B, so B = 2. Comparing linear coefficients, we see that −8 = −4A + C = −8 + C, so C = 0. Altogether,

 

The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system.

 उदाहरण ३ 

 

After long-division and factoring the denominator, we have

 

The partial fraction decomposition takes the form

 

Multiplying through by (x − 1)3(x2 + 1)2 we have the polynomial identity

 

Taking x = 1 gives 4 = 4C, so C = 1. Similarly, taking x = i gives 2 + 2i = (Fi + G)(2 + 2i), so Fi + G = 1, so F = 0 and G = 1 by equating real and imaginary parts. With C = G = 1 and F = 0, taking x = 0 we get A − B + 1 − E − 1 = 0, thus E = A − B.

We now have the identity

 

Expanding and sorting by exponents of x we get

 

We can now compare the coefficients and see that

 

with A = 2 − D and −A −3 D =−4 we get A = D = 1 and so B = 0, furthermore is C = 1, E = A − B = 1, F = 0 and G = 1.

The partial fraction decomposition of ƒ(x) is thus

 

Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at x=1 and at x=i in the above polynomial identity. (To this end, recall that the derivative at x=a of (x−a)mp(x) vanishes if m > 1 and it is just p(a) if m=1.)
Thus, for instance the first derivative at x=1 gives 
 
that is 8  =  4B + 8 so B=0.

उदाहरण ४ (residue method)

Thus, f(z) can be decomposed into rational functions whose denominators are z+1, z−1, z+i, z−i. Since each term is of power one, −1, 1, −i and i are simple poles.

Hence, the residues associated with each pole, given by
,
are 
,
respectively, and

.

उदाहरण ५ (limit method)

Limits can be used to find a partial fraction decomposition.

First, factor the denominator:

The decomposition takes the form of

As , the A term dominates, so the right-hand side approaches . Thus, we have

As , the right-hand side is

Thus, .

At , . Therefore, .

The decomposition is thus .

सन्दर्भ

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