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-

विधिसंपादित करें

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) :


उदाहरणसंपादित करें

उदाहरण १संपादित करें


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 २संपादित करें


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 ३संपादित करें


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 AB + 1 − E − 1 = 0, thus E = AB.

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 = AB = 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)संपादित करें


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




respectively, and


उदाहरण ५ (limit method)संपादित करें

Limits can be used to find a partial fraction decomposition.[1]


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  .

सन्दर्भसंपादित करें

  1. Bluman, George W. (1984). Problem Book for First Year Calculus. New York: Springer-Verlag. पपृ॰ 250–251.

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