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 "g" ("x") बहुपद हैं और ये "g"("x") के गुणखण्ड हैं। माना दिया हुआ भिन्न formula_5 है तो:जब दिये हुए भिन्न के हर को formula_6 जैसे रैखिक गुणनखण्ड हो सकें ; जहाँ n >=1
formula_7जब दिये हुए भिन्न के हर का रैखिक गुणनखण्ड न हो बल्कि formula_8 जैसे द्विघात गुणखण्ड हो (जहाँ n >= 1) :
formula_9
Here, the denominator splits into two distinct linear factors:
so we have the partial fraction decomposition
Multiplying through by "x" + 2"x" − 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
After long-division, we have
Since (−4) − 4×8 = −16 < 0, the factor "x" − 4"x" + 8 is irreducible, and the partial fraction decomposition over the reals has the shape
Multiplying through by "x" − 4"x" + 8"x", we have the polynomial identity
Taking "x" = 0, we see that 16 = 8"A", so "A" = 2. Comparing the "x" coefficients, we see that 4 = "A" + "B" = 2 + "B", so "B" = 2. Comparing linear coefficients, we see that −8 = −4"A" + "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.
After long-division and factoring the denominator, we have
The partial fraction decomposition takes the form
Multiplying through by ("x" − 1)("x" + 1) we have the polynomial identity
Taking "x" = 1 gives 4 = 4"C", so "C" = 1. Similarly, taking "x" = "i" gives 2 + 2"i" = ("Fi" + "G")(2 + 2"i"), 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)p(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".
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
Limits can be used to find a partial fraction decomposition.
First, factor the denominator:
The decomposition takes the form of
As formula_36, the "A" term dominates, so the right-hand side approaches formula_37. Thus, we have
As formula_40, the right-hand side is
Thus, formula_43.
At formula_44, formula_45. Therefore, formula_46.
The decomposition is thus formula_47.