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Adding 1 book for सत्यापनीयता) #IABot (v2.0.7) (GreenC bot

नया पृष्ठ

[[बीजगणित]] में, '''आंशिक भिन्न प्रसार''' (partial fraction expansion) एक विधि है जो किसी [[परिमेय भिन्न]] के [[अंश]] या [[भिन्न|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-

: <math> \frac{f(x)}{g(x)} </math>

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

: <math> \sum_j \frac{f_j(x)}{g_j(x)} </math>

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

;उदाहरण -

:<math>{25 \over (x+2)(x^2+1)^2} </math>को आंशिक भिन्नों में बदलकर निम्नलिखित प्रकार से भी लिखा जा सकता है-

:<math>\frac{1}{x+2} + \frac{-x+2}{x^2+1}+\frac{-5x+10}{(x^2+1)^2} \ . </math>

== विधि ==
माना दिया हुआ भिन्न <math>R(s)=\frac{P(s)}{Q(s)}</math> है तो:

;विधि 1

जब दिये हुए भिन्न के [[भिन्न|हर]] को <math>x-a</math> जैसे रैखिक गुणनखण्ड हो सकें ; जहाँ n >=1

<math>R(s)=\frac{P(s)}{(x-a)^n}=\frac{A1}{(x-a)}+\frac{A2}{(x-a)^2}+...+\frac{An}{(x-a)^n}</math>

;विधि 2

जब दिये हुए भिन्न के हर का रैखिक गुणनखण्ड न हो बल्कि <math>(x-a)^2+b^2</math> जैसे द्विघात गुणखण्ड हो (जहाँ n >= 1) :

<math>R(s)=\frac{P(s)}{[(x-a)^2+b^2]^n}=\frac{A1}{[(x-a)^2+b^2]}+\frac{A2}{[(x-a)^2+b^2]^2}+...+\frac{An}{[(x-a)^2+b^2]^n}</math>

==उदाहरण==
=== उदाहरण १ ===

: <math>f(x)=\frac{1}{x^2+2x-3}</math>

Here, the denominator splits into two distinct linear factors:

: <math>q(x)=x^2+2x-3=(x+3)(x-1)</math>

so we have the partial fraction decomposition

: <math>f(x)=\frac{1}{x^2+2x-3} =\frac{A}{x+3}+\frac{B}{x-1}</math>

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

: <math>1=A(x-1)+B(x+3)</math>

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

: <math>f(x) =\frac{1}{x^2+2x-3} =\frac{1}{4}\left(\frac{-1}{x+3}+\frac{1}{x-1}\right)</math>

'''

=== उदाहरण २ ===

: <math>f(x)=\frac{x^3+16}{x^3-4x^2+8x}</math>

After [[Polynomial long division|long-division]], we have

: <math>f(x)=1+\frac{4x^2-8x+16}{x^3-4x^2+8x}=1+\frac{4x^2-8x+16}{x(x^2-4x+8)}</math>

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

: <math>\frac{4x^2-8x+16}{x(x^2-4x+8)}=\frac{A}{x}+\frac{Bx+C}{x^2-4x+8}</math>

Multiplying through by ''x''3 − 4''x''2 + 8''x'', we have the polynomial identity

: <math>4x^2-8x+16 = A(x^2-4x+8)+(Bx+C)x</math>

Taking ''x'' = 0, we see that 16 = 8''A'', so ''A'' = 2. Comparing the ''x''2 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,

: <math>f(x)=1+2\left(\frac{1}{x}+\frac{x}{x^2-4x+8}\right)</math>

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

=== उदाहरण ३ ===

: <math>f(x)=\frac{x^9-2x^6+2x^5-7x^4+13x^3-11x^2+12x-4}{x^7-3x^6+5x^5-7x^4+7x^3-5x^2+3x-1}</math>

After [[Polynomial long division|long-division]] and [[polynomial factorization|factoring]] the denominator, we have

: <math>f(x)=x^2+3x+4+\frac{2x^6-4x^5+5x^4-3x^3+x^2+3x}{(x-1)^3(x^2+1)^2}</math>

The partial fraction decomposition takes the form

: <math>\frac{2x^6-4x^5+5x^4-3x^3+x^2+3x}{(x-1)^3(x^2+1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1)^2}</math>

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

: <math>

\begin{align}
& {} \quad 2x^6-4x^5+5x^4-3x^3+x^2+3x \\
& =A(x-1)^2(x^2+1)^2+B(x-1)(x^2+1)^2+C(x^2+1)^2+(Dx+E)(x-1)^3(x^2+1)+(Fx+G)(x-1)^3
\end{align}
</math>

Taking ''x'' = 1 gives 4 = 4''C'', so ''C'' = 1. Similarly, taking ''x'' = [[complex number|''i'']] gives 2 + 2''i'' = (''Fi'' + ''G'')(2 + 2''i''), so ''Fi'' + ''G'' = 1, so ''F'' = 0 and ''G'' = 1 by equating real and [[complex number|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

: <math>
\begin{align}
& {} 2x^6-4x^5+5x^4-3x^3+x^2+3x \\
& = A(x-1)^2(x^2+1)^2+B(x-1)(x^2+1)^2+(x^2+1)^2+(Dx+(A-B))(x-1)^3(x^2+1)+(x-1)^3 \\
& = A((x-1)^2(x^2+1)^2 + (x-1)^3(x^2+1)) + B((x-1)(x^2+1) - (x-1)^3(x^2+1)) + (x^2+1)^2 + Dx(x-1)^3(x^2+1)+(x-1)^3
\end{align}
</math>

Expanding and sorting by exponents of x we get

: <math>
\begin{align}
& {} 2 x^6 -4 x^5 +5 x^4 -3 x^3 + x^2 +3 x \\
& = (A + D) x^6 + (-A - 3D) x^5 + (2B + 4D + 1) x^4 + (-2B - 4D + 1) x^3 + (-A + 2B + 3D - 1) x^2 + (A - 2B - D + 3) x

\end{align}
</math>

We can now compare the coefficients and see that

: <math>
\begin{align}
A + D &=& 2 \\
-A - 3D &=& -4 \\
2B + 4D + 1 &=& 5 \\
-2B - 4D + 1 &=& -3 \\
-A + 2B + 3D - 1 &=& 1 \\
A - 2B - D + 3 &=& 3 ,
\end{align}
</math>

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

: <math>f(x)=x^2+3x+4+\frac{1}{(x-1)} + \frac{1}{(x - 1)^3} + \frac{x + 1}{x^2+1}+\frac{1}{(x^2+1)^2}.</math>

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
: <math> 2\cdot6-4\cdot5+5\cdot4-3\cdot3+2+3 = A\cdot(0+0) + B\cdot( 4+ 0) + 8 + D\cdot0 </math>
that is ''8 = 4B + 8'' so ''B=0''.

===उदाहरण ४ (residue method)===

:<math> f(z)=\frac{z^{2}-5}{(z^2-1)(z^2+1)}=\frac{z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}</math>

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
:<math>\frac{P(z_i)}{Q'(z_i)} = \frac{z_i^2 - 5}{4z_i^3}</math>,
are
:<math> 1, -1, \tfrac{3i}{2}, -\tfrac{3i}{2}</math>,
respectively, and

:<math> f(z)=\frac{1}{z+1}-\frac{1}{z-1}+\frac{3i}{2}\frac{1}{z+i}-\frac{3i}{2}\frac{1}{z-i}</math>.

===उदाहरण ५ (limit method)===

[[Limit (mathematics)|Limits]] can be used to find a partial fraction decomposition.<ref>{{cite book|last=Bluman|first=George W.|title=Problem Book for First Year Calculus|url=https://archive.org/details/problembookforfi00blum|year=1984|publisher=Springer-Verlag|location=New York|pages=[https://archive.org/details/problembookforfi00blum/page/n265 250]–251}}</ref>

:<math>f(x) = \frac{1}{x^3 - 1}</math>

First, factor the denominator:

:<math>f(x) = \frac{1}{(x - 1)(x^2 + x + 1)}</math>

The decomposition takes the form of

:<math>\frac{1}{(x-1)(x^2+x+1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}</math>

As <math>x \to 1</math>, the ''A'' term dominates, so the right-hand side approaches <math>\frac{A}{x - 1}</math>. Thus, we have

:<math>\frac{1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1}</math>
:<math>A = \lim_{x \to 1}{\frac{1}{x^2 + x + 1}} = \frac{1}{3}</math>

As <math>x \to \infty</math>, the right-hand side is

:<math>\lim_{x \to \infty}{\frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}} = \frac{A}{x} + \frac{Bx}{x^2} = \frac{A + B}{x}.</math>

:<math>\frac{A + B}{x} = \lim_{x \to \infty}{\frac{1}{x^3 - 1}} = 0</math>

Thus, <math>B = -\frac{1}{3}</math>.

At <math>x=0</math>, <math>-1 = -A + C</math>. Therefore, <math>C = -\frac{2}{3}</math>.

The decomposition is thus <math>\frac{\frac{1}{3}}{x - 1} + \frac{-\frac{1}{3}x - \frac{2}{3}}{x^2 + x + 1}</math>.

==सन्दर्भ==
{{टिप्पणीसूची}}

== बाहरी कड़ियाँ ==
* {{MathWorld |urlname=PartialFractionDecomposition |title=Partial Fraction Decomposition}}

[[श्रेणी:बीजगणित]]
[[श्रेणी:प्रारम्भिक बीजगणित]]

आंशिक भिन्न - अवतरण इतिहास

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