"आंशिक भिन्न": अवतरणों में अंतर
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अनुनाद सिंह (वार्ता | योगदान) |
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पंक्ति 30:
<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''<sup>2</sup> + 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)<sup>2</sup> − 4×8 = −16 < 0, the factor ''x''<sup>2</sup> − 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''<sup>3</sup> − 4''x''<sup>2</sup> + 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''<sup>2</sup> 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)<sup>3</sup>(''x''<sup>2</sup> + 1)<sup>2</sup> 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)<sup>m</sup>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
: <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|year=1984|publisher=Springer-Verlag|location=New York|pages=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>.
== बाहरी कड़ियाँ ==
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