Blog: `x/(16x^4 - 1)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Thursday, August 6, 2009

$\displaystyle\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}$

Let's factorize the denominator,

$\displaystyle{16}{{x}}^{{4}}-{1}={{\left({4}{{x}}^{{2}}\right)}}^{{2}}-{1}$

$\displaystyle={\left({4}{{x}}^{{2}}+{1}\right)}{\left({4}{{x}}^{{2}}-{1}\right)}$

$\displaystyle={\left({4}{{x}}^{{2}}+{1}\right)}{\left({2}{x}+{1}\right)}{\left({2}{x}-{1}\right)}$

Let $\displaystyle\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}=\frac{{A}}{{{2}{x}+{1}}}+\frac{{B}}{{{2}{x}-{1}}}+\frac{{{C}{x}+{D}}}{{{4}{{x}}^{{2}}+{1}}}$

$\displaystyle\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}=\frac{{{A}{\left({2}{x}-{1}\right)}{\left({4}{{x}}^{{2}}+{1}\right)}+{B}{\left({2}{x}+{1}\right)}{\left({4}{{x}}^{{2}}+{1}\right)}+{\left({C}{x}+{D}\right)}{\left({2}{x}+{1}\right)}{\left({2}{x}-{1}\right)}}}{{{\left({2}{x}+{1}\right)}{\left({2}{x}-{1}\right)}{\left({4}{{x}}^{{2}}+{1}\right)}}}$

$\displaystyle\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}=\frac{{{A}{\left({8}{{x}}^{{3}}+{2}{x}-{4}{{x}}^{{2}}-{1}\right)}+{B}{\left({8}{{x}}^{{3}}+{2}{x}+{4}{{x}}^{{2}}+{1}\right)}+{\left({C}{x}+{D}\right)}{\left({4}{{x}}^{{2}}-{1}\right)}}}{{{\left({2}{x}+{1}\right)}{\left({2}{x}-{1}\right)}{\left({4}{{x}}^{{2}}+{1}\right)}}}$

$\displaystyle\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}=\frac{{{A}{\left({8}{{x}}^{{3}}-{4}{{x}}^{{2}}+{2}{x}-{1}\right)}+{B}{\left({8}{{x}}^{{3}}+{4}{{x}}^{{2}}+{2}{x}+{1}\right)}+{4}{C}{{x}}^{{3}}-{C}{x}+{4}{D}{{x}}^{{2}}-{D}}}{{{\left({2}{x}+{1}\right)}{\left({2}{x}-{1}\right)}{\left({4}{{x}}^{{2}}+{1}\right)}}}$

$\displaystyle\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}=\frac{{{{x}}^{{3}}{\left({8}{A}+{8}{B}+{4}{C}\right)}+{{x}}^{{2}}{\left(-{4}{A}+{4}{B}+{4}{D}\right)}+{x}{\left({2}{A}+{2}{B}-{C}\right)}-{A}+{B}-{D}}}{{{\left({2}{x}+{1}\right)}{\left({2}{x}-{1}\right)}{\left({4}{{x}}^{{2}}+{1}\right)}}}$

$\displaystyle\therefore{x}={{x}}^{{3}}{\left({8}{A}+{8}{B}+{4}{C}\right)}+{{x}}^{{2}}{\left(-{4}{A}+{4}{B}+{4}{D}\right)}+{x}{\left({2}{A}+{2}{B}-{C}\right)}-{A}+{B}-{D}$

equating the coefficients of the like terms,

$\displaystyle{8}{A}+{8}{B}+{4}{C}={0}$ ----- equation 1

$\displaystyle-{4}{A}+{4}{B}+{4}{D}={0}$ ----- equation 2

$\displaystyle{2}{A}+{2}{B}-{C}={1}$ ----- equation 3

$\displaystyle-{A}+{B}-{D}={0}$ ------ equation 4

Now we have to solve the above four equations to find the solutions of A,B,C and D.

From equation 1,

$\displaystyle{4}{\left({2}{A}+{2}{B}+{C}\right)}={0}$

$\displaystyle{2}{A}+{2}{B}+{C}={0}$

Subtract equation 3 from the above equation,

$\displaystyle{\left({2}{A}+{2}{B}+{C}\right)}-{\left({2}{A}+{2}{B}-{C}\right)}={0}-{1}$

$\displaystyle{2}{C}=-{1}$

$\displaystyle{C}=-\frac{{1}}{{2}}$

From equation 2,

$\displaystyle{4}{\left(-{A}+{B}+{D}\right)}={0}$

$\displaystyle-{A}+{B}+{D}={0}$

Now subtract equation 4 from the above equation,

$\displaystyle{\left(-{A}+{B}+{D}\right)}-{\left(-{A}+{B}-{D}\right)}={0}$

$\displaystyle{2}{D}={0}$

$\displaystyle{D}={0}$

Now plug in the values of C in the equation 3,

$\displaystyle{2}{A}+{2}{B}-{\left(-\frac{{1}}{{2}}\right)}={1}$

$\displaystyle{2}{A}+{2}{B}+\frac{{1}}{{2}}={1}$

$\displaystyle{2}{A}+{2}{B}={1}-\frac{{1}}{{2}}$

$\displaystyle{2}{\left({A}+{B}\right)}=\frac{{1}}{{2}}$

$\displaystyle{A}+{B}=\frac{{1}}{{4}}$ ----- equation 5

Plug in the value of D in the equation 4,

$\displaystyle-{A}+{B}-{0}={0}$

$\displaystyle-{A}+{B}={0}$ ---- equation 6

Now add the equations 5 and 6,

$\displaystyle{2}{B}=\frac{{1}}{{4}}$

$\displaystyle{B}=\frac{{1}}{{8}}$

Plug in the value of B in the equation 6,

$\displaystyle-{A}+\frac{{1}}{{8}}-{0}$

$\displaystyle{A}=\frac{{1}}{{8}}$

$\displaystyle\therefore\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}=\frac{{\frac{{1}}{{8}}}}{{{2}{x}+{1}}}+\frac{{\frac{{1}}{{8}}}}{{{2}{x}-{1}}}+\frac{{{\left(-\frac{{1}}{{2}}\right)}{x}}}{{{4}{{x}}^{{2}}+{1}}}$

$\displaystyle\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}=\frac{{1}}{{{8}{\left({2}{x}+{1}\right)}}}+\frac{{1}}{{{8}{\left({2}{x}-{1}\right)}}}-\frac{{x}}{{{2}{\left({4}{{x}}^{{2}}+{1}\right)}}}$

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Thursday, August 6, 2009

$\displaystyle\frac{{x}}{{{16}{{x}}^{{4}}-{1}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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