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

Sunday, January 19, 2014

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

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

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

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

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,

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

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

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

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

Blog

Sunday, January 19, 2014

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

No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Sunday, January 19, 2014

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

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

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?