Question

2x Find a power series for the function f(x) and its interval of convergence. - 3.x2 5x – 2

Answer 1

$\small \frac{{2x}}{{3{x^2} - 5x - 2}} = \frac{{2x}}{{(3x + 1)(x - 2)}} = \frac{A}{{x - 2}} + \frac{B}{{3x + 1}}\\ \\ A = \mathop {\lim }\limits_{x \to 2} \left( {(x - 2) \times \frac{{2x}}{{3{x^2} - 5x - 2}}} \right) = \mathop {\lim }\limits_{x \to 2} \left( {(x - 2) \times \frac{{2x}}{{(3x + 1)(x - 2)}}} \right)\\ \\ = \mathop {\lim }\limits_{x \to 2} \frac{{2x}}{{(3x + 1)}} = \frac{4}{7}$

Also

$\small B = \mathop {\lim }\limits_{x \to \frac{{ - 1}}{3}} \left( {\left( {x + \frac{1}{3}} \right) \times \frac{{2x}}{{3{x^2} - 5x - 2}}} \right)\\ \\ = \mathop {\lim }\limits_{x \to \frac{{ - 1}}{3}} \left( {\left( {x + \frac{1}{3}} \right) \times \frac{{2x}}{{(3x + 1)(x - 2)}}} \right)\\ \\ = \mathop {\lim }\limits_{x \to \frac{{ - 1}}{3}} \left( {\left( {x + \frac{1}{3}} \right) \times \frac{{2x}}{{3\left( {x + \frac{1}{3}} \right)(x - 2)}}} \right)\\ \\ = \mathop {\lim }\limits_{x \to \frac{{ - 1}}{3}} \frac{{2x}}{{3(x - 2)}} = \frac{{\frac{2}{3}}}{{3\left( {\frac{1}{3} + 2} \right)}} = \frac{2}{7}$

That is .

2х 3х2 – 5х – 2 2r (3x+1)(х – 2) E 1 7 – 2 + 2 1 73x +1

Now $\small \frac{1}{{x - 2}} = \left( { - \frac{1}{2}} \right)\left( {\frac{1}{{1 - \frac{x}{2}}}} \right) = \left( { - \frac{1}{2}} \right)\sum\limits_{n = 0}^\infty {{{\left( {\frac{x}{2}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\left( {\frac{{ - 1}}{{{2^{n + 1}}}}} \right){x^n}}$ .....(i)

which is valid if
<13<2
. That is converges if  $\small x\in (-2,2)$.

Also $\small \frac{1}{{3x + 1}} = \frac{1}{{1 - ( - 3x)}} = \sum\limits_{n = 0}^\infty {{{\left( { - 3x} \right)}^n}} = \sum\limits_{n = 0}^\infty {{{\left( { - 3} \right)}^n}{x^n}}$ .....(ii)

which is valid if $\small |3x|<1\to |x|<\frac{1}{3}$ .Also interval of convergence is $\small \left (-\frac{1}{3},\frac{1}{3} \right )$ .

Now So both series valid is $\small |3x|<1\to |x|<\frac{1}{3}$ . Also interval of convergence is $\small \left (-\frac{1}{3},\frac{1}{3} \right )$ .

Combinding (i) and (ii) we get $\small \frac{4}{7}\frac{1}{{x - 2}} + \frac{2}{7}\frac{1}{{3x + 1}} = \frac{4}{7}\sum\limits_{n = 0}^\infty {\frac{{ - 1}}{{{{\left( 2 \right)}^{n + 1}}}}{x^n}} + \frac{2}{7}\sum\limits_{n = 0}^\infty {{{\left( { - 3} \right)}^n}{x^n}} = \sum\limits_{n = 0}^\infty {\frac{2}{7}\left( {\frac{{ - 1}}{{{{\left( 2 \right)}^n}}} + {{\left( { - 3} \right)}^n}} \right){x^n}}$ .

That is $\small \frac{{2x}}{{3{x^2} - 5x - 2}} = \sum\limits_{n = 0}^\infty {\frac{2}{7}\left( {\frac{{ - 1}}{{{{\left( 2 \right)}^n}}} + {{\left( { - 3} \right)}^n}} \right){x^n}}$ valid if $\small |x|<\frac{1}{3}$ . Also interval of convergence is $\small \left (-\frac{1}{3},\frac{1}{3} \right )$ .