Question

15.

Find a power series representation for the function. х f(x) (1 + 6x)2 f(x) = ( (-6).*- 1 nxt n = 0 x Determine the radius of convergence, R. R = 1/6

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16

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17

Answer 1

15)

Given

$f(x)=\frac{x}{(1+6x)^2}$

Power series of is given by 1 -

$\frac{1}{1-x} =\sum_{n=0}^\infty x^n$

$\text{ Differentiating with respect to x on both sides }$

d dir (1) - 1 1 – 1 d dc Σ n=0

$=> \frac{1}{(1-x)^2} = \sum_{n=0}^\infty \frac{\mathrm{d} }{\mathrm{d} x}( x^n)$

$=> \frac{1}{(1-x)^2} = \sum_{n=0}^\infty nx^{n-1}$

$\text{ Replacing } x\text{ by }-6x$

=> 1 (1 – (-6.x))2 Ση(-62)n-1 n=0

$=> \frac{1}{(1+6x)^2} = \sum_{n=0}^\infty n (-6)^{n-1}x^{n-1}$

$=> \frac{1}{(1+6x)^2} = \sum_{n=0}^\infty n (-1)^{n-1}\:\: 6^{n-1}x^{n-1}$

$\text{ Power series of }f(x)$

$f(x)= \frac{x}{(1+6x)^2} = x \left ( \sum_{n=0}^\infty n (-1)^{n-1}\:\: 6^{n-1}x^{n-1} \right )$

$=>f(x)= \sum_{n=0}^\infty n (-1)^{n-1}\:\: 6^{n-1}x^{n-1+1}$

$=>f(x)= \sum_{n=0}^\infty n (-1)^{n-1}\:\: 6^{n-1}x^{n}$

$\underline{\text{ Calculating the radius of convergence, R}}$

$\text{ nth term of series, }\:\:\:\:\:a_n= n (-1)^{n-1}\:\: 6^{n-1}x^{n}$

$\text{ (n+1)th term of series, }\:\:\:\:\:a_{n+1}= (n+1) (-1)^{n+1-1}\:\: 6^{n+1-1}x^{n+1}$

  

$\text{ Calculating, } L = \lim_{n\to\infty}\left | \frac{a_{n+1}}{a_n} \right |$

$L = \lim_{n\to\infty}\left | \frac{a_{n+1}}{a_n} \right |= \lim_{n\to\infty} \left | \frac{(n+1) (-1)^{n}\:\: 6^{n}x^{n+1}}{n (-1)^{n-1}\:\: 6^{n-1}x^{n}}\right |$

$=> L = \lim_{n\to\infty} \left | \frac{(n+1) (-1)\:\: 6x}{n } \right |=|6x| \lim_{n\to\infty} \left( \frac{n+1 }{n } \right )$

$=> L=|6x| \lim_{n\to\infty} \left(1+ \frac{1 }{n } \right ) =|6x| (1+0)$

$\text{ By ratio test, series will converge if }L<1$

$=>6|x|<1$

$=>|x|<\frac{1}{6}$

$=>\text{Radius of convergence, }R =\frac{1}{6}$

Hence,

$f(x)= \sum_{n=0}^\infty n (-1)^{n-1}\:\: 6^{n-1}x^{n}$

$\text{Radius of convergence, }R =\frac{1}{6}$