Question

(a) Use differentiation to find a power series representation forf(x) =1(6 + x)2.f(x) =∞ leftparen1.gif(−1)n(n+1)xn6n+2​ rightparen1.gifsum.gifn = 0What is the radius of convergence, R?R = (b) Use part (a) to find a power series forf(x) =1(6 + x)3.f(x) =∞ leftparen1.gif(−1)n(n+3)(n+1)xn6n+5​ rightparen1.gifsum.gifn = 0What is the radius of convergence, R?R = (c) Use part (b) to find a power series forf(x) =x2(6 + x)3.f(x) =∞ leftparen1.gif(−1)n(n+2)(n+1)xn+26n+3​ rightparen1.gifsum.gifn = 2What is the radius of convergence, R?R =

Answer 1

(a) Wild guess:

$f(x)=\dfrac1{(6+x)^2$

Recall the power series

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

With some manipulation, we can write

$\displaystyle\frac1{6+x}=\frac16\frac1{1-\left(-\frac x6\right)}=\frac16\sum_{n=0}^\infty\left(-\frac x6\right)^n=\sum_{n=0}^\infty\frac{(-x)^n}{6^{n+1}}$

Take the derivative and we get

$\displaystyle-\frac1{(6+x)^2}=-\sum_{n=0}^\infty\frac{n(-x)^{n-1}}{6^{n+1}}$

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

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

so we have

$\displaystyle\frac1{(6+x)^2}=\sum_{n=0}^\infty\frac{(n+1)(-x)^n}{6^{n+2}}$

By the ratio test, this series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+2)(-x)^{n+1}}{6^{n+3}}}{\frac{(n+1)(-x)^n}{6^{n+2}}}\right|=\left|\frac x6\right|\lim_{n\to\infty}\frac{n+2}{n+1}=\left|\frac x6\right|<1$

or $|x|<6$, so that the radius of convergence is $R=6$.

(b). If we take the second derivative, we get

$\displaystyle\frac2{(6+x)^3}=\sum_{n=0}^\infty\frac{n(n+1)(-x)^{n-1}}{6^{n+2}}$

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

$\displaystyle=\sum_{n=0}^\infty\frac{(n+1)(n+2)(-x)^n}{6^{n+3}}$

$\displaystyle\frac1{(6+x)^3}=\frac12\sum_{n=0}^\infty\frac{(n+1)(n+2)(-x)^n}{6^{n+3}}$

Apply the ratio test again and we get $R=6$.

(c) Multiply the previous series by $x^2$ and we get

$\displaystyle\frac{x^2}{(6+x)^3}=\frac12\sum_{n=0}^\infty\frac{(n+1)(n+2)(-x)^nx^2}{6^{n+3}}$

$\displaystyle=\frac12\sum_{n=0}^\infty\frac{(n+1)(n+2)(-1)^nx^{n+2}}{6^{n+3}}$

The ratio test yet again tells us $R=6$.