How do you find a power series converging to #f(x)=sinx/x# and determine the radius of convergence?

Question

Isabella Ackerman · Accepted Answer

#sinx/x = sum_(n=0)^oo (-1)^n x^(2n)/((2n+1)!)#

with radius of convergence #R=oo#.

Start from the MacLaurin series for #sin x#: #sinx = sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!)# We can divide by #x# term by term: #sinx/x = sum_(n=0)^oo (-1)^n x^(2n)/((2n+1)!)# and determine the radius of convergence using the ratio test: #lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs( (x^(2(n+1))/((2(n+1)+1)!))/( x^(2n)/((2n+1)!)# #lim_(n->oo) abs(a_(n+1)/a_n) = abs((x^(2n+2))/x^(2n)) ((2n+1)!)/((2n+3)!)# #lim_(n->oo) abs(a_(n+1)/a_n) = x^2/((2n+3)(2n+2)) = 0# So the series is convergent for every #x#.

Elijah Abram · Answer

undefined To find a power series converging to $ f(x) = \frac{\sin x}{x} $ and determine the radius of convergence, you can use Maclaurin series expansion. The Maclaurin series for $ \frac{\sin x}{x} $ is given by:

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

To determine the radius of convergence, use the ratio test:

$$ R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| $$

where $ a_{n} $ is the nth term of the series.

In this case, $ a_{n} = \frac{(-1)^n x^{2n}}{(2n + 1)!} $.

Apply the ratio test and find the limit as $ n $ approaches infinity:

$$ R = \lim_{n \to \infty} \left| \frac{(-1)^n x^{2n}}{(2n + 1)!} \cdot \frac{(2n - 1)!}{(-1)^{n+1} x^{2(n+1)}} \right| $$

$$ R = \lim_{n \to \infty} \left| \frac{x^2}{(2n + 1)(2n + 2)} \right| $$

$$ R = |x| \lim_{n \to \infty} \frac{1}{\sqrt{n(n + 1)}} $$

$$ R = |x| \cdot 0 $$

$$ R = 0 $$

Therefore, the radius of convergence is $ R = 0 $, which means the series converges only at $ x = 0 $.