How do you find a power series converging to #f(x)=e^(x/2)# and determine the radius of convergence?

Question

Ezra Adams · Accepted Answer

#e^(x/2) = sum_(n=0)^oo x^n/(2^n(n!))#

with radius of convergence #R=oo# Start from the MacLaurin expansion of #e^t#. Since: #d^n/(dt^n) e^t = e^t# #[d^n/(dt^n) e^t]_(t=0) = e^0 = 1# we have that: #e^t = sum_(n=0)^oo t^n/(n!)# Using the ratio test: #lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( ( t^(n+1)/((n+1)!))/(t^n/(n!))) = lim_(n->oo) abs t /(n+1) = 0# so the series is convergent for any #t in RR#, that is it has radius of convergence #R=oo# So, substituting #t=x/2#: #e^(x/2) = sum_(n=0)^oo (x/2)^n/(n!) = sum_(n=0)^oo x^n/(2^n(n!))# still with radius of convergence #R=oo#

Aurora Alvarado · Answer

undefined To find a power series converging to $ f(x) = e^{x/2} $ and determine the radius of convergence, we can use the Taylor series expansion of $ e^x $ centered at $ x = 0 $. Then, we'll substitute $ x/2 $ for $ x $. The Taylor series expansion of $ e^x $ is:

$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$

Now, substituting $ x/2 $ for $ x $, we get:

$$ e^{x/2} = \sum_{n=0}^{\infty} \frac{(x/2)^n}{n!} $$

This simplifies to:

$$ e^{x/2} = \sum_{n=0}^{\infty} \frac{x^n}{2^n \cdot n!} $$

This is the power series representation of $ f(x) = e^{x/2} $. To determine the radius of convergence, we'll use the ratio test. Let $ a_n = \frac{x^n}{2^n \cdot n!} $. Applying the ratio test:

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

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

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

This limit converges to zero for all $ x $, which means the radius of convergence is infinite, or in other words, the power series converges for all real numbers $ x $.