How do you find a power series converging to #f(x)=arcsin(x^3)# and determine the radius of convergence?

Question

Charles Anctil · Accepted Answer

#arcsin(x^3) = sum_(n=0)^oo ((2n)!)/(2^(2n)(n!)^2) x^(6n+3)/(2n+1)#

with radius of convergence #R=1# Start from the binomial series: #1/sqrt(1-t) = (1-t)^(-1/2) = sum_(n=0)^oo -1/2(-1/2-1)...(-1/2-n+1)(-t)^n/(n!)# simplifying: #1/sqrt(1-t) = (1-t)^(-1/2) = sum_(n=0)^oo -1/2(-3/2)...(-(2n-1)/2)(-t)^n/(n!)# #1/sqrt(1-t) = sum_(n=0)^oo (-1)^n (1*3* (2n-1))/(2^n(n!))(-t)^n# #1/sqrt(1-t) = sum_(n=0)^oo ((2n-1)!!)/(2^n(n!))t^n# Applying the ratio test we see that: #lim_(n->oo) abs(a_(n+1)/a_n ) = lim_(n->oo)abs ( (((2n+1)!!)/(2^(n+1)((n+1)!))t^(n+1))/( ((2n-1)!!)/(2^n(n!))t^n))# #lim_(n->oo) abs(a_(n+1)/a_n ) = lim_(n->oo) ((2n+1)/(2n+2)) abs t = abs(t)# so the series has radius of convergence #R=1#. Let now #t = u^2# and as #abs t < 1 => x^2 < 1# we have: #1/sqrt(1-u^2) = sum_(n=0)^oo ((2n+1)!!)/(2^n(n!))u^(2n)# still with #R=1#. Inside the interval #x in (-1,1)# we can then integrate term by term and obtain a series with the same radius of convergence: #int_0^v (du)/sqrt(1-u^2) = sum_(n=0)^oo ((2n+1)!!)/(2^n(n!))int_0^v u^(2n)du# #arcsinv= sum_(n=0)^oo ((2n-1)!!)/(2^n(n!)) v^(2n+1)/(2n+1)# Note that: # (2n-1)!! = ((2n-1)(2n-3)...3*1) = ((2n)(2n-1)(2n-2)(2n-3)...3*2*1)/((2n)(2n-2)...2)# # (2n-1)!! = ((2n)!)/(2^n(n(n-1)...1)) = ((2n)!)/(2^n(n!))# so we can write the series also as: #arcsinv= sum_(n=0)^oo ((2n)!)/(2^(2n)(n!)^2) v^(2n+1)/(2n+1)# Finally let: #v=x^3#. Again as #abs v<1 => abs(x^3) < 1# the radius of convergence does not change: #arcsin(x^3) = sum_(n=0)^oo ((2n)!)/(2^(2n)(n!)^2) (x^3)^(2n+1)/(2n+1)# #arcsin(x^3) = sum_(n=0)^oo ((2n)!)/(2^(2n)(n!)^2) x^(6n+3)/(2n+1)#

Aurora Alvarado · Answer

undefined To find a power series representation for $ f(x) = \arcsin(x^3) $, we can use the known power series representation for $ \arcsin(x) $:

$$ \arcsin(x) = \sum_{n=0}^{\infty} \frac{1}{2^n}\binom{2n}{n} \frac{x^{2n+1}}{2n+1} $$

First, we replace $ x $ with $ x^3 $ to get the power series for $ \arcsin(x^3) $:

$$ \arcsin(x^3) = \sum_{n=0}^{\infty} \frac{1}{2^n}\binom{2n}{n} \frac{(x^3)^{2n+1}}{2n+1} $$

$$ = \sum_{n=0}^{\infty} \frac{1}{2^n}\binom{2n}{n} \frac{x^{6n+3}}{2n+1} $$

Now we have the power series representation for $ f(x) = \arcsin(x^3) $:

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

The radius of convergence $ R $ for this power series can be found using the ratio test:

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

where $ a_n $ is the coefficient of $ x^n $ in the power series.

In this case, $ a_n = \frac{1}{2^n}\binom{2n}{n} \frac{1}{2n+1} $ for $ n = 6n+3 $. So,

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

Simplifying the expression inside the limit, we get:

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

$$ R = \lim_{n \to \infty} \left| \frac{(2n)!(n+1)!(2n+2)!}{(n!)^2(2n+1)!(2n+3)!} \frac{2(n+1)+1}{2n+1} \right| $$

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

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

$$ R = \lim_{n \to \infty} \left| \frac{2n+3}{2n+1} \right| $$

$$ R = 1 $$

So, the power series for $ f(x) = \arcsin(x^3) $ converges for $ |x| < 1 $, and the radius of convergence is $ R = 1 $.