Question 1

Use the binomial series to expand the function as a power series. Find the radius of convergence.

x/ â 9+ x^2

Question 2

Answer:

Explanation:

The given function is:

f(x) = (x)/(√(9+x^2))

Using the binomial series:

$= x(9+x^2)^(-1/2) \\ \\ = x *9^(-1/2)(1+(x^2)/(9))^(-1/2) \\ \\ = (x)/(3)(1+ (x^2)/(9))^(-1/2)$

$= (x)/(3) \sum \limits ^(\alpha )_(n=0)(^{-(1)/(2)}_n)((x^2)/(9))^n$

$\implies (x)/(3)\Bigg [ 1 + (-(1)/(2))*((x^2)/(9))+ ((-(1)/(2))(-(1)/(2)-1))/(2!)((x^2)/(9)) ^2 + ((-(1)/(2))(-(1)/(2)-1) (-(1)/(2)-2))/(3!) ) ((x^2)/(9)) ^3+ ... \Bigg ]$

$= (x)/(3)\Bigg [ 1 - (x^2)/(18)+ (3)/(5832)*(x^4)/(1)-(15)/(34992)x^6+... \Bigg ]$

$\mathbf{= (x)/(3)- (x^3)/(54)+ (1)/(5832)x^4 - (5)/(34992)x^7 + ...}$

To compute the radius of convergence:

$f(x) = (\lambda )/(3) \sum \limits ^(\alpha )_(n=0) (1+(x^2)/(9))^(-1/2)$

$f(x) = (\lambda )/(3) \sum \limits ^(\alpha )_(n=0) (^(-1/2) _n ) ((x^2)/(9))^n \\ \\ \implies (\lambda )/(3) \sum \limits ^(\alpha )_(n=0) (^(-1/2) _n ) ((x^2)/(9))^n \\ \\ \implies \sum \limits ^(\alpha)_(n=0) (^(-1/2) _n ) (1)/(3*9^n)*x^(2n) \\ \\ \implies \sum \limits ^(\alpha)_(n=0) (^(-1/2) _n ) (1)/(3^(2n+1))*x^(2n)$

Suppose
a_n = (^(-1/2)_(n))*(1)/(3^(2n+1))*x^(2n)

Then, rewriting the equation above as:

a_(n+1) = (^(-1/2)_(n+1))*(1)/(3^(2n+3))*x^(2n+2)

As such;

$\lim_(n \to x) \Big| (a_n+1)/(a_n) \Big| = \lim_(n \to x) \Bigg | ( (^(-1/2)_(n+1)) (x^(2n+2))/(3^(2n+3)) )/((^(-1/2)_(n) )(x^2)/(3^(2n+1)))} \Bigg |$

$\implies \lim_(n \to \alpha) \Bigg | ( (^(-1/2)_(n+1)) (x^(2))/(3^(2)) )/((^(-1/2)_(_n) )) \Bigg |$

$\implies \lim_(n \to \alpha) \Bigg | (((-1/2)!)/((-1/2-n -1)!(n+1)!)*(x^2)/(9) )/( ((-1/2!))/((-1/2-n)!(n!)) ) \Bigg| \\ \\ \\ \implies \lim_(n \to \alpha) \Bigg | ((-1/2 -n)! (n!) )/((-1/2 -n-1)! (n+1)! ) *(x^2)/(9) \Bigg| \\ \\ \\ \implies \lim_(n \to \alpha) \Bigg | ((-1/2 -n) (-1/2 -n-1)! \ n! )/((-1/2 -n-1)! (n+1) n! ) *(x^2)/(9) \Bigg|$