How do you find a power series representation for f(x) = ln(x^2+1) and what is the radius of convergence?

1 Answer
Jun 14, 2017

The Taylor approximation for f(x)=ln(1+x^2) centered around x=a^2, is ln(1+a)+sum_(n=1)^oo((x^2-a)^n(-1)^(n+1))/(n(1+a)^n), and its radius of convergence is sqrt(1+2a).

Explanation:

A power series is also called Taylor series. Taylor series are represented by sum_(n=0)^oo(f^n(a)(x-a)^n)/(n!), where the series is approximated around a value a.

Now, f(x)=ln(x^2+1) looks ridiculously similar to the function g(x)=ln(1+x), and for that function there is a well known Taylor expansion centered around a, which is:
g(x)=ln(1+a)+sum_(n=1)^oo((x-a)^n(-1)^(n+1))/(n(1+a)^n)

Letting h(x)=x^2, and recognising that f(x)=ln(x^2+1)-=g(h(x)), we can substitute h(x) into the series expansion for g(x).
f(x)=g(h(x))=ln(1+h(a))+sum_(n=1)^oo((h(x)-a)^n(-1)^(n+1))/(n(1+a)^n)=ln(1+a^2)+sum_(n=1)^oo((x^2-a)^n(-1)^(n+1))/(n(1+a)^n)

Therefore, the Taylor series (power series) approximation for f(x)=ln(x^2+1) centered around x=a is ln(1+a^2)+sum_(n=1)^oo((x^2-a)^n(-1)^(n+1))/(n(1+a)^n)

Now, we find its radius of convergence by applying the ratio test. The ratio test works by finding the ratio between the (n+1)th term and the nth term as n tends to oo. As a condition for convergence is that the ratio r has to satisfy absr<1 , we can limit the x for which the series will converge. I will show you how below:

Applying the ratio test:
lim_(k->oo)abs(u_(k+1)/u_k) =lim_(k->oo)abs((((x^2-a)^(k+1)(-1)^(k+1+1))/((k+1)(1+a)^(k+1)))/(((x^2-a)^k(-1)^(k+1))/(k(1+a)^k))) =lim_(k->oo)abs(((x^2-a)^(k+1))/((k+1)(1+a)^(k+1))*(k(1+a)^k)/((x^2-a)^k)) =lim_(k->oo)abs((x^2-a)/(1+a)*k/(k+1)) =abs((x^2-a)/(1+a))*lim_(k->oo)abs(k/(k+1)) =abs((x^2-a)/(1+a))*1

For the series to converge, the absolute ratio has to be less than 1, i.e.
abs((x^2-a)/(1+a))<1
abs(x^2-a)<1+a
abs(x^2)<1+2a
absx< sqrt(1+2a)
The radius of convergence is therefore sqrt(1+2a), which as we can see is dependent on the center of the series a.