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

Jun 14, 2017

The Taylor approximation for $f \left(x\right) = \ln \left(1 + {x}^{2}\right)$ centered around $x = {a}^{2}$, is $\ln \left(1 + a\right) + {\sum}_{n = 1}^{\infty} \frac{{\left({x}^{2} - a\right)}^{n} {\left(- 1\right)}^{n + 1}}{n {\left(1 + a\right)}^{n}}$, and its radius of convergence is $\sqrt{1 + 2 a}$.

#### 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, $f \left(x\right) = \ln \left({x}^{2} + 1\right)$ looks ridiculously similar to the function $g \left(x\right) = \ln \left(1 + x\right)$, and for that function there is a well known Taylor expansion centered around $a$, which is:
$g \left(x\right) = \ln \left(1 + a\right) + {\sum}_{n = 1}^{\infty} \frac{{\left(x - a\right)}^{n} {\left(- 1\right)}^{n + 1}}{n {\left(1 + a\right)}^{n}}$

Letting $h \left(x\right) = {x}^{2}$, and recognising that $f \left(x\right) = \ln \left({x}^{2} + 1\right) \equiv g \left(h \left(x\right)\right)$, we can substitute $h \left(x\right)$ into the series expansion for $g \left(x\right)$.
$f \left(x\right) = g \left(h \left(x\right)\right) = \ln \left(1 + h \left(a\right)\right) + {\sum}_{n = 1}^{\infty} \frac{{\left(h \left(x\right) - a\right)}^{n} {\left(- 1\right)}^{n + 1}}{n {\left(1 + a\right)}^{n}} = \ln \left(1 + {a}^{2}\right) + {\sum}_{n = 1}^{\infty} \frac{{\left({x}^{2} - a\right)}^{n} {\left(- 1\right)}^{n + 1}}{n {\left(1 + a\right)}^{n}}$

Therefore, the Taylor series (power series) approximation for $f \left(x\right) = \ln \left({x}^{2} + 1\right)$ centered around $x = a$ is $\ln \left(1 + {a}^{2}\right) + {\sum}_{n = 1}^{\infty} \frac{{\left({x}^{2} - a\right)}^{n} {\left(- 1\right)}^{n + 1}}{n {\left(1 + a\right)}^{n}}$

Now, we find its radius of convergence by applying the ratio test. The ratio test works by finding the ratio between the $\left(n + 1\right) t h$ term and the $n t h$ term as $n$ tends to $\infty$. As a condition for convergence is that the ratio $r$ has to satisfy $\left\mid r \right\mid < 1$ , 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.
$\left\mid \frac{{x}^{2} - a}{1 + a} \right\mid < 1$
$\left\mid {x}^{2} - a \right\mid < 1 + a$
$\left\mid {x}^{2} \right\mid < 1 + 2 a$
$\left\mid x \right\mid < \sqrt{1 + 2 a}$
The radius of convergence is therefore $\sqrt{1 + 2 a}$, which as we can see is dependent on the center of the series $a$.