Question

Question: Compute the Taylor series for f(x)= in(1+x²) about x₀= 0. [Hint:Multiply the series for (1+x²)^-1by 2xand integrate.]

Short answer

The required function is In(1+x²)=$\sum _{n-0}^{\infty }{(-1)}^n\frac{x2(n+1)}{n+1}$.

Step-by-step solution

Taylor series.

For a function ^f(x) the Taylor series expansion about a point ^$x_0$ is given by,

f(x-x₀) = f(x₀)+f'($x_0$)-(x-x₀)+f"(x₀)-$\frac{{\left(x-x_0\right)}^2}{2!}+f\text{'}\text{'}\text{'}\left(x_0\right)-\frac{{(x-x_0)}^3}{3!}+.....$

Multiply by 2x .

Use the known expansion:

$\frac{1}{1+x^2}=1-x^2+x^4-x^6+.....+{(-1)}^n\times x^{2n+...}$

Multiplying by 2x in the equation above gives:

$\frac{2x}{1+x^2}=2[x-x^3+x^5-x^7+...+(-1^n)x^{2n+1+...]}$

$=2\sum _{n-0}^{\infty }{(-1)}^n{x}^{2n+1}$

Integrating both sides yields:

${\int }_0^x\frac{2t}{1+t^2}dt={\int }_0^{t}2\sum _{n-0}^{\infty }{(-1)}^n{t}^{2n+1}dt$.............(1)

Integrate the left hand side.

The integral on the left hand side is:

Let, t = x

2tdt = du

u = 1+x²

Then,

${\int }_0^x\frac{2t}{1+t^2}dt={\int }_1^{1+x^2}\frac{du}{u}$

$=In{{\left|u\right|}_1}^{1+x^2}$

= In $\left|1+x^2\right|-In\left|1\right|$

=$\ln \left(1+x^2\right)$

Integrate the right hand side.

For the right hand side integral we obtain

= $2\sum _{n-0}^{\infty }{(-1)}^n{{\overline{)\frac{t^{2n+2}}{2n+2}}}^x}_0$

= $\sum _{n-0}^8{(-1)}^n\frac{x2n+2}{n+1}$

From the obtained results and equation (1), it follows that the series for the function f(x)=ln (1+x²) about x=0 is ln (1+x²) =