Question

Use the definition to find the Taylor series (centered at \(c\) ) for the function. $$ f(x)=\ln \left(x^{2}+1\right), \quad c=0 $$

Short answer

The Taylor series for the function \(f(x)=\ln\left(x^2+1\right)\) centered at \(c = 0\) starts as \(f(x) = 2x^2/2! + (f'''(0))x^3/3! + ...\). The remaining terms will depend on the third and higher derivatives of the function.

Step-by-step solution

Identify the Function to be Expanded

The function to be expanded into a Taylor series is \(f(x)=\ln\left(x^2+1\right)\). The Taylor series will be centered at \(c = 0\).

Find the Derivatives at the Point \(c\)

To formulate the Taylor series, we need to find the first several derivatives of the function at the point \(c = 0\). The 0th derivative is the function itself: \(f(x) = \ln\left(x^2+1\right)\). Therefore, \(f(0) = \ln(1) = 0\). The first derivative is \(f'(x) = 2x/(x^2+1)\). Therefore, \(f'(0) = 0\). The second derivative is \(f''(x) = 2(x^2 + 1 - 2x^2) / (x^2 + 1)^2 = 2 / (x^2 + 1)^2\). Therefore, \(f''(0) = 2\). The third and higher derivatives will follow a similar pattern, getting increasingly complex.

Formulate the Taylor Series

Now, we can formulate the Taylor series using the derivatives at \(c = 0\) and the formula for the Taylor series: \[ f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...\] Since \(f(0) = 0\) and \(f'(0) = 0\), the first two terms become 0 and the Taylor series simplifies to \[ f(x) = f''(0)x^2/2! + f'''(0)x^3/3! + ...\]. The remaining terms will depend on the third and higher derivatives, which we did not compute in this exercise.