Question

\(f(x)=\ln (x+1), \quad a=0\)

Short answer

The Taylor series expansion of \(f(x)=\ln (x+1)\) at \(a=0\) is \(f(x)=x-(x^2)/2+(x^3)/3!-...\)

Step-by-step solution

Calculate the first few derivatives of \(f(x)\) at \(a=0\)

The first derivative of \(f(x) = \ln(x + 1)\) is \(f'(x) = (1/(x+1))\). Substituting \(a=0\), we get: \(f'(0) = 1\). The second derivative of \(f(x)\) is \(f''(x)=-(1/(x+1)^2)\). Substituting \(a=0\), we get: \(f''(0) = -1\). Suppose we continue with this process, each time differentiating and then setting \(x=0\), alternately getting 1 and -1 for all higher order derivatives.

Substitute the derivatives into the Taylor Series expansion

The generic formula for a Taylor series is: \(f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... \). When we substitute \(a=0\) and the derivatives we calculated into this formula, we get: \(f(x)=\ln(1+0)+1*x-1*(x)^2/2+1*(x)^3/3!-... \). This performs the Taylor series expansion.

Simplify the series

We started with: \(f(x)=\ln(1+0)+1*x-1*(x)^2/2+1*(x)^3/3!-... \). The \ln(1) is just 0, so we can drop that, to get: \(f(x)=x-(x^2)/2+(x^3)/3!-...\). This is the Taylor series expansion of \(f(x)=\ln (x+1)\) at \(a=0\).