The general form of the Taylor series expansion:
\(f(x)\) \(=\) \(\int_{n=0}^{\infty}\) \(f^n(a)/n!(x-a)^n\)
So, the Taylor series of the function f(x) centered at 0:
\(f(x)\) \(=\) \(\int_{n=0}^{\infty}\) \(f^n(0)/n!(x^n)\)
\(=\) \(f(0)\) + \(f^1(0)/1!(x)\) + \(f^2(0)/2!(x^2)\) +\(f^3(0)/3!(x^3)\)+……+\(f^n(0)/n!(x^n)\)———–(1)
\(f(x) = 1/(1-x)\)
\(f(0) = 1\)
Now, find the value of the 1st, 2nd, 3rd….. nth derivatives of the function at x=0 are:
\(f^1(x)= 1/(1-x)^2\)
\(f^1(0)= 1\)
\(f^2(x)= 2/(1-x)^3\)
\(f^2(0)= 2\)
\(f^3(x)= 6/(1-x)^4\)
\(f^3(0)= 6\)
\(f^4(x)= 24/(1-x)^5\)
\(f^4(0)= 24\) ….
So, putting all the values in equation-1 we get the Taylor expansion of the function, f(x)=1/(1-x):
\(1/(1-x)= 1+ 1/1!(x^1) + 2/2!(x^2) + 6/3!(x^3) + 24/4!(x^4)+.....\) \(=\int_{n=0}^{\infty}x^n\)
\(f(x)=e^x\)
\(f(0)=e^0=1\)
\(f^1(x)=e^x\)
\(f^1(0)=e^0=1\)
\(f^2(x)=e^x\)
\(f^2(0)=e^0=1\)
\(f^3(x)=e^x\)
\(f^3(0)=e^0=1\)
\(f^4(x)=e^x\)
\(f^4(0)=e^0=1\) ……
Hence, putting all the values in equaion-1 we the Taylor series for the function \(f(x)=e^x\):
\(e^x=1+1/1!(x)+x^2/2!+x^3/3!+x^4/4!+....=\int_{n=0}^{\infty}x^n/n!\)
\(f(x)=ln(1+x)\) \(f(0)=ln(1+0)=0\)
\(f^1(x)=1/(1+x)\) \(f^1(0)=1/(1+0)=1\)
\(f^2(x)= -1/(1+x)^2\) \(f^2(0)= -1/(1+0)^2=-1\)
\(f^3(x)= 2/(1+x)^3\) \(f^3(0)= 2/(1+0)^3=2\)
\(f^4(x)= -6/(1+x)^4\) \(f^4(0)= -6/(1+0)^4=-6\) ………………..
Hence, the Taylor series expansion of ln(1+x):
\(ln(1+x)=0+1/1!x^1-1/2!x^2+2/3!x^3-6/4!x^4+....\)
\(=x-x^2/2+x^3/3-x^4/4+......\)
\(=\) \(\int_{n=0}^{\infty}(-1)^{n+1} x^n/n\)
Now,
\(f(x)=x^{1/2}\)
\(f(0)=0^{1/2}=0\)
\(f^1(x)=1/2x^{-1/2}\)
\(f^1(0)=1/2.0^{-1/2}=0\)
\(f^2(x)=-1/4x^{-3/2}\)
\(f^2(0)=-1/4.0^{-3/2}=0\)
\(f^3(x)=3/8x^{-5/2}\)
\(f^3(0)=3/8.0^{-5/2}=0\) ………………
It is found above that the function \(f(x)=x^{1/2}\) is not possible to be differentiated at 0. Therefore, this function cannot be approximated with the Taylor series expansion.