\[f(x)=1/(1-x)\] I started by finding the first few terms:
\[f(c)=1/(1-c)\] \[f'(c)=1/(1-c)^2\] \[f''(c)=2/(1-c)^3\] \[f'''(c)=6/(1-c)^4\] \[f''''(c)=24/(1-c)^5\] Then evaluated these initial terms at 0: \[f(0)=1/(1-(0))=1\] \[f'(0)=1/(1-0)^2=1\] \[f''(0)=2/(1-0)^3=2\] \[f'''(0)=6/(1-0)^4=6\] \[f''''(0)=24/(1-(0))^5=24\]
Then substituted these terms into the Taylor series like so:
\[f(x)=((1/((1-c)0!))(x-c)^0+(1/(((1-c)^2)1!))(x-c)^1+(2/(((1-c)^3)2!))(x-c)^2+(6/(((1-c)^4)3!)(x-c)^4+...+(1/(1-c)^(n+1))(x-c)^n\] Then evaluating the series at c=0: \[f(x)=(1/1)+(x/1!)+(2x^2/2!)+(6x^4/3!)+...\] And simiplified: \[f(x)=1+x+x^2+x^4+...\]
\[f(x)=e^x\] \(f(c)=e^c\) \(f'(c)=e^c\) \(f''(c)=e^c\) \(f'''(c)=e^c\)
\(f(0)=1\) \(f'(0)=1\) \(f''(0)=1\) \(f'''(0)=1\)
\[f(x)= (e^c/0!)(x-c)^0+(e^c/1!)(x-c)^1+(e^c/2!)(x-c)^2+(e^c/3!)(x-c)^3+...\] \[f(x)= 1+x+(x^2/2!)+((e^c/3!)(x-c)^3)x^3/3!)+...\]
\[f(x)=ln(1+x)\] \(f(c)=ln(1+c)\) \(f'(c)=1/(1+c)\) \(f''(c)=-1/(1+c)^2\) \(f'''(c)=2/(1+c)^3\) \(f''''(c)=6/(1+c)^4\)
\(f(0)=0\) \(f'(0)=1\) \(f''(0)=-1\) \(f'''(0)=2\) \(f''''(0)=6\)
\[f(x)=((ln(1+c))/0!)(x-c)^0+((1/(1+c)1!))(x-c)^1-(1/((1+c)^2!)(x-c)^2+(2/((1+c)^3)(3!))*x-c)^3+...\] \[f(x)= 0+x-(x^2/2)+(x^3/3)-(x^4/4)+...\]
\[f(x) = x^(1/2)\] \(f(c)=c^(1/2)\) \(f'(c)=1/2c^(1/2)\) \(f''(c)=-1/4c^(3/2)\) \(f'''(c)=3/8c^(5/2)\) \(f'''(c)=-15/16c^(7/2)\)
The derivatives of these terms cannot be defined at 0, and therefore, do not exist. Because of that, the Taylor series does not exist at c=0.