Problem 1

\[ f(x) = 1/(1-x) \]

first derivative \[ f^1(x)= 1/(1-x)^2 \]

Second derivative \[ f^2(x)= 2/(1-x)^3 \]

Third derivative \[ f^3(x)= 6/(1-x)^4 \]

Nth derivative \[ f^n(x)= n!/((1-x)^k] \] where k = n+1

\[\sum_{x=0}^{\infty} f^n(a)*(x-a)^n/n!\]

After substituting and canceling out terms we get at a=0

Taylor series of \(1/(1-x)\) is \[\sum_{x=0}^{\infty} x^n\]

where -1<x<1

Problem 2

\[ f(x) = e^x \]

first derivative \[ f^1(x)= e^x \]

Second derivative \[ f^2(x)= e^x \]

Third derivative \[ f^3(x)= e^x \]

Nth derivative \[ f^n(x)= e^x \]

Plugging it into the formula we have \[\sum_{x=0}^{\infty} (e^a)*(x-a)^n/n!\] when a=0 we can simplify as \[\sum_{x=0}^{\infty} (x)^n/n!\]

where x is for all values

Problem 3

\[ f(x) = ln(1+x) \]

first derivative \[ f^1(x)= \frac{1}{(1+x)} \]

Second derivative \[ f^2(x)= \frac{-1}{(1+x)^2} \]

Third derivative \[ f^3(x)= \frac{2}{(1+x)^3} \]

fourth derivative \[ f^3(x)= \frac{-6}{(1+x)^4} \]

Nth derivative \[ f^n(x)= \frac{(-1)^k*(n-1)!}{(1+x)^n} \] where k = n+1

We can plug this into the Taylor series formula and get \[\sum_{x=0}^{\infty} \frac{(-1)^k*(n-1)!*(x-a)^n}{(1+a)^n*n!}\] Afer simplifying at a=0 we get \[\sum_{x=0}^{\infty} \frac{(-1)^k*x^n}{n}\] where k=n+1