For this assignments we are finding the taylor expansions of the following functions:

1.

\(f\left( x \right) =\frac { 1 }{ 1-x }\)

The first few derivatives are:

\(f^{ ' }\left( x \right) =\frac { 1 }{ { (1-x) }^{ 2 } } ,\quad f^{ '' }\left( x \right) =\frac { 2 }{ { (1-x) }^{ 3 } } ,\quad f^{ ''' }\left( x \right) =\frac { 6 }{ { (1-x) }^{ 4 } } f^{ '''' }\left( x \right) =\frac { 24 }{ { (1-x) }^{ 5 } }\)
Evaluated at 0:

\(f^{ ' }\left( 0 \right) =1!,\quad f^{ '' }\left( 0 \right) =2!,\quad f^{ ''' }\left( 0 \right) =3!,\quad f^{ '''' }\left( 0 \right) =4!\)
The first few terms of the series:

\(1,\quad x,\quad \frac { 2! }{ 2! } { x }^{ 2 },\quad \frac { 3! }{ 3! } { x }^{ 3 },\quad \frac { 4! }{ 4! } { x }^{ 4 }\)
In summation notation:

\(\sum _{ n=0 }^{ \infty }{ { x }^{ n } }\)

2.

\(f\left( x \right) ={ e }^{ x }\)

The derivatives of this one are easy! They’re all \({ e }^{ x }\)
Evaluated at 0, they are therefore 1. The first few terms are:

\(1,\quad x,\quad \frac { { x }^{ 2 } }{ 2! } ,\quad \frac { { x }^{ 3 } }{ 3! } ,\quad \frac { { x }^{ 4 } }{ 4! }\)
In a more compact form:

\(\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n } }{ n! } }\)

3.

\(f(x)=\ln{(1+x)}\)

The derviatives:

\(f^{ ' }\left( x \right) =\frac { 1 }{ 1+x } ,\quad f^{ '' }\left( x \right) =-\frac { 1 }{ { (1+x) }^{ 2 } } ,\quad f^{ ''' }\left( x \right) =\frac { 2! }{ { (1+x) }^{ 3 } } ,\quad f^{ '''' }\left( x \right) =-\frac { 3! }{ { (1+x) }^{ 4 } }\)
Evaluated at 0:

\(f^{ ' }\left( 0 \right) =1,\quad f^{ '' }\left( 0 \right) =-1,\quad f^{ ''' }\left( 0 \right) =2,\quad f^{ '''' }\left( 0 \right) =-(3!)\)
The first few terms:

\(\quad 0,\quad x,\quad -\frac { { x }^{ 2 } }{ 2! } ,\quad \frac { 2!{ x }^{ 3 } }{ 3! } ,\quad -\frac { 3!{ x }^{ 4 } }{ 4! }\)
The factorials will cancel leaving n in the denominatitor, so in summation form we have:

\(\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 } } \frac { { x }^{ n } }{ n }\)