Perform the following Taylor Series Expansions:
1. \(f(x)=\frac { 1 }{ (1-x) }\)
\(\frac { 1 }{ (1-x) }\) is equivalent to \({ 1\cdot (1-x) }^{ -1 }\)
Taking the first derivative we get: \({ -(1-x) }^{ -2 }\cdot (-1)=\frac { 1 }{ { (1-x) }^{ 2 } }\)
The second derivative is: \({ -2(1-x) }^{ -3 }\cdot (-1)=\frac { 2 }{ { (1-x) }^{ 3 } }\)
The third derivative is: \({ -3\cdot 2\cdot { (1-x) }^{ -4 } }\cdot (-1)=\frac { 6 }{ { (1-x) }^{ 4 } }\)
The fourth derivative is: \({ -4\cdot 3\cdot 2\cdot { (1-x) }^{ -5 } }\cdot (-1)=\frac { 24 }{ { (1-x) }^{ 5 } }\)
Now, evaluating the above derivatives at \(x=0\), we get:
\({ f(0)=\frac { 1 }{ 1-0 } }=1=0!\)
\({ f'(0)=\frac { 1 }{ { (1-0) }^{ 2 } } }=1=1!\)
\({ f''(0)=\frac { 2 }{ { (1-0) }^{ 3 } } }=2=2!\)
\({ f'''(0)=\frac { 6 }{ { (1-0) }^{ 4 } } }=6=3!\)
\({ f''''(0)=\frac { 24 }{ { (1-0) }^{ 5 } } }=24=4!\)
So:
\(1+\frac { 1 }{ 1! } x+\frac { 2 }{ 2! } { x }^{ 2 }+\frac { 6 }{ 3! } { x }^{ 3 }+\frac { 24 }{ 4! } { x }^{ 4 }+...\)
\(=1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }...\)
\(=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } (x{ ) }^{ n } }\)
Since the above is a geometric series, it converges when \(|x|<1\)
2. \(f(x)={ e }^{ x }\)
Since the derivative of \({ e }^{ x }\) is always just \({ e }^{ x }\), our first through fourth derivatives will all result in \({ e }^{ x }\).
Again, since all the derivatives are the same, we’ll solve for where \(x=0\) and the result will be the same for all four derivatives:
\(f(0)={ e }^{ 0 }=1\)
Which leads us to:
\({ e }^{ x }=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } (x{ ) }^{ n } }\)
\(=1+\frac { 1 }{ 1! } x+\frac { 1 }{ 2! } { x }^{ 2 }+\frac { 1 }{ 3! } { x }^{ 3 }+\frac { 1 }{ 4! } { x }^{ 4 }+...\)
\(=\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n } }{ n! } }\)
3. \(f(x)=ln(1+x)\)
We know that the derivative of \(ln\) is \(\frac { 1 }{ x }\). So, taking the first derivative, we get:
\({ (1+x) }^{ -1 }\)
The second derivative would be:
\({ -1\cdot (1+x) }^{ -2 }\)
The third derivative would be:
\({ 2\cdot (1+x) }^{ -3 }\)
Finally, the fourth derivative would be:
\({ -3\cdot 2\cdot (1+x) }^{ -4 }\)
Next, we’ll evaluate each of the above derivatives at \(x=0\):
\(f(0)=ln(1+0)=ln(1)=0\)
\(f'(0)=(1+0{ ) }^{ -1 }=1\)
\(f''(0)=-1\cdot (1+0{ ) }^{ -2 }=-1\)
\(f'''(0)=2\cdot (1+0{ ) }^{ -3 }=2\)
\(f''''(0)=-3\cdot 2\cdot (1+0{ ) }^{ -4 }=-6\)
This leads us to:
\(ln(1+x)=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } (x{ ) }^{ n } }\)
\(=0+\sum _{ n=1 }^{ \infty }{ \frac { { (-1) }^{ (n+1) }(n-1)! }{ n! } (x{ ) }^{ n } }\)
\(=\sum _{ n=1 }^{ \infty }{ \frac { { (-1) }^{ (n+1) }({ x }^{ n }) }{ n } }\)