This week, we’ll work out some Taylor Series expansions of popular functions. • f(x) = 1 / (1−x) • f(x) = e^x • f(x) = ln(1 + x) For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as a R-Markdown document.
Solution:
The Taylor Series of f(x) centered at c is \(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(c) }{ n! } } { (x-c ) }^{ n }\) Let c = 0, this also gives us the Maclaurin Series of f(x), \(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } } { x }^{ n }\)
\(f\left( x \right) = \frac { 1 }{ (1-x) }\)
f(0) = 1
\(f'(x) = \frac { 1 }{ { (1-x) }^{ 2 } }\)
f’(0) = 1
\(f''(x) = \frac { 2 }{ { (1-x) }^{ 3 } }\)
f’’(0) = 2
\(f'''(x) = \frac { 6 }{ { (1-x) }^{ 4 } }\)
f’’’(0) = 6
\(f''''(x) = \frac { 24 }{ { (1-x) }^{ 5 } }\)
f’’’’(0) = 24
We can notice a pattern where the only term left is the n factorial of the nth derivative when we set that derivative equal to 0.
Plug the results into \(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } } { x }^{ n }\)
\(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } } { x }^{ n }\) =
\({ \frac { { f }^{ (0) }(0) }{ 0! } } { x }^{ 0 }+{ \frac { { f }^{ (1) }(0) }{ 1! } } { x }^{ 1 }+{ \frac { { f }^{ (2) }(0) }{ 2! } } { x }^{ 2 }+{ \frac { { f }^{ (3) }(0) }{ 3! } } { x }^{ 3 }+{ \frac { { f }^{ (4) }(0) }{ 4! } } { x }^{ 4 }+...\) =
\({ \frac { 1 }{ 1! } } { x }^{ 0 }+{ \frac { 1 }{ 1! } } { x }^{ 1 }+{ \frac { 2 }{ 2! } } { x }^{ 2 }+{ \frac { 6 }{ 3! } } { x }^{ 3 }+{ \frac { 24 }{ 4! } } { x }^{ 4 }+...\) =
\(1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+...\)
The Taylor Series expansion for \(f\left( x \right) =\frac { 1 }{ (1-x) }\) is \(1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+...\).
The formula for \(f\left( x \right) =\frac { 1 }{ (1-x) }\) is \(\sum _{ n=0 }^{ \infty }{ x }^{ n }\)
\(f\left( x \right) = { e }^{ x }\)
f(0) = 1
\(f'(x) = { e }^{ x }\)
f’(0) = 1
\(f''(x) = { e }^{ x }\)
f’’(0) = 1
\(f'''(x) = { e }^{ x }\)
f’’’(0) = 1
The derivative of \({ e }^{ x }\) is always \({ e }^{ x }\). That means we always get 1 when we set the derivative equal to 0.
Plug the results into \(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } } { x }^{ n }\)
\(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } } { x }^{ n }\) =
\({ \frac { { f }^{ (0) }(0) }{ 0! } } { x }^{ 0 }+{ \frac { { f }^{ (1) }(0) }{ 1! } } { x }^{ 1 }+{ \frac { { f }^{ (2) }(0) }{ 2! } } { x }^{ 2 }+{ \frac { { f }^{ (3) }(0) }{ 3! } } { x }^{ 3 }+...\) =
\({ \frac { { f }^{ (0) }(0) }{ 0! } } { x }^{ 0 }+{ \frac { { f }^{ (1) }(0) }{ 1! } } { x }^{ 1 }+{ \frac { { f }^{ (2) }(0) }{ 2! } } { x }^{ 2 }+{ \frac { { f }^{ (3) }(0) }{ 3! } } { x }^{ 3 }+...\) =
\({ \frac { 1 }{ 0! } } { x }^{ 0 }+{ \frac { 1 }{ 1! } } { x }^{ 1 }+{ \frac { 1 }{ 2! } } { x }^{ 2 }+{ \frac { 1 }{ 3! } } { x }^{ 3 }+...\) =
\(1+x+{ \frac { 1 }{ 2! } } { x }^{ 2 }+{ \frac { 1 }{ 3! } } { x }^{ 3 }+...\)
The Taylor Series expansion for \(f\left( x \right) ={ e }^{ x }\) is \(1+x+{ \frac { 1 }{ 2! } } { x }^{ 2 }+{ \frac { 1 }{ 3! } } { x }^{ 3 }+...\)
The formula for \(f\left( x \right) ={ e }^{ x }\) is \(\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } } { x }^{ n }\)
\(f\left( x \right)=ln(1+x)\)
f(0) = 0
\(f'(x) = \frac { 1 }{ (1+x) }\)
f’(0) = 1
\(f''(x) = -\frac { 1 }{ { (1+x) }^{ 2 } }\)
f’’(0) = -1
\(f'''(x) = \frac { 2 }{ { (1+x) }^{ 3 } }\)
f’’’(0) = 2
\(f''''(x) = -\frac { 6 }{ { (1+x) }^{ 4 } }\)
f’’’’(0) = -6
\(f'''''(x) = \frac { 24 }{ { (1+x) }^{ 5 } }\)
f’’’’’(0) = 24
We can see that from the second derivative onward, we get n - 1 factorial with alternating signs starting with a negative sign.
Plug the results into \(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } } { x }^{ n }\)
\(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(0) }{ n! } } { x }^{ n }\) =
\({ \frac { { f }^{ (0) }(0) }{ 0! } } { x }^{ 0 }+{ \frac { { f }^{ (1) }(0) }{ 1! } } { x }^{ 1 }+{ \frac { { f }^{ (2) }(0) }{ 2! } } { x }^{ 2 }+{ \frac { { f }^{ (3) }(0) }{ 3! } } { x }^{ 3 }+{ \frac { { f }^{ (4) }(0) }{ 4! } } { x }^{ 4 }+{ \frac { { f }^{ (5) }(0) }{ 5! } } { x }^{ 5 }+...\) =
\({ \frac { { 0 } }{ 0! } } { x }^{ 0 }+{ \frac { { 1 } }{ 1! } } { x }^{ 1 }-{ \frac { { 1 } }{ 2! } } { x }^{ 2 }+{ \frac { { 2 } }{ 3! } } { x }^{ 3 }-{ \frac { { 6 } }{ 4! } } { x }^{ 4 }+{ \frac { { 24 } }{ 5! } } { x }^{ 5 }+...\) =
\(x-{ \frac { { 1 } }{ 2! } } { x }^{ 2 }+{ \frac { { 2 } }{ 3! } } { x }^{ 3 }-{ \frac { { 6 } }{ 4! } } { x }^{ 4 }+{ \frac { { 24 } }{ 5! } } { x }^{ 5 }+...\) =
\(x-{ \frac { { 1 } }{ 2 } } { x }^{ 2 }+{ \frac { { 2 } }{ 6 } } { x }^{ 3 }-{ \frac { { 6 } }{ 24 } } { x }^{ 4 }+{ \frac { { 24 } }{ 120 } } { x }^{ 5 }+...\) =
\(x-{ \frac { { 1 } }{ 2 } } { x }^{ 2 }+{ \frac { { 1 } }{ 3 } } { x }^{ 3 }-{ \frac { { 1 } }{ 4 } } { x }^{ 4 }+{ \frac { { 1 } }{ 5 } } { x }^{ 5 }+...\)
The Taylor Series expansion for \(f\left( x \right)=ln(1+x)\) is \(x-{ \frac { { 1 } }{ 2 } } { x }^{ 2 }+{ \frac { { 1 } }{ 3 } } { x }^{ 3 }-{ \frac { { 1 } }{ 4 } } { x }^{ 4 }+{ \frac { { 1 } }{ 5 } } { x }^{ 5 }+...\)
The formula for \(f\left( x \right)=ln(1+x)\) is \(\sum _{ n=1 }^{ \infty }{ \frac { { (-1) }^{ n+1 } }{ n } } { X }^{ n }\)