Week 14 homework

Assignment 14 - Taylor Series

This week, we’ll work out some Taylor Series expansions of popular functions.

1. \(f(x) = (1-x)^{-1}\)
2. \(f(x) = e^x\)
3. \(f(x) = ln(1+x)\)
4. \(f(x) = x^\frac{1}{2}\)

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 an R- Markdown document.

a. \(f(x) = (1-x)^{-1}\)

We will work out the Taylor expansion at x = 0

Step 1: Find the first few derivatives:

The chain rule is shown for the first order derivatives. The next 3 higher order derivatives were calculated the same way. Also shown are the substitutions for x = 0 into the equations.

\[f(x) = (1-x)^{-1}~~~~~f(0) = 1\\ f(x) = x^{-1}, ~~g(x) = 1-x \\ f'(x) = -x^{-2}, ~~g'(-1) \\ f'(x) = f'(g(x)) * g'(x) \\ f'(x) = -(1-x)^{-2} * -1 \\ f'(x) = \frac{1}{(1-x)^{2}}~~~~~f'(0) = 1 \\ f''(x) = \frac{2}{(1-x)^{3}}~~~~~f''(0) = 2 \\ f'''(x) = \frac{6}{(1-x)^{4}}~~~~~f'''(0) = 6 \\ f'''(x) = \frac{24}{(1-x)^{5}}~~~~~f'''(0) = 24 \] Step 2: Substituting the derivatives into the general Taylor expansion formula of, where c = 0

\[\sum^{\inf}_{n=0} \frac{f^n(C)}{n!}(x-C)^n \\ \frac{1}{0!}*(x)^0 ~+~ \frac{1}{1!}*(x)^1 ~+~ \frac{2}{2!}*(x)^2 ~+~ \frac{6}{3!}*(x)^3 ~+~ \frac{24}{4!}*(x)^4 \\ 1 ~+~ x ~+~ x^2 ~+~ x^3 ~+~ x^4 ...\\ \]

The Taylor expansion of \(f(x) = (1-x)^{-1}\) is \(1 ~+~ x ~+~ x^2 ~+~ x^3 ~+~ x^4\)

b.\(f(x) = e^x\)

Step 1: Find the first few derivatives:

The derivative of \(e^x\) will always be \(e^x\) and thus the derivative at x = 0 will always be 1

Step 2: Substituting the derivatives into the general Taylor expansion formula of, where c = 0

\[\sum^{\inf}_{n=0} \frac{f^n(C)}{n!}(x-C)^n \\ \frac{1}{0!}*(x)^0 ~+~ \frac{1}{1!}*(x)^1 ~+~ \frac{1}{2!}*(x)^2 ~+~ \frac{1}{3!}*(x)^3 ~+~ \frac{1}{4!}*(x)^4 \\ 1 ~+~ x ~+~ \frac{x^2}{2!} ~+~ \frac{x^3}{3!} ~+~ \frac{x^4}{4!} ...\\ \]

The Taylor expansion of \(f(x) = e^x\) is \(1 ~+~ x ~+~ \frac{x^2}{2!} ~+~ \frac{x^3}{3!} ~+~ \frac{x^4}{4!}...\)

c.\(f(x) = ln(x+1)\)

Step 1: Find the first few derivatives:

The derivatives and substitutions for x = 0 into the equations are shown below.

\[f(x) = ln(1+x)^{-1}~~~~~f(0) = 0 \\ f'(x) = \frac{1}{1+x}~~~~~f'(0) = 1\\ f''(x) = -\frac{1}{(1+x)^{2}}~~~~~f''(0) = -1\\ f'''(x) = \frac{2}{(1+x)^{3}}~~~~~f'''(0) = 2 \\ f''''(x) = -\frac{6}{(1+x)^{4}}~~~~~f''''(0) = -6 \\ \]

Step 2: Substituting the derivatives into the general Taylor expansion formula of, where c = 0

\[\sum^{\inf}_{n=0} \frac{f^n(C)}{n!}(x-C)^n \\ \frac{0}{0!}*(x)^0 ~+~ \frac{1}{1!}*(x)^1 ~+~ -\frac{1}{2!}*(x)^2 ~+~ \frac{2}{3!}*(x)^3 ~+~ -\frac{6}{4!}*(x)^4 \\ 0 ~+~ x ~-~ \frac{x^2}{2!} ~+~ \frac{2x^3}{3!} ~-~ \frac{6x^4}{4!} \\ x ~-~ \frac{x^2}{2} ~+~ \frac{x^3}{3} ~-~ \frac{x^4}{4} ...\\ \]

The Taylor expansion of \(ln(1+x)\) is \(x ~-~ \frac{x^2}{2} ~+~ \frac{x^3}{3} ~-~ \frac{x^4}{4} ...\)

d. \(f(x) = x^\frac{1}{2}\)

Step 1: Find the first few derivatives:

The derivatives and substitutions for x = 1 into the equations are shown below.

\[f(x) = x^\frac{1}{2}~~~~~f(1) = 0 \\ f'(x) = \frac{1}{2}x^{-\frac{1}{2}}~~~~~f'(1) = \frac{1}{2}\\ f''(x) = -\frac{1}{4}x^{-\frac{3}{2}}~~~~~f''(1) = -\frac{1}{4}\\ f'''(x) = \frac{3}{8}x^{-\frac{5}{2}}~~~~~f'''(1) = \frac{3}{8}\\ f''(x) = -\frac{15}{16}x^{-\frac{7}{2}}~~~~~f''(1) = -\frac{5}{128}\\ \] Step 2: Substituting the derivatives into the general Taylor expansion formula of, where c = 1

\[\sum^{\inf}_{n=0} \frac{f^n(C)}{n!}(x-C)^n \\ \frac{1}{0!}*(x-1)^0 ~+~ \frac{\frac{1}{2}}{2}*(x-1)^1 ~+~ \frac{-\frac{1}{4}}{2!}*(x-1)^2 ~+~ \frac{\frac{3}{8}}{3!}*(x-1)^3 ~+~ \frac{-\frac{15}{16}}{4!}*(x-1)^4 \\ 1 ~+~ \frac{1}{2}*(x-1) ~-~ \frac{1}{8}*(x-1)^2 ~+~ \frac{1}{16}*(x-1)^3 ~-~ \frac{5}{128}*(x-1)^4 \\ \]

The Taylor expansion of \(x^\frac{1}{2}\) is \(1 ~+~ \frac{1}{2}*(x-1) ~-~ \frac{1}{8}*(x-1)^2 ~+~ \frac{1}{16}*(x-1)^3 ~-~ \frac{5}{128}*(x-1)^4...\)