This week, we’ll work out some Taylor Series expansions of popular functions.
\(\bullet \boldsymbol{f(x) = \frac{1}{(1-x)}}\)
\(\bullet \boldsymbol{f(x) = e^x}\)
\(\bullet \boldsymbol{f(x) = \ln(1+x)}\)
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.
\(\boldsymbol{f(x) = \frac{1}{(1-x)}}\)
This function is not defined at \(x = 1\).
| Derivatives | Evaluation at \(x = 0\) |
|---|---|
| \(f(x) = \frac{1}{(1-x)}\) | \(f(0) = 1\) |
| \(f'(x) = \frac{1}{(1-x)^2}\) | \(f(0) = 1\) |
| \(f''(x) = \frac{2}{(1-x)^3}\) | \(f(0) = 2\) |
| \(f'''(x) = \frac{6}{(1-x)^4}\) | \(f(0) = 6\) |
| \(f^{(4)}(x) = \frac{24}{(1-x)^5}\) | \(f(0) = 24\) |
| \(f^{(5)}(x) = \frac{120}{(1-x)^6}\) | \(f(0) = 120\) |
Insert into McClaurin Series formula:
\(1 + \frac{1}{1!}x^1 + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 + \frac{120}{5!}x^5 \dots\)
Simplify to:
\(1 + x + x^2 + x^3 + x^4 + x^5 \dots x^n\)
In summation form:
\(\sum\limits_{n=0}^{\infty} x^n\)
\(\boldsymbol{f(x) = e^x}\)
| Derivatives | Evaluation at \(x = 0\) |
|---|---|
| \(f(x) = e^x\) | \(f(0) = 1\) |
| \(f'(x) = e^x\) | \(f(0) = 1\) |
| \(f''(x) = e^x\) | \(f(0) = 1\) |
| \(f'''(x) = e^x\) | \(f(0) = 1\) |
| \(f^{(4)}(x) = e^x\) | \(f(0) = 1\) |
| \(f^{(5)}(x) = e^x\) | \(f(0) = 1\) |
Insert into McClaurin Series formula:
\(1 + \frac{1}{1!}x^1 + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \frac{1}{5!}x^5 \dots\)
Simplify to:
\(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} \dots \frac{x^n}{n!}\)
In summation form:
\(\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}\)
\(\boldsymbol{f(x) = \ln(1+x)}\)
| Derivatives | Evaluation at \(x = 0\) |
|---|---|
| \(f(x) = \ln(1+x)\) | \(f(0) = 0\) |
| \(f'(x) = \frac{1}{x+1}\) | \(f(0) = 1\) |
| \(f''(x) = -\frac{1}{(x+1)^2}\) | \(f(0) = -1\) |
| \(f'''(x) = \frac{2}{(x+1)^3}\) | \(f(0) = 2\) |
| \(f^{(4)}(x) = -\frac{6}{(x+1)^4}\) | \(f(0) = -6\) |
| \(f^{(5)}(x) = \frac{24}{(x+1)^5}\) | \(f(0) = 24\) |
Insert into McClaurin Series formula:
\(0 + \frac{1}{1!}x^1 - \frac{1}{2!}x^2 + \frac{2}{3!}x^3 - \frac{6}{4!}x^4 + \frac{24}{5!}x^5 \dots\)
Simplify to:
\(x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 \dots (-1)^{n+1}\frac{1}{n}x^n\)
In summation form:
\(\sum\limits_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n}x^n\)