This week, we’ll work out some Taylor Series expansions of popular functions.
• \(f (x) = \frac{1}{(1−x)}\)
• \(f (x) = e^x\)
• \(f (x) = ln(1 + x)\)
• \(f(x)=x^{1/2}\)
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.
\[ f(x) = \frac{1}{1-x} = \sum_{n=0}^{\infty} = \frac{f^n ( \alpha)}{n!} (x- \alpha) ^n \\ \] Suppose, for simplicity, the general formula for the taylor series of f(x) centered at \(\alpha = 0\)
\[ \begin{align} f(x) = \frac{1}{1-x}, && f(0) = 1 \\ f^1(x) = \frac{1}{(1-x)^2}, && f^2(0) =1 \\ f^2(x) = \frac{2}{(1-x)^3}, && f^2(0) =2 \\ f^3(x) = \frac{6}{(1-x)^4}, && f^2(0) =6 \\ f^4(x) = \frac{24}{(1-x)^5}, && f^2(0) =24 \\ \end{align} \]
and the Taylor Series becomes,
\[ \frac{1}{1-x} = \sum_{n=0}^\infty x^n = 1+x + x^2+ x^3+ x^4 + \ ... \] Now, we will use the ratio test to check for convergence
\[ \lim_{n \rightarrow 0} \Big| \frac{x^{n+1}}{x^n} \Big| = |x| \]
So, this series converges for \(|x| <1\)
For x = -1,
\[ \sum_{n=0}^\infty (-1)^n = 1 - 1+ 1 -1 + \ ... - ... \]
neither diverges nor convergence since we dont know the nth partial sum of this series
For x = 1,
\[ \sum_{n=0}^\infty (-1)^n = 1 + 1+ 1+ \ ... - ... \]
As n goes to infinity, it diverges to positive infinity.
So, the valid range as stated in the notes is the open interval (-1,1).
\[ f(x) = e^x = \sum_{n=0}^{\infty} = \frac{f^n ( \alpha)}{n!} (x- \alpha) ^n \\ \]
Suppose, for simplicity, the general formula for the taylor series of f(x) centered at \(\alpha = 0\)
\[ \begin{align} f(x) = e^x, && f(0) = 1 \\ f^1(x) = e^x, && f^2(0) =1 \\ f^2(x) = e^x, && f^2(0) =1 \\ f^3(x) = e^x, && f^2(0) =1 \\ f^4(x) = e^x, && f^2(0) =1 \\ \end{align} \]
and the Taylor Series becomes,
\[ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1+x + \frac{x^2}{2!}+ \frac{x^3}{3!}+ \frac{x^4}{4!} + \ ... \] Now, we will use the ratio test to check for convergence
\[ \lim_{n \rightarrow \infty} \Big| \frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}} \Big| = \lim_{n \rightarrow \infty} \Big| \frac{x}{n} \Big| = 0 \]
So, this series interval of convergence is \((-\infty, \infty)\)
\[ f(x) = ln(1+x) = \sum_{n=0}^{\infty} = \frac{f^n ( \alpha)}{n!} (x- \alpha) ^n \\ \]
Suppose, for simplicity, the general formula for the taylor series of f(x) centered at \(\alpha = 0\)
\[ \begin{align} f(x) = ln(1+x), && f(0) = 0 \\ f^1(x) = \frac{1}{(1-x)}, && f^2(0) =1 \\ f^2(x) = \frac{1}{(1-x)^2}, && f^2(0) =1 \\ f^3(x) = \frac{2}{(1-x)^3}, && f^2(0) =2 \\ f^4(x) = \frac{6}{(1-x)^4}, && f^2(0) =6 \\ \end{align} \]
and the Taylor Series becomes,
\[ ln(1+x) = \sum_{n=0}^\infty (-1)^n \frac{x^n}{n} = x - \frac{x^2}{2}+ \frac{x^3}{3}- \frac{x^4}{4} + \ ... \ - \ ... \] Now, we will use the ratio test to check for convergence
\[ \lim_{n \rightarrow \infty} \Big| \frac{(-1)^{n+1}\frac{x^{n+1}}{(n+1)}}{(-1)^n\frac{x^n}{n}} \Big| = \lim_{n \rightarrow \infty} \Big| -(x) \Big| = |x| \]
So, this series interval of convergence is \((-1, 1)\).
\[ f(x) = x^{1/2} = \sum_{n=0}^{\infty} = \frac{f^n ( \alpha)}{n!} (x- \alpha) ^n \\ \]
Suppose, for simplicity, the general formula for the taylor series of f(x) centered at \(\alpha = 0\)
\[ \begin{align} f(x) = x^{1/2}, && f(0) = 0 \\ f^1(x) = \frac{1}{2x^{1/2}}, && f^2(0) = DNE\\ f^2(x) = -\frac{1}{4x^{3/2}}, && f^2(0) =DNE \\ f^3(x) = \frac{1}{2x^{1/2}}, && f^2(0) =DNE \\ f^4(x) = -\frac{15}{16x^{7/2}}, && f^2(0) =DNE \\ \end{align} \]
The taylor series for \(f(x) = x^{1/2}\) is not defined at \(\alpha = 0\) since f(x) is not differentiable at 0.