This week, we’ll work out some Taylor Series expansions of popular functions.
\(f (x) = \frac{1}{(1−x)}\)
\(f (x) = e^x\)
\(f (x) = ln(1 + x)\)
\(f(x)=x^{\frac{1}{2}}\)
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.
This function cannot take \(x=1\) so its valid range is \(\mathbb{R} \setminus \{1\}\) or \((- \infty , 1) \cup (1, \infty)\)
So the Taylor series expansion of this function at \(x=0\) is
\(\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\)
The valid range for this function is all real numbers \(\mathbb{R}\) or \((- \infty, \infty)\)
So the Taylor series expansion of this function at \(x=0\) is
\(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\)
“ln” function is only defined on only posituve real numbers \(\mathbb{R^+}\) so
\[ \begin{align} 1+x&>0\\ x&>-1\\ \end{align} \]
So the valid range of \(ln(1+x)\) is \((-1, \infty)\)
We know that \(ln(x) = \sum_{n=0} ^{\infty} (-1)^{n+1} \frac{(x-1)^n}{n}\)
\(\Rightarrow\) to find the Taylor series for \(ln(1+x)\) at \(x=0\), we substitute \((1+x)\) for \(x\):
\[ \begin{align} ln(1+x) &= \sum _{n=0} ^ {\infty} (-1)^{n+1} \frac{((1+x)-1)^n}{n}\\ &= \sum _{n=0} ^ {\infty} (-1)^{n+1} \frac{x^n}{n}\\ \end{align} \]
This function can be also written as \(f(x) = \sqrt x\) which is defined on the positive real numbers including 0 $ _{} $ but we are going to find the Taylor series expansion at \(x=1\) since we cannot evaluate the derivatives at \(x=0\) since it will be undefined:
\[ \begin{align} f(x) &= \sqrt x \qquad \Rightarrow f(1) = 1 \\ f'(x) &= \frac{1}{2 \sqrt{x}} \quad \Rightarrow f'(1) = \frac{1}{2} \\ f''(x) &= \frac{-1}{4x^{\frac{3}{2}}} \quad \Rightarrow f''(1) = \frac{-1}{4} \\ f'''(x) &= \frac{3}{8 x ^ {\frac{5}{2}}} \quad \Rightarrow f'''(1) = \frac{3}{8}\\ f^{(4)} (x) &= \frac{-15}{16 x ^\frac{7}{2}} \quad \Rightarrow f^{(4)} (1) = \frac{-15}{16}\\ f^{(5)} (x) &= \frac{105}{32 x^{\frac{9}{2}}} \quad \Rightarrow f^{(5)} (1) = \frac{105}{32} \end{align} \]
So the Taylor series Expansion of \(x^{\frac{1}{2}}\) is:
\(x^{\frac{1}{2}} = 1 + \frac{1}{2} (x-1) - \frac{1}{8} (x-1)^2 + \frac {1}{48} (x-1)^3 - \frac{5}{384} (x-1)^4 + \frac{35}{1536} (x-1)^5\)
We also know that \((1+x)^k = \sum_{n=0}^{\infty} \frac{k(k-1)...(k-(n-1))}{n!} x^n\)
For \(k=\frac{1}{2}\) we get:
\[ \begin{align} (1+x)^{\frac{1}{2}} &= \sum_{n=0}^{\infty} \frac{\frac{1}{2}(\frac{1}{2}-1)...(\frac{1}{2}-(n-1))}{n!} x^n\\ &= \sum_{n=0}^{\infty} \frac{\frac{1}{2}(\frac{1}{2}-1)...(\frac{2n-3}{2})}{n!} x^n\\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n-1} (2n-3)!!}{2^nn!} x^n\\ \end{align} \]