• f (x) =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.
\[ f(x) = \frac{1}{1-x} \]
Starting with the derivatives of \(f(x)\):
\[ f'(x) = \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2} \]
\[ f''(x) = \frac{d}{dx} \left( \frac{1}{(1-x)^2} \right) = \frac{2}{(1-x)^3} \]
General formula for the \(n\)-th derivative, \[ f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}} \]
\[ f^{(n)}(0) = n! \]
Substituting these derivatives into the Taylor series formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty} x^n \]
This series is valid for \(|x| < 1\)
\[ f(x) = e^x \]
The derivatives: \[ f(x) = e^x \] \[ f'(x) = e^x \] \[ f''(x) = e^x \] \[ f^{(n)}(x) = e^x \quad \text{for all } n \]
Evaluating derivatives at 0: \[ f(0) = e^0 = 1 \] \[ f'(0) = e^0 = 1 \] \[ f''(0) = e^0 = 1 \] \[ f^{(n)}(0) = e^0 = 1 \quad \text{for all } n \]
Given that all derivatives at \(x=0\) are 1, the Taylor series of \(f(x)\) is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty} \frac{1}{n!} x^n \]
This series is valid for all real numbers \(x\)
\[ f(x) = \ln(1 + x) \]
The derivatives:} \[ f(x) = \ln(1 + x) \] \[ f'(x) = \frac{1}{1 + x} \] \[ f''(x) = -\frac{1}{(1 + x)^2} \] \[ f'''(x) = \frac{2}{(1 + x)^3} \]
Evaluating Derivatives at \(x=0\): \[ f'(0) = 1 \] \[ f''(0) = -1 \] \[ f'''(0) = 2 \]
Using the pattern observed from the derivatives, the general derivative \(f^{(n)}(x)\) evaluated at zero is: \[ f^{(n)}(x) = (-1)^{n-1} \frac{(n-1)!}{(1 + x)^n} \Rightarrow f^{(n)}(0) = (-1)^{n-1} (n-1)! \] The Taylor series of \(f(x)\) at \(x = 0\) can thus be written as: \[ f(x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} \]
This series is valid for \(-1 < x \leq 1\)
\[ f(x) = x^{1/2} \]
We want to find the Taylor series of \(f(x)\) around the point \(x=1\) which is is valid for \(x \geq 0\). To find the Taylor series at \(x=1\), we begin by shifting our function to: \[ g(x) = f(x+1) = (x+1)^{1/2} \]
We do this shifting because f(x) = x^{1/2} is not defined at x=0. We can use g(x) as an alternative.
The derivatives of \(g(x)\):
\[ g(x) = (x+1)^{1/2} \] \[ g'(x) = \frac{1}{2}(x+1)^{-1/2} \] \[ g''(x) = -\frac{1}{4}(x+1)^{-3/2} \] \[ g'''(x) = \frac{3}{8}(x+1)^{-5/2} \]
Evaluating Derivatives at \(x=0\): \[ g(0) = 1 \] \[ g'(0) = \frac{1}{2} \] \[ g''(0) = -\frac{1}{4} \] \[ g'''(0) = \frac{3}{8} \]
The Taylor series of \(g(x)\) at \(x = 0\) is: \[ g(x) = \sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n!} x^n \] \[ f(x) = g(x-1) = \sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n!} (x-1)^n \]
This series provides an approximation valid near \(x=1\).