Notes on Derivatives and Taylor Series Calculations

Overview of Derivatives and Taylor Series

Finding Derivatives
  • Objective: To compute the first several derivatives of a given function at a specific point, then evaluate these derivatives.
  • Step 1: Compute the first, second, third, fourth, and potentially fifth derivatives of the function. This process is often applied to various functions, including trigonometric, exponential, and polynomial functions.
  • Evaluation Point: Derivatives are evaluated at a specified point (e.g., x=0x = 0 or x=1x = 1), which is crucial for constructing Taylor series.
Example: Evaluation of Derivatives

Let's consider the function f(x)=3sin(x)f(x) = 3\text{sin}(x) and evaluate its first four derivatives at x=0x = 0.

  • Function: f(x)=3sin(x)f(x) = 3\text{sin}(x)

  • First Derivative:

    • Calculation: f(x)=ddx(3sin(x))=3cos(x)f'(x) = \frac{d}{dx}(3\text{sin}(x)) = 3\text{cos}(x)
    • Evaluation at zero: f(0)=3cos(0)=31=3f'(0) = 3\text{cos}(0) = 3 \cdot 1 = 3
  • Second Derivative:

    • Calculation: f(x)=ddx(3cos(x))=3sin(x)f''(x) = \frac{d}{dx}(3\text{cos}(x)) = -3\text{sin}(x)
    • Evaluation at zero: f(0)=3sin(0)=30=0f''(0) = -3\text{sin}(0) = -3 \cdot 0 = 0
  • Third Derivative:

    • Calculation: f(x)=ddx(3sin(x))=3cos(x)f'''(x) = \frac{d}{dx}(-3\text{sin}(x)) = -3\text{cos}(x)
    • Evaluation at zero: f(0)=3cos(0)=31=3f'''(0) = -3\text{cos}(0) = -3 \cdot 1 = -3
  • Fourth Derivative:

    • Calculation: f(x)=ddx(3cos(x))=3sin(x)f''''(x) = \frac{d}{dx}(-3\text{cos}(x)) = 3\text{sin}(x)
    • Evaluation at zero: f(0)=3sin(0)=30=0f''''(0) = 3\text{sin}(0) = 3 \cdot 0 = 0
  • Note: The number of derivatives needed (up to five or more) depends on the desired precision for the function approximation.

Taylor Series Construction
  • Definition: A Taylor series approximates a function, f(x)f(x), as an infinite sum of terms, where each term is calculated from the values of the function's derivatives at a single point, aa (often called the center of the series).

  • General form: The Taylor series for f(x)f(x) expanded around point aa is given by:
    f(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3++f(n)(a)n!(xa)n+f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \dots

  • Summation Form: This can be concisely written using summation notation as:
    f(x)=n=0f(n)(a)n!(xa)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

    • For this discussion, the point of expansion is referred to as aa or cc. For example, if c=1c = 1, the terms will involve (x1)(x-1) powers.
  • Specific Calculation Steps (for constructing terms):

    1. Determine the expansion point, aa.
    2. Calculate the function value f(a)f(a).
    3. Calculate the first derivative f(x)f'(x) and evaluate f(a)f'(a).
    4. Calculate higher-order derivatives f(x),f(x),,f(n)(x)f''(x), f'''(x), \dots, f^{(n)}(x) and evaluate them at aa.
    5. Substitute these values into the Taylor series formula. For example, a term might look like f(n)(a)n!(xa)n\frac{f^{(n)}(a)}{n!}(x-a)^n.
Example for Taylor Series Calculation
  • Given: Taylor series for f(x)=ln(x)f(x) = \text{ln}(x) expanded around c=1c = 1. The general term for the Taylor series of ln(x)\text{ln}(x) around c=1c=1 is given by: n=1(1)n+1n(x1)n=(x1)12(x1)2+13(x1)3\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \dots
    • Note that for ln(x)\text{ln}(x), f(1)=ln(1)=0f(1) = \text{ln}(1) = 0, so the series starts from n=1n = 1.
Derivative Notation and Understanding
  • Notation: Various notations are used to represent derivatives:
    • Leibniz's notation: dydx\frac{dy}{dx}, ddxf(x)\frac{d}{dx}f(x), d2ydx2\frac{d^2y}{dx^2} for second derivative.
    • Lagrange's notation (Prime notation): f(x)f'(x), f(x)f''(x), f(x)f'''(x), f(n)(x)f^{(n)}(x).
    • Newton's notation (Dot notation): y˙,y¨\dot{y}, \ddot{y} (primarily used in physics for derivatives with respect to time).
    • Euler's notation (D-operator notation): D<em>xf(x)D<em>xf(x), $$D