91Ó°ÊÓ

Derivatives are a fundamental concept in calculus. They describe the rate at which a function changes. When working with a Taylor series, finding derivatives is the first important step. By calculating derivatives, we can find the necessary coefficients for the series expansion.
For example, given the function \( f(x) = \sqrt{x+1} \), we start by finding its derivatives:

• The first derivative, \( f'(x) \), tells us how the function's slope changes at each point. It's \( f'(x) = \frac{1}{2\sqrt{x+1}} \).
• The second derivative, \( f''(x) \), gives us information about the curve's concavity. It's \( f''(x) = -\frac{1}{4(x+1)^{3/2}} \).
• The third derivative, \( f'''(x) \), continues this pattern, providing further insight into the function's shape. It's \( f'''(x) = \frac{3}{8(x+1)^{5/2}} \).

Derivatives tell us a lot about the function's behavior around a particular point. Knowing this allows us to build a Taylor series.

A series expansion is a way to express complex functions as infinite sums of simpler terms. The Taylor series is one such expansion. It uses derivatives of a function at a single point to approximate the function. This method allows us to understand and compute values of functions more easily.
In this exercise, expanding \( f(x) = \sqrt{x+1} \) into a Taylor series is the goal. With derivatives calculated, we use the Taylor series formula:

• \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]

This formula shows that each term of the series involves a derivative, evaluated at \( a \), multiplied by a power of \( (x-a) \). With each additional term, our approximation of the function becomes more accurate.
The ideas behind series expansion help us simplify and solve difficult problems by breaking them into manageable parts.

Once we've found the derivatives of a function, the next step is evaluating these derivatives at a specific point. In our exercise, we evaluate them at \( x = 0 \). This process helps us to pinpoint the specific coefficients needed to construct the Taylor series.
Let's see it in action:

• The zeroth derivative, or the function itself, here results in \( f(0) = 1 \).
• The first derivative \( f'(0) = \frac{1}{2} \).
• The second derivative \( f''(0) = -\frac{1}{4} \).
• The third derivative \( f'''(0) = \frac{3}{8} \).

Substituting these values into the Taylor series formula gives us the approximation of \( \sqrt{x+1} \) around \( x = 0 \), ultimately leading to the series:

• \[ f(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \ldots \]

This evaluated series gives a close approximation of our function around \( x = a = 0 \), a crucial step in solving real-world problems with calculus.