Problem 9 - Calculus

Question

Find the Taylor polynomials of degree \(n\) approximating the functions for \(x\) near \(0 .\) (Assume \(p\) is a constant. \()\) $$\frac{1}{\sqrt{1+x}}, \quad n=2,3,4$$

Step-by-Step Solution

Answer

Taylor polynomials: Degree 2: \(1 - \frac{1}{2}x + \frac{3}{4}x^2\), Degree 3: \(1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3\), Degree 4: \(1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3 + \frac{105}{16}x^4\).

Step 1: Recall Taylor Series Formula

The Taylor series expansion of a function \(f(x)\) around \(x = a\) is given by: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \cdots\]For this exercise, \(a=0\) and we need the expansions up to \(n=4\).

Step 2: Derive the Function and Its Derivatives

The function given is \(f(x) = \frac{1}{\sqrt{1+x}}\). First, calculate the derivatives at \(x = 0\):- \(f(x) = (1+x)^{-1/2}\).- \(f'(x) = -\frac{1}{2}(1+x)^{-3/2}\).- \(f''(x) = \frac{3}{4}(1+x)^{-5/2}\).- \(f^{(3)}(x) = -\frac{15}{8}(1+x)^{-7/2}\).- \(f^{(4)}(x) = \frac{105}{16}(1+x)^{-9/2}\).Evaluate these derivatives at \(x=0\).

Step 3: Simplify the Derivatives at x = 0

Evaluate the derivatives calculated in the previous step at \(x=0\):- \(f(0) = 1\).- \(f'(0) = -\frac{1}{2}\).- \(f''(0) = \frac{3}{4}\).- \(f^{(3)}(0) = -\frac{15}{8}\).- \(f^{(4)}(0) = \frac{105}{16}\).

Step 4: Construct Taylor Polynomials

Using the derivatives at \(x=0\), construct the Taylor polynomials for different degree values:- Degree 2: \[T_2(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2\]- Degree 3: \[T_3(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3\]- Degree 4: \[T_4(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3 + \frac{105}{16}x^4\]

Step 5: Present the Final Polynomials

The final Taylor polynomials approximating \(\frac{1}{\sqrt{1+x}}\) near \(x=0\) are:- Degree 2: \(T_2(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2\).- Degree 3: \(T_3(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3\).- Degree 4: \(T_4(x) = 1 - \frac{1}{2}x + \frac{3}{4}x^2 - \frac{15}{8}x^3 + \frac{105}{16}x^4\).

Key Concepts

Taylor Series

The Taylor series is a powerful tool in calculus for approximating functions. It allows a function to be expressed as an infinite sum of terms. Each term is calculated from the derivatives of the function at a given point. The formula for a Taylor series expansion of a function \(f(x)\) around the value \(x = a\) is:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \cdots\]The series provides an approximation that becomes closer to the real function as more terms are added.

• The point \(a\) is where the series is centered.
• Each term involves higher order derivatives of the function.
• In practical use, only a finite number of terms are computed for a desired degree of accuracy.

For the present exercise, the series is centered at \(x = 0\) to a degree of 4. Taylor series make it easier to handle complex functions by using polynomial expressions.

Derivatives

A derivative represents the rate at which a function changes at a particular point. It's a central concept in calculus and fundamental to the formation of Taylor series. Calculating derivatives is necessary to determine each polynomial term in a Taylor series.

• The first derivative \(f'(x)\) indicates the slope at any point on the function.
• The second derivative \(f''(x)\) shows the rate of change of the slope.
• Higher-order derivatives, such as third \(f^{(3)}(x)\) and fourth \(f^{(4)}(x)\), provide further information about changes in the function's behavior.

In our exercise:

• We determined the derivatives of \(f(x) = \frac{1}{\sqrt{1+x}}\).
• Calculated these derivatives at \(x = 0\) to construct the Taylor polynomial.
• As each derivative is evaluated, it contributes to a specific term in the polynomial.

Understanding derivatives is crucial for constructing Taylor polynomials and interpreting how a function behaves around a certain point.

Polynomial Approximation

Polynomial approximation is an approach used to estimate more complex mathematical functions using simpler polynomials. Taylor polynomials are a type of polynomial approximation derived from the Taylor series. They offer a useful way to get close to the actual values of a function, especially near the point where the series is centered.

• Short polynomial expressions replace a potentially infinite series.
• Each additional term in the Taylor polynomial provides a more accurate approximation.
• For computations or predictions, several terms usually provide enough precision.

In this exercise, the function \(\frac{1}{\sqrt{1+x}}\) is approximated using Taylor polynomials of degrees 2, 3, and 4:

• These polynomials approximate the function near \(x = 0\).
• As more terms are added (higher degrees), accuracy improves.
• Taylor polynomials are especially useful when dealing with functions near their center point, \(x = a\).

By recognized practices, these polynomials help simplify complex function calculations.

Mathematical Functions

A mathematical function maps a set of inputs (domain) with corresponding outputs (range). In calculus, functions are the basis for determining behavior over intervals or specific points. A function like \(f(x) = \frac{1}{\sqrt{1+x}}\) reveals how \(f\) changes as \(x\) changes.

• Functions can exhibit complex behavior that requires approximation for analysis.
• Polynomial and Taylor series approximations provide simpler ways to study functions.
• They help calculate values over an input domain and understand local behavior around specific values.

For our problem, we're focused on approximating the mathematical behavior near \(x = 0\).

• This involves understanding how the original function behaves in this region.
• The calculated Taylor polynomials give insight into predicted values close to the center.
• This foundational tool in calculus supports deeper exploration of changes in rate and extent in mathematical functions.

Understanding mathematical functions is essential to leverage Taylor series and polynomial approximations effectively.