Problem 11 - Calculus

Question

Find the Taylor series about 0 for the functions including the general term. $$(1+x)^{3}$$

Step-by-Step Solution

Answer

The Taylor series is \((1 + x)^3 = 1 + 3x + 3x^2 + x^3\). The general term is \(\frac{3!}{n!(3-n)!}x^n\).

Step 1: Identify the Function

The given function is \((1 + x)^3\). We need to expand this function as a Taylor series about 0.

Step 2: Determine the Derivatives

To find the Taylor series, we need the derivatives of \((1 + x)^3\) evaluated at \(x = 0\). Calculate the first few derivatives:- \(f(x) = (1 + x)^3\)- \(f'(x) = 3(1 + x)^2\)- \(f''(x) = 6(1 + x)\)- \(f'''(x) = 6\)- Higher-order derivatives are zero since the third derivative is constant.

Step 3: Evaluate the Derivatives at Zero

Evaluate each derivative at \(x = 0\):- \(f(0) = (1 + 0)^3 = 1\)- \(f'(0) = 3(1 + 0)^2 = 3\)- \(f''(0) = 6(1 + 0) = 6\)- \(f'''(0) = 6\)- Higher-order derivatives are zero.

Step 4: Write the Taylor Series Expression

The Taylor series for a function \(f(x)\) about 0 is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]Using the values obtained:- Constant term: \(1\)- Linear term: \(3x\)- Quadratic term: \(\frac{6}{2!}x^2 = 3x^2\)- Cubic term: \(\frac{6}{3!}x^3 = x^3\)Hence, the Taylor series is: \((1 + x)^3 = 1 + 3x + 3x^2 + x^3\).

Step 5: Express the General Term

In the computed series, it is clear that the general term of the function \((1 + x)^3\) is \[ \frac{3!}{n!(3-n)!} x^n \] for \(n = 0, 1, 2, 3\). This is derived from the binomial theorem for integer powers.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies continuous change. It provides tools to understand rates of change and how quantities accumulate. Two main concepts in calculus are differentiation and integration. Differentiation helps in determining the rate at which quantities change, represented by derivatives. Integration, on the other hand, helps in finding the total accumulation of quantities, such as areas under curves. These tools are essential for solving real-world problems in physics, engineering, and economics. In the context of a Taylor series, calculus is used to find derivatives, enabling the expression of functions as infinite sums of terms calculated from the values of its derivatives at a single point. The ability to approximate functions using Taylor series showcases the power of calculus in simplifying complex calculations.

Binomial Theorem

The Binomial Theorem is a formula that provides a way to expand powers of a binomial. A binomial is an algebraic expression with two terms, such as (1 + x). The theorem states that for any positive integer n , (1 + x)^n can be expanded into a sum involving terms of the form \\( \binom{n}{k} x^k \)\, where \( \binom{n}{k} \)\ is the binomial coefficient, representing the number of ways to choose k elements from n elements. This expansion simplifies the multiplication of binomials raised to large powers without needing to manually expand the equation multiple times. In the given problem, (1 + x)^3 is expanded using concepts from the Binomial Theorem. The coefficients of each term in the expanded form correspond to the binomial coefficients \( \binom{3}{k} \). This is integrated into the Taylor series to offer insight into how the powers of x relate to the sequence of terms generated.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable. It is a core concept in calculus and is used to explore the behavior of functions. Derivatives are essential for the Taylor series because they help approximate functions at a specific point by using the function's rate of change information. To find the Taylor series for (1 + x)^3 , the derivatives up to the third degree were calculated:

• First derivative: f'(x) = 3(1 + x)^2
• Second derivative: f''(x) = 6(1 + x)
• Third derivative: f'''(x) = 6

These derivatives allow the creation of polynomial expressions that sum up to approximate the function around a center point, in this case, x = 0 . The derivatives evaluated at this point are crucial to forming each term in the Taylor series.

Polynomial Expansion

Polynomial expansion involves expressing a function as a series of terms involving powers of a variable. This is crucial for Taylor series, where functions are written as infinite sums of polynomial terms. For (1 + x)^3 , polynomial expansion breaks down the expression into a sum of polynomial terms of increasing degree. The Taylor series specifically centers around a point, frequently around x = 0 , simplifying calculations by focusing on the behavior near this point. In our example, the expansion resulted in: (1 + x)^3 = 1 + 3x + 3x^2 + x^3 , where each term represents the evaluated derivatives at this point. Polynomial expansion through Taylor series is especially helpful in approximating complex functions using simpler polynomial terms, making calculations more manageable.