Question

Let \(1-\sum_{1}^{\infty} c_{n} t^{n}\) be the Maclaurin series for \((1-t)^{1 / 2}\). a. The series converges absolutely and uniformly on compact subsets of \((-1,1)\), as does the termwise differentiated series \(-\sum_{1}^{\infty} n c_{n} t^{n-1}\). Thus, if \(f(t)=\) \(1-\sum_{1}^{\infty} c_{n} t^{n}\), then \(f^{\prime}(t)=-\sum_{1}^{\infty} n c_{n} t^{n-1}\). b. By explicit calculation, \(f(t)=-2(1-t) f^{\prime}(t)\). from which it follows that \((1-t)^{-1 / 2} f(t)\) is constant. Since \(f(0)=1, f(t)=(1-t)^{1 / 2}\).

Short answer

The function is \(f(t) = (1-t)^{1/2}\).

Step-by-step solution

Identify the Maclaurin Series Function

We are given that \( f(t) = 1 - \sum_{n=1}^{\infty} c_n t^n \) is the Maclaurin series for \((1-t)^{1/2}\). This implies that this series is equivalent to the function \((1-t)^{1/2}\) expanded around \(t = 0\).

Differentiate the Series Termwise

The differentiated series is given as \(-\sum_{n=1}^{\infty} n c_n t^{n-1}\). This represents the derivative of the function \(f(t)\), denoted as \(f'(t)\). Since the series converges absolutely and uniformly on compact subsets of \((-1,1)\), the differentiated series also converges within these bounds.

Use the Relationship between f(t) and its Derivative

The problem states that \( f(t) = -2(1-t) f'(t) \). Substitute the series expressions for \(f(t)\) and \(f'(t)\) from steps 1 and 2 to establish this relationship: \(1 - \sum_{n=1}^{\infty} c_n t^n = -2(1-t)(-\sum_{n=1}^{\infty} n c_n t^{n-1})\).

Express f(t) as a Function of t

Recognizing from step 3 that the equation simplifies the relationship between \(f(t)\) and its derivative, express \(f(t)\) in terms of \((1-t)\). Since the series matches \((1-t)^{1/2}\), given \(f(0) = 1\), infer that \(f(t) = (1-t)^{1/2}\).

Establish the Constant Function

Since \( (1-t)^{-1/2} f(t) \) is constant, verify this by examining that \(((1-t)^{-1/2})(1-t)^{1/2} = 1\). It confirms that \((1-t)^{-1/2} f(t) = 1\) is constant, thus supporting the conclusion that \(f(t) = (1-t)^{1/2}\).

Key concepts

Convergence of Series

Understanding the convergence of a series is crucial, especially when dealing with infinite series, such as the Maclaurin series. In essence, convergence refers to whether the series approaches a finite limit as the number of terms increases. In this exercise, the Maclaurin series \[ f(t) = 1 - \sum_{n=1}^{\infty} c_n t^n \]and its termwise differentiated series \[- \sum_{n=1}^{\infty} n c_n t^{n-1} \]converge absolutely and uniformly within compact subsets of the interval \((-1, 1)\).

• Absolute convergence: This implies that if you take the absolute value of each term in the series, and the series still converges, the series is said to converge absolutely.
• Uniform convergence: This means the series converges at the same rate for all points within the interval. It ensures that the limit of the series is a continuous function if each series component is continuous.

These properties are important because they allow us to differentiate and integrate the series term by term within the radius of convergence.

Termwise Differentiation

Termwise differentiation is the process of differentiating each term in a series independently and is valid under certain convergence conditions. In the exercise, this technique is applied to the series \[f(t) = 1 - \sum_{n=1}^{\infty} c_n t^n \]to find its derivative. The termwise differentiation results in the series:\[ f'(t) = -\sum_{n=1}^{\infty} n c_n t^{n-1} \]This means that the derivative of the original series function is also a series, where each term of the original series \( c_n t^n \) is differentiated yielding \( n c_n t^{n-1} \).
For termwise differentiation to be valid:

• The original series should converge absolutely and uniformly within its interval.
• Each function in the series is differentiable.

These conditions ensure that differentiating term by term does not affect the convergence of the series.

Function Derivation

Function derivation involves finding the derivative of a given function, which tells us the rate at which the function is changing at any given point. In the provided exercise, the derivation of the function \[ f(t) = 1 - \sum_{n=1}^{\infty} c_n t^n \]is achieved through termwise differentiation, resulting in:\[ f'(t) = -\sum_{n=1}^{\infty} n c_n t^{n-1} \]Function derivation is vital because it helps establish the relationship between a function and its gradient, often giving insights into solutions of differential equations or real-world problems modeled by these functions.A core relationship demonstrated in the exercise is \( f(t) = -2(1-t) f'(t) \), which provides a deeper understanding of the behavior of the function in relation to its derivative. Recognizing such relationships is key for solving real-world problems analytically.

Uniform Convergence

Uniform convergence is a powerful concept in mathematical analysis that indicates the rate at which different series or functions approach their respective limits. If a sequence of functions converges uniformly to a limit function within a particular interval, it ensures several favorable properties, such as continuity and differentiability, can be preserved while passing to the limit.In our exercise, both the original Maclaurin series and its termwise differentiated series converge uniformly on compact subsets of \((-1, 1)\). This convergence means no matter how close we get to a limit, every point in the interval will get equally close, making analysis much more straightforward.

• Uniform convergence prevents unexpected behavior by having consistent function values through a domain.
• This property provides a foundation for manipulating the series, such as ensuring valid termwise differentiation or integration.

Understanding how uniform convergence applies to series helps in studying stability and predictability across mathematical models.