Question

Given the function \(\phi(x)=e^{2 x}:\) (a) Write the polynomial part \(P_{n}\) of its Maclaurin series. (b) Write the Lagrange form of the remainder \(R_{n}\). Determine whether \(R_{n} \rightarrow 0\) as \(n \rightarrow \infty,\) that is, whether the series is convergent to \(\phi(x)\) (c) If convergent, so that \(\phi(x)\) may be expressed as an infinite series, write out this series.

Short answer

(a) \( P_n(x) = 1 + 2x + \frac{4x^2}{2!} + \cdots + \frac{2^n x^n}{n!} \); (b) \( R_n(x) \rightarrow 0 \), series converges; (c) \( \phi(x) = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!} \).

Step-by-step solution

Maclaurin Series Definition

The Maclaurin series of a function \( f(x) \) is given by the series:\[S_n(x) = f(0) + \frac{f'(0)x}{1!} + \frac{f''(0)x^2}{2!} + \ldots + \frac{f^{(n)}(0)x^n}{n!}\]Here, \( \phi(x) = e^{2x} \) and its derivatives are needed to obtain the coefficients.

Compute Derivatives

Calculate derivatives of \( \phi(x) = e^{2x} \):- \( \phi'(x) = 2e^{2x} \)- \( \phi''(x) = 4e^{2x} \)- \( \phi^{(n)}(x) = 2^n e^{2x} \)Each derivative at \( x=0 \) results in \( \phi^{(n)}(0) = 2^n \).

Write Polynomial Part of the Maclaurin Series

Using the derivatives:\[P_n(x) = 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \cdots + \frac{2^nx^n}{n!}\]

Write Lagrange Form of Remainder

The Lagrange form of the remainder is given by:\[R_n(x) = \frac{\phi^{(n+1)}(\zeta)x^{n+1}}{(n+1)!}\]where \( \zeta \) is some value in \( (0, x) \). Since \( \phi^{(n+1)}(x) = 2^{n+1} e^{2x} \), substitute:\[R_n(x) = \frac{2^{n+1}e^{2\zeta}x^{n+1}}{(n+1)!}\]

Check Convergence of Remainder

As \( n \rightarrow \infty \), \( \frac{2^{n+1}e^{2\zeta}x^{n+1}}{(n+1)!} \rightarrow 0 \) for any finite \( x \) because of the factorial in the denominator growing much faster than the exponential.Therefore, \( R_n(x) \rightarrow 0 \). The series is convergent.

Write the Infinite Series for \( \phi(x) \)

Since convergent, we can express \( \phi(x) = e^{2x} \) as:\[\sum_{n=0}^{\infty} \frac{2^n x^n}{n!} = 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \cdots\]

Key concepts

Lagrange Remainder

When we deal with series like the Maclaurin series, precision is crucial. Often, we can't compute all the infinite terms in practice, so we need a way to understand how accurate our partial sum is. This is where the **Lagrange remainder** comes in handy. For a function, the Lagrange remainder provides an estimate of the error made when we approximate the function by its Maclaurin series up to the nth term. It's expressed as:\[R_n(x) = \frac{f^{(n+1)}(\zeta)x^{n+1}}{(n+1)!}\]Here, \(\zeta\) is a value between 0 and \(x\). This formula tells us that the part of the function that's not captured by the polynomial terms is tied to its next derivative. In our exercise, since the exponential function's growth is ultimately checked by the factorial, \(R_n(x)\) becomes negligible as \(n\) increases. By understanding the behavior of \(R_n(x)\), we gain confidence in how close our polynomial approximation is to the true function value.

Convergence of Series

Understanding when a series converges is key to using series expansions confidently. A series converges if its remainder, as expressed by the Lagrange remainder, tends to zero as \(n\) approaches infinity. Let's break this down:

• **Convergence** means that as we include more terms from a series, the approximation of the function becomes more accurate.
• The Lagrange remainder helps us determine this behavior by showing how the error behaves as more terms are added.
• In the case of the exponential function \(e^{2x}\), the Lagrange remainder goes to zero rapidly because the factorial denominator grows faster than the exponential growth in the numerator.

For example, with \(R_n(x) = \frac{2^{n+1}e^{2\zeta}x^{n+1}}{(n+1)!}\), the factorial \((n+1)!\) grows much faster than the exponential \(2^{n+1}\), ensuring that as more terms are included, the error practically vanishes. This implies the series is convergent, and indeed, accurate.

Exponential Function

The **exponential function** \(e^{x}\) is one of the most important functions in mathematics. It has a unique property of having a growth rate proportional to its value. Here are some key features:

• The derivative of an exponential function \(e^{kx}\) is \(ke^{kx}\). For \(e^{2x}\), its derivatives continue to be multiples of \(e^{2x}\).
• The Maclaurin series for \(e^{2x}\) can be constructed by assessing the function and its derivatives at \(x = 0\). They all bear a form of \(2^n\).
• When we express the exponential function as a series, \(e^{2x} \approx 1 + 2x + \frac{4x^2}{2!} + \ldots\), it reflects how rapidly it encompasses new values of \(x\).

This potential for expansion into an infinite series highlights both its simplicity and universality in mathematics. Whether calculating compound interest, population growth, or using in calculus, understanding these expansions helps predict behaviors effectively.