Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(y=y(x)=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\ldots\). Show that \(y\) satisfies the differential equation \(y^{\prime \prime}+y=0\) with the conditions \(y(0)=0\) and \(y^{\prime}(0)=1\). From this, guess a simple formula for \(y\).

Step-by-Step Solution

Answer

The function is y(x) = sin(x), satisfying the differential equation and conditions.

Step 1: Identify the Function

The function \( y = y(x) \) is presented as an infinite series:\[ y = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots \] This is an alternating series where the power of \( x \) increases by 2 in each term.

Step 2: Differentiate the Function

Differentiate the function \( y \) term by term with respect to \( x \) to find \( y' \):\[ y' = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \] Simplifying gives the series for \( y' \).

Step 3: Differentiate Again to Find Second Derivative

Differentiate \( y' \) to find \( y'' \):\[ y'' = -x + \frac{x^3}{3!} - \frac{x^5}{5!} + \ldots \] Simplifying gives the series for \( y'' \).

Step 4: Substitute into the Differential Equation

Substitute \( y \) and \( y'' \) into the differential equation \( y'' + y = 0 \):\[ (-x + \frac{x^3}{3!} - \frac{x^5}{5!} + \ldots) + (x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots) = 0 \] The terms correctly cancel out, verifying the equation.

Step 5: Verify Initial Conditions

Check \( y(0) \) and \( y'(0) \): \[ y(0) = 0 \] (because all terms have \( x \) in them) and \[ y'(0) = 1 \] (from the definition of the series for \( y' \)). These satisfy the given conditions.

Step 6: Identify a Possible Simple Formula

The series \( y \) resembles the Maclaurin series for \( \sin(x) \), which is given by \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots \). Thus, the function \( y(x) = \sin(x) \).

Key Concepts

Infinite Series

An infinite series is a sum of an infinite sequence of terms. In mathematics, it's often used to represent functions rather than evaluate an endless addition. For our exercise, the function \(y(x)\) is expressed as an infinite series:

• Each term in the series alternates in sign, which indicates that it is an alternating series.
• The power of \(x\) increases by 2 in each subsequent term, showcasing a pattern typical of series expansions for trigonometric functions.

This kind of series allows us to closely approximate complicated functions using polynomials, which can be differentiated and integrated term by term, simplifying many calculus operations.

Maclaurin Series

The Maclaurin series is a specific kind of Taylor series, centered at zero. It provides a polynomial expression for functions that can be used to approximate a function near this central point. In our exercise, the representation of \(y(x)\) is reminiscent of the Maclaurin series:

• Maclaurin series expand functions in powers of \(x\), highlighting the polynomial nature of the series.
• The terms of our function \(y(x)\) match those of the sine function's Maclaurin series: \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \).
• This connection allows us to identify the function \(y(x)\) with a known trigonometric function, facilitating further analysis and solution of differential equations.

By utilizing the Maclaurin series, complex trigonometric functions like sine and cosine can be easily handled in calculus problems through simple polynomial calculations.

Initial Conditions

Initial conditions are crucial when solving differential equations as they provide specific values the solution must satisfy. They often determine a unique solution from the general solution of a differential equation. In our case, the initial conditions given were:

• \(y(0) = 0\): since all terms of \(y(x)\) include \(x\), evaluating at zero results in zero.
• \(y'(0) = 1\): derived from the first derivative of the series, this condition confirms that the slope at the starting point (or the velocity in a physical context) is 1.

These initial conditions were verified in the sequence of calculations to ensure the differential equation setup was consistent with a specific solution for \(y(x)\).

Trigonometric Functions

Trigonometric functions like sine and cosine have patterns that can be represented using infinite series. For instance, the exercise uses the series to represent the sine function. This representation includes:

• The characteristic pattern of alternation in sign, reflecting the sine function's oscillating nature.
• The increasing odd powers of \(x\), capturing the periodicity and smooth, wave-like behavior of the function.

The connection in this exercise highlights the utility of these trigonometric series in solving differential equations. By recognizing the series as that of a sine function, we not only find a solution but also see the underlying connection between calculus and trigonometry. This provides an insight into how series can be used mathematically to simplify otherwise complex expressions and solve equations efficiently.