Problem 34 - Essential Calculus Early Transcendentals

Question

Show that the function $$ f(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !} $$ is a solution of the differential equation $$ f^{\prime \prime}(x)+f(x)=0 $$

Step-by-Step Solution

Answer

The function \( f(x) = \cos(x) \) is a solution to the differential equation \( f''(x) + f(x) = 0 \).

Step 1: Recall the Expansion for Cosine

The function given can be recognized as the Taylor series expansion for the cosine function, \[ f(x) = \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \].We need to prove that \( f(x) = \cos(x) \) satisfies the equation \( f''(x) + f(x) = 0 \).

Step 2: Compute the First Derivative

Differentiate the function \( f(x) = \cos(x) \) with respect to \( x \). The derivative of \( \cos(x) \) is \( -\sin(x) \), so\[ f'(x) = -\sin(x) \].

Step 3: Compute the Second Derivative

Differentiate the first derivative \( f'(x) = -\sin(x) \) with respect to \( x \). The derivative of \( -\sin(x) \) is \( -\cos(x) \), so \[ f''(x) = -\cos(x) \].

Step 4: Substitute into the Differential Equation

Substitute \( f''(x) = -\cos(x) \) and \( f(x) = \cos(x) \) into the differential equation:\[ f''(x) + f(x) = -\cos(x) + \cos(x) = 0 \].

Step 5: Conclude the Solution

Since \( f''(x) + f(x) = 0 \), the function \( f(x) = \cos(x) \) is indeed a solution of the differential equation \( f''(x) + f(x) = 0 \).

Key Concepts

Taylor Series

The Taylor series is a way to represent functions as infinite sums of terms calculated from the values of their derivatives at a single point. It is an incredibly powerful tool in calculus and differential equations. In simpler terms, it allows us to express complex functions as series so that we can analyze and solve them more easily.
In the context of the exercise, the given function:

• \( f(x)= \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !} \)

is recognized as the Taylor series expansion of the cosine function. This series form helps in differentiating and substituting into equations, crucial for verifying that it solves the original differential equation provided.

Cosine Function

The cosine function, commonly notated as \(\cos(x)\), is a fundamental trigonometric function. It demonstrates periodic oscillations that are symmetrical about the y-axis. Understanding the cosine function involves knowing its basic properties.

• It has a period of \(2\pi\), meaning it repeats every \(2\pi\) intervals.
• The amplitude, or the height from the centerline to the peak, is 1.

In simpler terms, the cosine function can tell us about angles and rotations on a plane. When working with differential equations, recognizing that our function matches the behavior of \(\cos(x)\) allows identifying it as a solution for the given equation.

Second Derivative

The second derivative provides information about the curvature or concavity of a function. When you compute the second derivative of a function, you obtain a deeper understanding of how the function behaves.
For the function \(\cos(x)\), the process of differentiation involves:

• First Derivative: \(f'(x) = -\sin(x)\)
• Second Derivative: \(f''(x) = -\cos(x)\)

Notice how the second derivative, \(-\cos(x)\), helps in returning to the original form. This relationship is crucial for fulfilling the differential equation \(f''(x) + f(x) = 0\), which ultimately verifies whether the given function is a solution.

Solution Verification

Verification involves checking that the proposed function really satisfies the differential equation. For this exercise, we need to ensure that substituting into the equation returns zero.
We have:

• Second derivative, \(f''(x) = -\cos(x)\)
• Function itself, \(f(x) = \cos(x)\)

Substituting these into the equation yields:

• \(f''(x) + f(x) = -\cos(x) + \cos(x) = 0\)

Since both terms cancel each other, resulting in zero, the verification process shows that the function \(f(x) = \cos(x)\) is indeed a solution to the differential equation \(f''(x) + f(x) = 0\). This confirms that our understanding of derivatives and trigonometric functions has been applied correctly.